8

YEAR

CambridgeMATHS NSW SYLLABUS FOR THE AUSTRALIAN CURRICULUM

>> Additional resources online STUART PALMER | DAVID GREENWOOD BRYN HUMBERSTONE | JUSTIN ROBINSON JENNY GOODMAN | JENNIFER VAUGHAN © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

Cambridge University Press

477 Williamstown Road, Port Melbourne, VIC 3207, Australia Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.edu.au Information on this title: www.cambridge.org/9781107671812 © Stuart Palmer, David Greenwood, Bryn Humberstone, Justin Robinson, Jenny Goodman, Jennifer Vaughan 2014 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2014 Reprinted 2014 Cover designed by Sardine Design Typeset by Aptara Corp. Printed in Singapore by C.O.S Printers Pte Ltd A Cataloguing-in-Publication entry is available from the catalogue of the National Library of Australia at www.nla.gov.au ISBN 978-1-107-67181-2 Paperback Additional resources for this publication at www.cambridge.edu.au/GO Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this publication, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact: Copyright Agency Limited Level 15, 233 Castlereagh Street Sydney NSW 2000 Telephone: (02) 9394 7600 Facsimile: (02) 9394 7601 Email:[emailprotected]Reproduction and communication for other purposes Except as permitted under the Act (for example a fair dealing for the purposes of study, research, criticism or review) no part of this publication may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher at the address above Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables and other factual information given in this work is correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter.

© David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

Cambridge University Press

table of Contents strands, substrands and outcomes

About the authors Introduction and guide to this book Acknowledgements

1

algebraic techniques 2 and indices 1A 1B 1C 1D 1E 1F 1G 1H 1I 1J 1K

Pre-test The language of algebra revision Substitution and equivalence Adding and subtracting terms revision Multiplying and dividing terms revision Adding and subtracting algebraic fractions extension Multiplying and dividing algebraic fractions extension Expanding brackets Factorising expressions Applying algebra Index laws for multiplication and division The zero index and power of a power Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

viii ix xi

2 4 5 10 14 19

Number and algebra Algebraic Techniques MA4–8NA, MA4–9NA

23 28 33 38 42 46 51 55 57 58 59 59 61

iii © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

Cambridge University Press

2

Equations 2 Pre-test Reviewing equations REVISION Equivalent equations REVISION Equations with fractions Equations with pronumerals on both sides Equations with brackets Solving simple quadratic equations Formulas and relationships EXTENSION Applications EXTENSION Inequalities EXTENSION Solving inequalities EXTENSION Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

64 65 70 75 80 85 89 92 96 101 105 110 112 113 114 115 117

Measurement and Pythagoras’ theorem

118

2A 2B 2C 2D 2E 2F 2G 2H 2I 2J

3

62

3A 3B 3C 3D 3E 3F 3G 3H 3I 3J 3K 3L 3M

Pre-test Length and perimeter REVISION Circumference of circles REVISION Area REVISION Area of special quadrilaterals Area of circles Area of sectors and composite figures Surface area of prisms EXTENSION Volume and capacity Volume of prisms and cylinders Time REVISION Introducing Pythagoras’ theorem Using Pythagoras’ theorem Calculating the length of a shorter side Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

120 121 127 132 139 144 149 155 160 165 170 178 183 188 193 195 196 197 198 200

Number and Algebra Equations MA4–10NA

Measurement and Geometry Length Area Volume Time Right-angled Triangles (Pythagoras) MA4–12MG, MA4–13MG, MA4–14MG, MA4–15MG, MA4–16MG

iv © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

Cambridge University Press

4

Fractions, decimals, percentages and financial mathematics 4A 4B 4C 4D 4E 4F 4G 4H 4I 4J 4K

5

Pre-test Equivalent fractions REVISION Computation with fractions REVISION Decimal place value and fraction/decimal conversions REVISION Computation with decimals REVISION Terminating decimals, recurring decimals and rounding REVISION Converting fractions, decimals and percentages REVISION Finding a percentage and expressing as a percentage Decreasing and increasing by a percentage The Goods and Services Tax (GST) Calculating percentage change, profit and loss Solving percentage problems with the unitary method and equations Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response question

Ratios and rates 5A 5B 5C 5D 5E 5F 5G 5H 5I

Pre-test Introducing ratios Simplifying ratios Dividing a quantity in a given ratio Scale drawings Introducing rates Ratios and rates and the unitary method Solving rate problems Speed Distance/time graphs Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response question

Semester review 1

Number and Algebra

202 204 205 211

Fractions, Decimals and Percentages Financial Mathematics MA4–5NA, MA4–6NA

220 226 233 239 247 253 259 265 270 275 277 278 281 281 283

284 286 287 292 297 303 310 315 320 325 331 342 343 344 345 346 349

350

Number and Algebra Ratios and Rates Financial Mathematics MA4–7NA, MA4–6NA

© David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

v Cambridge University Press

6

Angle relationships and properties of geometrical figures 1 6A 6B 6C 6D 6E 6F 6G

7

Pre-test The language, notation and conventions of angles REVISION Transversal lines and parallel lines REVISION Triangles Quadrilaterals Polygons EXTENSION Line symmetry and rotational symmetry Euler’s formula for three-dimensional solids FRINGE Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

Linear relationships 1 7A 7B 7C 7D 7E 7F 7G 7H 7I

Pre-test The Cartesian plane Using rules, tables and graphs to explore linear relationships Finding the rule using a table of values Gradient EXTENSION Gradient–intercept form EXTENSION The x-intercept EXTENSION Solving linear equations using graphical techniques Applying linear graphs EXTENSION Non-linear graphs Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

360

Measurement and Geometry

362

Angle Relationships

363 371 379 386 392 397

MA4–18MG, MA4–17MG

Properties of Geometrical Figures

402 408 410 411 412 413 415

416 418 419

Number and Algebra Linear Relationships MA4–11NA

423 427 433 440 448 453 462 468 474 476 477 478 479 482

vi © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

Cambridge University Press

8

Transformations and congruence

8A 8B 8C 8D 8E 8F 8G 8H

9

Pre-test Reflection Translation Rotation Congruent figures Congruent triangles Similar figures EXTENSION Similar triangles EXTENSION Using congruent triangles to establish properties of quadrilaterals Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response question

484 486 487 493 498 504 509 515 521

9A 9B 9C 9D 9E 9F 9G 9H 9I 9J

Semester review 2

Linear Relationships Properties of Geometrical Figures MA4–11NA, MA4–17MG

528 533 535 536 537 538 541

Data collection, representation and analysis 542 Pre-test Types of data Dot plots and column graphs Line graphs Sector graphs and divided bar graphs Frequency distribution tables Frequency histograms and frequency polygons Mean, median, mode and range Interquartile range EXTENSION Stem-and-leaf plots Surveying and sampling Investigation Puzzles and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

Number and Algebra Measurement and Geometry

544 545 551 561 568 574 580 588 594 599 606 612 613 614 615 617 619

Statistics and Probability Data Collection and Representation Single Variable Data Analysis MA4–19SP, MA4–20SP

620

Answers 628 Index 693 Note: Students who require additional revision of Computation with integers may find Appendix 1 useful. This can be accessed online at www.cambridge.edu.au/GO. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

vii Cambridge University Press

Table of About theContents authors Stuart Palmer was born and educated in NSW. He is a high school mathematics teacher with more than 25 years’ experience teaching students from all walks of life in a variety of schools. Stuart has taught all the current NSW Mathematics courses in Stages 4, 5 and 6 numerous times. He has been a head of department in two schools and is now an educational consultant who conducts professional development workshops for teachers all over NSW and beyond. He also works with pre-service teachers at The University of Sydney and The University of Western Sydney. David Greenwood is the Head of Mathematics at Trinity Grammar School in Melbourne

and has 20 years’ experience teaching mathematics from Years 7 to 12. He has run numerous workshops within Australia and overseas regarding the implementation of the Australian Curriculum and the use of technology for the teaching of mathematics. He has written more than 20 mathematics titles and has a particular interest in the sequencing of curriculum content and working with the Australian Curriculum proficiency strands. Bryn Humberstone graduated from University of Melbourne with an Honours degree in

Pure Mathematics, and is currently teaching both junior and senior mathematics in Victoria. Bryn is particularly passionate about writing engaging mathematical investigations and effective assessment tasks for students with a variety of backgrounds and ability levels. Justin Robinson is Head of Positive Education and a mathematics teacher at Geelong Grammar School. Prior to this, he spent 20 years teaching mathematics and was a key writer of in-house maths material. He has a keen interest in engaging all students through a wide variety of effective teaching methods and materials.

Jenny Goodman has worked for 20 years in comprehensive State and selective high schools

in NSW and has a keen interest in teaching students of differing ability levels. She was awarded the Jones medal for education at Sydney University and the Bourke prize for Mathematics. She has written for Cambridge NSW and was involved in the Spectrum and Spectrum Gold series.

viii

Jennifer Vaughan has taught secondary mathematics for over 30 years in NSW, WA, Queensland and New Zealand and has tutored and lectured in mathematics at Queensland University of Technology. She is passionate about providing students of all ability levels with opportunities to understand and to have success in using mathematics. She has taught special needs students and has had extensive experience in developing resources that make mathematical concepts more accessible; hence, facilitating student confidence, achievement and an enjoyment of maths. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

Cambridge University Press

Introduction and guide to this book This resource has been developed from an analysis of the NSW Syllabus for the Australian Curriculum and the Australian Curriculum: Mathematics. It is structured on a detailed teaching program for the implementation of the NSW Syllabus, and a comprehensive copy of the teaching program can be found on the companion website. The chapters are based on a logical teaching and learning sequence for the syllabus topic concerned, so that chapter sections can be used as ready-prepared lessons. Exercises have questions graded by level of difficulty, as indicated in the teaching program, and are grouped according to the Working Mathematically components of the NSW Syllabus, as indicated by badges in the margin of the exercises. This facilitates the management of differentiated learning and reporting on students’ achievement. For certain topics the prerequisite knowledge has been given in sections marked as REVISION, whereas EXTENSION marks a few sections that go beyond the Syllabus. Similarly, the word FRINGE is used to mark a few topics treated in a way that lies at the edge of the Syllabus requirements, but which provide variety and stimulus. Apart from these, all topics are aligned exactly to the NSW Syllabus, as indicated at the start of each chapter and in the teaching program. In Stage 5, separate textbooks are provided for 5.1/5.2 and 5.1/5.2/5.3. In addition the NSW Syllabus allocates topics to certain pathways, which are designated 5.2◊, 5.3§ and 5.3#. These categories, together with Stage 4, 5.1, 5.2 and 5.3, are indicated for each chapter section by ‘ladder icons’ in the Year 9 and 10 textbooks and teaching programs.

Guide to this book

2

Chapter 1 Algebraic techniques 2 and indices

Number and Algebra

3

NSW Syllabus

Features:

for the Australian Curriculum Strand: Number and Algebra

Substrand: ALGEBRAIC TECHNIQUES

NSW Syllabus for the Australian Curriculum: strands, substrands and content outcomes for chapter (see teaching program for more detail)

Outcomes A student generalises number properties to operate with algebraic expressions.

1

Chapter

What you will learn: an overview of chapter contents

A student operates with positive-integer and zero indices of numerical bases.

Algebraic techniques 2 and indices

(MA4–9NA)

Avatar algebra Computer gaming is a billion-dollar industry that employs a range of computer specialists including game programmers. To create a virtual threedimensional world on a two-dimensional screen requires the continual variation of thousands of numbers (or coordinates); for example, if an avatar leaps up, the position of its shadow must be changed. In the two-dimensional image, the change in any one measurement results in many, many other measurement changes in order to produce a realistic image. It would be annoying for programmers if every time they changed one measurement they had to write a separate program instruction for hundreds of changes! If their avatar jumps into a doorway, the door’s dimensions, the light and shadows, the size and movement of enemies, the viewing angle etc. must all be recalculated. However, a programmer avoids such tedious work by using algebra. For example, an algebraic rule making a door height equal twice the avatar’s height can be written into a game’s program.

What you will learn 1A 1B 1C 1D 1E 1F 1G 1H 1I 1J 1K

Chapter introduction: use to set a context for students 4

The language of algebra REVISION Substitution and equivalence Adding and subtracting terms REVISION Multiplying and dividing terms REVISION Adding and subtracting algebraic fractions Multiplying and dividing algebraic fractions Expanding brackets Factorising expressions Applying algebra Index laws for multiplication and division The zero index and power of a power

EXTENSION EXTENSION

Algebraic rules linking many varying but related quantities are programmed into computer games. Other examples of related variables include how fast the avatar runs and how fast the background goes past, the avatar’s direction of movement and the route that its enemies follow. Computer-game programmers deal with many more complex and sophisticated issues of input and output. Avatars, programmed with artificial intelligence, can make decisions, react unpredictably, interact with the terrain and try to outwit their human enemy. Expertise in mathematics, physics, logic, problemsolving and attention to detail are all essential skills for the creation of realistic and exciting computer games.

Chapter 1 Algebraic techniques 2 and indices

Pre-test

Pre-test: establishes prior knowledge (also available as a printable worksheet)

(MA4–8NA)

1 Evaluate: a 8+4×6 c 12 − (6 + 2) + 8

b d

4×5−2×3 3(6 + 4)

2 Evaluate: a 10 + 6 − 12 c −3 − 8

b d

4−7 −1 − 1 − 1

b d

7 less than m half of w

3 Write an expression for: a 5 more than x c the product of x and y e double the sum of p and q

4 If y = 2x + 5, find the value of y when x = 1.2. Complete the tables using the given equations. Cambridge University Press © David Greenwood et al.5 2014 ISBN: 9781107671812 a M = 2A + 3 Photocopying is restricted under law and this material must not be transferred to another party

ix

Guide to this book (continued)

10

Chapter 1 Algebraic techniques 2 and indices

1A The language of algebra

Topic introduction: use to relate the topic to mathematics in the wider world

RE V I SI ON

A pronumeral is a letter that can represent one or more

1B Substitution and equivalence numbers. For instance, x could represent the number of

MA

T

Key ideas

Key ideas: summarises the knowledge and skills for the lesson Examples: solutions with explanations and descriptive titles to aid searches (digital versions also available for use with IWB)

c Write 7 a +another 2b + 5expression = 7( 3) + 2( 2with ( 4 ) +three 5 terms.

This soccer player scored x goals last year.

= 7b 21 51 Using 2 The expression 5a + ++ c8 −+ 3ab + 6 has terms. of algebra Example thefive language a State the constant term. the individual terms in the expression 4a + b − 12c + 5. = 34 a List ■ In algebra, a letter can be used to represent one or more numbers. These letters are called b ■State theexpressions coefficient of: Two are equivalent if they equal regardless of d. the number that is b In the expression 4a + b − 12chave + 5 state the values coefficients of a, b, c and pronumerals. A variable is a letter that is used to represent more than one number. i substituted a Whatpronumeral. is the constant The term 4a +of b −arithmetic 12c + 5? help to determine equivalence. forceach laws a in ■ a × b is written ab and a ÷ the b iscoeffi written d State cient of b.in the expression 3a + 4ab + 5b2 + 7b. ii b ■ The commutative laws of arithmetic b tell us that a + b = b + a and a × b = b × a for any values of 2 ■ a × a is written a . iii acand b. S O L U T IO N EX P L A N AT IO N ■ cAnWrite expression a combination offinumbers another is expression that has ve terms. and pronumerals combined with mathematical

Key ideas

Example 1c,d

The associative laws of arithmetic tell us that a + (b + c) = (a + b) + c and

Each part of an expression is a term. Terms get added There are four terms: 4a, b, 12c and 5. Chapter 1 Algebraic 2 and8indices 1A operations, e.g.techniques 3x +a 2yz and ÷ (3a − 2b) + 41 are expressions. Number and Algebra ■

8

MA

T

LLLLY Y

Y

LL

LL

L L L LY Y

MA

LL

Y

T

T

LL

MA

bordering rectangular pools that are 4 m in breadth and x m long.

Y

T AT M AM

Puzzles and challenges

T

MA

Chapter summary: mind map of key concepts & interconnections

Y

T

Puzzles and challenges

LL

MA

Investigations: inquiry-based activities

Y

T SIZE M

Questions are linked to examples

b 7 seasons of Proof by Induction 2 Semester reviews per book

4 The following six expressions could appear on either side of an equation. Using just two of the

5 For each of Algebraic the following expressions, state the coefficient of b. expressions, create an equation that has of no tiles solution. 6 a Use your algebraic rule to form equations for each of the following total numbers when Enrichment: alphabet 5 seasons Number and Algebra ac59 3a + 2b + cof both shows b 3a + b + 2c two rows of flat tiles are used for pool edging. 2x 3x + 1 7x + 4 4(x + 7) 2 + 3(x + 1) 2(3 + x) − 1 Chapter 1 Algebraic techniques 2 and indices Semester review 1 350 all+7 9b seasons if the naleach is−halved purchased in a sale expression terms, onefifor letter of+the alphabet. It starts i 96 tiles ii 120 tiles iii 136 tiles iv 248 tiles between 100 kg and 150 kg, but it allows more than16 An cd 4a + 2ccontains +ofdeach26show, dprice 3a 2b f when 5 A certain pair of scales only registers weights Multiple-choice b By manually solving each equation, determineone the person lengths that use thesequestions numbers to of getthe on pools at a time. aeA+plumber +2a9c+charges +4techniques 16d + a25e + call-out … b4b1:+Algebraic fthen2a + 5c Chapter 2$70 and indices 11 fee and $90 per hour. Write an expression for Algebraic terms Pronumeral: a letter which stands 2 Consider the expression − 3bthe + 8.fi Which one ofweighs the following true? a If three people weigh 1themselves in pairs5aand rst pair 117statements kg, theis second pair one or more numbers offortiles. Multiple-choice questions a What is the coefficient of f ? Concise form 5x 2y –2ab 2

Language

A The coefficient of a is 5.

Expanded form 5xxy –2abb

B

It has 5 terms.

weighs 120 kg and the third weighs kg, whatD are individual C Thepair constant term is 127 8. Thetheir coefficient of b is 3. weights?

Semester review 1

Chapter summary

58

9

subtracted) to make an expression. × (bof× the c) =following (a × b) ×worded c for any values of a and b. (or mathematical 3 Matchaeach statements with the correct expression. Numberand anddivision, Algebra 7 ■ A term is a part of an expression with only pronumerals, numbers, multiplication The coefficient is the number in front of a pronumeral. a The sum of x and3 xb7 The coefficient of a isA4. 3 − x e.g. 9a, 10cd and are terms. For b the coefficient is 1 because b is the same as 1 × b. Theall coeffi cient of b is 1. x b 3 less than x KINING G 5 The coefficient of c isB−12. Ois ORRKbeing For c the coeffi cient is −12 12 because this term 12. W W 12Write A mobile phone call costs 20of cents connection centsisper minute. an expression each the following. Example 2c,d 6■ 3fee and then A coeffi cient thefor number in front of a pronumeral. If the50term being subtracted, the coefficient UU FF Example 3 isby Substituting values subtracted. For d the coefficient is 0 because there The coefficient of d isC0. x − 3 caa x7Write is divided CC are an expression for ifthe total cost (in cents) a call t minutes. more than y 2 and b 3of less thanlasting xcoeffi RR PS is a negative number, is 1. For PS termscient with d. Exercise 1A REV Ithere S I O Nis no number in front, theno HHE RKIN A A OMA TTI CIGC b xThe Write anxof expression total cost (in dollars) ofproduct aexpressions. call lasting t minutes. Substitute =a3and and y for = 6the to evaluate the following sum dx isThe of 4 and WEM A dcthe is tripled D of 3x expression 3x + cyb −57z, the coefficient 3, the coeffi cient of term yp is is1 any andterm the that coeffi cient A constant does not contain U Fa 2 dollars) Write for the cost (in ofthird a calloflasting t hours. eac Half ofan q isexpression subtracted from 4total fx +One r is added to 10 5x b 5x + 2y x C expression 3a + 2b + 5c has three terms. Example 1a,b 1 The pronumeral. −7. eof xz is subtracted from 3 E R PS HE The the sumterms. of b and c multiplied by 2 h2 The sum of b and twice the value of c A ag List A T I C of ■ f A term that does notd contain any variables is+called a constant term. 7 Although there is a 4 in front of ab and a 5 in M front x is divided by 3 F x 7 L U Tproduct I O Ncoeffi E X P of L A2aNadded AT I O N The ofcient a, b and j A quarter tothese half isofa bterm containing just b,ORso bi S OState the of: c divided by 7 KINthey b , neither of G W ■ The sum of a and b is a + b. U F x and 2y l The difference of afalse. and half of b aaquotient should ignored. 13 kThe If ixThe isdifference positive of number, as true orbe C Remember that 5(3) is another way of writing 5 × 3. ■ of a andclassify b is a −the b. following statements a 5 x = 5 ( 3 ) I NG RORKPS ma iiThe of k andthan itself n The square of w H EW x isb product alwaysof smaller A ■ The product a and bexpressions: is 2a ××x.b. UM A T FI C =of15 4 For the following b iiieach x isc always smaller than x +expressions 2. Number and Algebra C 9 7■ Describe each ofof the following in words. R PS The quotient a and b is a ÷ b. 2 i x isstate howsmaller many termsxwith there are terms. ii list the terms HE c c Write another expression . three M AT I C A 2 than Replace aThe 3 +square xalways c 4all × bthe × cpronumerals by their values 2 b a+b ■ a x 22b + 2+ycof + xa =isless 5a(3.)than + 24(6−)x.+ (3) b 7a b 19y − 52x + 32 15−+ Example 2(4Creating from dd 2a + xb is always × 2fiexpressions f a 4description − 2b the order in which to evaluate remember expression 5a + 7b + c −e3ab +−6b)has ve terms. and Example 1c,d 2 The ce ax+−2b d 7u − 3v + 2a + 123c (9) + 12 number. +3 3 is always=a5positive RK I N WO RKING Write an expression for each of theand following. a mobile State the constant term.20 (multiplication 12 A phone call costs cents connection minute.addition). ef 10f f 9fee − 2b +then 4c +50 d +cents e per before WUO F G x ++x2be − 1 is always aThe positive = 45 12 sum + 3 ofnumber. a + 3 and k b The product of m and 7 UC F b Write State 2the coefficient of: for the cost (inh cents) a call lasting t minutes. yan+expression 4abc − 2nk ab +of 2bc + 3cd +The 4de 5 − x buys 8 agMarcela 7 plants from thetotal local nursery. R CPS c 5 is added to one half of k d sum of a and b is doubled = 6 0 Chapter 2 Equations 2 112 H R PS A 14b If Write number, the following statements true ort false. Give a brief reason. ib isaa negative E expression for classify the totalwrite cost (in dollars) of a callaslasting minutes. H EM A T I C A a If the an cost is $x for each plant, an M AT I C 5 cFor the following the coeffiofcient oflasting b. P L A Nt AT a Write b −b 4of must be negative. Chapter 2 Equations 2 iieach Ofor L Uexpressions, Tthe IO Ntotal EX IO N an expression coststate (in dollars) a call hours. expression for theStotal cost in dollars. ab iii 3a 2b + c b 3a + b + 2c b ++ 2 could be negative. c The word ‘sum’ means +. 3 + kis decreased by b If the cost of eachaplant 1 Find the unknown value in the following puzzles. cc 4a + 2c be +expression dpositive. that has five terms. d 3a − 2b + f b ×+ 29b could another a A number is increased by 2, then doubled, then increased by 3 and then tripled. The result is 99. c Write The word ‘product’ means ×. b m ×an7 or 7m $3 during a sale, write expression for: R K I NG WO ed bb++2a + 4 be negative. f 2a + 5c b must b A number is doubled and then one third of the number is subtracted. The result is 5 larger than Tiling a pool edge U F i aeach the new cost per plant in dollars 3 If Match of the following worded with the correct mathematical expression. Example 2a,b 13 1classify k statements 1 the original number. x is positive number, the following statements as true C g 7 − 54c + d h 5a − 6b + c c k + 5 or + 5 One or halffalse. of k can be written × k (because ‘of’ 15a What the difference between andA2(a3 +− 5)? an expression in words to describe each ofR PS 2in dollars 2 The Sunny Swimming Pool Company constructs rectangular pools each m will in breadth various c In five years’ time4Alf be twicewith as old as he was two years ago. How old is Alf now? iiis is the new sum of xtotal and cost 7 than x Give 2 HE ai xThe always smaller 2 × x.2a2 +of5the A k 4a − b + c + d j 2a + 4b − 12b M means ×), or because k is being divided by two. AT I C d The of aused shirt for is increased by 10% for GST and then decreased by 10% on a sale. The new lengths. There are non-slip square tiles, 50 cm by 50 cm, thatprice can be the external edging them and describe how the grouping symbols change the meaning. x 7 plants 2 b x is always smaller than x + 2. Bl 8a + c − 3b + d k 37aless − bthan +cx price is $44. What was the original price? around the pool perimeter where swimmers walk. 3 2 e One-third of a number is subtracted from 10 and then the result is tripled, giving the original . × her x is always than (a +xb) 2 or job. 2(a +She b)C x − 3 The values of a and b are being added and the result is 9 ccFrancine earnssmaller $p 2perd week for x is divided by 1 Draw a neat diagram illustrating the pool edge with one row back of flat tiles bordering the perimeter of number again. dworks 1 − for x is 48 always less than 4 −Write x. multiplied by 2. Grouping symbols (the brackets) are weeks each year. an a rectangular pool 4 m in breadth and 5 m long. is 3tripled D 3x 2 Consider the following ‘proof’ that 0 = 1. required to multiply the whole result by two and not Enrichment: Algebraic alphabet edexpression xx − is always positive number. for the aamount she earns: x just the value of b. 2 Develop a table showing the dimensions of rectangular pools 2 + 5each 2x = 3xwith + 5 breadth 4 m and ranging in fea xin+ − 1 is always a positive number. isaxsubtracted from 3 E fortnightcontains 26 terms, one for each2letter of the alphabet. It starts 16 An expression −5 for each pool when length from 5 m to 10 m. Add a column for the total−5 number of tiles required in4b year 2 = 3x 2x fbab+xis isaone divided bynumber, 3+ 25e +classify F x +statements 7 one row of flat tiles borders the outside edge of the pool. + 9c + 16d … 14 If negative the following as true or false. Give a brief reason. ÷ ÷x ÷ ÷x year her wage is is the of increased f? aca binWhat −one 4 must beifcoefficient negative. 3 Develop an algebraic rule for the total number of tiles, T, required 2 = 3 for bordering the perimeter of week after she has already What isper thebe coefficient of z? rectangular pools that are 4 m in breadth and x m long. −2 R K I NG b b bby + $20 2 could negative. −2 WO 0=1 30 weeks inhas the c bworked Which pronumeral ayear coefficient of 400? U F × 2 could be positive. 4 cFor each of the following expressions: 4 a Use your algebraic rule to form equations for each of the following total number of tiles when a C d bi One term is removed and now the coefficient of k is zero. What was the term? R PS d + b must be negative. single row of flat tiles is used for pool edging. a Which step caused the problem in this proof? (Hint: Consider the actual solution to the state manyDVDs terms of there areTV ii list the terms HE 10 Jon likes to how purchase some A equation.) e Another expression containing 26 terms starts a + 2b + 4c + 8d + 16e + … What is the sum of allM A T I C i 64 tiles ii 72 tiles iii 80 tiles iv 200 tiles ashows. 7a +One 2b +show, c b 19y − 52x + 32 Numbers, costs $a per b Prove that 0 = 1 is equivalent to the equation 22 = 50 by adding, subtracting, multiplying and15 What is the difference between 2a + 5 and 2(a + 5)? Give an expression in words to describe each of b By manually solving each equation, determine the lengths of the various pools that use each of the coefficients? cseason, a + 2b d 7u − 3v + 2a + 123c dividing both sides. another show, by Induction, them andand describe how the Proof grouping symbols change the meaning. the above numbers of tiles. ecosts 10f$b + per 2be season. Write an expression for f the 9 −cost 2b +of:4c + d + e 3 Find all the values of x that would make both these inequalities false. 5 Develop an algebraic rule for the total number of tiles, T,

Exercise questions categorised by the working mathematically components and enrichment (see next page)

Investigation

LL

Let’s start: an activity (which can often be done in groups) to start the lesson

Y

goals a particular soccer player scored last year. Or p could Onethe common thing to doofwith algebraic expressions is to replace the pronumerals (also known as represent price (in dollars) a book. If a pronumeral can variables)values with it numbers. This is referred to as substitution, or evaluation. In the expression 4 + x we can take different is also called a variable. substitute x = 3 to get the result 7. Two expressions are called equivalent if they always give the same result when a number is substituted. For example, 4 + x and x + 4 are equivalent, because no matter what Let’sthestart: valueAlgebra of x, 4 + xsort and x + 4 will be equal numbers. Consider the four expressions x + 2, x × 2, x − 2 and x ÷ 2. Number and Algebra 7 start: algebra • IfLet’s you know thatAFL x is 10, sort the four values In Australian Rules football, the final team score is given by 6x + y, where x is the number of goals and from lowest to highest. y is an theexample number of of abehinds • Give value ofscored. x that would • State make x × 2the lessscore than if x +x = 2. 3 and y = 4. Exercise 1A REVISION R K I NG • different If the score is 29, areif the • Try values of xwhat to see youvalues can: of x and y? Try to list all the possibilities. WO U F Ifexpression y =x ÷ 9 and thethan is2 has a two-digit number, what are the possible values of x? make 2 less C 3a +score 2b x+−5c three terms. Example 1a,b 1 – •The R PS HE – a make + terms. 2 less than x − 2 List xthe M AT I C A Statexthe cient xof: – b make × 2coeffi less than ÷2 ■i To a evaluate an expression or to substitute values means to replace each pronumeral in an with a number to obtain a final value. ii expression b 6 Chapter 1 Algebraic techniques 2 and indices iii For c example, if a = 3 and b = 4, then we can evaluate the expression 7a + 2b + 5:

HOTmaths icons: links to interactive online content via the topic number, 1A in this case (see next page for more)

110

5

Number and Algebra

7ppqqq q 7 Determine an algebraic rule for 7pthe total number ofb tiles, T, required for required forcientn ofrows flatand get weights of 108 kg, 118 kg and 130 kg, E The coeffi a2 isin 10.of If another three people weigh themselves pairs 4abb 4ab – – + – × ÷ sum difference product rectangular quotient tiles bordering pools 5cthat are5ccc4 m in breadth and m inindividual length. 2 Half the sum of double x and 3 can be written as: what arextheir weights? more than less than times divide 2 3 2 3

added increased

minus decreased

double (2×) twice (2×)

one third one half

1

2( x − 3)

2x + 3

2x + 6

× 2x + 3 B 6 D E c A group of four childrenAwho thanC50x +kg, weigh themselves in 2groups of three, 2 all weigh less 2 2

Adding and subtracting

Like terms

like terms triple (3×) quarter Variables have identical for each of the following pools, and then manually solve 8 Use this algebraic rule to form equations • Count ‘how many’ expanded form. 3 kg, If n +128 2 is ankg, odd 125 integer, next odd be written getting the weights 122 kgtheand 135integer kg. can How muchas:do they each weigh? • Don’t change variables. m 6aam ×÷ +4a –3 +9a –12a A n+3 B n+4 C n+5 D n+1 E n eachan expression equation to determine the–2a6alength of each–6apool. m –2aam not used in terms 2

2

2

-7xy

5x

5ma 2

algebraic expressions

–8

Pool

Breadth of pool 4 m

A

4

5 is the coefficient of x constant term is –8 –7 is the coefficient of xy Substitution ‘evaluate’

B

‘substitute’

C

4

Algebraic techniques 2 and indices

Length of pool x m

Distributive law a(b +c) = ab + ac 5(2a +m) = 5(2a) +5(m) =10a + 5m 7(k – 3a ) = 7(k ) – 7(3a) = 7k – 21a

a stays the same

Number of layers n

a 2 stays the same

5(2a –3) – 7(4 + a) = 5(2a) – 5(3) – 7(4) + –7(a) = 10a – 15 – 28 – 7a = 3a – 43

a(b –c) = ab – ac Factorising

Equivalent expressions x=2 x = 10

2

sign in front belongs to term = –6a – 12a + 4a 2 + 9a 2 – 3 = –18a + 13a 2 – 3

Expanding brackets 3(x+y) = x + y + x + y + x + y = 3x + 3y

4

replace variables with numbers and calculate answer 7a + 3(a + b) + 4b 2 4 a=8b=2 7 × 8 + 3 × (8 + 2) + 4 × 2 × 2 = 4 = 14 + 30 + 16 = 60

5maa = 5aam

6a 2m, –2a 2m and 5ma 2 are like terms ab = ba

12a 2m + 8am 2 = 12aam + 8amm HCF = 4am = 4am x 3a + 4am x 2m = 4am(3a + 2m)

12x + 6a (HCF = 6) = 6 × 2x + 6 × a = 6(2x + a)

7 − 3x 2 − 3x + 5 7−6=1 2−6+5=1 7 − 30 = –23 2 − 30 + 5 = –23 7 − 3x = 2 – 3x + 5

Index laws

Index notation

Examples × 35 = 37

4 Find the value of 5 − 4a2 given a = 2. A 3 numberB of −3tiles T C 21 Total

3

5 3 × x × y is equivalent to: 228 A 3x + y B xy

4

6

5

12ab can be simplifi 288 ed to: 24 a 2 2a A 2ab B 500 b

C 3+x+y

9 57 × 54 is equal to: A 2511 B 528

E 13 E xy + 2xy

3

4

−1

7

7

12

2

5 Which answer is NOT equivalent to (m × n) ÷ (p ( × q)? A

mn pq

B

m×

n pq

C

m n × p q

b d

the product of p and 3 the sum of x and y, divided by 2

D

mnq p

■

Short-answer questions

C

b 2a

7 The expanded form of 2x(3 + 5y) is: A 6x + 5y B 3x + 5y C 6x + 5xy 8 Simplifying 3a ÷ 6b gives: a A 2 B b

D −11 D 3x + 3y

Textbooks also include:

the gtotal 7 −cost 54cof+calling d h 5a − 6b + c a plumber out for x hours. 8 × 8 is the same as: bi A 8What of z? D 8 C 8 4a − bis+Bthe c64+coefficient d j 2a + 4b2 − 12b 2 4x + 5 + 3xx is the same as: ck A 7xWhich pronumeral has a coefficient of l400?8a + c − 3b + d 7a + 5 − b +B c12x 12x C 12 + x D 2x 2 + 12 ■ 3d 12mOne + 18 factorises to: is removed and now the coefficient of k is zero. What was the term? term A 2(6m – 9) B −6(2m – 3) C 6(3 – 2m) D 6(2m + 3) e Another expression containing 26 terms starts a + 2b + 4c + 8d + 16e + … What is the sum of all 4 5a + 5 – 4a – 4 – a – 1 equals: A a–1 B 0 C 2–a D a+1 ■ the coefficients? 1

C

2a b

C 253

D

ab 2

D 6 + 10y

D

ab 2

E

b 2

E 6x + 10xy

E

a 2b

D 53

E 511

D 6a(a – b)

E 3(a2 - 2ab)

1

Write an expression for: a the sum of p and q c half the square of m

2 If a = 6, b = 4 and c = −1, evaluate: a a+b+c b ab – c d 3a2 + 2b

Complete answers Index Using technology activities

c a(b2 – c)

e abc

f

3 Simplify each algebraic expression. a 4 × 6kk b a+a+a c e 3ab + 2 + 4ab f 7x + 9 – 6xx – 10 g

a×a×a 18xy ÷ 9xx

4 Simplify: 5 xy a 5

w w + 5 2

ab c d 7p 7 ÷ 14 h m + n – 3m + n

Chapter reviews with multiple-choice, short-answer and extended-response questions a n = a × a × a × .... × a

Algebraic fractions

n ‘lots’ of a

base

Adding and subtracting

Multiplying

3a + 7m = 3a x 5 + 7m x 2 2 x 5 5 x 2 2 5 = 15a + 14m 10 10 = 15a + 14m 10

‘unit’ is tenths

4 1 7a 12m × a 31 a a

=1

1

÷a

+a ×a

a

a2

–a Dividing

3a ÷ 2 4 3 = 3a × 3 4 2 = 9a 8

O

15x ÷ y 20x 1 = 15x y × 20x 3 1 = 15x × 1 y 420x1 =3 4y

a ÷ a = a × 1a = a a =1

2 x 3

1

÷x 3

+x 3 ×x 3 6 x x3 –x 3 O

2.5m ÷ 5n =

5m = 5m–n 5n

3. (5m) n = 5m

1.

32

5

2. 122 = 123 12 3. (23) 4 = 212

10 The factorised form of 3a2 − 6ab is: A 3a2(1 – 2b) B 3a(a – 2b) C 3a(a – b)

Short-answer questions

2a

= 28 m

1.

5m × 5n = 5m+n

1 State whether each of the following is true or false. a The constant term in the expression 5x + 7 is 5. b 16xy and 5yx are like terms. c The coefficient of d in the expression 6d 2 + 7d + 8abd + 3 is 7. d The highest common factor of 12abc and 16c is 2c. e The coefficient of xy in the expression 3x + 2y is 0. 2 For the expression 6xy + 2x − 4y2 + 3, state: a the coefficient of x b the constant term c the number of terms d the coefficient of xy

5 Simplify: m 5 a × 5 6

b

3x 2 x − 7 7

b

c

ab 1 ÷ 7 7

d

c

3a +

a 2

m n mn × ÷ 3 2 4

6 Expand, and simplify where necessary. a 6(2m – 3) b 10 + 2(m – 3)

c 5(A 5( + 2) + 4(A 4( – 1)

7 Factorise: a 18a – 12

c −8m2 – 16mn

b 6m2 + 6m

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Working Mathematically badges All exercises are divided into sections marked by Working Mathematically badges, such as this example: Fluency & Problem-solving & Reasoning & Understanding & Communicating Communicating Communicating Communicating

M AT I C A

R

HE

C

F PS

Y

Y

PS

R K I N G

LL

HE

C

U

T

M AT I C A

R

WO

F LL

PS

Y

F

R K I N G

MA

HE

C

U

T

M AT I C A

R

WO

MA

PS

R K I N G

LL

U

T

T

HE

C

WO

F MA

MA

R

R K I N G

Y

U

LL

WO

M AT I C A

The letters U (Understanding), F (Fluency), PS (Problem-solving), R (Reasoning) and C (Communication) are highlighted in colour to indicate which of these components apply mainly to the questions in that section. Naturally there is some overlap between the components.

Acknowledgments The authors would like to thank the following sources for permission to reproduce material: Cover: Shutterstock.com / ollyy Images: Photo by Andargor / Wikimedia Commons, p.178; © Commonwealth of Australia. Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Australia License, p.269(b); © Bidgee. Creative Commons Attribution-ShareAlike 3.0 Unported License, p.548(t); © Jastrow. Creative Commons Attribution 3.0 Unported , p.127; Photo by Mark A. Wilson, Department of Geology, The College of Wooster, p.363; © Mike Young. Creative Commons Attribution-ShareAlike 3.0 Unported License, p.329(b); 2013 Used under license from Shutterstock. com / Shawn Pecor, pp.5, 525 / Galina Barskaya, p.8 / Kotomiti, p.17 / Andresr, pp.18, 19, 242 / Vartanov Anatoly, p.22 / Flashgun, p.27, 451 / corepics , p.31 / Gl0ck, p.32 / Jacek Chabraszewski, p.36 / Sergielev, p.40 / Racheal Grazias, p.42 / Radu Razvan, pp.43, 68(b) / erwinova, p.44 / Dmitriy Bryndin, p.45 / Christophe Rolland, p.50 / Alin B., p.54 / Orla, p.55 / Robert Cumming, p.61 / Natural Sports, pp.62-63 / Roobcio, p.67 / Lasse Kristensen , p.68(t) / corepics, p.78 / Radist, p.79 / Stephen Coburn, p.83 / liubomir, p.84 / Slavo Valigursky, p.85 / mangostock, p.88 / Oleksiy Mark, p .89 / Committer, p.91 / Neil Balderson, p.94 / Giuseppe_R, p.95(t) / Lucky Business, p.95(b) / Yarek Gora, p.96 / kurhan, p.99(t) , 263(t) / Mark Snelson, p.99(b) / Vlue, p.100 / courtyardpix, p.103 / Gemenacom, p.104 / egd, p.105 / Neale Cousland, pp.108, 291, 509 , 559 / corgarashu, p.109 / Robert Naratham, p.111 / Xavier Pironet, p.116 / racorn, p.117 / saiko3, pp.118-119 / SF photo, p.121 / Jean Frooms, p (t) / Vasyl Helevachuk, p.125(b) / Bork, p.130 / Eder, p.132 / Leigh Prather, p.137(l) / BelleMedia, p.139 / Muellek Josef, p.142 / szefei , p.143 / Yuriy Kulyk, p.148 / Alex Kosev, p.149 / frog-traveller, p.154 / Dean Mitchell, p.155 / sarah johnson, p.156 / J. Helgason, p.159 / Rob , p.160 / Faraways, p.163(t) / Wire_ man, p.163(b) / Valerie Potapova, p.170 / Diego Cervo, p.177 / Jiri Foltyn, p.181 / John Teate, p. 182 / Phillip Minnis, p.187 / Pakhnyushcha, p.186 / s_oleg, p.190 / federicophoto, p.201 / schankz, pp.202-203 / Pete Nielsen, p.205 / Monkey Business Images, p.210 / Suzanne Tucker, p.209 / Muelleck Joseph, p.211 / Mitya, p.217(l) / Sophos

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Design, p.217(r) / Robyn Mackenzie, p.218(t) / Bogdan Shahanski, p.218(b) / Denys Kurbatov, p.220 / Adrian Britton, p.224 / Josek Bosak, p.226 / Brian A Jackson, p.228 / Chee-Onn Leong, p.230 / Maridav, p.231(t) / Ramona Heim, p.231(b) / James Cornejo, p.232 / Dan Collier, p.233 / Benis Arapovic, p.237 / Yellowj, p.238 / London Bikers, p.245 / c., p.246(t) / KamiGami, p.246(b) / magicphoto, p.247 / RTimages, p.251 / oliveromg, p.252 / Mike Flippo, p.253 / ImageryMajestic, p.254 / Ingvar Bjork, p.256(t) / iDesign, p.256(b) / gorillaimages, p.257(t) / Margo Harrison, p.257(b) / Most Photos, p.263(b) / Constantinex, p.263(c-l) / Globe Turner, p.263(c) / Anatoly Tiplyashin, p.263(c-r) / Anna Chelnokova, p. 264 / Lipsky, p.265(l) / Olga Popova, p.265(r) / Karen Givens, p.268(t) / Stephen Coburn, p.268(b) / Lauren Cameo, p.269(t) / evv, p.270 / OlegDoroshin, p.271 / photobac, p.272 / Tatiana Belova, p.273 / Tomasz Trojanowski, p.274 / benicce, p.276 / gielmichal, pp.284-285 / H. Brauer , p.287 / Glue Stock, p.289 / Deyan Georgiev, p.292(t) / aslysun, p.292(c) / Nazarenko Andril, p.292(b) / irabel8, p.295 / Dennis Donohue, p.296 / Bailey Image, p.297 / Bogdan Vasilescu, p.298 / Sven Mare, p.301(t) / Steve Cukrov, p.301(b) / Christopher Meder Photography, p.302 / spe, p. 305 / Like, p.306 / Oliveromg, p.308(b) / Ashley Whitworth, p.309, 513 / Appalachian Trail, p.310 / Idiz, p.312 / Christine F, p.313 / Vladimir Daragan, p .314 / Losevsky Pavel, p.315 / Polina Lobanova, p.318 / great_photos, p.319 / Dmitry Shironosov, p.320 / Erik E. Cardona, p.323(t) / IgorGolovniov, p.323(b) / Vaclav Volrab, pp.325, 601 / GoodMood Photo, p.329(t) / Laurin Rinder, p.330(t) / Jason Grower, p.330(c) / Elina Pasok, p.330(b) / Igor Stepovik, p.336 / Ljupco Smokovsky, p.337(t) / Tropinina Olga, p.337(b) / Ilia Torlin, p.339 / Ehrman Photographic, p.341 / Tania A, p.342 / Olaru Radian -Alexandru, p.349 / MQ Naufal, p.359 / StudioSmart, pp.360-361 / ssguy, p.367 / Poznyakov, p.369 / Igumnova Irina, p.371(t) / Khafizov Ivan, p.371 (b) / Tanor, p.375 / Andrei Merkulov, p.378 / Chris Howey, p.379 / TranceDrummer, p.384 / Christian Musat, p.386 / Frontpage, p.392 / jopelka, p.397(l ) / Ian 2010, p.397(r) / Alex Yeung, p.402(l) / Zelenskaya, p.402(r) / Lance Bellers, p.403 / Ragma Images, p.408(l) / Doroshin Oleg , p.408(r) / abutyrin, pp.416-417 / Goodlight, p.419 / Luis Stortini Taste of the Square, p.422 / Bojan, p.423 / Pressmaster, p.427 / djgis, p.432 / Elena Elisseeva, p.433 / Flockholl, p.434 / Hood, p.437 / agsandrew, p.443 / Timothy R. Nichols, p.452 / Moreno Soppelsa, p.462 / fritz16, p.465 / Daryl Marquardt, p.466 / Kalim, p.467 / Kharidehal Abhirama Ashwin, p.468 / Vladitto, p.469 / Chaikin, p.476 / slowfish, p.482 / Angelo, p.483 / Phant, pp.484-485 / doglikehorse, p.487 / PeterG, p.493 / Tom Reichner, p.497 / Terry Davis, p.498 / rooster photo, p.503 / Michele Perbellini, p.504 / Michael G. Mill, p.515 / Coprid, p.520 / Oleg Zabielin, p.521 / Anyka, p.526 / Gustavo Miguel, p.532 / Chernetskiy, p.535 / nmedia, pp.542-543 / wavebreakmedia, p.545 / Ulrich Mueller, p.547 / Monticello, p.548(c) / Miles1987, p.548(b) / Amy Myers, p.555 / jcjgphotography, p.556 / Cloudia Newland, p.557(l) / Can Balcioglu, p.557(r ) / Ronnie Howard, p.560 / WilleeCole, p.564 / Yegor Korzh, p.565 / Bertrand Benoit, p.566 / Albert H. Teich, p.567 / Eric Isselee, p.572 / Sportslibrary, p.557 (t) / Sinitar, p.577(b) / GG Pro Photo, p.579 / David Arts (graph) & Faith Kocylidir (camera), p.580 / Poznyakov, p.582 / carrot theater, p.585 / oleandra , p.584 / Yuri Arcurs, p.588 / sportgraphic, p.591 / Jeff Davies, p.593 / Palych1378, p.594 / afaizal, p.597 / Jinlide, p.602 / Tsian, p.603 / Anna Dickie, p.604 / Tomasz Parys, p.605 / EM Karuna, p.606 / velefante, p.608 / arteretum, p.609 / mama_mia, p.611 / Vacclav, p.613 / Kristina Postnikova, p.619 . . . . All curriculum material taken from NSW Mathematics 7–10 Syllabus © Board of Studies NSW for and on behalf of the Crown in right of the State of New South Wales, 2012. Every effort has been made to trace and acknowledge copyright. The publisher apologizes for any accidental infringement and welcomes information that would remedy this situation.

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2

Chapter 1 Algebraic techniques 2 and indices

1

Chapter

Algebraic techniques 2 and indices

What you will learn 1A 1B 1C 1D 1E 1F 1G 1H 1I 1J 1K

The language of algebra REVISION Substitution and equivalence Adding and subtracting terms REVISION Multiplying and dividing terms REVISION Adding and subtracting algebraic fractions Multiplying and dividing algebraic fractions Expanding brackets Factorising expressions Applying algebra Index laws for multiplication and division The zero index and power of a power

EXTENSION EXTENSION

Cambridge University Press

Number and Algebra

NSW Syllabus

for the Australian Curriculum Strand: Number and Algebra

Substrand: ALGEBRAIC TECHNIQUES

Outcomes A student generalises number properties to operate with algebraic expressions. (MA4–8NA) A student operates with positive-integer and zero indices of numerical bases. (MA4–9NA)

Avatar algebra Computer gaming is a billion-dollar industry that employs a range of computer specialists including game programmers. To create a virtual threedimensional world on a two-dimensional screen requires the continual variation of thousands of numbers (or coordinates); for example, if an avatar leaps up, the position of its shadow must be changed. In the two-dimensional image, the change in any one measurement results in many, many other measurement changes in order to produce a realistic image. It would be annoying for programmers if every time they changed one measurement they had to write a separate program instruction for hundreds of changes! If their avatar jumps into a doorway, the door’s dimensions, the light and shadows, the size and movement of enemies, the viewing angle etc. must all be recalculated. However, a programmer avoids such tedious work by using algebra. For example, an algebraic rule making a door height equal twice the avatar’s height can be written into a game’s program.

Algebraic rules linking many varying but related quantities are programmed into computer games. Other examples of related variables include how fast the avatar runs and how fast the background goes past, the avatar’s direction of movement and the route that its enemies follow. Computer-game programmers deal with many more complex and sophisticated issues of input and output. Avatars, programmed with artiﬁ cial intelligence, can make decisions, react unpredictably, interact with the terrain and try to outwit their human enemy. Expertise in mathematics, physics, logic, problemsolving and attention to detail are all essential skills for the creation of realistic and exciting computer games.

Cambridge University Press

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Chapter 1 Algebraic techniques 2 and indices

Pre-test

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1 Evaluate: a 8+4×6 c 12 − (6 + 2) + 8

b d

4×5−2×3 3(6 + 4)

2 Evaluate: a 10 + 6 − 12 c −3 − 8

b d

4−7 −1 − 1 − 1

b d

7 less than m half of w

3 Write an expression for: a 5 more than x c the product of x and y e double the sum of p and q

4 If y = 2x + 5, find the value of y when x = 1.2. 5 Complete the tables using the given equations. a M = 2A + 3 0

A

3

7

10

11

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b y=

1 (x + 1) 2

x

1

3

y

6 Substitute x = 6 and y = −2 into each expression and then evaluate. a x+y b xy c 3x − y d 2x + 3y 7 Write these numbers in expanded form. a 52 b 24

c

33

d (−8)3

8 Evaluate: a 53

c

(−4)2

d 33 − 22

b

6+6+6+6+6+6=

d

9×9×9×8×8=9

b 42 + 52

9 Complete: a 5+5+5+5=4× c 10 ÷ 5 =

5 e 117 × 21 = 117 × 20 + 117 ×

10 Write down the HCF (highest common factor) of: a 24 and 36 b 15 and 36

×6 ×8

c 48 and 96

Cambridge University Press

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Number and Algebra

1A The language of algebra

R E V I S I ON

A pronumeral is a letter that can represent one or more numbers. For instance, x could represent the number of goals a particular soccer player scored last year. Or p could represent the price (in dollars) of a book. If a pronumeral can take different values it is also called a variable.

Let’s start: Algebra sort Consider the four expressions x + 2, x × 2, x − 2 and x ÷ 2. • If you know that x is 10, sort the four values from lowest to highest. • Give an example of a value of x that would make x × 2 less than x + 2. • Try different values of x to see if you can: – make x ÷ 2 less than x − 2 – make x + 2 less than x − 2 – make x × 2 less than x ÷ 2

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In algebra, a letter can be used to represent one or more numbers. These letters are called pronumerals. A variable is a letter that is used to represent more than one number. a × b is written ab and a ÷ b is written a . b a × a is written a2. An expression is a combination of numbers and pronumerals combined with mathematical operations, e.g. 3x + 2yz and 8 ÷ (3a − 2b) + 41 are expressions. A term is a part of an expression with only pronumerals, numbers, multiplication and division, 3x e.g. 9a, 10cd and are all terms. 5 A coefficient is the number in front of a pronumeral. If the term is being subtracted, the coefficient is a negative number, and if there is no number in front, the coefficient is 1. For the expression 3x + y − 7z, the coefficient of x is 3, the coefficient of y is 1 and the coefficient of z is −7. A term that does not contain any variables is called a constant term. The sum of a and b is a + b. The difference of a and b is a − b. The product of a and b is a × b. The quotient of a and b is a ÷ b. The square of a is a2.

Cambridge University Press

Key ideas

This soccer player scored x goals last year. x x goals last year.

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Chapter 1 Algebraic techniques 2 and indices

Example 1 Using the language of algebra a b c d

List the individual terms in the expression 4a + b − 12c + 5. In the expression 4a + b − 12c + 5 state the coefficients of a, b, c and d. What is the constant term in 4a + b − 12c + 5? State the coefficient of b in the expression 3a + 4ab + 5b2 + 7b.

SOLUTION

EXPLANATION

a There are four terms: 4a, b, 12c and 5.

Each part of an expression is a term. Terms get added (or subtracted) to make an expression.

b The coefficient of a is 4. The coefficient of of b is 1. The coefficient of c is −12. 12. The coefficient of d is 0.

The coefficient is the number in front of a pronumeral. For b the coefficient is 1 because b is the same as 1 × b. For c the coefficient is −12 12 because this term is being subtracted. For d the coefficient is 0 because there are no terms with d.

c 5

A constant term is any term that does not contain a pronumeral.

d 7

Although there is a 4 in front of ab and a 5 in front of b2, neither of these is a term containing just b, so they should be ignored.

Example 2 Creating expressions from a description Write an expression for each of the following. a The sum of 3 and k b The product of m and 7 c 5 is added to one half of k d The sum of a and b is doubled SOLUTION

EXPLANATION

a 3+k

The word ‘sum’ means +.

b m × 7 or 7m

The word ‘product’ means ×.

c

1 k k + 5 or + 5 2 2

d (a + b) × 2 or 2(a + b)

1 One half of k can be written × k (because ‘of’ 2 means ×), or k because k is being divided by two. 2 The values of a and b are being added and the result is multiplied by 2. Grouping symbols (the brackets) are required to multiply the whole result by two and not just the value of b.

Cambridge University Press

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Number and Algebra

1 The expression 3a + 2b + 5c has three terms. a List the terms. b State the coefficient of: i a ii b iii c c Write another expression with three terms.

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2 The expression 5a + 7b + c − 3ab + 6 has five terms. a State the constant term. b State the coefficient of: i a ii b iii c c Write another expression that has five terms.

Example 2a,b

3 Match each of the following worded statements with the correct mathematical expression. a The sum of x and 7 A 3−x x b 3 less than x B 3 c x is divided by 2 C x−3 D

e x is subtracted from 3

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f

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x is divided by 3

3x x 2 x+7

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7 Describe each of the following expressions in words. a 3+x b a + b d 2a + b e (4 − b) × 2

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3 less than x The product of 4 and p One third of r is added to 10 The sum of b and twice the value of c A quarter of a added to half of b The difference of a and half of b The square of w

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6 Write an expression for each of the following. a 7 more than y b c The sum of a and b d e Half of q is subtracted from 4 f g The sum of b and c multiplied by 2 h i The product of a, b and c divided by 7 j k The quotient of x and 2y l m The product of k and itself n

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8 Marcela buys 7 plants from the local nursery. a If the cost is $x for each plant, write an expression for the total cost in dollars. b If the cost of each plant is decreased by $3 during a sale, write an expression for: i the new cost per plant in dollars ii the new total cost in dollars of the 7 plants

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9 Francine earns $p per week for her job. She works for 48 weeks each year. Write an expression for the amount she earns: a in a fortnight b in one year c in one year if her wage is increased by $20 per week after she has already worked 30 weeks in the year 10 Jon likes to purchase DVDs of some TV shows. One show, Numbers, costs $a per season, and another show, Proof by Induction, costs $b per season. Write an expression for the cost of: a 4 seasons of Numbers b 7 seasons of Proof by Induction c 5 seasons of both shows d all 7 seasons of each show, if the final price is halved when purchased in a sale 11 A plumber charges a $70 call-out fee and then $90 per hour. Write an expression for the total cost of calling a plumber out for x hours.

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Chapter 1 Algebraic techniques 2 and indices

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15 What is the difference between 2a + 5 and 2(a + 5)? Give an expression in words to describe each of them and describe how the grouping symbols change the meaning.

Enrichment: Algebraic alphabet 16 An expression contains 26 terms, one for each letter of the alphabet. It starts a + 4b + 9c + 16d + 25e + … a What is the coefficient of f ? b What is the coefficient of z? c Which pronumeral has a coefficient of 400? d One term is removed and now the coefficient of k is zero. What was the term? e Another expression containing 26 terms starts a + 2b + 4c + 8d + 16e + … What is the sum of all the coefficients?

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14 If b is a negative number, classify the following statements as true or false. Give a brief reason. a b − 4 must be negative. b b + 2 could be negative. c b × 2 could be positive. d b + b must be negative.

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13 If x is a positive number, classify the following statements as true or false. a x is always smaller than 2 × x. b x is always smaller than x + 2. c x is always smaller than x2. d 1 − x is always less than 4 − x. e x − 3 is always a positive number. f x + x − 1 is always a positive number.

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12 A mobile phone call costs 20 cents connection fee and then 50 cents per minute. a Write an expression for the total cost (in cents) of a call lasting t minutes. b Write an expression for the total cost (in dollars) of a call lasting t minutes. c Write an expression for the total cost (in dollars) of a call lasting t hours.

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Chapter 1 Algebraic techniques 2 and indices

1B Substitution and equivalence One common thing to do with algebraic expressions is to replace the pronumerals (also known as variables) with numbers. This is referred to as substitution, or evaluation. In the expression 4 + x we can substitute x = 3 to get the result 7. Two expressions are called equivalent if they always give the same result when a number is substituted. For example, 4 + x and x + 4 are equivalent, because no matter what the value of x, 4 + x and x + 4 will be equal numbers.

Let’s start: AFL algebra

Key ideas

In Australian Rules football, the final team score is given by 6x + y, where x is the number of goals and y is the number of behinds scored. • State the score if x = 3 and y = 4. • If the score is 29, what are the values of x and y? Try to list all the possibilities. • If y = 9 and the score is a two-digit number, what are the possible values of x?

■

To evaluate an expression or to substitute values means to replace each pronumeral in an expression with a number to obtain a final value. For example, if a = 3 and b = 4, then we can evaluate the expression 7a + 2b + 5: 7a + 2b + 5 = 7(3 77(3) (3)) + 22(4)) + 5 = 21 + 8 + 5

■

■

■

= 34 Two expressions are equivalent if they have equal values regardless of the number that is substituted for each pronumeral. The laws of arithmetic help to determine equivalence. The commutative laws of arithmetic tell us that a + b = b + a and a × b = b × a for any values of a and b. The associative laws of arithmetic tell us that a + (b + c) = (a + b) + c and a × (b × c) = (a × b) × c for any values of a and b.

Example 3 Substituting values Substitute x = 3 and y = 6 to evaluate the following expressions. a 5x b 5x2 + 2y + x SOLUTION

EXPLANATION

a

5 x = 5( 3) = 15

Remember that 5(3) is another way of writing 5 × 3.

b

5 x 2 + 2 y + x = 5(3)2 + 2(6) + (3) = 5(9) + 12 + 3 = 45 + 12 + 3

Replace all the pronumerals by their values and remember the order in which to evaluate (multiplication before addition).

= 60 © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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11

Number and Algebra

Example 4 Deciding if expressions are equivalent a Are x − 3 and 3 − x equivalent expressions? b Are a + b and b + 2a − a equivalent expressions?

a No.

The two expressions are equal if x = 3 (both equal zero). But if x = 7 then x − 3 = 4 and 3 − x = −4. Because they are not equal for every single value of x, they are not equivalent.

b Yes.

Regardless of the values of a and b substituted, the two expressions are equal. It is not possible to check every single number but we can check a few to be reasonably sure they are equivalent. For instance, if a = 3 and b = 5, then a + b = 8 and b + 2a − a = 8. If a = 17 and b = −2 then a + b = 15 and b + 2a − a = 15.

Exercise 1B

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3 What is the value of a + 2b if a and b both equal 10? 4 a State the value of 4 + 2x if x = 5. b State the value of 40 − 2x if x = 5. c Are 4 + 2x and 40 − 2x equivalent expressions?

6 Substitute a = 4 and b = −3 into each of the following. a 5a + 4 b 3b d ab − 4 + b e 2 × (3a + 2b) g 12 + 6 a b j a2 + b

h ab + b 3 k 5 × (b + 6)2

c a+b f 100 − (10a + 10b) i l

100 a+b a − 4b

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Example 3b

5 Substitute the following values of x into the expression 7x + 2. a 4 b 5 c 2 d 8 e 0 f −6 g −9 h −3

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x = 1 and y = −10 x = −10 and y = −6 x = −2 and y = 9

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8 Evaluate the expression 4ab − 2b + 6c when: a a = 4 and b = 3 and c = 9 b a = −8 and b = −2 and c = 9 c a = −1 and b = −8 and c = −4 d a = 9 and b = −2 and c = 5 e a = −8 and b = −3 and c = 5 f a = −1 and b = −3 and c = 6 9 For the following state whether they are equivalent (E) or not (N). a x + y and y + x b 3 × x and x + 2x c 4a + b and 4b + a d 7 − x and 4 − x + 3 e 4(a + b) and 4a + b f 4 + 2x and 2 + 4x g 1 × a and a h 3 + 6y and 3(2y + 1) 2 2 10 For each of the following, two of the three expressions are equivalent. State the odd one out. a 4x, 3 + x and 3x + x b 2 − a, a − 2 and a + 1 − 3 c 5t − 2t, 2t + t and 4t − 2t d 8u − 3, 3u − 8 and 3u − 3 + 5u 11 Copy and complete the following table. 1

2

3

x + 3

4

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6

4

5

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100

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12 Give three expressions that are equivalent to 2x + 4y + 5.

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4x + 2

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4 – 3x

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2x - 4

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0.25 6

−2 8

14 a Evaluate x2 for the following values of x. Recall (−3)2 = −3 × −3. i 3 ii −3 iii 7 2 2 b Given that 25 = 625, what is the value of (−25) ? c Explain why the square of a number is the square of its negative. d Why is (−3)2 = 9 but −32 = −9? What effect do the brackets have?

iv −7

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15 a Explain with an example why a ÷ (b × c) is not equivalent to (a ÷ b) × c. b Does this contradict the associative law (see Key ideas)? Justify your answer. c Is a ÷ (b ÷ c) equivalent to (a ÷ b) ÷ c? Why or why not?

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16 a Is it possible to substitute values of x and y so that x + y and x + 2y are equal? Try to describe all possible solutions. b Does this imply that x + y and x + 2y are equivalent? Give reasons. 17 a

By substituting a range of numbers for a and b, determine whether (ab)2 is equivalent to a2b2.

b Is (a + b)2 equivalent to a2 + b2? Why/why not? a × b ? Why/why not?

c Is

ab equivalent to

d Is

a + b equivalent to

a + b ? Why/why not?

e For pairs of expressions in a−d that are not equivalent, find a few values for a and b that make them equal. 18 Sometimes when two expressions are equivalent you can explain why they are equivalent. For example, x + y is equivalent to y + x because ‘the order in which you add numbers does not matter’, or ‘because addition is commutative’. For each of the following pairs of expressions, try to describe why they are equivalent. a x × y and y × x b x + x and 2x 1 c y − y and 0 d × x and x ÷ 2 2 e a × 3a and 3a2 f k2 and (−k)2

Enrichment: Missing values 19 Find the missing values in the table below. a

5

b

2

a+b

−20

8 1 10

10

a + 2b a–b

17 1

−19

7 0

11 13

a – 2b

−29

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Chapter 1 Algebraic techniques 2 and indices

1C Adding and subtracting terms

R E V I S I ON

Recall from Year 7 that an expression such as 3x + 5x can be simplified to 8x, but an expression such as 3x + 5y cannot be simplified. The reason is that 3x and 5x are like terms — they have exactly the same pronumerals. The terms 3x and 5y are not like terms. Also, 4ab and 7ba are like terms because ab and ba are equivalent, as multiplication is commutative. However, a2b, ab and ab2 are all unlike terms, since a2b means a × a × b, which is different from a × b and a × b × b.

Let’s start: Like terms

Key ideas

• Put these terms into groups of like terms. 4a 5b 2ab 3ba 2a 7b2 5a2b 9aba • What is the sum of each group? • Ephraim groups 5a2b and 2ab as like terms, so he simplifies 5a2b + 2ab to 7ab. How could you demonstrate to him that 5a2b + 2ab is not equivalent to 7ab?

Like terms contain exactly the same pronumerals with the same powers; the pronumerals do not need to be in the same order, e.g. 4ab and 7ba are like terms. ■ Like terms can be combined when they are added or subtracted to simplify an expression, e.g. 3xy + 5xy = 8xy. ■ A subtraction sign stays in front of a term even when it is moved. − sign stays with following term e.g. 3x + 7y − 2x + x 3y + x − 4y = 3x − 2x + x + 7y + 3y − 4y = 2x + 6y ■

Example 5 Identifying like terms a Are 5abc and −8abc like terms? b Are 12xy 12 2 and 4y2x like terms? c Are 3ab2 and 7a2b like terms? SOLUTION

EXPLANATION

a Yes.

Both terms have exactly the same pronumerals: a, b and c.

b Yes.

When written out in full 12 12xy2 is 12 × x × y × y and 4y2x is 4 × y × y × x. Both terms include one x and two occurrences of y being multiplied, and the order of multiplication does not matter.

c No.

3ab2 = 3 × a × b × b and 7a2b = 7 × a × a × b.. They are not like terms because the first includes only one a and the second includes two.

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Number and Algebra

Example 6 Simplifying by combining like terms Simplify the following by combining like terms. a 7t + 2t − 3t b 4x + 3y + 2x 2 + 7y

c

7ac + 3b − 2ca + 4b − 5b

SOLUTION

EXPLANATION

a 7t + 2t − 3t = 6t

These are like terms, so they can be combined: 7 + 2 − 3 = 6.

b 4x + 3y + 2x 2 + 7y = 4x + 2x 2 + 3y + 7y = 6x + 10y

Move the like terms next to each other.

c 7ac + 3b − 2ca + 4b − 5b = 7ac − 2ca + 3b + 4b − 5b = 5ac + 2b

Move the like terms together. Recall that the subtraction sign stays in front of 2ca even when it is moved. 7 − 2 = 5 and 3 + 4 − 5 = 2

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Combine the pairs of like terms.

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2 a If x = 3 and y = 4, evaluate 5x + 2y. b If x = 3 and y = 4, evaluate 7xy. c 5x + 2y is equivalent to 7xy. True or false? 3 a Substitute x = 4 into the expression 10x − 5x + 2x. b Substitute x = 4 into: i 3x ii 5x iii 7x c Which one of the expressions in part b is equivalent to 10x − 5x + 2x?

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4 Classify the following pairs as like terms (L) or not like terms (N). a 3a and 5a b 7x and −12x c 2y and 7y d 4a and −3b e 7xy and 3y f 12ab and 4ba g 3cd and −8c h 2x and 4xy

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7 Simplify the following by combining like terms. a 7f + 2f + 8 + 4 b 10x + 3x + 5y + 3y c 2a + 5a + 13b − 2b d 10a + 5b + 3a + 4b e 10 + 5x + 2 + 7x f 10a + 3 + 4b − 2a − b g 10x + 31y − y + 4x h 11a + 4 − 2a + 12a i 7x2y + 5x + 10yx2 j 12xy − 3yx + 5xy − yx 2 2 k −4x + 3x l −2a + 4b − 7ab + 4a m 10 + 7q − 3r + 2q − r n 11b − 3b2 + 5b2 − 2b

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6 Simplify the following by combining like terms. a 3x + 2x b 7a + 12a c 15x − 6x d 4xy + 3xy e 16uv − 3uv f 10ab + 4ba g 11ab − 5ba + ab h 3k + 15k – 2k i 15k – 2k – 3k

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5 Classify the following pairs as like terms (L) or not like terms (N). a −3x2y and 5x2y b 12ab2 and 10b2a 2 2 c 2ab and 10ba d 7qrs and −10rqs 2 e 11q r and 10rq f −15ab2c and −10cba2

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8 For each expression choose an equivalent expression from the options listed. a 7x + 2x A 10y + 3x b 12y + 3x − 2y B 9xy c 3x + 3y C 9x d 8y − 2x + 6y − x D 3y + 3x e 4xy + 5yx E 14y − 3x

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9 Write expressions for the perimeters of the following shapes in simplest form. a b c 7x

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10 Towels cost $c each at a shop. a John buys 3 towels, Mary buys 6 towels and Naomi buys 4 towels. Write a fully simplified expression for the total amount spent on towels. b On another occasion, Chris buys n towels, David buys twice as many as Chris and Edward buys 3 times as many as David. Write a simplified expression for the total amount they spent on towels.

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11 State the missing numbers to make the following equivalences true. a 10x + 6y − x+ y = 3x + 8y b 5a − 7b + c

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a2b+

a+ d+

b = 11a = 4c + 2d + 1 + 3c + 7d + 4

b2a + 2a2b + b2a = 7b2a + 10a2b

12 Add the missing expressions to the puzzle to make all six equations true. 5x

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= +

+

=

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= +

7x

= =

7x + 3y

13 In how many ways could the blanks below be filled if all the coefficients must be positive integers? a+ b+ a = 10a + 7b 14 Simplify a − 2a + 3a − 4a + 5a − 6a + … + 99a − 100a.

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15 Prove that 10x + 5y + 7x + 2y is equivalent to 20x − 3x + 10y − 3y. Hint: Simplify both expressions.

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Chapter 1 Algebraic techniques 2 and indices

Enrichment: Missing expressions 18 a

Fill in the missing expressions to make all eight equations true. 5a

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+2b

+ +

+

+

4+a

=

+ +

3a + 8

= −3a

+ =

= +

10a + 3 +

+

= +

=

+

+ +

4a

1 =

=

10a + 2b + 8

b Design your own ‘missing values’ puzzle like the one above. It should only have one possible solution.

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1D Multiplying and dividing terms

R E V I S I ON

Recall that a term written as 4ab is shorthand for 4 × a × b. Observing this helps to see how we can multiply terms. 4 ab × 3c 3c = 4 × a × b × 3 × c = 4 × 3× a ×b × c = 12 abc Division is written as a fraction so 12 ab means (12ab) ÷ (9ad). To simplify a division we look for 9 ad common factors: 4 3

12 × a × b 4 b = 9 × a × d 3d

a ÷ a = 1 for any value of a except 0 a cancels to 1. so a

Let’s start: Multiple ways

■ ■ ■ ■

12abc means 12 × a × b × c. When multiplying, the order is not important: 2 × a × 4 × b = 2 × 4 × a × b. x2 means x × x and x3 means x × x × x. When dividing, cancel any common factors. For example:

3

15 x y

4

20 y z

=

3x4z

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Key ideas

Multiplying 4a × 6b × c gives you 24abc. • In how many ways can positive integers fill the blanks in a× b× c = 24abc? • In how many other ways can you multiply three terms to get 24abc? For example, 12ab × 2 × c. You should assume the coefficients are all integers.

Chapter 1 Algebraic techniques 2 and indices

Example 7 Multiplying and dividing terms a Simplify 7a × 2bc × 3d. b Simplify 3xy × 5xz. c Simplify 10 ab . 15bc d Simplify

18 x 2 and . 8xz

SOLUTION

EXPLANATION

a 7a × 2bc × 3d = 7×a×2×b×c×3×d = 7×2×3×a×b×c×d

Write the expression with multiplication signs and bring the numbers to the front.

= 42abcd

Simplify: 7 × 2 × 3 = 42 and a × b × c × d = abcd

b 3xy × 5xz

d

Write the expression with multiplication signs and bring the numbers to the front.

= 15x2yz

Simplify, remembering that x × x = x2.

10 ab 2 10 × a × b = 15bc 3 15 × b × c 2a = 3c

Write the numerator and denominator in full, with multiplication signs. Cancel any common factors and remove the multiplication signs.

18 x 2 y 9 18 × x × x × y = 4 8 xz 8 × x ×z 9 xy = 4z

Write the numerator and denominator in full, remembering that x2 is x × x. Cancel any common factors and remove the multiplication signs.

Exercise 1D

REVISION

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12 8

D3a2b

d

15 25

3 Which one of these is equivalent to a × b × a × b × b? A 5ab B a2b3 C a3b2

D (ab)5

4 Write these without multiplication signs. a 3×x×y b 5×a×b×c c

d 4×a×c×c×c

12 × a × b × b

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1 Which is the correct way to write 3 × a × b × b? A 3ab B 3ab2 C ab3

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6 Simplify the following. a 8ab × 3c d 5d × 2d × e g 4xy × 2xz j −3xz × (− 2z)

b e h k

c 3 × 12x f 4a × 6de × 2b

a×a 7x × 2y × x 4abc × 2abd −5xy × 2yz

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b 5a × 2b e 3a × 10bc × 2d

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5 Simplify the following. a 7d × 9 d 4a × 2b × cd

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3d x d 5xy x 2x 9ab x 2a2 4xy2 x 4y

7 Write the following terms as a product of prime numbers and pronumerals. a 18xy b 12abc c 15x2 d 51ab2 8 Simplify the following divisions by cancelling any common factors. 5a 10 a

b

7x14 y.l

c

e

7 xyz 21yz

f

2 12x

g

i

−4 a 2 8ab

j

21 p − 3q

k

10 xy 12 y −5 x 10 yz

2

− 21 p −3 p

d

ab 4b

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12 and 2 −18 and

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4x

d

e

9x

f

a

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9 9y

4 4y

x

2x y

3y 10 Fill in the missing terms to make the following equivalences true. a 3x × d

4 years

× z = 6xyz = 7s

b 4a × e

2ab

= 12ab2 = 4b

c −2q × f

14 xy

× 4s = 16qs = −2y

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9 Write a simplified expression for the area of the following shapes. Recall that rectangle area = length × breadth. a b c 2 2y 4b 6x

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14 Simplify the following expressions, remembering that you can combine like terms when adding or subtracting. a

2ab × 3bc 3bc × 4cd 4a × 3bc × 2d

b

12 a 2 b + 4 a 2 b 4 b + 2b

c

7 x 2 y − 55yx 2 12 xy

d

8 a 2 b + ( 44aa × 2ba ) 3ba − 22ba

e

10 abc + 5 cba + 5 a × bc 4 c × 1100 ab

f

10 x 2 y − ( 4 x × 6 xxyy ) 7 xy 2

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12 A square has a side length of x cm. a State its area in terms of x. b State its perimeter in terms of x. c Prove that its area divided by its perimeter is equal to a quarter of its side length. 2a 2 2 13 Joanne claims that the following three expressions are equivalent: , × a, . 5 5 5a a Is she right? Try different values of a. b Which two expressions are equivalent? c There are two values of a that make all three expressions equal. What are they?

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1E Adding and subtracting algebraic fractions

E X T E N S I ON

An algebraic fraction is a fraction that could include any algebraic expression in the numerator or the denominator. 2 3

5 7

20 3

2x5

2x − 4y 7 a + 9b

3 7a + 4

Algebraic fractions

Fractions

The rules for working with algebraic fractions are the same as the rules for normal fractions. For example, two fractions with the same denominator can be added or subtracted easily. Normal fractions

Algebraic fractions 5x

2 7 9 + = 13 13 13 8 11

−

2 11

=

13

6

5x

11

11

−

3x

+

13

2y11

=

=

8x13

5x − 2 y 11

If two fractions do not have the same denominator, they must be converted to have the lowest common denominator (LCD) before adding or subtracting. Normal fractions

Algebraic fractions

2 1 10 3 + = + 3 5 15 15 13 = 15

2a 3

+

b 5

= =

10a

+

3b

15 15 10a + 3b 15

Let’s start: Adding thirds and halves x x x 5x + . Dallas gets and Casey gets . 3 2 5 6 • Which of the two students has the correct answer? You could try substituting different numbers for x. Dallas and Casey attempt to simplify

• How can you prove that the other student is incorrect? x x • What do you think + is equivalent to? Compare your answers with others in the class. 3 4

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Key ideas

Chapter 1 Algebraic techniques 2 and indices

■

■

■

An algebraic fraction is a fraction with an algebraic expression as the numerator or the denominator. The lowest common denominator (or LCD) of two algebraic fractions is the smallest number that both denominators are factors of. Adding and subtracting algebraic fractions requires that they both have the same denominator, e.g. 2 x + 4 y = 2 x + 4 y . 5 5 5

Example 8 Working with denominators 3x 2y and . 10 15 2x b Convert to an equivalent algebraic fraction with the denominator 21. 7 a Find the lowest common denominator of

SOLUTION

EXPLANATION

a 30

The multiples of 10 are 10, 20, 30, 40, 50, 60 etc. The multiples of 15 are 15, 30, 45, 60, 75, 90 etc. The smallest number in both lists is 30.

b

2x 3 × 2x = 7 3× 7 =

Multiply the numerator and denominator by 3, so that the denominator is 21.

6 x 21

Simplify the numerator: 3 × 2x 2x is 6x.

Example 9 Adding and subtracting algebraic fractions Simplify the following expressions. a

3x 5 x + 11 11

b

SOLUTION a

4 a 2a + 3 5

c

6 k 3k − 5 10

d

a b − 6 9

EXPLANATION

3x 5x 3x + 5x + = 11 11 11 =

8x11

The two fractions have the same denominator, so the two numerators are added. 3xx and 5x are like terms, so they are combined to 8x.

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Number and Algebra

4 a 2a 5 × 4 a 3 × 2a + = + 3 5 15 15

The LCD = 15, so both fractions are converted to have 15 as the denominator.

=

20 a 6a + 15 15

Simplify the numerators.

=

26a 15

Combine: 20a + 6a is 26a.

6k 3k 12k 3k − = − 5 10 10 10

LCD = 10, so convert the first fraction (multiplying numerator and denominator by 2).

=

12k − 3k 10

Combine the numerators.

=

9k 10

Simplify: 12k − 3k = 9k.

a b 3a 2b − = − 6 9 18 18

d

=

LCD = 18, so convert both fractions to have 18 as a denominator.

3a − 2b 18

Combine the numerators. Note this cannot be further simplified since 3a and 2b are not like terms.

Exercise 1E

EXTENSION

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a

1 1 + 4 3

b

2 1 + 7 5

c

1 1 + 10 5

d

2 1 − 5 4

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1 Complete these sentences. 2 a For the fraction the numerator is 2 and the denominator is _____. 7 4 b For the numerator is _____ and the denominator is _____. 9 12 x c The expression is an example of an __________ fraction. 5 5x + 3 d The denominator of is _____. 7 2 Find the lowest common denominator (LCD) of the following pairs of fractions. 1 2 1 1 3 5 2 1 a and b and c and d and 3 5 4 5 7 6 3 6

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x = 5 10

b

2a = 7 21

Example 9d

x y and 4 5

d

x y and 12 6

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4z = 5 20

d

3k = 10 50

7 Simplify the following sums. a

x 2x + 4 4

b

5a 2a + 3 3

c

2b b + 5 5

d

4k k + 3 3

e

a a + 2 3

f

a a + 4 5

g

p p + 2 5

h

q q + 4 2

k

7p 2p + 6 5

l

x 3x + 4 8

2k 3k 2m 2m + j + 5 7 5 3 8 Simplify the following differences. i

Example 9c

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6 Copy and complete the following, to make each equation true. a

Example 9a,b

b

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3y y − 5 5

b

7p 2p − 13 13

c

10 r 2 r − 7 7

d

8q 2q − 5 5

e

p p − 2 3

f

2t t − 5 3

g

9u and − 11 2

h

8y 5y − 3 6

i

r r − 3 2

j

6u 7u − 7 6

k

9u 3u − 1 4

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5p 7p − 12 11

9 Simplify the following expressions, giving your final answer as an algebraic fraction. Hint: 4x is the 4x same as . 1 x x a a 4x + b 3x + c + 2a 3 2 5 d

8p − 2p 3

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2t +

7p2

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10u 3v + 3 10

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7y 2x − 10 5

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x −y 3

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5−

2x 7 WO

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1 10 Cedric earns an unknown amount $x every week. He spends of his income on rent 3 1 and on groceries. 4 a Write an algebraic fraction for the amount of money he spends on rent. b Write an algebraic fraction for the amount of money he spends on groceries. c Write a simplified algebraic fraction for the total amount of money he spends on rent and groceries.

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11 Egan fills the bathtub so it is a quarter full and then adds half a bucket of water. A full bathtub can contain T litres and a bucket contains B litres. a Write the total amount of water in the bathtub as the sum of two algebraic fractions. b Simplify the expression in part a to get a single algebraic fraction. c If a full bathtub contains 1000 litres and the bucket contains 2 litres, how many litres of water are in the bathtub?

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12 Afshin’s bank account is halved in value and then $20 is removed. If it initially had $A in it, write an algebraic fraction for the amount left.

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x x 2x + is not equivalent to . 4 5 9 x x x c Is + equivalent to x − ? Explain why or why not. 2 5 3 14 a Simplify: b Demonstrate that

i

x x − 2 3

ii

x x − 3 4

iii x − x 4 5

iv

x x − 5 6

b What patterns did you notice in the above results? c Write a difference of two algebraic fractions that simplifies to

x . 110

Enrichment: Equivalent sums and differences 15 For each of the following expressions, find a single equivalent algebraic fraction. a

z z z + + 4 3 12

b

2x x x + − 5 2 5

c

7u 3u 5u + − 2 4 8

d

8k k 5k + − 3 6 12

e

p p + −3 4 2

f

u u u + + 3 4 5

g

5j j − +2 12 3

h

7t t 2r − + 5 3 15

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x x 5x Demonstrate that + is equivalent to by substituting at least three different 2 3 6 values for x.

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Chapter 1 Algebraic techniques 2 and indices

1F Multiplying and dividing algebraic fractions As with fractions, it is generally easier to multiply and divide algebraic fractions than it is to add or subtract them. 3 2 6 × = 5 7 35

3×2 5×7

4 x 2 y 8 xy × = 7 11 77

Fractions

Algebraic fractions

Dividing is done by multiplying by the reciprocal of the second fraction. 4 1 4 3 ÷ = × 5 3 5 1 12 = 5

2 x 3y 2 x 7 ÷ = × 5 7 5 3y 14 x = 15 y

Fractions

Algebraic fractions

Let’s start: Always the same One of these four expressions always gives the same answer, no matter what the value of x is. x x + 2 3

x x − 2 3

x x × 2 3

x x ÷ 2 3

Key ideas

• Which of the four expressions always has the same value? • Can you explain why this is the case? • Try to find an expression involving two algebraic fractions that is equivalent to 3 . 8

■

To multiply two algebraic fractions, multiply the numerators and the denominators separately. Then cancel any common factors in the numerator and the denominator. 2 x 10 y 4 20 xy × = 3 5 3 15 =

■

4 xy 3

The reciprocal of an algebraic fraction is formed by swapping the numerator and denominator. 2a 3b 2a 4 ÷ = × 5 4 5 3b 8a = 15b

■

The reciprocal of

4 3b is . 3b 4

To divide algebraic fractions, take the reciprocal of the second fraction and then multiply.

Cambridge University Press

E X T E N S I ON

Number and Algebra

Example 10 Multiplying algebraic fractions Simplify the following products. a

2 a 3b × 5 7

b

SOLUTION a

4 x 3y × 15 2 EXPLANATION 2a × 3b = 6ab and 5 × 7 = 35.

2 a 3b 6 ab × = 5 7 35

b Method 1: 4x × 3y = 12xy 12 15 × 2 = 30

4 x 3y 12 xy × = 15 2 30 =

2 xy 5

Divide by a common factor of 6 to simplify.

Method 2: 2 5

First divide by any common factors in the numerators and denominators: 4xx and 2 have a common factor of 2. Also 3y and 15 have a common factor of 3.

4 x 1 3 y 2 xy × 1 = 5 15 2

Example 11 Dividing algebraic fractions Simplify the following divisions. a

3a b ÷ 8 5

b

SOLUTION a

EXPLANATION

3a b 3a 5 ÷ = × 8 5 8 b =

b

u 15 p ÷ 4 2

15a 8b

u 15 p u 2 ÷ = × 4 2 4 15 p

Take the reciprocal of

b 5 , which is . 5 b

Multiply as before: 3a × 5 = 15a, 8 × b = 8b.

Take the reciprocal of

2 15 p , which is . 15 p 2

=

2 and 60 p

Multiply as before: u × 2 = 2u and 4 × 15p 15 = 60p 60 .

=

u 30 p

Cancel the common factor of 2.

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EXTENSION

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3 5 15 × = 4 8 2 4 2 Which one of the following shows the correct first step in calculating ÷ ? 3 5 2 4 3 5 2 4 2 3 2 4 2 5 2 4 3 4 A B C D ÷ = × ÷ = × ÷ = × ÷ = × 3 5 2 4 3 5 4 5 3 5 3 4 3 5 2 5 c

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3 Calculate the following by hand, remembering to simplify your result. b f

1 10 × 5 11 1 2 ÷ 4 3

c g

1 6 × 4 11 4 1 ÷ 17 3

d h

5 6 × 12 10 2 1 ÷ 9 2 WO

d

c f

2bc 5bc

2 4a × 3 5 3a 7a × 2 5

Example 10b

6 Simplify the following products, remembering to cancel any common factors. 6x 7y 2b 7d 8a 3b a × b × c × 5 6 5 6 5 4c 4 3k 9d 4 e 3x 1 × d × e × f 9k 2 2 7 2 6x

Example 11

7 Simplify the following divisions, cancelling any common factors. 3a 1 2x 3 9a 1 a ÷ b ÷ c ÷ d 4 5 5 7 10 4 4 2y 1 2 4a 2 e ÷ f ÷ g ÷ h 5 3 7 x 7 5 2y 3 2x 4 y 5 7x ÷ i ÷ j k ÷ l x y 5 3 12 x 2 8 Simplify the following. (Recall that 3 = a d

4x ×3 5 a 4× 3

b e

4x ÷3 5 7 5× 10 x

2 4x ÷ 3 7 4 b 2c ÷ 7 5 4 a 2b ÷ 5 7a

3 .) 1 c f

x 5 x 1÷ 2 2÷

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Example 10a

4 Simplify the following algebraic fractions. 5 xy 2x a b c 2y 6 5 Simplify the following products. x 2 1 a a × b × 3 5 7 9 4c 1 4 a 2b d × e × 5 5 3 5

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2 7 × 3 10 2 4 ÷ 3 5

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9 Helen’s family goes to dinner with Tess’ family. The bill comes to a total of $x and each family pays half. a Write an algebraic fraction for the amount Helen’s family pays. b Helen says that she will pay for one third of her family’s bill. Write an algebraic fraction for the amount she pays.

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10 The rectangular field below has length x metres and breadth y metres. xm ym a Write an expression for the area of the field. 1 3 b A smaller section is fenced off. It is the length and the breadth. 2 4 i Write an expression for the length of the smaller section. ii Write an expression for the breadth of the smaller section. iii Hence, write an expression for the area of the smaller section. c To find the proportion of the field that is fenced off, you can divide the fenced area by the total area. Use this to find the proportion of the field that has been fenced. 11 Write an algebraic fraction for the result of the following operations. a A number q is halved and then the result is tripled. 2 1 b A number x is multiplied by and the result is divided by 1 . 3 3 a b c The fraction is multiplied by its reciprocal . b a d The number x is reduced by 25% and then halved.

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x 12 Recall that any value x can be thought of as the fraction . 1 1 a Simplify x × . x

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b Simplify x ÷ 1 . x c Show that x ÷ 3 is equivalent to

1 × x by writing them both as algebraic fractions. 3

a ÷c. b b e Simplify a ÷ . c d Simplify

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b Which one of the expressions above will always have the same value regardless of x? 14 Assume that a and b are any two whole numbers. a a a Prove that 1 ÷ is the same as the reciprocal of the fraction . b b a b Find the reciprocal of the reciprocal of by evaluating 1 ÷ 1 ÷ b

a . b

Enrichment: Irrational squares 15 Consider a square with side length x. a Write an expression for the area of the square. b The length of each side is now halved. Give an expression for the area of the new square. 3 c If each side of the original square is multiplied by , show that the resulting 5 area is less than half the original area.

x

d If each side of the original square is multiplied by 0.7, find an expression for the area of the square. 7 Recall that 0.7 = . 10 e Each side of the square is multiplied by some amount, which results in the square’s area being halved. Try to find the amount by which they were multiplied correct to three decimal places.

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1G Expanding brackets Two expressions might look quite different when in fact they are equivalent. For example, 2(3 − 7b) is equivalent to 4b + 6(1 − 3b) even though they look quite different. One use for expanding brackets is that it allows us quite easily to convert between equivalent expressions.

Let’s start: Room plans An architect has prepared ﬂoor plans for a house but some numbers are missing. Four students have attempted to describe the total area of the plans shown. a 10

■

5

b

To expand brackets also known as removing the grouping symbols, you use the distributive law, which states that: – a(b + c) = a × b + a × c = ab + ac

■

– a(b − c) = a × b − a × c = ab – ac e.g. 4(2x + 5) = 8x + 20 and 3(5 − 2y) = 15 − 6y, we say the expansion of 4(2x + 5) = 8x + 20 The distributive law can be illustrated by considering rectangle areas. b c a

■

a×b

a×c

Area = a(b + c) Area = ab + ac

The distributive law is used in arithmetic, e.g. 5 × 31 = 5(30 + 1) = 5(30) + 5(1) = 155

Cambridge University Press

Key ideas

Alice says it is 5a + 50 + ab. Brendan says it is 5(a + 10) + ab. Charles says it is a(5 + b) + 50. David says it is (5 + b)(a + 10) − 10b. • Discuss which of the students is correct. • How do you think each student worked out their answer? • The architect later told them that a = 4 and b = 2. What value would each of the four students get for the area?

Chapter 1 Algebraic techniques 2 and indices

Example 12 Expanding using the distributive law Expand the following. a 3(2x 3(2 + 5)

b

−8(7 + 2y)

c

4x(2x (2 − y) (2x

SOLUTION

EXPLANATION

and 3(2x 3(2 + 5) = 3(2x 3(2 ) + 3(5) = 6x + 15

Distributive law: 3(2x 3(2 + 5) = 3(2x 3(2 ) + 3(5) Simplify the result.

b −8(7 8(7 + 22yy) = −8(7) 8(7) + ((−8)(2 8)(2yy) = −56 + (−16y) = −56 − 16y

Distributive law: −8(7 8(7 + 22yy) = −8(7) 8(7) + ((−8(2 8(2yy)) Simplify the result. Adding −16y is the same as subtracting positive 16y.

c 4x(2x (2 − y) = 4x(2x (2x (2 ) − 4x(y) (2x = 8x2 – 4xy

Distributive law: 4x(2x (2 − y) = 4x(2x (2x (2 ) − 4x(y) (2x Simplify the result, remembering 4x × 2x 2 = 8x2.

Example 13 Expanding and combining like terms Expand the brackets in each expression and then combine like terms. a 3(2b + 5) + 3b b 12xy 12 + 7x(2 − y) SOLUTION

EXPLANATION

a 3(2b + 5) + 3b = 3(2b) + 3(5) + 3b = 6b + 15 + 3b = 9b + 15

Use the distributive law. Simplify the result. Combine the like terms.

b 12xy 12 + 7x(2 − y) = 12xy 12 + 7x(2) − 7x(y) = 12xy 12 + 14x − 7xy = 5xy + 14x

Use the distributive law. Simplify the result and combine the like terms.

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c 5(3a − 2) = 15a −

7

a 6

D 12x + 28

b

3(2x + 4y) = 6x +

d

7(4x − 2y) = 28x −

and and

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Example 13

7 Use the distributive law to remove the grouping symbols. a 8z(k – h) b 6j(k + a) c 4u(r – 2q) e m(10a + v) f −2s(s + 5g) g −3g(8q + g) i −8u(u + 10t) j −s(t + 5s) k m(2h – 9m)

d 3(8 – v) h −8(5 + h) l 2(c – 8) d 2p(3c – v) h −f (n + 4f) l 4a(5w – 10a)

8 Simplify the following by expanding and then collecting like terms. a 7(9f + 10) + 2f b 8(2 + 5x) + 4x c 4(2a + 8) + 7a d 6(3v + 10) + 6v e 7(10a + 10) + 6a f 6(3q – 5) + 2q g 6(4m – 5) + 8m h 4(8 + 7m) – 6m 9 The distributive law also allows expansion with more than two terms in the brackets, for instance 3(2x − 4y + 5) = 6x − 12y + 15. Use this fact to simplify the following. a 2(3x + 2y + 4z) b 7a(2 − 3b + 4y) c 2q(4z + 2a + 5) d −3(2 + 4k + 2p) e −5(1 + 5q − 2r) f −7k(r + m + s) 10 Simplify the following by expanding and then collecting like terms. a 3(3 + 5d ) + 4(10d + 7) b 10(4 + 8f ) + 7(5f + 2) c 2(9 + 10j ) + 4(3j + 3) d 2(9 + 6d ) + 7(2 + 9d) e 6(10 − 6j ) + 4(10j − 5) f 8(5 + 10g) + 3(4 − 4g)

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6 Use the distributive law to expand these expressions. Example 12a,b a 9(a + 7) b 2(2 + t) c 8(m – 10) e −5(9 + g) f −7(5b + 4) g −9(u + 9) i 5(6 – j) j 6(2 – m) k 3(10 – b)

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4 Which of the following is the correct expansion of 4(3x + 7)? A 12x + 7 B 4x + 28 C 3x + 28

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3 Consider this combined rectangle, which has an area of 6(a + 7). a What is the area of the purple rectangle? b What is the area of the orange rectangle? c Write an expression for the sum of these two areas. d Write the expanded form for 6(a + 7).

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2 The expression 3(2 + 7x) is equivalent to 2 + 7x + 2 + 7x + 2 + 7x. Simplify this expression by combining like terms.

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11 Write an expression for each of the following and then expand it. a A number t has 4 added to it and the result is multiplied by 3. b A number u has 3 subtracted from it and the result is doubled. c A number v is doubled, and then 5 is added. The result is tripled. d A number w is tripled, and then 2 is subtracted. The result is doubled.

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12 Match each operation on the left with an equivalent one on the right. (Hint: First convert the descriptions into algebraic expressions.) a The number x is doubled and 6 is added. b The number x is reduced by 5 and the result is doubled. c The number x is added to double the value of x. d The number x is halved, then 3 is added and the result is doubled. e 2 is subtracted from one third of x and the result is tripled.

A x is doubled and reduced by 10. B The number x is tripled. C x is decreased by 6. D x is increased by 3 and the result is doubled. E x is increased by 6.

13 The number of boys in a classroom is b and the number of girls is g. Each boy has 5 pencils and each girl has 3 pencils. a Write an expression for the total number of pencils in the class. b If the pencils cost $2 each, write and expand an expression for the total cost of all the pencils in the room. c Each boy and girl also has one pencil case, costing $4 each. Write a simplified and expanded expression for the total cost of all pencils and cases in the room. d If there are 10 boys and 8 girls in the room, what is the total cost for all the pencils and cases in the room? 14 a When expanded, 4(2a + 6b) gives 8a + 24b. Find two other expressions that expand to give 8a + 24b. b Give an expression that expands to 4x + 8y. c Give an expression that expands to 12a − 8b. d Give an expression that expands to 18ab + 12ac.

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16 Prove that 4a(2 + b) + 2ab is equivalent to a(6b + 4) + 4a by expanding both expressions. 17 Find an expanded expression for (x + y)(x + 2y) by considering the diagram below. Ensure your answer is simplified by combining any like terms. x

y

y

x y 18 Prove that the following sequence of operations has the same effect as doubling a number. 1 Take a number, add 2. 2 Multiply by 6. 3 Subtract 6. 4 Multiply this result by 1 . 3 5 Subtract 2.

Enrichment: Expanding algebraic fractions x+5 x + change both fractions to have a common denominator of 6, giving 3 2 2( x + 5) 3 x . Then expand to finish off the simplification: 2 x + 1100 3 x 5 x + 110 . + + = 6 6 6 6 6 Use this method to simplify the following sums.

19 To simplify

a

x +1 x + 3 2

b

d

x + 2 x +1 + 4 3

e

x+5 x + 5 3 2 x + 1 3x + 1 + 5 10

c

3x x − 1 + 8 4

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2 x − 1 3x + 2 + 7 5

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15 The distributive law is often used in multiplication of whole numbers. For example, C R PS HE 17 × 102 = 17 × (100 + 2) = 17(100) + 17(2) = 1734. M AT I C A a Use the distributive law to find the value of 9 × 204. Start with 9 × 204 = 9 × (200 + 4). b Use the distributive law to find the value of 204 × 9. Start with 204 × 9 = 204 × (10 − 1). c Given that a × 11 = a × (10 + 1) = 10a + a, find the value of these multiplications. i 14 × 11 ii 32 × 11 iii 57 × 11 iv 79 × 11 2 d It is known that (x + 1)(x − 1) expands to x − 1. For example, if x = 7 this tells you that 8 × 6 = 49 − 1 = 48. Use this fact to find the value of: i 7×5 ii 21 × 19 iii 13 × 11 iv 201 × 199 e Using a calculator, or otherwise, evaluate 152, 252 and 352. Describe how these relate to the fact that (10n + 5)(10n + 5) is equivalent to 100n(n + 1) + 25.

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Chapter 1 Algebraic techniques 2 and indices

1H Factorising expressions Factorising and expanding are opposite procedures. Factorising allows us to simplify expressions and solve harder mathematical problems. Because 3(2x + 5) expands to 6x + 15, this means that the factorisation of 6x + 15 is 3(2x + 5). The aim in factorising is to write expressions as the product of two or more factors, just as with numbers we can factorise 30 and write 30 = 2 × 3 × 5.

Let’s start: Expanding gaps

Key ideas

• Try to fill in the gaps to make the following equivalence true: ( + ) = 12x + 18xy. • In how many ways can this be done? Try to find as many ways as possible. • If the aim is to make the term outside the brackets as large as possible, what is the best possible solution to the puzzle?

■

■

■

The highest common factor (HCF) of a set of terms is the largest factor that divides into each term. e.g. HCF of 15x and 21y is 3. HCF of 10a and 20c is 10. HCF of 12x and 18xy is 6x. e.g. 10 10x + 15y To factorise an expression, first take the HCF of the HCF = 5 terms outside the brackets and divide each term by it, Result 5(2 5(2x + 3y) leaving the result in brackets. To check your answer, expand the factorised form, HCF 10x ÷ 5 15y ÷ 5 10 e.g. 5(2x + 3y) = 10x + 15y

Example 14 Finding the highest common factor (HCF) Find the highest common factor (HCF) of: a 20 and 35 b

18a and 24ab

SOLUTION

EXPLANATION

a 5

5 is the largest number that divides into 20 and 35.

b 6a

6 is the largest number that divides into 18 and 24, and a divides into both terms.

c 3x

3 divides into both 12 and 15, and x divides into both terms.

c 12x 12x and 15x2

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Number and Algebra

Example 15 Factorising expressions Factorise the following expressions. a 6x + 15 b 12a + 18ab d −10a − 20 e x2 + 4x

c 21x − 14y

a 6x + 15 = 3(2x 3(2 + 5)

HCF of 6xx and 15 is 3. 6x ÷ 3 = 2x 2x and 15 ÷ 3 = 5

b 12a + 18ab = 6a(2 + 3b)

HCF of 12a and 18ab is 6a. 12a ÷ 6a = 2 and 18ab ÷ 6a = 3b

c 21 21x − 14 14y = 7(3 7(3x − 22y))

HCF of 21x 21x and 14y is 7. 21x 21 ÷ 7 = 33xx and 14y ÷ 7 = 22y The subtraction sign is included as in the original expression.

d −10a − 20 = −10(a + 2)

HCF of −10a and −20 is 10 but because both terms are negative we bring −10 out the front. −10a ÷ −10 = a and −20 ÷ −10 = 2.

e x2 + 4x = x(x + 4)

HCF of x2 and 4xx is x. x2 ÷ x = x and 4x ÷ x = 4.

Exercise 1H

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2 The factors of 14 are 1, 2, 7 and 14. The factors of 26 are 1, 2, 13 and 26. What is the highest factor that these two numbers have in common? 3 Find the highest common factor of the following pairs of numbers. a 12 and 18 b 15 and 25 c 40 and 60 d 24 and 10 4 Fill in the blanks to make these expansions correct. x+3 b 5(7 − 2x) = − 10x a 3(4x + 1) = c 6(2 + 5y) = + y d 7(2a − 3b) = − e 3(2a + ) = 6a + 21 f 4( − 2y) = 12 − 8y g 7( + ) = 14 + 7q h (2x + 3y) = 8x + 12y 5 Verify that 5x + 15 and 5(x + 3) are equivalent by copying and completing the table below. x

2

7

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−3

−6

5x + 15 5(x + 3)

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7 Factorise the following by first finding the highest common factor. Check your answers by expanding them. a 3x + 6 b 8v + 40 c 15x + 35 d 10z + 25 e 40 + 4w f 5j − 20 g 9b − 15 h 12 − 16f i 5d − 30 8 Factorise the following. a 10cn + 12n b 24y + 8ry e 10h + 4z f 30u − 20n i 21hm − 9mx j 49u − 21bu

c g k

14jn + 10n 40y + 56ay 28u − 42bu

9 Find the factorisation of the following expressions. a −3x − 6 b −5x − 20 c −10a − 15 2 2 e x + 6x f k + 4k g 2m2 + 5m

d 24g + 20gj h 12d + 9dz l 21p − 6c d −30p − 40 h 6p2 + 3p WO

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11 A group of students lines up for a photo. They are in 6 rows with x students in each row. Another 18 students join the photo. a Write an expression for the total number of students in the photo. b Factorise the expression above. c How many students would be in each of the 6 rows now? Write an expression. d If the photographer wanted just 3 rows, how many students would be in each row? Write an expression. e If the photographer wanted just 2 rows, how many students would be in each row? Write an expression.

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6 Find the highest common factor (HCF) of the following pairs of terms. a 15 and 10x b 20a and 12b c 27a and 9b d 7xy and 14x e −2yz and 4xy f 11xy and −33xy g 8qr and −4r h −3a and 6a2 i 14p and 25pq

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13 Consider the diagram shown to the right. What is the factorised form of xy + 3x + 2y + 6?

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12 a Expand 2(x + 1) + 5(x + 1) and simplify. b Factorise your result. c Make a prediction about the equivalent of 3(x + 1) + 25(x + 1) if it is expanded, simplified and then factorised. d Check your prediction by expanding and factorising 3(x + 1) + 25(x + 1). y

3

x

xxy

3x

2

2 2y

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Enrichment: Factorising fractions 5 x + 110 can be simplified 7 x + 114

5 ( x + 2) 5 by first factorising the numerator and the denominator = . Factorise and then simplify the 7 ( x + 2) 7 following fractions as much as possible. 2x + 4 7x − 7 3ac + 55a a b c 5 x + 110 2x − 2 a + 2ab 5 q − 15 1 7 p + 1144 pq 4 a + 2b d e f 3q − 9 9 p + 1188 pq 8c + 1100 d g

7a − 221 2a − 6

h

12 p 8 p + 2 pq

i

100 − 10 x 20 − 2 x

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14 In English, people often convert between ‘factorised’ and ‘expanded’ sentences. For instance, ‘I like John and Mary’ is equivalent in meaning to ‘I like John and I like Mary’. The first form is factorised with the common factor that I like them. The second form is expanded. a Expand the following sentences. i I eat fruit and vegetables. ii Rohan likes maths and English. iii Petra has a computer and a television. iv Hayden and Anthony play tennis and chess. b Factorise the following sentences. i I like sewing and I like cooking. ii Olivia likes ice-cream and Mary likes ice-cream. iii Brodrick eats chocolate and Brodrick eats fruit. iv Adrien likes chocolate and Adrien likes soft drinks, and Ben likes chocolate and Ben likes soft drinks.

15 Factorising can be used to simplify algebraic fractions. For example,

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Chapter 1 Algebraic techniques 2 and indices

1I Applying algebra The skills of algebra can be applied to many situations within other parts of mathematics as well as to other fields such as engineering, sciences and economics.

Let’s start: Carnival conundrum

Key ideas

Alwin, Bryson and Calvin have each been offered special deals for the local carnival. – Alwin can pay $50 to go on all the rides all day. Algebra can be applied to both the engineering of the ride and the price of the tickets. – Bryson can pay $20 to enter the carnival and then pay $2 per ride. – Calvin can enter the carnival at no cost and then pay $5 per ride. • Which of them has the best deal? • In the end, each of them decides that they were happiest with the deal they had and would not have swapped. How many rides did they each go on? Compare your different answers.

■ ■

Different situations can be modelled with algebraic expressions. To apply a rule, the variables should first be clearly defined. Then known values are substituted for the variables.

Example 16 Writing expressions from descriptions Write an expression for the following situations. a The total cost of k bottles if each bottle costs $4 b The area of a rectangle if its breadth is 2 cm more than its length and its length is x cm c The total cost of hiring a plumber for n hours if he charges $40 call-out fee and $70 per hour SOLUTION

EXPLANATION

a 4 × k = 4k

Each bottle costs $4 so the total cost is $4 multiplied by the number of bottles purchased.

b x × ((x + 2) = x((x + 2)

Length = x so breadth = x + 2. The area is length × breadth.

c 40 + 70n

$70 per hour means that the cost to hire the plumber would be 70 × n. Additionally $40 is added for the call-out fee, which is charged regardless of how long the plumber stays.

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x = −3

3 Consider the isosceles triangle shown. a Write an expression for the perimeter of the triangle. b Find the perimeter when x = 3 and y = 2.

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6 An electrician charges a call-out fee of $30 and $90 per hour. Which of the following represents the total cost for x hours? A x(30 + 90) B 30x + 90 C 30 + 90x D 120x

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Write an expression for the total area of the shape shown. b If x = 9, what is the area?

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7 a Give an expression for the perimeter of this regular pentagon. b If each side length were doubled, what would the perimeter be? c If each side length were increased by 3, write a new expression for the perimeter.

x

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Example 16b

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8 An indoor soccer pitch costs $40 per hour to hire plus a $30 booking fee. a Write an expression for the cost of hiring the pitch for x hours. b Hence, find the cost of hiring the pitch for a round-robin tournament that lasts 8 hours.

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10 A repairman says the cost in dollars to hire his services for x hours is 20(3 + 4x). a How much would it cost to hire him for 1 hour? b Expand the expression he has given you. c Hence, state: i his call-out fee ii the amount he charges per hour. 11 Three deals are available at a fair. Deal 1: Pay $10, rides cost $4/each. Deal 2: Pay $20, rides cost $1/each. Deal 3: Pay $30, all rides are free. a Write an expression for the total cost of n rides using deal 1. (The total cost includes the entry fee of $10.) b Write an expression for the total cost of n rides using deal 2. c Write an expression for the total cost of n rides using deal 3. d Which of the three deals is best for someone going on just two rides? e Which of the three deals is best for someone going on 20 rides? f Fill in the gaps: i Deal 1 is best for people wanting up to _____ rides. ii Deal 2 is best for people wanting between _____ and _____ rides. iii Deal 3 is best for people wanting more than _____ rides.

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9 A plumber says that the cost in dollars to hire her for x hours is 50 + 60x. a What is her call-out fee? b How much does she charge per hour? c If you had $200, what is the longest period you could hire the plumber?

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y−4 x−3

14 Tamir notes that whenever he hires an electrician, they charge a call-out fee $F and an hourly rate of $H per hour. a Write an expression for the cost of hiring an electrician for one hour. b Write an expression for the cost of hiring an electrician for two hours. c Write an expression for the cost of hiring an electrician for 30 minutes. d How much does it cost to hire an electrician for t hours?

Enrichment: Ticket sales 15 At a carnival there are six different deals available to reward loyal customers. Deal

Entry cost ($)

Cost per ride ($)

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The queue consists of 100 customers. The first customer knows they will go on one ride, the second will go on two rides, and the pattern continues, with the 100th customer wanting to go on 100 rides. Assuming that each customer can work out their best deal, how many of each deal will be sold? © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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13 Roberto draws a rectangle with unknown dimensions. He notes that the area is (x − 3)(y − 4). a If x = 5 and y = 7, what is the area? b What is the value of (x − 3)(y − 4) if x = 1 and y = 1? c Roberto claims that this proves that if x = 1 and y = 1 then his rectangle has an area of 6. What is wrong with his claim? (Hint: Try to work out the rectangle’s perimeter.)

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12 In a particular city, taxis charge $4 to pick someone up (ﬂagfall) and then $2 per minute of travel. Three drivers have different ways of calculating the total fare. Russell adds 2 to the number of minutes travelled and doubles the result. Jessie doubles the number of minutes travelled and then adds 4. Arash halves the number of minutes travelled, adds 1 and then quadruples the result. a Write an expression for the total cost of travelling x minutes in: i Russell’s taxi ii Jessie’s taxi iii Arash’s taxi b Prove that all three expressions are equivalent by expanding them. c A fourth driver starts by multiplying the number of minutes travelled by 4 and then adding 8. What should he do to this result to calculate the correct fare?

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Chapter 1 Algebraic techniques 2 and indices

1J Index laws for multiplication and division Recall that x2 means x × x and x3 means x × x × x. Index notation provides a convenient way to describe repeated multiplication. index or exponent 35 = 3 × 3 × 3 × 3 × 3 base Notice that 35 × 32 = 3 × 3 × 3 × 3 × 3 × 3 × 3 35 which means that 35 × 32 = 37.

32

Similarly it can be shown that 26 × 25 = 211. When dividing, note that: 510 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 = 5×5×5×5×5×5×5 57 =5×5×5

Index notation has wide application, particularly in modelling growth and decay, in science, economics and computer applications.

So 510 ÷ 57 = 53. These observations are generalised in index laws.

Let’s start: Comparing powers

Key ideas

• Arrange these numbers from smallest to largest. 23, 32, 25, 43, 34, 24, 42, 52, 120 • Did you notice any patterns? • If all the bases were negative, how would that change your arrangement from smallest to largest? For example, 23 becomes (−2)3.

index or exponent

■

5n = 5 × 5 × … × 5 base

■ ■

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The number 5 appears n times

e.g. 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64 An expression such as 4 × 53 can be written in expanded form as 4 × 5 × 5 × 5. The index law for multiplying terms with the same base: 3m × 3n = 3m + n e.g. 34 × 32 = 36. 3m The index law for dividing terms with the same base: 3m ÷ 3n = n = 3m − n 3 e.g. 38 ÷ 35 = 33.

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Number and Algebra

Example 17 Using the index law for multiplication Simplify the following using the index law for multiplication. a 53 × 57 × 52 b 53 × 54 c 65 × 6 × 63 SOLUTION

EXPLANATION

a 53 × 57 × 52 = 512

3 + 7 + 2 = 12 and the first index law applies (using a = 5).

b 53 × 54 = 57

Using the first index law, 3 + 4 = 7, so 53 × 54 = 57.

c 65 × 6 × 63 = 65 × 61 × 63 = 69

Write a as a1 so the index law can be used. 5 + 1 + 3 = 9, so the final result is 69.

Example 18 Using the index law for division Simplify the following using the index law for division. a

57

b

53

SOLUTION a

57 3

5

620

10 × 56

c

65

4 × 52

EXPLANATION

= 54

Considering

57 3

5

=

5×5×5×5× 5 × 5 × 5 5×5×5

= 54 or just 7 − 3 = 4. b

c

620 65

Use the second index law, so 20 − 5 = 15.

= 615

10 × 56 4 × 52 =

5 54 × 2 1

=

55 2

=

10 56 × 4 52

First separate the numbers into a separate fraction. Cancel the common factor of 2 and use the second index law. Combine the result as a single fraction.

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3 Which of the following is the same as 4 × 4 × 4? A 44 B 43 C 34 D 4×3 4 a

Calculate the value of: i 22 iii 25 b Is 22 × 23 equal to 25 or 26?

ii 23 iv 26

5 a Calculate the value of: i 56 iii 53 b Is 56 ÷ 52 equal to 53 or 54?

ii 52 iv 54

Write 83 in expanded form. Write 84 in expanded form. Write the result of multiplying 83 × 84 in expanded form. Which of the following is the same as 83 × 84? A 812 B 85 7 C 8 D 81

8 Simplify the following using the index law for multiplication. a 23 × 24 b 52 × 54 c 710 × 73 d 37 × 32 2 4 3 e 3 ×3 ×3 f 72 × 74 × 73 g 22 × 23 × 24 h 210 × 212 × 214 3 i 10 × 10 j 62 × 6 h 54 × 53 × 5 l 22 × 2 × 2

Example 17b,c

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310 35

c g

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23 × 510 2 2 × 54

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24 × 53 22

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3700 × 3300 ? 31000

11 A student tries to simplify 32 × 34 and gets the result 96. a Use a calculator to verify this is incorrect. b Write out 32 × 34 in expanded form, and explain why it is not the same as 96. c Explain the mistake they have made in attempting to apply the first index law.

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iv (−2)5

53 in index form. 53 53 b Given that 53 = 125, what is the numerical value of 3 ? 5 c According to this, what is the value of 50? Check whether this is also the result your calculator gives. d What is the value of 120? Check with a calculator.

13 a Use the index law for division to write

3a = 9 , what does this tell you about the value of a and b? 3b 2a 5a b Given that b = 8, find the value of b . 2 5

14 a If

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Enrichment: Simplifying larger expressions with non-numerical bases 15 Simplify the following. a

53 x 4 y 7 × 510 xy 3 52 xy × 55xx 3 y 7 × 57

b

a 2 a 4 a6 a8 a10 a1a3 a 5 a 7 a 9

c

ab 2 c3 d 4 e 5 bc 2 d 3 e 4

d

1q × 2q2 × 3q3 × 4q4 × 5q5

16 Simplify the following expressions involving algebraic fractions. a

53 24 × 2 2 x 5 52 x

3 5 c 10 x y ÷

1 −10 x 4 y3

b

7a3 b 2 49a 2 b ÷ 3 220 2c 5 2 c

d

12 x 2 y 5 7 x 4 y × 21x 3 3y6

Paris – the Renaissance mathematician Nicolas Chuquet developed a form of index notation here in the ﬁ fteenth century, but our present index notation was not widely used before Rene Descartes’ writing spread it in the seventeenth century.

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51

Number and Algebra

1K The zero index and power of a power Consider what the expanded form of (53)4 would be: (53 )4 = 53 × 53 × 53 × 53 = 5×5×5 × 5×5×5 × 5×5×5 × 5×5×5 = 51122 Similarly:

(6 ) 4

2

= 64 × 64 = 6×6×6×6 × 6×6×6×6 = 68

This leads us to an index law: (am)n = amn. When a number (other than zero) is raised to the power of 0 the result is 1. This can be seen by the pattern in this table. 44

43

42

41

40

256

64

16

4

1

÷4

÷4

÷4

A pictorial representation of (43)3 . Each of the 43 green cubes in the top ﬁ gure is made up of 43 tiny blue cubes shown magniﬁ ed in the lower ﬁ gure. How many blue cubes are there in total?

÷4

Let’s start: How many factors?

■

■

■

When a number (other than zero) is raised to the power of 0 the result is 1. This gives the law a0 = 1 (where a is any number other than 0). A power of a power can be simplified by multiplying indices: (am)n = amn, e.g. (62)5 = 610. Expressions involving powers can be expanded, so (3 × 5)4 = 34 × 54 and (2 × 5)10 = 210 × 510.

Cambridge University Press

Key ideas

The number 7 has two factors (1 and 7) and the number 72 has three factors (1, 7 and 49). • Which of these has the most factors? 710 75 72 × 73 (72)3 76 10 7 • Which has more factors: 7 or 10 ? Compare your answers with others in your class.

Chapter 1 Algebraic techniques 2 and indices

Example 19 Working with the zero index Simplify the following expressions using the index laws. a 40 × 8 b 40 × 80 SOLUTION

EXPLANATION

a 40 × 8 = 1 × 8 =8

Any number to the power of 0 equals 1, so 40 = 1.

b 40 × 80 = 1 × 1 =1

40 = 1 and 80 = 1.

Example 20 Using the index law for power of a power Simplify the following expressions using the index laws. a (23)5 b (52)4 × (53)2 EXPLANATION

a (23)5 = 215

3 × 5 = 15, so we can apply the index law easily.

b (52)4 × (53)2 = 58 × 56 = 514

Apply the index law with 2 × 4 = 8 and 3 × 2 = 6. Apply the first index law: 8 + 6 = 14.

Exercise 1K

WO

U

2 Which of the following is equivalent to (3x)2? A 3×x B 3×x×x C 3×x×3×x

D 3×3×x

3 a Copy and complete the table below. 35

34

243

81

33

32

31

30

b What is the pattern in the bottom row from one number to the next? 4 a Calculate the value of: i (42)3 ii 45 2 3 5 6 b Is (4 ) equal to 4 , 4 or 423?

iii 46

R

HE

T

D 5×3×2

MA

1 Which one of the following is equivalent to (53)2? A 5×5×5 B 53 × 53 C 52 × 52

iv 423

Cambridge University Press

R K I N G

C

F PS

Y

SOLUTION

LL

52

M AT I C A

53

Number and Algebra

6 Simplify the following. a (23)4 b (32)8 3 3 d (6 ) e (78)3

c f

(74)9 (25)10

7 Simplify the following using the index laws. b (52)6 × (53)2 a (23)2 × (25)3 3 6 2 2 d (3 ) × (3 ) e (23)2 × (24)3 h

52 (22 )10

k

( 23 ) 2

c (74)2 × (75)3 f (112)6 × (112)3

(117 )2

i

(113 )2 320

l

(33 )5

(115 )3 112 (112 )10 (113 )6 WO

(32)3 × 3 = 311

c

(

9 a Use the fact that (52 )3 = 56 to simplify (52 )3

(

b Simplify (53 )4

)

5

)

4

d

(74)

× (73)2 = 714

HE

T

b (2 )4 = 212

R

MA

= 515

R K I N G

U

8 Find the missing value that would make the following simplifications correct. a (53)

PS

M AT I C A

C

F PS

Y

j

(53 )4

R

HE

F

LL

g

C

LL

30 + 50 30 × 50 (35 × 53)0

R K I N G

M AT I C A

.

.

c Put the following numbers into ascending order. (You do not need to calculate the actual values.) 2100, (27)10, ((25)6)7, ((23)4)5 10 a How many zeroes do the following numbers have? i 102 ii 105 iii 106 5 6 7 3 b How many zeroes does the number (10 × 10 × 10 ) have? 11 a Simplify x3 × x4. b Simplify (x3)4. c Find the two values of x that make x3 × x4 and (x3)4 equal. U

R

HE

T

12 For this question you will be demonstrating why should equal 1 for any value of a other than zero. 52 a State the value of 2 . 5 52 b Use the index law for division to write 2 as a power of 5. 5 c Use this method to prove that 30 = 1. d Use this method to prove that 1000 = 1. e Explain why you cannot use this method to prove that 00 = 1.

MA

a0

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R K I N G

C

F PS

Y

WO

LL

Example 20b

c f i

T

Example 20a

5 Simplify the following. a 30 b 50 0 d 3+5 e (3 + 5)0 g (3 × 5)0 h 3 × 50

MA

Example 19

U

Y

WO

M AT I C A

1K

WO

13 Ramy is using his calculator and notices that (23)4 = (26)2. a Explain why this is the case. b Which of the following are also equal to (23)4? A (24)3 B (22)6 C (42)3

R K I N G

U

MA

R

T

HE

D (43)2 × (62)2

According to the index laws, what number is equal to (90.5)2? What positive number makes the equation x2 = 9 true? What should 90.5 equal according to this? Check on a calculator. Use this observation to predict the value of 360.5.

15 Alexis notices that

(a 3 ) 2 × a 4 is always equal to one, regardless of the value of a. (a 2 ) 5

a Simplify the expression above. b Give an example of two other expressions that will always equal 1 because of the index laws.

Enrichment: Simplifying expressions with non-numerical bases 16 Simplify the following using the index laws. a

( 5 x 2 )3 × ( 5 x 3 ) 4 ( 5 x 6 )3

b

( x 2 )4 x7 3 × x ( x 2 )2

c

( x 2 y 3 )4 × ( x 3 y 2 )5 ( xy )7 × ( x 2 y )6

d

(a 2 b3 c 4 )10 a3 ÷ a10 b 20 c30 b 2

e

( x 20 y10 )5 ( x10 y 20 )2

f

(78 )9 (710 )7

g

( 76 ) 5 ( 7 5 )6

h

511 × 513 (52 )111

i

100 20 100012

The number of hydrogen atoms (the most abundant element) in the observable universe is estimated to be between 1079 and 1081.

Cambridge University Press

F PS

M AT I C A

c Freddy claims that (25)6 can be written in the form (4 ) . Find one way to fill in the two missing values. 14 a b c d

C

Y

Chapter 1 Algebraic techniques 2 and indices

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54

Number and Algebra

Investigation

Card pyramids Using a pack of playing cards, build some pyramids on your desk like the ones illustrated below.

Pyramid 3 (Six-triangle pyramids) (Fifteen cards)

Pyramid 2 (Three-triangle pyramids) (Seven cards)

Pyramid 1 (One-triangle pyramid) (Two cards)

1 Copy and complete this table: Number of triangle pyramids on base Total number of triangle pyramids Total number of cards required

1

2

1

3

2

3

5

45

15

55

100

2 Describe the number of pyramids in, and the number of cards required for, pyramid 20 (20 pyramids on the base). How did you get your answer? 3 If you had 10 decks of playing cards, what is the largest tower you could make? Describe how you obtained your answer.

Number pyramids Number pyramids with a base of three consecutive numbers

16

1 Can you explain how this number pyramid is constructed? 2 Draw a similar number pyramid starting with the number 4 on the left of the base. 3 Draw a similar number pyramid that has 44 as its top number. Remember the base of the pyramid must be consecutive numbers. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

7

3

9

4

55

5

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56

Chapter 1 Algebraic techniques 2 and indices

4 Can you draw a similar number pyramid that has 48 as its top number? Explain your answer. 5 Draw several of these pyramids to investigate how the top number is related to the starting value. • Set up a table showing starting values and top numbers. • Can you work out an algebraic rule that calculates top numbers given the starting number? 6 Draw a number pyramid that has a base row of n, n + 1 and n + 2. What is the algebraic expression for the top number? Check this formula using some other number pyramids. 7 What is the sum of all the numbers in a pyramid with base row −10, −9, −8? 8 Determine an expression for the sum of all the numbers in a pyramid starting with n on the base.

Number pyramids with four consecutive numbers on the base 1 Copy and complete the following number pyramids. 60

5

2

17

4

6

8

2 Investigate how the top number is related to the starting number. Can you show this relationship using algebra? 3 Write the sequence of all the possible top numbers less than 100. 4 What patterns can you see in this sequence of top numbers? Can you find some ways of showing these patterns using algebraic expressions? Let the bottom row start with n. (In the examples above n = 6 and n = 2.)

Number pyramids with many consecutive numbers on the base 1 Determine the algebraic rule for the value of the top number for a pyramid with a base of six consecutive numbers starting with n. 2 List out the algebraic expressions for the first number of each row for several different sized pyramids all starting with n. What patterns can you see occurring in these expressions for: • the coefficients of n? • the constants? 3 What is the top number in a pyramid with a base of 40 consecutive numbers starting with 5? 4 Write an expression for the top number if the base had 1001 consecutive numbers starting with n.

Cambridge University Press

1 Find the values of A, B and C so that the rows and columns add up correctly. A

2b

4B + A

Sum = 2

2a

C + A

3BD

Sum = 44

3A

C + B

BD

Sum = 30

Sum = 42

Sum = 11

Sum = 23

2 Fill in the missing expressions to make the six equivalences true. +

3x +

=

7x − 3y + 1

+ +

=

+ =

=

2y + 3x

+

= =

3x

3 Finding the largest value a If m can be any number, what is the largest value that 10 − m(m + 5) could have? b If x + y evaluates to 15, what is the largest value that x × y could have? c If a and b are chosen so that a2 + b2 is equal to (a + b)2, what is the largest value of a × b? 4 Simplify these algebraic expressions. a

a a +1 a + − 5 6 2

b

x − 1 2x − 3 x − + 3 7 6

5 The following three expressions all evaluate to numbers between 1 and 100, but most calculators cannot evaluate them. Find their values using the index laws. a

21001 × 2 2002 (2150 )20

b

51000 × 31001 15 999

c

8 50 × 4100 × 2 200 (2 250 )2 × 2 48

a

a

6 Consider the following pattern.

a

a

a

a

a

n=1

n=2

a n=3

a n=4

a n=5

The perimeter for the shape when n = 1 is given by the expression 4a and the area is a2. a Give expressions for the perimeter and area of the other shapes shown above and try to find a pattern. b If a = 6 and n = 1000, state the perimeter and give the approximate area.

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57

Puzzles and challenges

Number and Algebra

Chapter summary

58

Chapter 1 Algebraic techniques 2 and indices

Algebraic terms

Pronumeral: a letter which stands for one or more numbers

Concise form 5x 2y –2ab 2 7p 2q 3 4ab 2 – 3 5c

Language + sum more than added increased

– difference less than minus decreased

× product times double (2×) twice (2×) triple (3×)

an expression

-7xy

–8

6a 2m –2a 2m 5ma 2

‘substitute’

replace variables with numbers and calculate answer 7a + 3(a + b) + 4b 2 4 a=8b=2 7 × 8 + 3 × (8 + 2) + 4 × 2 × 2 = 4 = 14 + 30 + 16 = 60

Algebraic techniques 2 and indices

7 − 3x = 2 – 3x + 5

Index notation a n = a × a × a × .... × a

Algebraic fractions

base

n ‘lots’ of a

Multiplying

‘unit’ is tenths

4 1 7a 12m × a 31

=1

5(2a –3) – 7(4 + a) = 5(2a) – 5(3) – 7(4) + –7(a) = 10a – 15 – 28 – 7a = 3a – 43

12a 2m + 8am 2 = 12aam + 8amm HCF = 4am = 4am x 3a + 4am x 2m = 4am(3a + 2m)

Index laws 1. 5m × 5n = 5m+n 5m 2. 5m ÷ 5n = n = 5m –n 5 3. (5m ) n = 5mn

Examples 1. 32 × 35 = 37 5 2. 122 = 123 12 3. (23) 4 = 212

2a

= 28m a a

a 2 stays the same

a(b –c) = ab – ac

12x + 6a (HCF = 6) = 6 × 2x + 6 × a = 6(2x + a)

7 − 3x 2 − 3x + 5 7−6=1 2−6+5=1 7 − 30 = –23 2 − 30 + 5 = –23

3a + 7m = 3a x 5 + 7m x 2 2 x 5 5 x 2 2 5 = 15a + 14m 10 10 = 15a + 14m 10

Distributive law a(b +c) = ab + ac 5(2a +m) = 5(2a) +5(m) =10a + 5m 7(k – 3a ) = 7(k ) – 7(3a) = 7k – 21a

a stays the same

Factorising

Equivalent expressions

Adding and subtracting

sign in front belongs to term = –6a – 12a + 4a 2 + 9a 2 – 3 = –18a + 13a 2 – 3

Expanding brackets 3(x+y) = x + y + x + y + x + y = 3x + 3y

Substitution

x=2 x = 10

6aam -2aam 5maa = 5aam

6a 2m, –2a 2m and 5ma 2 are like terms ab = ba

5 is the coefficient of x constant term is –8 –7 is the coefficient of xy

‘evaluate’

Adding and subtracting like terms • Count ‘how many’ • Don’t change variables. –6a +4a 2 –3 +9a 2 –12a

Like terms Variables have identical expanded form.

×÷ not used in algebraic expressions

terms 5x

÷ quotient divide one third one half quarter

Expanded form 5xxy –2abb 7ppqqq 4abb – 5ccc

1

÷a

+a ×a

a

a2

–a Dividing

3a ÷ 2 4 3 = 3a × 3 4 2 = 9a 8

O

15x ÷ y 20x 1 = 15x y × 20x 3 1 15x × 1 = y 420x1 =3 4y

a ÷ a = a × 1a = a a =1

2 x 3

1

÷x 3

+x 3 ×x 3

x3

x6

–x 3 O

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Number and Algebra

Multiple-choice questions 1 Consider the expression 5a2 − 3b + 8. Which one of the following statements is true? A The coefficient of a is 5. B It has 5 terms. C The constant term is 8. D The coefficient of b is 3. E The coefficient of a2 is 10. 2 Half the sum of double x and 3 can be written as: 1 2x + 6 A × 2x + 3 B C x+6 2 2

D

2x + 3 2

E

2( x − 3) 2

3 If n + 2 is an odd integer, the next odd integer can be written as: A n+3 B n+4 C n+5 D n+1

E n

4 Find the value of 5 − 4a2 given a = 2. A 3 B −3 C 21

D −11

E 13

5 3 × x × y is equivalent to: A 3x + y B xy

D 3x + 3y

E xy + 2xy

6

12ab can be simplified to: 24 a 2 2a A 2ab B b

C 3+x+y

C

b 2a

7 The expanded form of 2x(3 + 5y) is: A 6x + 5y B 3x + 5y C 6x + 5xy 8 Simplifying 3a ÷ 6b gives: a A 2 B b 9 57 × 54 is equal to: A 2511 B 528

C

2a b

C 253

10 The factorised form of 3a2 − 6ab is: A 3a2(1 – 2b) B 3a(a – 2b) C 3a(a – b)

D

ab 2

D 6 + 10y

D

ab 2

E

b 2

E 6x + 10xy

E

a 2b

D 53

E 511

D 6a(a – b)

E 3(a2 - 2ab)

Short-answer questions 1 State whether each of the following is true or false. a The constant term in the expression 5x + 7 is 5. b 16xy and 5yx are like terms. c The coefficient of d in the expression 6d 2 + 7d + 8abd + 3 is 7. d The highest common factor of 12abc and 16c is 2c. e The coefficient of xy in the expression 3x + 2y is 0. 2 For the expression 6xy + 2x − 4y2 + 3, state: a the coefficient of x b the constant term c the number of terms d the coefficient of xy

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59

60

Chapter 1 Algebraic techniques 2 and indices

3 Substitute the following values of a to evaluate the expression 8 − a + 2a2. a −1 b 2 c −3 d 0 4 Substitute x = 2 and y = −3 into each of the following. a 2y + 3 b 3x + y 5 xy +2 6 5 For what value of x is 3 − x equal to x − 3? d 4x − 2y

e

c xy + y −3 x + 2 y f x+y

6 Simplify each of these expressions by collecting like terms. a 7m + 9m b 3a + 5b – a c x2 – x + x2 + 1 d 5x + 3y + 2x + 4y e 7x – 4x2 + 5x2 + 2x f −8m + 7m + 6n – 18n 7 Simplify: a 9a × 4b

b 30 × x × y ÷ 2

c −8x × 4y ÷ (−2)

8 Copy and complete the following equivalences. a 3x + 4y = 8x − + 4y b 3ab × 3x c = d 9a2b ÷ 4 20

= −12abc = 3a

9 Express each of the following in their simplest form. 5x x 2a b 6 x 15 a − b + c × 12 6 5 15 5 2x2

d

4 a 8a ÷ 7 21b

10 Expand and simplify when necessary. a 3(x – 4) b −2(5 + x) e 7 – 3(x – 2) f 10(1 – 2x)

c g

k(3 – 4m) d 2(x – 3y) + 5x 4(3x – 2) + 2(3x + 5)

11 Factorise fully. a 2x + 6

c

12x + 3xy

b 24 – 16g

d 7a2 + 14ab

12 By factorising first, simplify the following fractions. a

5a + 110 5

b

12 x − 24 x−2

c

16 p 64 p + 48 pq

13 Find the missing values. a 75 × 72 = 7

b 54 ÷ 5 = 5

c

(32)3 = 3

d 30 =

14 Use the index laws to simplify each of the following expressions. a 52 × 55 d

56 55

15 Simplify: a 6 × 50

75

b 67 × 6

c

e (23)4

f

b (6 × 5)0

c 60 × 5

73 (62)3

Cambridge University Press

Number and Algebra

Extended-response questions 1 Two bus companies have different pricing structures. Company A: $120 call-out fee, plus $80 per hour Company B: $80 call-out fee, plus $100 per hour a Write an expression for the total cost $A of travelling n hours with company A. b Write an expression for the total cost $B of travelling for n hours with company B. c Hence state the cost of travelling for three hours with each company. d For how long would you need to hire a bus to make company A the cheaper option? e In the end, a school principal cannot decide which bus company to choose and hires three buses from company A and two buses from company B. Give an expanded expression to find the total cost for the school to hire the five buses for n hours. f If the trip lasts for five hours, how much does it cost to hire the five buses for this period of time?

2 Consider the ﬂoor plan shown. a Write an expanded expression for the ﬂoor’s area in terms of x and y. b Hence find the ﬂoor’s area if x = 6 metres and y = 7 metres. c Write an expression for the ﬂoor’s perimeter in terms of x and y. d Hence find the ﬂoor’s perimeter if x = 6 metres and y = 7 metres. e Another ﬂoor plan is shown at right. Write an expression for the ﬂoor’s area and an expression for its perimeter. f i By how much does the area differ in the two ﬂoor plans? ii By how much does the perimeter differ in the two ﬂoor plans?

y

x

y

x

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61

2

Chapter

Equations 2

What you will learn

2A 2B 2C 2D 2E 2F 2G 2H 2I 2J

Reviewing equations REVISION Equivalent equations REVISION Equations with fractions Equations with pronumerals on both sides Equations with brackets Solving simple quadratic equations Formulas and relationships EXTENSION Applications EXTENSION Inequalities EXTENSION Solving inequalities EXTENSION

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63

NSW Syllabus

for the Australian Curriculum Strand: Number and Algebra Substrand: EQUATIONS

Outcome A student uses algebraic techniques to solve simple linear and quadratic equations. (MA4–10NA)

Protecting sea turtles Loggerhead sea turtles are magniﬁ cent marine predators that feed on shellﬁ sh, crabs, sea urchins and jellyﬁ sh. Sadly, Australian loggerhead turtles have lost more than 50% of their nesting females in the last 10 years. These diminishing numbers mean that loggerhead turtles are now an endangered species. They are under threat for a number of reasons, including human activity on the beaches where the females lay eggs and accidental death in ﬁ shing nets. In order to best work out how to save the loggerhead sea turtle, scientists need to work out what effect various actions, such as closing beaches, will have on the loggerhead turtle population. To do this scientists use mathematics! With a combination of data and equations, they use computer models to predict population numbers of future generations of the loggerhead turtle.

For example, this simple equation describes the number of turtles that there will be next year: F = C (1 + B − D) The pronumerals used in this equation are: • F = future population • C = current population • B = birth rate • D = death rate By mathematically predicting the future of the loggerhead turtle population, environmental scientists can advise governments of the best decisions to help save the loggerhead sea turtle from extinction.

Cambridge University Press

Chapter 2 Equations 2

Pre-test

64

1 Simplifythesealgebraicexpressions. a 9m+2m b 4a−3a d 8a +2a−10 e 4x−7x

c 7n+3n−n f 4p×3q

2 Expandthesealgebraicexpressionsusingthedistributivelaw. a 3(m+4) b 2(a+b) c 3(x+7) d −2(x−3) 3 Expandandsimplify. a 3(m+5)+2m

b 7(2x+3)+3x

4 Simplify: a m+3−m

c 3x÷3

b a+6−6

d −m×(−1)

5 Ithinkofanumber,doubleit,andthenaddthreetoget27.Whatisthenumber? 6 Findthemissingnumbertomakethefollowingequationstrue. b 6× =24 c 6− a 4+ =12

=8

7 Completeeachstatement. a 6+ =0 c 4a÷ =a

b 8× =1 x d × =x 3 8 Solveeachofthefollowingequationsbyinspectionorusingguessandcheck. a x+8=12 b 4x=32 c m−6=−2 d 3m=18 9 Copyandcompletethisworkingtofindthesolution. a 2x− 1=15 +1 +1 2x=_____ ÷2 x=_____

÷2

x +1 b =5 3 ×3 x+1=_____

x=_____

5x − 3 = 7 2 +3 +3 5 x =_____ 2

d 2(x+2)=8

÷2 x+2=_____

5x=_____

x=_____

c

×3

÷2

x=_____

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65

Number and Algebra

2A Reviewing equations

R E V I S I ON

Anequationisastatementthattwothingsareequal,suchas: 2+2=4 7×5=30+5 4+x=10+y. Itconsistsoftwoexpressionsseparatedbytheequalssign(=),anditisconsideredtrueiftheleft-hand sideandright-handsideareequal.Trueequationsinclude7+10=20−3and8=4+4;examplesof falseequationsare2+2=7and10×5=13. Ifanequationhasapronumeralinit,suchas3+x=7,thenasolutiontotheequationisavalueto substituteforthepronumeraltoformatrueequation.Inthiscaseasolutionisx=4because3+4=7isa trueequation.

Let’s start: Solving the equations

■

■ ■

■ ■

■

Anequationisamathematicalstatementthattwoexpressionsareequal,suchas4+x=32.It couldbetrue(e.g.4+28=32)orfalse(e.g.4+29=32). Inanequationthepronumeralissometimescalledanunknown. Afalseequationcanbemadeintoatruestatementbyusingthe≠sign.Forinstance,4+29≠32is atruestatement. Anequationhasaleft-handside(LHS)andaright-handside(RHS). Asolutiontoanequationisavaluethatmakesanequationtrue.Theprocessoffindinga solutioniscalledsolving. Anequationcouldhavenosolutionsoritcouldhaveoneormoresolutions.

Example 1 Classifying equations as true or false Foreachofthefollowingequations,statewhethertheyaretrueorfalse. a 3+8=15−4 b 7×3=20+5 c x+20=3×x,ifx=10 SOLUTION

EXPLANATION

a True

Left-handside(LHS)is3+8,whichis11. Right-handside(RHS)is15−4,whichisalso11. SinceLHSequalsRHS,theequationistrue.

b False

LHS=7×3=21 RHS=20+5=25 SinceLHSandRHSaredifferent,theequationisfalse.

c True

Ifx=10thenLHS=10+20=30. Ifx=10thenRHS=3×10=30. LHSequalsRHS,sotheequationistrue.

Cambridge University Press

Key ideas

• Findanumberthatwouldmaketheequation25=b×(10−b)true. • Howcanyouprovethatthisvalueisasolution? • Trytofindasolutiontotheequation11×b=11+b.

66

Chapter 2 Equations 2

Example 2 Stating a solution to an equation Stateasolutiontoeachofthefollowingequations. a 4+x=25 b 5y=45

c26=3z+5

SOLUTION

EXPLANATION

a x=21

Weneedtofindavalueofxthatmakestheequationtrue. Since4+21=25isatrueequation,x=21isasolution.

b y=9

Ify=9then5y =5×9=45,sotheequationistrue.

c z=7

Ifz=7then3z+5=3×7+5 =21+5 =26 Note:Thefactthatzisontheright-handsideofthe equationdoesnotchangetheprocedure.

Example 3 Writing equations from a description Writeequationsforthefollowingscenarios. a Thenumberkisdoubled,thenthreeisaddedandtheresultis52. b Akiraworksnhours,earning$12perhour.Thetotalsheearnedwas$156.

to 2k+3=52

Thenumberkisdoubled,givingk×2.Thisisthesame as2k. Since3isadded,theleft-handsideis2k+3,whichmust beequalto52accordingtothedescription.

b 12n=156

IfAkiraworksnhoursat$12perhour,thetotalamount earnedis12×n,or12n.

Exercise 2A

U

c 5+3=16÷2 f 2=8−3−3

R

HE

T

1 Classifytheseequationsastrueorfalse. a 5×3=15 b 7+2=12+3 d 8−6=6 e 4×3=12×1

WO

MA

Example 1a, b

REVISION

Cambridge University Press

R K I N G

C

F PS

Y

EXPLANATION

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SOLUTION

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3 Statethevalueofthemissingnumbertomakethefollowingequationstrue. a 5+

=12

b 10×

=90

c

−3=12

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2 Ifthevalueofxis3,whatisthevalueofthefollowing? a 10+x b 3x c 5−x

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4 Considertheequation15+2x=x×x. a Ifx=5,findthevalueof15+2x. b Ifx=5,findthevalueofx×x. c Isx=5asolutiontotheequation15+2x=x×x? d Giveanexampleofanotherequationthathasx=5asasolution.

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6 Ifa=3,statewhetherthefollowingequationsaretrueorfalse. a 7+a=10 b 2a+4=12 c 8−a=5 d 4a−3=9 e 7a+2=8a f a=6−a 7 Someonehasattemptedtosolvethefollowingequations.Statewhetherthesolutioniscorrect(C) orincorrect(I). a 5+2x=4x−1,proposedsolution:x=3 b 4+q=3+2q,proposedsolution:q=10 c 13−2a=a+1,proposedsolution:a=4 d b×(b+3)=4,proposedsolution,b=−4 Example 2

8 Stateasolutiontoeachofthefollowingequations. a 5+x=12 b 3=x−10 d 17=p−2 e 10x=20 g 4u+1=29 h 7k=77

c 4v+2=14 f 16−x=x i 3+a=2a

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5 Ifx=2,statewhetherthefollowingequationsaretrueorfalse. a 7x=8+3x b 10−x=4x c 3x=5−x d x+4=5x e 10x=40÷x f 12x+2=15x

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10 Findthevalueofthenumberinthefollowingstatements. a Anumberistripledtoobtaintheresult21. b Halfofanumberis21. c Sixlessthananumberis7. d Anumberisdoubledandtheresultis−16. e Three-quartersofanumberis30. f Sixmorethananumberis−7. 11 Berkeleybuysxkgoforangesat$3.20perkg.Hespendsatotal of$9.60. a Writeanequationinvolvingx todescribethissituation. b Stateasolutiontothisequation. 12 Emily’sagein10years’timewillbetriplehercurrentage.Sheis currentlyEyearsold. a WriteanequationinvolvingEtodescribethissituation. b Findasolutiontothisequation. c HowoldisEmilynow? d Howmanyyearswillshehavetowaituntilsheisfourtimeshercurrentage? 13 Findtwopossiblevaluesoftthatmaketheequation“t(10−t)=21true”. 14 a b c d e

Whyisx=3asolutiontox2=9? Whyisx=−3asolutiontox2=9? Findthetwosolutionstox2=64(hint:oneisnegative). Explainwhyx2=0hasonlyonesolutionbutx2=1hastwo. Explainwhyx2=−9hasnosolutions.(Hint:considerpositiveandnegativemultiplication)

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9 Writeequationsforeachofthefollowingproblems.Youdonotneedtosolvetheequations. a Anumberxisdoubledandthen7isadded.Theresultis10. b Thesumofxandhalfofxis12. c Aston’sageisa.Hisfather,whois25yearsolder,istwiceas oldasAston. d Fel’sheightishcmandherbrotherPatis30cmtaller.Pat’s heightis147cm. e Coffeecosts$cpercupandteacosts$t.Fourcupsofcoffee andthreecupsofteacostatotalof$21. f Chairscost$ceach.Topurchase8chairsanda$2000table costsatotalof$3600.

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Example 3

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16 a The equation p × (p + 2) = 3 has two solutions. State the two solutions. Hint: One of them is negative. b How many solutions are there for the equation p + (p + 2) = 3? c Try to find an equation that has three solutions.

Enrichment: More than one unknown 17 a There are six equations in the square below. Find the values of a, b, c, d and e to make all six equations true. a

+

× 2

=

÷ ×

= d

12

b =

÷

e

22 −

=

c =

=

10

b Find the value of f that makes the equation a × b × e = c × d × f true. 18 For each of the following pairs of equations, find values of c and d that make both equations true. More than one answer may be possible. a c + d = 10 and cd = 24 b c − d = 8 and c + d = 14 c c ÷ d = 4 and c + d = 30 d cd = 0 and c − d = 7

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Explain why the equation x + 3 = x has no solutions. Explain why the equation x + 2 = 2 + x is true, regardless of the value of x. Show that the equation x + 3 = 10 is sometimes true and sometimes false. Classify the following equations as always true (A), sometimes true (S) or never true (N). i x + 2 = 10 ii 5 − q = q iii 5 + y = y iv 10 + b = 10 v 2 × b = b + b vi 3 − c = 10 vii 3 + 2z = 2z + 1 viii 10p = p ix 2 + b + b = (b + 1) × 2 e Give a new example of another equation that is always true.

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2B Equivalent equations

R E V I S I ON

Ifwehaveanequation,wecanobtainanequivalentequationbyperformingthesameoperationtoboth sides.Forexample,ifwehave2x+4=20,wecanadd3tobothsidestoobtain2x+7=23. Thenewequationwillbetrueforexactlythesamevaluesofxastheoldequation.Thisobservation helpsustosolveequationssystematically.Forexample,2x+4=20isequivalentto2x=16(subtract4 frombothsides),andthisisequivalenttox=8(dividebothsidesby2).Thebottomequationisonlytrue ifxhasthevalue8,sothismeansthesolutiontotheequation2x+4=20isx=8.Wewritethisas: 2 +4=20 2x −4

−4

2 = 16 2x ÷2

÷2 x =8

Let’s start: Attempted solutions Belowarethreeattemptsatsolvingtheequation4x−8=40.Eachhasaproblem. Attempt 3 Attempt 1 Attempt 2 4x −8=40 +8

+8

+8

4x =48 −4

−8

÷4

÷4

÷4 x −8=10

4x =32 −4

x =44

4x −8=40

4x −8=40

÷4

+8

x =8

+8 x =18

Key ideas

• Canyouprovethattheseresultsarenotthecorrectsolutionstotheequationabove? • Foreachone,findthemistakethatwasmade. • Canyousolve4x−8=40systematically?

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Twoequationsareequivalentifyoucangetfromonetotheotherbyrepeatedly: – addinganumbertobothsides – subtractinganumberfrombothsides – multiplyingbothsidesbyanumberotherthanzero – dividingbothsidesbyanumberotherthanzero – swappingtheleft-handsideandright-handsideoftheequation. Tosolveanequationsystematically,youshouldrepeatedlyfindanequivalentequationthatis simpler.Forexample: 5x +2=32 −2 −2 5x =30 ÷5 ÷5 x =6 Tocheckasolutionsubstitutetheunknown’svalueintoseeiftheequationistrue, e.g.LHS=5(6)+2andRHS=32.

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Number and Algebra

Example 4 Finding equivalent equations Showtheresultofapplyingthegivenoperationtobothsidesoftheseequations. a 8y=40 [÷8] b 10+2x=36 [−10] c 5a−3=12 [+3] SOLUTION

EXPLANATION

8y = 40

a ÷8

Writetheequationoutandthendividebothsidesby8. 40÷8is5and8y÷8isy

÷8 y =5 10+ 2x 2 =36

b −10

−10 2 = 26 2x 5a − 3 = 12

c +3

Writetheequationoutandthensubtract10fromboth sides. 36−10is26 10+2x−10is2x Writetheequationoutandthenadd3tobothsides. 12+3is15 5a−3+3is5a

+3 5a = 15

Example 5 Solving equations systematically Solvethefollowingequationsandcheckthesolutionbysubstituting. a x−4=16 b 2u+7=17 c 40−3x=22 SOLUTION a

EXPLANATION

x −4=16 +4

+4 x =20 Sothesolutionisx=20. 2u +7=17

b −7

2u =10 ÷2

u =5 Sothesolutionisu=5. c

Check:2(5)+7=10+7=17

40−3x =22 −40

−40 −3x = −18

÷ −3

Check:20−4=16 Toremovethe+7,wesubtract7frombothsides. Finallywedivideby2toreversethe2u.Rememberthat 2umeans2×u.

−7

÷2

Byadding4tobothsidesoftheequation,wegetan equivalentequation.

÷ −3 x =6 Sothesolutionisx=6.

Wesubtract40frombothsidestoremovethe40atthe startoftheLHS. Since−3×x=−18,wedivideby−3togetthefinal solution. Check:40−3(6)=40−18=22

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2 Foreachequationfillintheblanktogetanequivalentequation. a b 5x =10 10−2x 2 =20 2x +2 +2 +5 +5 5x +2= __ 15−2x 2 = __ 2x c d 3q +4=__ 7z +12=4z +10 −4 −4 −10 3q = 12 7z +2= __

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1 Foreachofthefollowingequationsadd4tobothsidestoobtainanequivalentequation. a 3x=10 b 7+k=14 c 5=2x

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3 Considertheequation4x=32. a Copyandcompletethefollowingworking. 4x =32 ÷4 ÷4 x = __ b Whatisthesolutiontotheequation4x=32? 4 Tosolvetheequation10x+5=45,whichofthefollowingoperationswouldyoufirstapplytoboth sides? A Divideby5 B Subtract5 C Divideby10 D Subtract45

6 Copyandcompletethefollowingtosolvethegivenequationsystematically. a b 10 =30 10x q +5=2 ÷10 ÷10 −5 −5 x = __ __= __ 4x +2=22

c

30=4p 4 +2 4p

d −2

−2

−2

÷4

e

__ =__

÷4

__ =__

20−4x =8 −20

−20 −4x = −12

÷ −4

x =__

−2 __=__

4x =__

f

p ÷3+6=8 −6

−6 p ÷3= 2

÷ −4 __ =__

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5 Foreachequation,showtheresultofapplyingthegivenoperationtobothsides. a 10+2x=30 [−10] b 4+q=12 [−2] c 13=12−q [+5] d 4x=8 [×3] e 7p=2p+4 [+6] f 3q+1=2q+1 [−1]

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8 Solvethefollowingequationssystematically.Checkyoursolutionsusingsubstitution. a 5+9h=32 b 9u−6=30 c 13=5s−2 d −18=6−3w e −12=5x+8 f −44=10w+6 g 8=−8+8a h 4y−8=−40

Example 5c

9 Solvethefollowingequationssystematicallyandcheckyoursolutions. a 20−4d=8 b 34=4−5j c 21−7a=7 d 6=12−3y e 13−8k=45 f 44=23−3n g 13=−3b+4 h −22=14−9b

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d 11=k+2 h g÷3=−3

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7 Solvethefollowingequationssystematically. a a+5=8 b t×2=14 c 7=q−2 e 19=x+9 f −30=3h g −36=9

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10 Thefollowingequationsdonotallhavewholenumbersolutions.Solvethefollowingequations systematically,givingeachsolutionasafraction. a 2x+3=10 b 5+3q=6 c 12=10b+7 d 15=10+2x e 15=10−2x f 13+2p=−10 g 22=9+5y h 12−2y=15

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11 Foreachofthefollowing,writeanequationandsolveitsystematically. a Thesumofpand8is15. b Theproductofqand−3is12. c 4issubtractedfromdoublethevalueofkandtheresultis18. d Whenr istripledand4isadded,theresultis34. e Whenxissubtractedfrom10,theresultis6. f Whentripleyissubtractedfrom10,theresultis16.

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12 Solvethefollowingequationssystematically.Morethantwostepsareinvolved. a 14×(4x+2)=140 b 8=(10x−4)÷2 c −12=(3−x)×4 13 Thefollowingshapesarerectangles.Bysolvingequationssystematically,findthevalueofthe variables.Someoftheanswerswillbefractions. a b 13x 10 5 +3 5y

17 2x − 4

c

4x 2x

Perimeter = 40

10x + 5

25 y d

25 4q − 1

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70 − 3p 3 14 Sidneyworksfor10hoursatthenormalrateofpay($xperhour)andthenthenextthreehours atdoublethatrate.Ifheearnsatotalof$194.88,writeanequationandsolveittofindhisnormal hourlyrate.

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Example 5a

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7x +4=39

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15 a Prove that 7x+4=39and−2x+13=3areequivalentby fillinginthemissingsteps.

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−4 7x =35

÷7

÷7 ___=___

× −2 +13

× −2 ___=___ +13

−2x 2 +13=3 2x

b Provethat10k+4=24and3k−1=5areequivalent. 16 a P rovethat4x+3=11and2x=4areequivalent.Trytousejusttwostepstogetfromone equationtotheother. b Aretheequations5x+2=17andx=5equivalent? c Provethat10−2x=13and14x+7=20arenotequivalent,nomatterhowmanystepsareused. 17 Astudenthastakentheequationx=5andperformedsomeoperationstobothsides: x =5 ×4

×4 4x =20

+3 ×2 a b c d

+3

4x +3=23 (4x +3)×2=46

×2

Solve(4x+3)×2=46systematically. Describehowthestepsyouusedinyoursolutioncomparewiththestepsthestudentused. Giveanexampleofanotherequationthathasx=5asitssolution. Explainwhythereareinfinitelymanydifferentequationswiththesolutionx=5.

Enrichment: Dividing whole expressions 18 Itispossibletosolve2x+4=20byfirstdividingbothsidesby2,aslongaseverytermisdivided by2.Soyoucouldsolveitineitherofthesefashions. 2 +4=20 2x −4 2 =16 2x ÷2

2 +4=20 2x −4 ÷2

x =8

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÷2

÷2 x + 2=10

−2

−2 x =8

Notethat2x+4dividedby2isx+2,notx+4.Usethismethodofdividingfirsttosolvethe followingequationsandthencheckthatyougetthesameanswerasifyousubtractedfirst. a 2x+6=12 b 4x+12=16 c 10x+30=50 d 2x+5=13 e 5x+4=19 f 3+2x=5 g 7=2x+4 h 10=4x+10

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2C Equations with fractions x x Recallfromalgebrathatafractionsuchas representsx÷3.Thismeansthattosolveanequationwith 3 3 ononeside,weshouldfirstmultiplybothsidesby3.Forexample: x = 10 3 ×3 ×3 x =30

Let’s start: Practising with fractions • Ifxisanumbergreaterthan1,evaluatetheseexpressionsandputthemintoascendingorder(smallest tolargest): 2x + 1 2

x 2 + 1 2

2 + 2x 2 2

2 x +1

1 2 x + 2

■ ■

a meansa÷b. b Tosolveanequationwithafractionononeside,multiplybothsidesbythedenominator. q = 12 4 ×4 ×4 q =48

Example 6 Solving equations with fractions Solvethefollowingequationssystematically. 4x = 8 a b 3 5x = 29 c 4+ d 2 SOLUTION a ×3

7−

2x = 5 3

EXPLANATION 4x =8 3

×3

4x =24 ÷4

4 y + 115 =3 9

x =6

Multiplyingbothsidesby3removesthedenominator of3. Bothsidesaredividedby4tosolvetheequation.

÷4

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Key ideas

• Investigatehowtheordermightchangeifxcouldbeanumberlessthan1.

Chapter 2 Equations 2

4 y + 115 =3 9

b ×9

×9

Multiplyingbothsidesby9removesthedenominator of9.

−15

Theequation4y+15=27issolvedintheusualfashion (subtract15,divideby4).

4y +15=27 −15 4y =12 ÷4

c

y =3

÷4

5x = 29 2 −4 5x = 25 2 ×2

4+ −4 ×2

Wemustsubtract4firstbecausewedonothaveafraction byitselfontheleft-handside.Oncethereisafractionby itself,multiplybythedenominator(2).

5x = 50 5

÷5 x =10 d

2x =5 3 2x − =−2 3 2x =2 3 7−

−7 × −1 ×3

−7

Subtract7firsttogetafraction.Whenbothsidesare negative,multiplying(ordividing)by−1makesthemboth positive.

× −1 ×3

2 =6 2x x =3

÷2

Exercise 2C

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1 a Ifx=4,findthevalueof

m =2 5 m = __ p =7 10 p = __

×5

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d h l

k =3 6 5v = 15 4 −6 f = − 24 5

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a =3 5 3s = −9 2 −7 j =7 5

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5 Solvethefollowingequationssystematically.Checkyoursolutionsbysubstituting. c−7 h + 10 a + 12 a t − 8 = − 10 b c d = 4 =2 = −5 3 5 2 2 5j + 6 s−2 4n 7v e −1 = f =2 g 3= h + 10 −6=−2 8 8 12 9 4r 3 − 112 f − 15 i 7q + 112 = − 6 j −4 = k 15 = l 9− =5 7 3 5 5 5k + 4 3 + 113b m − 6 = 5x − 8 n 5u − 7 = − 2 o p 20 = = −3 −8 −7 −7 −4 7m g−3 −7y p − 15 q t − 12 = − 5 r 4 + = −1 = −3 s 4 = 12 5 8 −3

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7 Theaverageoftwonumberscanbefoundbyaddingthemandthendividingtheresultby2. x+5 a Iftheaverageofxand5is12,whatisx?Solvetheequation = 12 tofindout. 2 b Theaverageof7andpis−3.Findpbywritingandsolvinganequation. c Theaverageofanumberanddoublethatnumberis18.Whatisthatnumber? d Theaverageof4xand6is19.Whatistheaverageof6xand4?(Hint:Findxfirst.)

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6 Forthefollowingpuzzles,writeanequationandsolveittofindtheunknownnumber. a Anumberxisdividedby5andtheresultis7. b Halfofyis−12. c Anumberpisdoubledandthendividedby7.Theresultis4. d Fourisaddedtox.Thisishalvedtogetaresultof10. e xishalvedandthen4isaddedtogetaresultof10. f Anumberkisdoubledandthen6isadded.Thisresultishalvedtoobtain−10.

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Example 6b,c,d

4 Solvethefollowingequationssystematically. b g a =4 =2 b c 5 10 2 7w =−7 e =8 f g 5 10 2n 3m i =6 j =4 k 7 7

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Example 6a

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3 Matcheachoftheseequationswiththecorrectfirststeptosolveit. x x x−4 x a =7 b c d +4=3 =5 −4=7 4 4 2 2 A Multiplybothsidesby2. B Add4tobothsides. C Multiplybothsidesby4. D Subtract4frombothsides.

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x 2 = 10 anddividedbothsidesby2. 3 x a Solve 2 = 10 . 3

2x b Isthesolutionthesameasthesolutionfor = 10 ifbothsidesarefirstmultipliedby3? 3 147q c Solve = 1470byfirst: 13 i multiplyingbothsidesby13 ii dividingbothsidesby147 d Whatisoneadvantageindividingfirstratherthanmultiplying? e Solvethefollowingequations. −4 p 20 p 123r 13q =4 = 40 ii iii iv = 246 i = − 39 77 14 17 27 10 Tosolveanequationwithavariableonthedenominatorwecanfirstmultiplybothsidesbythat variable. 30 = 10 x ×x ×x ÷10

30 = 10 x

÷10

3= x

Usethismethodtosolvetheequations. 12 −15 a =2 b = −5 x x d

4+

20 = 14 x

e

16 +1 = 3 x

c

1 +3= 4 x

f 5 =

−10 +3 x

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8 Arestaurantbillof$100istobepaid.Blakeputsinone−thirdoftheamountinhiswallet, leaving$60tobepaidbytheotherpeopleatthetable. a Writeanequationtodescribethissituation,ifbrepresentstheamountinBlake’swallet beforehepays. b Solvetheequationsystematically,andhencestatehowmuchmoneyBlakehasinhis wallet.

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Enrichment: Fractional solutions 11 Solvethefollowingequations.Notethatthesolutionsshouldbegivenasfractions. x 1 10 − 3 x 4x + 3 8 + 3x a b c 7 = + d 2 = = 12 =6 4 3 4 5 5 12 RecallfromSectionIE(Addingandsubtractingalgebraicfractions)thatalgebraicfractionscanbe combinedbyfindingacommondenominator,forexample: 2 x 5 x 8 x 15 x + = + 3 4 12 12 23 x = 12 Usethissimplificationtosolvethefollowingequations. a

2 x 5x += 46 3 4

b

x x + = 22 5 6

c 10 =

d

4=

x x − 2 3

e

6x 2x + = 28 5 3

f 4 =

x x + 2 3

3x x − 7 3

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2D Equations with pronumerals on both sides Sofaralltheequationswehaveconsideredinvolvedapronumeraleitherontheleft-handside, e.g.2x +3=11orontherightside,e.g.15=10−2x.Buthowcanyousolveanequationwithpronumerals onbothsides,e.g.12+5x=16+3x?Theideaistolookforanequivalentequationwithpronumeralson justoneside. Theequation12+5x=16+3xcanbethoughtofasbalancingscales. x x x

x x

12

16

x x

x

Then3xcanberemovedfrombothsidesofthisequationtoget:

12

x x

16

Theequation12+2x=16isstraightforwardtosolve.

Let’s start: Moving pronumerals

Key ideas

Youaregiventheequation11+5x=7+3x. • Canyoufindanequivalentequationwithxjustontheleft-handside? • Canyoufindanequivalentequationwithxjustontheright-handside? • Trytofindanequivalentequationwith9xontheleft-handside. • Doalloftheseequationshavethesamesolution?Trytofindit.

Ifbothsidesofanequationhaveapronumeraladdedorsubtracted,thenewequationwillbe equivalent totheoriginalequation. Ifpronumeralsareonbothsidesofanequation,addorsubtractitsothatthepronumeralappears ononlyoneside.Forexample:

■

■

-2a

10 + 5a = 13 + 2a 10 + 3a = 13

■

-2a

+3b

4b +12=89−3b

+3b

7b +12 = 89

Sometimesitiswisetoswaptheleft-handsideandright-handside.

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Example 7 Solving equations with pronumerals on both sides Solvethefollowingequations. a 7t+4=5t+10

b 6x+4=22−3x

SOLUTION a

c 2u=7u−20

EXPLANATION

7t +4=5t +10 −5t

Pronumeralsareonbothsidesoftheequation,so subtract5tfrombothsides.

−5t 2t + 4=10

−4

−4

Once5tissubtracted,theusualprocedureisappliedfor solvingequations. LHS=7 × 3+4 RHS=5×3+10 =25 =25

2t = 6 ÷2

÷2 t =3

6x +4=22−3x

b +3x

Pronumeralsareonbothsides.Toremove3x,weadd3x tobothsidesoftheequation. OncepronumeralsarejustontheLHS,theusual procedureisappliedforsolvingequations. Alternatively,6xcouldhavebeensubtractedfromboth sidesoftheequationtoget4=22−9x. LHS=6 × 2+4 RHS=22−3×2 =16 =16

+3x 9x + 4=22 −4

−4

9x = 18

÷9

÷9 x =2

2u =7u −20

c

Choosetoremove2ubysubtractingit.

−2u

−2u 0 = 5u −20

Notethat2u−2uisequalto0,sotheLHSofthenew equationis0. LHS=2 × 4 RHS=7×4−20 =8 =8

+20

+20 20=5th ÷5

÷5

4= u ∴ u=4

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3 +9=5−2p 3p 2 2p +2p 2 2p ___=___

9q +5=12q +21 −9q ___=3q +21

___=8 +2p 2 2p

d

+7k

15k +12 = 13−7k

+7k

___=___

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2 Fillintheblanksfortheseequivalentequations. a b 5x +3=2x 2 +8 2x −9q −2x 2 2x −2x 2 2x

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1 Ifx=3,arethefollowingequationstrueorfalse? a 5+2x=4x−1 b 7x=6x+5 c 2+8x=12x d 9x−7=3x+11

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5 Solvethefollowingequationssystematically,checkingyoursolutionsusingsubstitution. a 9+4t=7t+15 b 2c−2=4c−6 c 6t−3=7t−8 d 7z−1=8z−4 e 8t−24=2t−6 f 2q−5=3q−3 g 5x+8=6x−1 h 8w−15=6w+3 i 6j+4=5j−1

Example 7c

6 Solvethefollowingequationssystematically.Yoursolutionsshouldbecheckedusingsubstitution. a 1−4a=7−6a b 6−7g=2−5g c 12−8n=8−10n d 2+8u=37+3u e 21−3h=6−6h f 37−4j=7−10j g 13−7c=8c−2 h 10+4n=4−2n i 10a+32=2a j 10v+14=8v k 18+8c=2c l 2t+7=22−3t m 6n−47=9−8n n 3n=15+8n o 38−10=10+4

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4 Solvethefollowingequationssystematicallyandcheckyoursolutions. a 10f+3=23+6f b 10y+5=26+3y c 7s+7=19+3s d 9j+4=4j+14 e 2t+8=8t+20 f 4+3n=10n+39 g 4+8y=10y+14 h 5+3t=6t+17 i 7+5q=19+9q

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7 Solvethefollowingequationssystematically.Youranswersshouldbegivenasfractions. a 3x+5=x+6 b 5k−2=2k c 3+m=6+3m d 9j+4=5j+14 e 3−j=4+j f 2z+3=4z−8

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8 Writeanequationandsolveitsystematicallytofindtheunknownnumberintheseproblems. a Doublingxandadding3isthesameastriplingxandadding1. b Ifzisincreasedby9,thisisthesameasdoublingthevalueofz. c Theproductof7andyisthesameasthesumofyand12. d Whenanumberisincreasedby10,thishasthesameeffectastriplingthenumberand subtracting6.

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10 Findtheareaandtheperimeterofthisrectangle. 8x − 21 2 −4 2y

y+8 4x + 7

11 Atanewsagency,Preetabought4pensanda $1.50newspaper,whileherhusbandLevybought 2pensanda$4.90magazine.Totheirsurprise thecostwasthesame. a Writeanequationtodescribethis,usingpforthe costofasinglepen. b Solvetheequationtofindthecostofpens. c IfFredhasa$20note,whatisthemaximum numberofpensthathecanpurchase?

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14 a Trytosolvetheequation4x+3=10+4x. b Thistellsyouthattheequationyouaretryingtosolvehasnosolutions(because10=3isnever true).Provethat2x+3=7+2x hasnosolutions. c Giveanexampleofanotherequationthathasnosolutions.

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Enrichment: Geometric equations 15 Findthevaluesofthepronumeralsinthefollowinggeometricdiagrams. a

(6x − 5)°

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(2z + 7) 7)° (9x)°

(4x + 35)°

(3y)°

( + 10)° (x c

(6k − 32)°

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(3k + 4)°

(6b + 5)° e

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f a° (100 − 2x)° (2b)°

(26x + 2)°

(3c)°

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2E Equations with brackets InChapter1itwasnotedthatexpressionswithbrackets couldbeexpandedbyconsideringrectangleareas.

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4 × 2 Area = 4(x 4( + 2) = 8 Area = 4x + 8

4 × x = 4x

So4(x+2)and4x+8areequivalent.Thisbecomes quitehelpfulwhensolvinganequationlike4(x+2)=5x+1. Wejustsolve4x+8=5x+1usingthetechniques fromtheprevioussection.

Let’s start: Architect’s dilemma

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Toexpandbrackets,usethedistributive law,whichstatesthat: – a(b+c)=ab+ac.Forexample,3(x+4)=3x+12.

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– a(b−c)=ab−ac.Forexample,4(b−2)=4b−8. b Like termsaretermsthatcontainexactlythesamepronumeralsandcanbecollectedtosimplify expressions.Forexample,5x+10+7xcanbesimplifiedto12x +10. Equationsinvolvingbracketscanbesolvedbyfirstexpandingbracketsandcollectinglike terms.

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Key ideas

Inthefollowinghouseplans,thekitchenand diningroomareseparatedbyadividingdoor. • Ifthebreadthofthedividerisx,whatistheareaofthe kitchen?Whatistheareaofthediningroom? • Trytofindthebreadthofthedivideriftheareasof thetworoomsareequal. • Isiteasiertosolve3(7+x)=4(x +4)or 21+3x=4x+16?Whichofthesedidyousolvewhen tryingtofindthebreadthofthedivider?

Chapter 2 Equations 2

Example 8 Solving equations with brackets Solvethefollowingequationsbyfirstexpandinganybrackets. a 3(p+4)=18 b −12(3q+5)=−132 c 4(2x−5)+3x=57 d 2(3k+1)=5(2k−6) EXPLANATION

a

Usethedistributivelawtoexpandthebrackets. Solvetheequationbyperformingthesameoperationsto bothsides.

3(p+4)=18 3 +12=18 3p −12 −12 3 =6 3p ÷3 ÷3 p =2

b

−12(3q+5)=−132 −36q+(−60)=−132 −36q −60= −132 +60 +60 −36q = −72 ÷ −36 ÷ −36 q =2

Usethedistributivelawtoexpandthebrackets. Simplify−36q+(−60)to−36q−60. Solvetheequationbyperformingthesameoperationsto bothsides.

c

4(2x−5)+3x=57 8x−20+3x=57 11x −20=57 +20 +20 11x =77 ÷11 ÷11 x =7

Usethedistributivelawtoexpandthebrackets. Combinetheliketerms:8x+3x=11x. Solvetheequationbyperformingthesameoperationsto bothsides.

d

2(3k+1)=5(2k−6) 6k +2=10k −30 −6k −6k 2 =4k −30 +30 +30 32=4k ÷4 8= k ÷4

Usethedistributivelawonbothsidestoexpandthe brackets. Solvetheequationbyperformingthesameoperationsto bothsides.

Exercise 2E

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b Whatisthecorrectexpansionof4(x+3)? 4 Simplifythefollowingexpressionsbycollectingliketerms. a 4x+3x b 7p+2p+3 c 8x−2x+4 d 2k+4+5k e 3x+6−2x f 7k+21−2k

Example 8b

6 Solvethefollowingequationsinvolvingnegativenumbers. a −6(4p+4)=24 b −2(4u−5)=34 c −2(3v−4)=38 d 28=−4(3r+5) e −3(2b−2)=48 f −6=−3(2d−4)

Example 8c

7 Solvethefollowingequationsbyexpandingandcombiningliketerms. a 4(3y+2)+2y=50 b 5(2−5)+3=1 c 4(5+3w)+5=49 d 49=5(3c+5)−3c e 28=4(3d+3)−4d f 58=4(2w−5)+5w g 23=4(2p−3)+3 h 44=5(3k+2)+2k i 49=3(2c−5)+4

Example 8d

8 Solvethefollowingequationsbyexpandingbracketsonbothsides. a 5(4x−4)=5(3x+3) b 6(4+2r)=3(5r+3) c 5(5f−2)=5(3f+4) d 4(4p−3)=2(4+3p) e 2(5h+4)=3(4+3h) f 4(4r−5)=2(5+5r) g 4(3r−2)=4(2r+3) h 2(2p+4)=2(3p−2) i 3(2a+1)=11(a−2)

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5 Solvethefollowingequationsbyfirstexpandingthebrackets. a 2(4u+2)=52 b 3(3j−4)=15 c 5(2p−4)=40 d 15=5(2m−5) e 2(5n+5)=60 f 26=2(3a+4)

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9 Solvethefollowingequationssystematically. a 2(3+5r)+6=4(2r+5)+6 b 3(2+2)+18=4(4+3)−8 c 2(3x−5)+16=3+5(2x−5) d 3(4s+3)−3=3(3s+5)+15 e 4(4y+5)−4=6(3y−3)+20 f 3(4h+5)+2=14+3(5h−2)

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10 Desmondnotesthatin4years’timehisagewhendoubledwillgivethenumber50. Desmond’scurrentageisd. a WriteanexpressionforDesmond’sagein4years’time. b Writeanexpressionfordoublehisagein4years’time. c Writeanequationtodescribethesituationdescribedabove. d Solvetheequationtofindhiscurrentage.

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11 Rahda’susualhourlywageis$w.Sheworksfor5hoursatthis wageandthen3morehoursatanincreasedwageof$(w+4). a WriteanexpressionforthetotalamountRahdaearnsforthe 8hours. b Rahdaearns$104forthe8hours.Writeandsolveanequation tofindherusualhourlywage.

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12 Kate’sage5yearsago,whendoubled,isequaltotripleherage 10yearsago. a Writeanequationtodescribethis,usingkforKate’scurrent age. b SolvetheequationtofindKate’scurrentage.

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15 Ajithclaimsthatthreetimeshisage5yearsagoisthesameasninetimeshowoldhewillbe nextyear.ProvethatwhatAjithissayingcannotbetrue. 16 Acommonmistakewhenexpandingistowrite2(n+3)as2n+3.Thesearenotequivalent, since,forexample,2(5+3)=16and2×5+3=13. a Provethattheyareneverequalbytryingtosolve2(n+3)=2n+3. b Provethat4(2x+3)isneverequalto8x+3butitissometimesequalto4x+12.

Enrichment: Challenging expansions 17 Solvethefollowingequations.Notethat,ingeneral,youranswerswillnotbeintegers. a 2(3x+4)+5(6x+7)=64x+1 b −5(3p+2)+5(2p+3)=−31 c −10(n+1)+20(2n+13)=7 d 4(2q+1)−5(3q+1)=11q−1 e x+2(x+1)+3(x+2)=11x f m−2(m+1)−3(m−1)=2(1−4m)

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14 Abrahamisaskedhowmanypeopleareintheroomnextdoor.Heanswersthatifthreemore peoplewalkedinandthentheroom’spopulationwasdoubled,thiswouldhavethesameeffect asquadruplingthepopulationandthen11peopleleaving.ProvethatwhatAbrahamsaid cannotbetrue.

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2F Solving simple quadratic equations Mostoftheequationsyouhaveworked withsofararecalledlinearequations like2x−3=7and5(a+2)=7(a−1) wherethepowerofthepronumeralis 1andthereisusuallyasinglesolution. Anothertypeofequationisoftheform x2=candthisisanexampleofasimple quadraticequation.Notethatthepower ofthepronumeralxis2.Dependingon thevalueofc,xcanhavezero,oneor twosolutions.Thesetypesofequations appearfrequentlyinmathematicsand inproblemsinvolvingdistance,area, graphsandmotion.

Let’s start: How many solutions?

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x2=cisasimplequadratic equation. – Ifc>0,thentherearetwosolutions:x = c andx=− c . Forexample:Thesolutionsforx2=16arex=4orx=−4(thiscanbewrittenasx=±4) because(−4)2=16and42=16. – Ifc=0,thenthereisonesolution,x=0. – Ifc<0,thentherearenosolutionsbecausex2willalwaysbepositive.

Example 9 Solving x 2 = c if c > 0 Solvethefollowingequations.Roundto2decimalplacesinpartbbyusingacalculatortoassist. a x2=81 b x2=23 SOLUTION

EXPLANATION

a x=9orx=−9

Since81isapositivenumber,theequationhastwo solutions.Both9and−9squaretogive81.

b x= 23 =4.80(to2decimalplaces) orx=− 23 =−4.80(to2decimalplaces)

Thenumber23isnotaperfectsquareso 23 can beroundedifrequired.

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Key ideas

Considertheequationx2=c.Howmanyvaluesofxcanyouthinkofthatsatisfytheequationif: • c=0? • c=9? • c=−4? Whatconclusionscanyoucometoregardingthenumberofsolutionsforxdependingonthevalueofc?

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Example 10 Stating the number of solutions c x2=7

EXPLANATION

a 0solutions

Inx2=c,ifc<0therearenosolutionsbecauseany numbersquaredispositiveorzero.

b 1solution

x=0istheonlysolutiontox2=0

c 2solutions

Both 7 and− 7 squaretogive7.

Exercise 2F

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2 a Useacalculatortomultiplythesenumbersbythemselves.Recallthataneg×neg=pos. i −3 ii 7 iii 13 iv −8 b Didyouobtainanynegativenumbersinparta? 3 Writeinthemissingnumbers. a (−3)2=_____and32=9soifx2=9thenx=_____orx=_____ b (5)2=_____and(−5)2=25soifx2=25thenx =_____orx=_____ c (11)2=121and(−11)2=____soifx2=121thenx=_____orx=_____

Example 10

c x2=100 g x2=36 k x2=900

d x2=64 h x2=121 l x2=10000

5 Solvethefollowingandroundto2decimalplaces. a x2=6 b x2=12 c x2=37 e x2=104 f x2=317 g x2=390

d x2=41 h x2=694

6 Statethenumberofsolutionsfortheseequations. a x2=10 b x2=4 c x2 =3917 e x2=−94 f x2=0 g a2=0

d x2=−4 h y2=1

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4 Solvethefollowingequations. a x2=4 b x2=49 e x2=1 f x2=144 2 i x =169 j x2=256

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9 Byfirstdividingbothsidesbythecoefficientofx2,solve thesesimplequadraticequations. a 2x2=8 b 3x2=3 c 5x2=45 d −3x2=−12 2 e −2x =−50 f 7x2=0 g −6x2=−216 h −10x2=−1000

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11 Solvetheequationswiththegivenconditions. a x2=16ifx>0 b x2=25ifx<0 2 c x =49ifx<0 d x2=196ifx>0 12 Theexactvaluesolutionstox2=5,forexample,arewrittenasx= 5 or− 5 .Alternatively wecanwritex=± 5 . Writetheexactvaluesolutionstotheseequations. a x2=11 b x2=17 c x2=33

d x2 =156

13 Thetriad(a,b,c)=(3,4,5)satisfiestheequationa2+b2=c2because32+42=52. a Decideifthefollowingtriadsalsosatisfytheequationa2+b2=c2. i (6,8,10) ii (5,12,13) iii (1,2,3) iv (−3,−4,−5) v (−2,−3,−6) vi (−8,15,17) b Canyoufindanytriads,inwhicha,b,carepositiveintegers,thatsatisfya3+b3=c3?

Enrichment: Solving more complex linear equations 14 Comparethislinearandquadraticequationsolution. 3x−1=11 11 3x 3 2−1=11 +1 +11 +1 +1 3x=12 12 33x2=12 ÷3 ÷3 x=4 ÷3 x2=4 ÷3 x=±2 Nowsolvethesequadraticequations. a 2x2+1=9 b 5x2−2=3 2 d −x +1=0 e −2x2+8=0 g 4−x2=0 h 27−3x2=0

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2G Formulas and relationships

E X T E N S I ON

Someequationsarecalledformulas.Aformulashowstherelationship A=×b b betweentwoormorevariables.Forexample,youknowfrommeasurement P = 2 + 2b thattheareaofarectangleisrelatedtoitslengthandbreadth,givenbythe formulaA=×banditsperimeterisgivenbyP=2+2b. Althoughtheseareoftenusedasadefinitionfortheareaandadefinition fortheperimeter,theyarealsojustequations−twoexpressionswrittenoneithersideofanequalssign.

Let’s start: Rectangular dimensions

Key ideas

YouknowthattheareaandperimeterofarectanglearegivenbyA=×bandP=2+2b. • If=10andb=7,findtheperimeterandthearea. • If=2andb=8,findtheperimeterandthearea. • Noticethatsometimestheareavalueisbiggerthantheperimetervalueandsometimestheareavalueis lessthantheperimetervalue.If=10,isitpossibletomaketheareaandtheperimetervaluesequal? • If=2,canyoumaketheareaandtheperimeterequal?Discuss.

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Aformulaorruleisanequationcontainingtwoormorepronumerals,oneofwhichisthe subjectoftheequation. Thesubjectofaformulaisapronumeralthatoccursbyitselfontheleft-handside, e.g.VisthesubjectofV=3x+2y. Touseaformulasubstitutealltheknownvaluesandthensolvetheequationtofindtheunknown value.

Example 11 Applying a formula Applytheformulaforarectangle’sperimeterP =2 +2btofind: a Pwhen =4andb =7 b whenP =40andb =3 SOLUTION

EXPLANATION

a P =2 +2b P =2× 4+2× 7 P =22

Writetheformula. Substituteinthevaluesforandb. Simplifytheresult.

b P =2 +2b 40=2 +2× 3

Writetheformula. SubstituteinthevaluesforPandbtoobtainanequation.

40=2 +6 −66 ÷2

Solvetheequationtoobtainthevalueof. −66

34 =2

÷2

17= © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Substitutex=4intotheexpressionx+7. Substitutea=2intotheexpression3a. Substitutep=5intotheexpression2p−3. Substituter=−4intotheexpression7r.

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2 IfyousubstituteP=10andx=2intotheformulaP=3m+x,whichofthefollowingequations wouldyouget? A 10=6+x B 10=3m+2 C 2=3m+10 D P=30+2 3 Ifyousubstitutek=10andL=12intotheformulaL=4k+Q,whichofthefollowingequations wouldyouget? A 12=40+Q B L=40+12 C 12=410+Q D 10=48+Q

c FindAifp=−2. Example 11b

b FindAifp=11. 13 d FindAifp= . 2

5 ConsidertheruleU=8a+4. a FindaifU=44.Setupandsolveanequation. b FindaifU=92.Setupandsolveanequation. c IfU=−12,findthevalueofa. 6 Considertherelationshipy=2x+4. a Findyifx =3. b Bysolvinganappropriateequation,findthevalueofx thatmakesy=16. c Findthevalueofxify=0. 7 UsetheformulaP=mvtofindthevalueofmwhenP=22andv=4. 8 Assumethatxandyarerelatedbytheequation4x+3y=24. a Ifx=3,findybysolvinganequation. b Ifx=0,findthevalueofy. c Ify=2,findxbysolvinganequation. d Ify =0,findthevalueofx. 9 ConsidertheformulaG=k(2a+p)+a. a Ifk=3,a=7andp=−2,findthevalueofG. b IfG=78,k=3andp=5,findthevalueofa.

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4 ConsidertheruleA=4p+7. a FindAifp=3.

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10 Thecost$CtohireataxiforatripoflengthdkmisC=3+2d. a Findthecostofa10kmtrip(i.e.ford=10). b Atriphasatotalcostof$161. i SetupanequationbysubstitutingC=161. ii Solvetheequationsystematically. iii Howfardidthetaxitravel?(Giveyouranswerinkm.)

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a Findtheareaofthetrapeziumshowntotheright.

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b FindthevalueofhifA=20,a=3andb=7.

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c Findthemissingvalueinthetrapeziumshowntotheright.

7 Area = 72

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a WhatisapossiblerulebetweenVandT? b Useyourruletofindthevolumeatatemperatureof27°C. (V − 10)2 c Provethattherule T = + 10 wouldalsoworkforKaty’sresults. 20 14 ConsidertheruleG=120−4p. a Ifpisbetween7and11,whatisthelargestvalueofG? b IfpandGareequal,whatvaluedotheyhave? © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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13 Katyisascientistwhotriestoworkouttherelationshipbetweenthevolumeofagas, V mL,anditstemperatureT°C.Shemakesafewmeasurements.

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Whichoneormoreofthefollowingrulesare consistentwiththeexperiment’sresults? 3V B T =V+2A A T= A C T=17− A 16 TemperaturesindegreesFahrenheitandCelsiusarerelatedbytherule F=1.8C+32. a BysubstitutingF=xandC=x,findavaluesuchthatthetemperatureinFahrenheitandthe temperatureinCelsiusareequal. b BysubstitutingF=2xandC=x,findatemperatureinCelsiusthatdoublestogivethetemperature inFahrenheit. c ProvethattherearenoCelsiustemperaturesthatcanbemultipliedby1.8togivethetemperature inFahrenheit.

Enrichment: Mobile phone plans 17 Twocompanieshavemobilephoneplansthatfactorinthenumberofminutesspenttalkingeach month(t)andthetotalnumberofcallsmade(c). CompanyA’scostincents:A=20t+15c+300 CompanyB’scostincents:B=30t+10c a Inonemonth12callsweremade,totalling50minutesonthephone.Findthecostindollarsthat companyAandcompanyBwouldhavecharged. b Inanothermonth,acompanyAuserwascharged$15(1500cents)formaking20calls.Howlong werethesecallsintotal? c Inanothermonth,acompanyBusertalkedfor60minutes intotalandwascharged$21.Whatwastheaveragelength ofthesecalls? d Brionynoticesonemonththatforhervaluesoftandc,the twocompaniescostthesame.Findapossiblevalueoftand cthatwouldmakethishappen. e Brionyrevealsthatshemadeexactly20callsforthemonth inwhichthetwocompanies’chargeswouldbethesame. Howmuchtimedidshespendtalking?

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15 MarieisascientistwhoistryingtodiscovertherelationshipbetweenthevolumeofagasV, itstemperatureTanditstransparencyA.Shemakesafewmeasurements.

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E X T E N S I ON

Althoughknowinghowtosolveequationsisuseful,it isimportanttobeabletorecognisewhenreal-world situationscanbethoughtofasequations.Thisisthecase wheneveritisknownthattwovaluesareequal.

Let’s start: Sibling sum

Key ideas

Johnandhiseldersisterare4yearsapartintheirages. • Ifthesumoftheiragesis26,describehowyoucould workouthowoldtheyare. • Couldyouwriteanequationtodescribethesituation above,ifxisusedforJohn’sage? • Howwouldtheequationchangeifx isusedforJohn’s sister’sageinstead?

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The difference in two people’s ages can be expressed as an equation.

Anequationcanbeusedtodescribeanysituationinwhichtwovaluesareequal. Tosolveaproblemfollowthesesteps. 1 Definepronumeralstostandforunknownnumbers(e.g.letj=John’scurrentage). 2 Writeanequationtodescribethesituation. 3 Solvetheequationsystematically,ifpossible,orbyinspection. 4 Ensureyouanswertheoriginalquestion,includingthecorrectunits(e.g.dollars,years,cm). 5 Checkthatyoursolutionisreasonableandmakessense.

Example 12 Solving a problem using equations Theweightof6identicalbooksis1.2kg.Whatistheweightofonebook? SOLUTION

EXPLANATION

Letb =weightofonebook. 6b =1.2

Step1:Defineapronumeraltostandfortheunknownnumber.

6b = 1.2 ÷ 6

÷6

Step2:Writeanequationtodescribethesituation. Step3:Solvetheequation.

Step4:Answertheoriginalquestion.Itisnotenoughtogive b =0.2 afinalansweras0.2;thisisnottheweightofabook,itisjust Thebooksweigh0.2kgeach,or200g anumber. each.

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Number and Algebra

Example 13 Solving a harder problem using equations Purchasing5applesanda$2.40mangocoststhesameaspurchasing7applesandamandarinthatcosts 60cents.Whatisthecostofeachapple? SOLUTION

EXPLANATION

Letc =costofoneappleindollars. 5c +2.4=7c +0.6

Step1:Defineapronumeraltostandfortheunknownnumber. Step2:Writeanequationtodescribethesituation.Notethat 60centsmustbeconvertedto$0.6tokeeptheunitsthesame throughouttheequation. Step3:Solvetheequation.

−5c

5c +2.4=7c +0.6

−5c

2.4 = 2c +0.6 −0.6

−0.6 1.8=2c ÷2

÷2 0.9= c

Applescost90centseach.

Step4:Answertheoriginalquestion.Itisnotenoughtogive afinalansweras0.9;thisisnotthecostofanapple,itisjusta number.

Example 14 Solving problems with two related unknowns JaneandLukehaveacombinedageof60.GiventhatJaneistwiceasoldasLuke,findtheagesof LukeandJane. SOLUTION

EXPLANATION

Let =Luke’sage.

Step1:Defineapronumeralfortheunknownvalue.Once Luke’sageisfound,wecandoubleittofindJane’sage. Step2:Writeanequationtodescribethesituation.Note thatJane’sageis2becausesheistwiceasoldasLuke. Step3:Solvetheequationbyfirstcombiningliketerms.

+2 =60 3 =60 ÷3

÷3 =20

Lukeis20yearsoldandJaneis40years old.

Step4:Answertheoriginalquestion.

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1 Matcheachofthesewordeddescriptionswithanappropriateexpression. a Thesumofx and3 A 2x b Thecostof2applesiftheycost$xeach B x+1 c Thecostofxorangesiftheycost$1.50each C 3x d Triplethevalueofx D x+3 e Onemorethanx E 1.5x

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2 Forthefollowingproblemschoosetheequationtodescribethem. a Thesumofxand5is11. A 5x=11 B x+5=11 C x−5=11 D 11−5 b Thecostof4pensis$12.Eachpencosts$p. A 4=p B 12p C 4p=12 D 12p=4 c Josh’sagenextyearis10.Hiscurrentageisj. A j+1=10 B j=10 C 9 D j−1=10 d Thecostofnpencilsis$10.Eachpencilcosts$2. A n÷10=2 B 5 C 10n=2 D 2n=10. d 10=3a+1 WO

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5 Acombinationof6chairsandatablecosts$3000.Thetablealonecosts$1740. a Defineapronumeralforthecostofonechair. b Writeanequationtodescribetheproblem. c Solvetheequationsystematically. d Hencestatethecostofonechair. 6 Theperimeterofthisrectangleis72cm. a Writeanequationtodescribetheproblem,usingbforthebreadth. b Solvetheequationsystematically. c Hencestatethebreadthoftherectangle.

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9 Asquarehasaperimeterof26cm. a Solveanequationtofinditssidelength. b Hencestatetheareaofthesquare. Example 14

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10 AlisonandFlynn’scombinedageis40.GiventhatFlynnis4yearsolderthanAlison,writean equationanduseittofindAlison’sage. 11 Recallthatinaquadrilateralthesumofallanglesis360°. Findthevaluesofxandyinthediagrambelow.

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12 Thesumofthreeconsecutivenumbersis357. a Useanequationtofindthesmallestofthethreenumbers. b Whatistheaverageofthesethreenumbers? c Ifthesumofthreeconsecutivenumbersis 38064,whatistheiraverage? 13 Thebreadthofarectangularpoolis5metres longerthanthelength.Theperimeterofthepool is58metres. a Drawadiagramofthissituation. b Useanequationtofindthepool’slength. c Hencestatetheareaofthepool.

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8 Anumberistripled,then2isadded.Thisgivesthesameresultasifthenumberwere quadrupled.Setupandsolveanequationtofindtheoriginalnumber.

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7 Aplumberchargesa$70call-outfeeand$52perhour.Thetotal costofaparticularvisitwas$252. a Defineapronumeraltostandforthelengthofthevisitinhours. b Writeanequationtodescribetheproblem. c Solvetheequationsystematically. d Statethelengthoftheplumber’svisit,givingyouranswerin minutes.

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14 Theaverageoftwonumberscanbefoundbyaddingthemanddividingby2. a Iftheaverageofxand10is30,whatisthevalueofx? b Iftheaverageofxand10is2,whatisthevalueofx? c Iftheaverageofxand10issomenumberR,createaformulaforthevalueofx.

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15 Sometimesyouaregivenanequationtosolveapuzzle,butthesolutionoftheequationisnotactually possibleforthesituation.Considerthesefiveequations. A 10x=50 B 8+x=10 C 10+x=8 D 10x=8 E 3x+5=x+5 a Youaretoldthatthenumberofpeopleinaroomcanbedeterminedbysolvinganequation.Which oftheseequationscouldbeusedtogiveareasonableanswer? b Ifthelengthofaninsectisgivenbythevariablex cm,whichoftheequationscouldbesolvedto giveareasonablevalueofx? c ExplainwhyequationDcouldnotbeusedtofindthenumberofpeopleinaroombutcouldbe usedtofindthelengthofaninsect. d GiveanexampleofapuzzlethatwouldmakeequationCreasonable.

Enrichment: Unknown numbers 16 Findtheunknownnumberusingequations.Theanswersmightnotbewholenumbers. a Theaverageofanumberanddoublethenumberis25.5. b Adding3totwiceanumberisthesameassubtracting9fromhalfthenumber. c Theaverageofanumberanddoublethenumbergivesthesameresultasaddingonetotheoriginal numberandthenmultiplyingbyone-third. d Theproductof5andanumberisthesameasthesumoffourandtwicetheoriginalnumber. e Theaverageof5numbersis7.Whenonemorenumberisadded,theaveragebecomes10.What numberwasadded?

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Aninequalityislikeanequationbut,insteadofindicatingthattwoexpressionsareequal,itindicates whichofthetwohasthegreatervalue.Forexample,2+4<7,3×5≥15andx≤10areallinequalities. Thefirsttwoaretrue,andthelastonecouldbetrueorfalsedependingonthevalueofx.Forinstance,the numbers9.8,8.45,7and−120allmakethisinequalitytrue. Wecouldrepresentallthevaluesofxthatmakex≤10atruestatement. x 8 9 10 11

Let’s start: Small sums

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Aninequalityisastatementoftheform: – LHS>RHS(greaterthan),e.g.5>2 – LHS≥RHS(greaterthanorequal),e.g.7≥7or10≥7 – LHS

Example 15 Representing inequalities on a number line Representthefollowinginequalitiesonanumberline. a x ≥4 b x <6 c 1< x ≤5 SOLUTION a

3

EXPLANATION 4

5

6

x

Acircleisplacedat4andthenthearrowpointstotheright, towardsallnumbersgreaterthan4. Thecircleisfilled(closed)because4isincludedintheset.

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Twopositivewholenumbersarechosen:xandy.Youaretoldthatx+y≤5. • Howmanypossiblepairsofnumbersmakethistrue?Forexample,x=2andy=1isonepairanditis differentfromx=1andy=2. • Ifx+y≤10,howmanypairsarepossible?Trytofindapatternratherthanlistingthemall. • Ifallyouknowaboutxandyisthatx+y>10,howmanypairsofnumberscouldtherebe?

Chapter 2 Equations 2

b x <6 3

4

5

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Acircleisplacedat6andthenthearrowpointstothe left,towardsallnumberslessthan6. Thecircleishollow(open)because6isnotincludedin theset.

x

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Circlesareplacedat1and5,andalinegoesbetween themtoindicatethatallnumbersinbetweenareincluded. Thecircleat1isopenbecausetheinequalityis

Example 16 Using inequalities to describe real-life situations Describethefollowingsituationsasaninequality,usingxtostandfortheunknownquantity. x xtostandfortheunknownquantity. a Fredisshorterthan160cm. b JohnisatleastasoldasMaria,whois10. c Rose’stestscoreisbetween40and50inclusive. SOLUTION

EX PLANATION

a x <160

UsingxtostandforFred’sheight, x xtostandforFred’sheight, x xmustbelessthan160.

b x ≥10

Johnisatleast10,sohisagesgreaterthanorequalto10.

c 40≤ x ≤50

x xisbetween40and50.Theword‘inclusive’tellsusthat 40and50arebothincluded,so≤isused(ratherthan<).

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2 Matcheachoftheseinequalitieswiththeappropriatedescription. a x>5 b x<5 c x≥3 d x≤ 3 A Thenumberxislessthan5. B Thenumberxisgreaterthanorequalto3. C Thenumberx islessthanorequalto3. D Thenumberxisgreaterthan5. 3 Foreachofthefollowing,statewhethertheymaketheinequalityx>4trueorfalse. a x=5 b x=−2 c x=4 d x=27 4 Ifx=12,classifythefollowinginequalitiesastrueorfalse. a x>2 b x<11 c x≥13

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5 Representthefollowinginequalitiesonseparatenumberlines. Example 15a,b a x>3 b x<10 c x≥2 e x≥1 f x≥4 g x<−5 i x≤2 j x<−6 k x≥−3 m 10>x n 2

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6 a Listwhichofthefollowingnumbersmaketheinequality2≤x<7true. 8, 1, 3, 4, 6, 4.5, 5, 2.1, 7, 6.8, 2 b Representtheinequality2≤x<7onanumberline. 7 Representthefollowinginequalitiesonseparatenumberlines. a 1≤x≤6 b 4≤x<11 c −2

9 ItisknownthatTim’sageisbetween20and25inclusive,andNick’sage isbetween23and27inclusive. a Ift=Tim’sageandn=Nick’sage,writetwoinequalitiestorepresent thesefacts. b Representbothinequalitiesonthesamenumberline. c NickandTimaretwins.Whatisthepossiblerangeoftheirages? Representthisonanumberline. 10 Atacertainschoolthefollowinggradesareawardedfordifferentscores. Score

x ≥ 80

60 ≤ x < 80

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20 ≤ x < 40

x < 20

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a Convertthefollowingscoresintogrades. i 15 ii 79 iii 80 iv 60 v 30 b EmmagotaBononetest,buthersisterRebeccagotanAwithjust7moremarks.Whatisthe possiblerangeforEmma’sscore? c Hugh’smarkearnedhimaC.Ifhehadscoredhalfthismark,whatgradewouldhehaveearned? d AlfredandReubenearnedaDandaCrespectively.Iftheirscoreswereaddedtogether,what gradeorgradescouldtheyearn? e MichaelearnedaDandwastoldthatifhedoubledhismarkhewouldhaveaB.Whatgradeor gradescouldheearnifhegotanextra10marks?

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8 Foreachofthefollowingdescriptions,chooseanappropriateinequalityfromA–Hbelow. a Johnismorethan12yearsold. b Marikaisshorterthan150cm. c Matthewisatleast5yearsoldbutheisyoungerthan10. d Thetemperatureoutsideisbetween−12°Cand10°Cinclusive. A x<150 B x<12 C x>12 D x≤150 E 10≤x≤−12 F −12≤x≤10 G 5≤x<10 H 5

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11 Sometimesmultipleinequalitiescanbecombinedtoasimplerinequality. a Explainwhythecombinationx≥5,x≥7isequivalenttotheinequalityx≥7. b Simplifythefollowingpairsofinequalitiestoasingleinequality. i x>5,x≥2 ii x<7,x<3 iii x≥1,x>1 iv x≤10,x<10 v x>3,x<10 vi x>7,x≤10 c Simplifythefollowingpairsofinequalitiestoasingleinequality. i 3

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Enrichment: Working within boundaries 13 Ifitisknownthat0≤x≤10and0≤y≤10,whichofthefollowinginequalitiesmustbetrue?Justify youranswers. a x+y≤30 b 2x≤20 c 10≤2y≤20 d x ×y≤100 e 0≤ x−y≤10 f x+5y≤100

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12 Someinequalities,whencombined,havenosolutions;somehaveonesolutionandsomehave infinitelymanysolutions.Labeleachofthefollowingpairsusing0,1or∞(infinity)tosayhowmany solutionstheyhave. a x≥5andx≤5 b x>3andx<10 c x≥3andx<4 d x>3andx<2 e −2

14 Ifitisknownthat0≤a≤10,0≤b≤10and0≤c≤10,whatisthelargestvaluethatthefollowing expressionscouldhave? a a+b+c b ab+c c a(b +c) d a ×b ×c e a−b−c f a−(b−c) g 3a+4 h a−bc

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2J Solving inequalities

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Sometimesaproblemarisesinwhichaninequalityismorecomplicatedthansomethingsuchasx>5or y≤40.Forinstance,youcouldhavetheinequality2x+4>100.Tosolveaninequalitymeanstofind allthevaluesthatmakeittrue.Fortheinequalityabove,x=50,x=90andx=10000areallpartofthe solution,butthesolutionisbestdescribedasx>48,becauseanynumbergreaterthan48willmakethe inequalitytrueandanyothernumbermakesitfalse. Therulesforsolvinginequalitiesareverysimilartothoseforequations:performthesameoperation tobothsides.Theoneexceptionoccurswhenmultiplyingordividingbyanegativenumber.Wecando this,butwemustﬂipthesignbecauseofthefollowingobservation. 5>2

5>2

× −1

× −1

× −1

−5 > −2 Incorrectmethod

−5 < −2

× −1

Correctmethod

Let’s start: Limousine costing

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Givenaninequality,anequivalentinequalitycanbeobtainedby: – addingorsubtractinganexpressionfrombothsides – multiplyingordividingbothsidesbyanypositivenumber – multiplyingordividingbothsidesbyanegativenumberandreversingtheinequality symbol – swappingleft-handsideandright-handsideandreversingtheinequalitysymbol.

−4 ÷2

2 +4<10 2x 2 <6 2x x <3

−4 ÷2

−2

−4x +2<6

−2

−4xx <4 ÷ −4

÷ −4 x > −1 Signisreversed

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Alimousineishiredforawedding.Thechargeisa $50hiringfeeplus$200perhour. • Ifthetotalhiretimewasmorethan3hours,what canyousayaboutthetotalcost? • Ifthetotalcostislessthan$850butmorethan $450,whatcanyousayaboutthetotaltimethe limousinewashired?

Chapter 2 Equations 2

Example 17 Solving inequalities Solvethefollowinginequalities. a 5x +2<47

b

SOLUTION a

5x +2<47 −2 5x <45

b ×9

c 15−2x 2 >1 2x

EX PLANATION

−2 ÷5

3 + 4x ≥3 9

Theinequalityissolvedinthesamewayasanequationis solved:2issubtractedfromeachsideandthenbothsides aredividedby5.Thesigndoesnotchangethroughout.

÷5 x <9 3 + 4x ≥3 9

×9

3+4x ≥27 −3

Theinequalityissolvedinthesamewayasanequationis solved.Bothsidesaremultipliedby9firsttoeliminate9 fromthedenominator.

−3 4x ≥24

÷4

÷4 x ≥6 15−2x 2 >1 2x

c −15

15issubtractedfromeachside. −15

−2x 2 > −14 2x ÷ −2 x <7

Exercise 2J

Bothsidesaredividedby−2.Becausethisisanegative number,theinequalityisreversedfrom>to<.

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1 Ifx=3,classifythefollowinginequalitiesastrueorfalse. a x+4>2 b 5x≥10 c 10−x<5

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6 Solvethefollowinginequalitiesinvolvingfractions. y+4 x−3 a d − 9 > 10 b ≤7 c > 2 2 4 2 2x + 4 7 + 3h 4 + 6p e >6 f < 5 g ≥4 3 2 4

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7 Solvethefollowinginequalitiesinvolvingnegativenumbers.Remembertoreversetheinequality whenmultiplyingordividingbyanegativenumber. a 6−2x<4 b 24−6s≥12 c 43−4n>23 d 34−2j<14 e 2−9v≤20 f 2−7j≤37 g 48−8c≥32 h 42−8h≤42 i 7−8s>31 j 6−8v>22 k 10−4v≥18 l 4−5v<29 8 Matchthefollowinginequalitieswiththeirsolutionsdepictedonanumberline. x +1 a 5x+2≥17 b c 9(x+4)<45 d 5−2x<3 >3 6

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9 Kartikbuys4cartonsofmilkanda$20phonecard.Thetotalcostofhisshoppingwas greaterthan$25. a Ifcisthecostofacartonofmilk,writeaninequalitytodescribethesituationabove. b Solvetheinequalitytofindthepossiblevaluesofc. c Ifthemilk’scostisamultipleof5cents,whatistheminimumpriceitcouldbe?

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10 InAFLfootballthescoreisgivenby6g+bwhere g isthenumberofgoalsandbisthenumberof behinds.Ateamscored4behindsandtheirscore waslessthanorequalto36. a Writeaninequalitytodescribethissituation. b Solvetheinequality. c Giventhatthenumberofgoalsmustbeawhole number,whatisthemaximumnumberofgoals thattheycouldhavescored?

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11 RecallthattoconvertdegreesCelsiustoFahrenheittheruleisF=1.8C+32.Pippainformsyouthat thetemperatureisbetween59°and68°Fahrenheitinclusive. a Solve1.8C+32≥59. b Solve1.8C+32≤68. c Hencestatethesolutionto59≤1.8C+32≤68,givingyouranswerasasingleinequality. d Pippalaterrealisedthatthetemperaturesshegaveyoushouldhavebeendoubled−therangewas actually118°to136°Fahrenheit.StatetherangeoftemperaturesinCelsius,givingyouransweras aninequality.

Provethatif5x−2ispositivethenx ispositive. Provethatif2x+6ispositivethenx+5ispositive. Isitpossiblethat10−xispositiveand10−2xispositivebut10−3xisnegative?Explain. Isitpossiblethat10−xispositiveand10−3xispositivebut10−2xisnegative?Explain.

14 Apuzzleisgivenbelowwithfourclues. ClueA:3x>12 ClueB:5−x≤4 ClueC:4x+2≤42 ClueD:3x+5<36 a Twoofthecluesareunnecessary.Statewhichtwocluesarenotneeded. b Giventhatxisawholenumberdivisibleby4,whatisthesolutiontothepuzzle?

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12 Tosaythatanumberx ispositiveistosaythatx>0. a If10x−40ispositive,findallthepossiblevaluesofx.Thatis,solve10x−40>0. b Findallkvaluesthatmake2k−6positive. c If3a+6isnegativeand10−2aispositive,whatarethepossiblevaluesofa? d If5a+10isnegativeand10a+30ispositive,whatarethepossiblevaluesofa?

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Enrichment: Pronumerals on both sides 16 Thismethodforsolvinginequalitiesallowsbothsidestohaveanyexpressionsubtractedfromthem. Thisallowsustosolveinequalitieswithpronumeralsonbothsides.Forexample: −10x 10 10x

12 +5≤10x 12x 10 +11 10x

−10x 10 10x

2 +5≤11 2x whichcanthenbesolvedasusual.Ifweendupwithapronumeralontheright-handside,such as5

c 5−2b≥3b−35 f 9g+40

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15 Multiplyingordividingbyanegativenumbercanbeavoidedbyaddingthepronumeral totheothersideoftheequation.Forexample:

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Chapter 2 Equations 2

Tiling a pool edge TheSunnySwimmingPoolCompanyconstructsrectangularpoolseach4minbreadthwithvarious lengths.Therearenon-slipsquaretiles,50cmby50cm,thatcanbeusedfortheexternaledging aroundthepoolperimeterwhereswimmerswalk. 1 Drawaneatdiagramillustratingthepooledgewithonerowofﬂattilesborderingtheperimeterof arectangularpool4minbreadthand5mlong. 2 Developatableshowingthedimensionsofrectangularpoolseachwithbreadth4mandrangingin lengthfrom5mto10m.Addacolumnforthetotalnumberoftilesrequiredforeachpoolwhen onerowofﬂattilesborderstheoutsideedgeofthepool. 3 Developanalgebraicruleforthetotalnumberoftiles,T,requiredforborderingtheperimeterof rectangularpoolsthatare4minbreadthandxmlong. 4 a U seyouralgebraicruletoformequationsforeachofthefollowingtotalnumberoftileswhena singlerowofﬂattilesisusedforpooledging. i 64tiles ii 72tiles iii 80tiles iv 200tiles b Bymanuallysolvingeachequation,determinethelengthsofthevariouspoolsthatuseeachof theabovenumbersoftiles. 5 Developanalgebraicruleforthetotalnumberoftiles,T,requiredfortworowsofﬂattiles borderingrectangularpoolsthatare4minbreadthandxmlong. 6 a Useyouralgebraicruletoformequationsforeachofthefollowingtotalnumbersoftileswhen tworowsofﬂattilesareusedforpooledging. i 96tiles ii 120tiles iii 136tiles iv 248tiles b Bymanuallysolvingeachequation,determinethelengthsofthepoolsthatusethesenumbers oftiles. 7 Determineanalgebraicruleforthetotalnumberoftiles,T,requiredforrequiredfornrowsofﬂat tilesborderingrectangularpoolsthatare4minbreadthandxminlength. 8 Usethisalgebraicruletoformequationsforeachofthefollowingpools,andthenmanuallysolve eachequationtodeterminethelengthofeachpool. Pool

Breadth of pool 4 m

A

4

3

228

B

4

4

288

C

4

5

500

Length of pool x m

Number of layers n

Total number of tiles T

Cambridge University Press

Number and Algebra

Hiring a tuk-tuk taxi Thecostofhiringatuk-tuktaxiinThailandis50bahthiringcost(ﬂagfall)plus15bahtevery2km.

2 Ongraphpaper,drawagraphofthehiringcost (verticalaxis)versusthedistance(horizontalaxis) travelled.Usethegridatrightasaguide. 3 Onyourgraph,rulelinestoshowthedistance travelledforhiringcostsof125bahtand170baht. 4 WriteanalgebraicformulaforthehiringcostCof anxkmrideinatuk-tuktaxi.

Cost (baht)

1 Untable,showthecostofhiringatuk-tuktaxifor2,4,6,8,10and20km. 350 300 250 200 150 100 50 0

0 5 10 15 20 5 a U seyouralgebraicformulatoformequations Distance (km) foreachofthefollowingtripcosts: i 320baht ii 800baht iii 1070baht b Useyourequation-solvingprocedurestocalculatethedistancetravelledforeachtrip. 6 AtouristwaschargedA$100forthe116kmtripfromPattayaBeachtotheBangkok Internationalairport.If1bahtcanbeboughtforA$0.03318,doyouthinkthatthiswasafair price?Justifyyouranswerwithcalculations.

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111

Puzzles and challenges

112

Chapter 2 Equations 2

1 Findtheunknownvalueinthefollowingpuzzles. a Anumberisincreasedby2,thendoubled,thenincreasedby3andthentripled.Theresultis99. b Anumberisdoubledandthenonethirdofthenumberissubtracted.Theresultis5largerthan theoriginalnumber. c Infiveyears’timeAlfwillbetwiceasoldashewastwoyearsago.HowoldisAlfnow? d Thepriceofashirtisincreasedby10%forGSTandthendecreasedby10%onasale.Thenew priceis$44.Whatwastheoriginalprice? e One-thirdofanumberissubtractedfrom10andthentheresultistripled,givingtheoriginal numberbackagain. 2 Considerthefollowing‘proof’that0=1. 2 +5=3x +5 2x −5

−5 2 =3x 2x ÷ ÷x

2=3

−2

÷ ÷x −2

0=1 a Whichstepcausedtheprobleminthisproof?(Hint:Considertheactualsolutiontothe equation.) b Provethat0=1isequivalenttotheequation22=50byadding,subtracting,multiplyingand dividingbothsides. 3 Findallthevaluesofxthatwouldmakeboththeseinequalitiesfalse. 19−2x<5and20+x>4x+2 4 Thefollowingsixexpressionscouldappearoneithersideofanequation.Usingjusttwoofthe expressions,createanequationthathasnosolution. 2x 3x+1 7x+4 4(x+7) 2+3(x+1) 2(3+x)−1 5 Acertainpairofscalesonlyregistersweightsbetween100kgand150kg,butitallowsmorethan onepersontogetonatatime. a Ifthreepeopleweighthemselvesinpairsandthefirstpairweighs117kg,thesecondpair weighs120kgandthethirdpairweighs127kg,whataretheirindividualweights? b Ifanotherthreepeopleweighthemselvesinpairsandgetweightsof108kg,118kgand130kg, whataretheirindividualweights? c Agroupoffourchildrenwhoallweighlessthan50kg,weighthemselvesingroupsofthree, gettingtheweights122kg,128kg,125kgand135kg.Howmuchdotheyeachweigh?

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Equivalent equations stay balanced

Unknown –5 ÷2

2x + 5 = 8 2x = 3 x = 32

Equation –5 Solving ÷2

–x –1 ÷4

Solution

5x + 1 = x + 9 4x + 1 = 9 4x = 8

–x –1 ÷4

x=2

Solving simple quadratic equations If x2 = c then: • If c > 0, x = √ c or x = – √ c e.g x2 = 16 gives x = ±4 • If c = 0 then x = 0 with one solution. • If c < 0 then there are no solutions for x.

Equations 2

× –1 –5x = –2 5x = 2 ÷5 x = 25

P = 2 + 2w

–3 3 – a = 5 –a = 2 × –1 a = –2

× –3 –6 ÷ –2

6-2a-3

=8

6 – 2a = –24 –2a = –30 a = 15

÷5

–3 × –1

=4

5x = 12 x = 12 5

×3

–9

÷5

Problem solving with equations • Pronumerals: use words to explain • Rule: starts with word or pronumeral, has = sign • Solve: state solution • Sentence answer with units

=6 9 – 2a 3 = –3 –2a 3

×3

–2a = –9 a = 92

÷ –2

3k – 2 = 5

×5

–2

3k – 2 = –10 3k = –8 k = – 83

–9 ×3 ÷ –2

×5 +2 ÷3

Equations with brackets Expand brackets Collect like terms Distributive law

× –1

× –1

÷5

5x 3

÷3

subject pronumerals –x = –3 × –1 x=3

×3

+2

Formulas rules, relationships Pronumerals with negative coefficients

Equations with fractions

a (b + c) = ab + ac

5(2x – 3) + 8 = 6x – 19 10x – 15 + 8 = 6x – 19 –6x 10x – 7 = 6x – 19 –6x 4x – 7 = –19 +7 +7 4x = –12 ÷4 ÷4 x = –3

An orange costs 5c more than an apple. An orange and an apple together cost $1.15. Determine the cost of each. Cost of apple = x Cost of orange = x + 5 Total = x + x + 5 2x + 5 = 115 2x = 110 x = 55 Apples cost 55 cents each. Oranges cost 60 cents each.

× –3 –6 ÷ –2

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113

Chapter summary

Number and Algebra

114

Chapter 2 Equations 2

Multiple-choice questions 1 Whichoneofthefollowingequationsistrue? A 20÷(2×5)=20÷10 B 12−8=2 1 C 5÷5×5= D 15+5×4=20+4 5 E 10−2×4=4×2−5 2 Whichoneofthefollowingequationsdoesnothavethesolutionx=9? x A 4x=36 B x+7=16 C =3 3 D x+9=0 E 14−x=5 3 Thesolutiontotheequation3a+8=29is: 1 C a=7 D a=18 E a=3 A a=21 B a=12 3 4 ‘Threelessthanhalfanumberisthesameasfourmorethanthenumber’canbeexpressedasan equationby: x x ( x − 3) A −3=4x B C −3=x + 4 = x + 4 2 2 2 x x D +3=x+4 E −3+4 2 2 5 Thesolutiontotheequation−3(m+4)=9is: A m=7 B m =−7 C m =−1 D m=1 E m=−3 6 If12+2x=4x−6thenxequals: A 8 B 9 C 12

D 15

E 23

7 WhichequationdoesNOThavethesamesolutionastheothers? A 2x+4=0

B 2x=−4

C 0=4+2x

D 2x=4

And 2(x+2)=0

8 Whichoneofthefollowingequationshasthesolutionn=10? A 4−n=6 B 2n+4=3n+5 C 50−4n=90 D 2(n+5)=3(n+1) E 70−6n=n 9 Malcolmsolvesanequationasfollows: 5−2x+4=11 line1 1−2x=11 line2 −2x=10 line3 x=−5 line4 Choosethecorrectstatement. A Theonlymistakewasmadeinmovingfromline1toline2. B Theonlymistakewasmadeinmovingfromline2toline3. C Theonlymistakewasmadeinmovingfromline3toline4. D Amistakewasmadeinmovingfromline1toline2andfromline3toline4. E Nomistakesweremade.

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Number and Algebra

10 Thevalueofxinthisisoscelestriangleis: A 30 B 45 C 62.5 D 65 E 130

50° (2x + 5°)

Short-answer questions 1 Classifyeachofthefollowingastrueorfalse. a If3x=6thenx=3. b Ifa=21thena+ a−a=a. c 5×4=10+10 2 Findthesolutionstotheseequations. m a 4m=16 b =−4 3 c 9−a=10 d 10m=2 e 2m+6=36 f a+a=12 3 Writeanequationtorepresenteachofthefollowingstatements.Youdonotneedtosolvethe equations. a Doublemplus3equalsthreetimesm. b Thesumofnandfourismultipliedbyfive;theansweris20. c Thesumoftwoconsecutiveevennumbers,thefirstbeingx,is74. 4 Foreachequationbelow,statethefirstoperationyouwouldapplytobothsides. x c 3a+3=2a+10 a 15+2x=45 b −5=6 2 5 Solve: a a+8=12 b 6−y=15 c 2x−1=−9 d 5+3x=17 e 20−4x=12 f 8a−8=0 6 Solvethefollowingequationssystematically. m 5x a =−2 b = 20 3 2 −2 y k+3 = 12 c d −5 = 3 11 e

8 − 2w = 4 3

f 13 = 2a − 8 6

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115

116

Chapter 2 Equations 2

7 Solvetheseequationsbyfirstfindingacommondenominator. x x x 1 − = 10 a 2= + b 5 x − 3 x = 1 c 2 4 8 4 3 2 8 Solvetheseequations. a 8a −3=7a+5 c 3x+5=7x−11 e 2x+8=x

b 2−3m=m d 10−4a=a f 4x+9=5x

9 Solvethefollowingequationsbyfirstexpandingthebrackets. a 2(x+5)=16 b 3(x+1)=−9 c 18=−2(2x−1) d 3(2a+1)=27 e 5(a+4)=3(a+2) f 2(3m+5)=16+3(m+2) g 8(3−a)+16=64 h 2x +10=4(x−6) i 4(2x−1)−3(x−2)=10x−3 10 a T hesumofthreeconsecutivenumbersis39.Firstwriteanequationandthenfindthevalueofthe smallestnumber. b Fourtimesanumberless5isthesameasdoublethenumberplus3.Writeanequationtofindthe number. c Thedifferencebetweenanumberandthreetimesthatnumberis17.Whatisthenumber? 11 Solvethefollowingsimplequadraticequations.Roundto2decimalplaceswherenecessary. a x2=4 b x2=100 c a2=49 2 2 d x =8 e y =39 f b2=914 12 Statethenumberofsolutionsforxintheseequations a 2x=6 b x2=16 2 d 5x =0 e 4x−1=3

c x2=0 f x2=−3

13 Writeaninequalitytorepresentthesesituations. a Theprofitofacompanyisatleast$100000. b Thecostofanewcarcannotexceed$6700. c Torideontheroller-coaster,aperson’sheightmustbe between1.54mand1.9minclusive.

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Number and Algebra

Extended-response questions 1 Touploadanadvertisementtothewww.searches.com.auwebsitecosts$20andthen12cents wheneversomeoneclicksonit. a Writeaformularelatingthetotalcost($S )andthenumberofclicks(n)ontheadvertisement. b Ifthetotalcostis$23.60,writeandsolveanequationtofindouthowmanytimesthe advertisementhasbeenclickedon. c Touploadtothewww.yousearch.com.auwebsitecosts$15initiallyandthen20centsfor everyclick.Writeaformulaforthetotalcost$Ywhentheadvertisementhasbeenclickedn times. d Ifapersonhasatmost$20tospend,whatisthemaximumnumberofclickstheycanafford ontheiradvertisementatyousearch.com.au? e Setupandsolveanequationtofindtheminimumnumberofclicksforwhichthetotalcostof postinganadvertisementtosearches.com.auislessthanthecostofpostingtoyousearch.com.au. 2 Mahniplanstospendthenext12weekssavingsomeofherincome.Shewillsave$xaweekfor thefirst6weeksandthen$(2x−30)aweekforthefollowing6weeks. a Writeanexpressionforthetotalamountsavedoverthe12weeks. b Ifshemanagedtosave$213inthefirstsixweeks,howmuchdidshesave: i inthefirstweek? ii inthe7thweek? iii intotaloverthe12weeks? c IfMahniwantstosaveatotalof$270,writeandsolveanequationtofindouthowmuchshe wouldhavetosaveinthefirstweek. d IfMahniwantstosavethesameamountinthefirst6weeksasinthelast6weeks,howmuch wouldshesaveeachweek? e IntheendMahnidecidesthatshedoesnotmindexactlyhowmuchshesavesbutwantsitto bebetween$360and$450.Statetherangeofxvaluesthatwouldachievethisgoal,giving youranswerasaninequality.

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118

Chapter 3 Measurement and Pythagoras’ theorem

Chapter

3

Measurement and Pythagoras’ theorem

What you will learn

3A 3B 3C 3D 3E 3F 3G 3H 3I 3J 3K 3L 3M

Length and perimeter REVISION Circumference of circles REVISION Area REVISION Area of special quadrilaterals Area of circles Area of sectors and composite figures Surface area of prisms EXTENSION Volume and capacity Volume of prisms and cylinders Time REVISION Introducing Pythagoras’ theorem Using Pythagoras’ theorem Calculating the length of a shorter side

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119

NSW Syllabus

for the Australian Curriculum

Strand: Measurement and Geometry Substrand:

LENGTH AREA VOLUME TIME RIGHT-ANGLED TRIANGLES (PYTHAGORAS)

Outcomes Length A student calculates the perimeters of plane shapes and the circumferences of circles. (MA4–12MG) Area

A student uses formulas to calculate the areas of quadrilaterals and circles, and converts between units of area. (MA4–13MG)

The wheels are turning Civilisations in ancient and modern times have used measurement to better understand the world in which they live and work. The circle, for example, in the form of a wheel helped civilisations gain mobility, and modern society to develop machines. For thousands of years mathematicians have studied the properties of the wheel or circle shape including such measurements as its circumference. The ancient civilisations knew of the existence of a special number (which we know as pi) that links a circle’s radius with its circumference and area. It was the key to understanding the precise geometry of a circle, but they could only guess its value. We now know that pi is a special number that has an infinite number of decimal places with no repeated pattern. From a measurement perspective, pi is the distance a wheel with diameter 1 unit will travel in one full turn.

Volume A student uses formulas to calculate the volumes of prisms and cylinders, and converts between units of volume. (MA4–14MG) Time

A student performs calculations of time that involve mixed units, and interprets time zones. (MA4–15MG)

Right-angled triangles (Pythagoras)

A student applies Pythagoras’ theorem to calculate side lengths in right-angled triangles, and solves related problems. (MA4–16MG)

Cambridge University Press

Chapter 3 Measurement and Pythagoras’ theorem

Pre-test

120

1 Convert these measurements to the units shown in the brackets. a 3 m (cm) b 20 cm (mm) c 1.8 km (m) d 0.25 m (cm) e 35 mm (cm) f 4200 m (km) g 500 cm (m) h 100 mm (m) i 2 minutes (seconds) j 3 L (mL) k 4000 mL (L) l 3000 g (kg) 2 Name these shapes. a b

e

f

c

d

g

h

3 Find the area of these rectangles and triangles. 1 Remember: Area (rectangle) = × b and Area (triangle) = bh 2 a b c 3 cm 5 cm 4 cm 10 cm

d 8 cm

8 cm 5 cm 4 Find the perimeter of these shapes. a b

c

6 cm

2.5 cm

12 m

9 cm

10 cm

3m 5 Evaluate the following. 1 b 1 (2 + 7) × 6 c 52 d 112 ×5×4 2 2 1 e g h (22 + 17) × 3 f 36 81 144 2 6 Using V = × b × h, find the volume of these rectangular prisms. a

a

2 cm 3 cm

b

c 10 m

2 cm

5m 5m

20 cm

30 cm 7 cm

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121

Measurement and Geometry

3A Length and perimeter For thousands of years, civilisations have found ways to measure length. The Egyptians, for example, used the cubit (length of an arm from the elbow to the tip of the middle finger), the Romans used the pace (5 feet) and the English developed their imperial system using inches, feet, yards and miles. The modern-day system used in Australia (and most other countries) is the metric system, which was developed in France in the 1790s and is based on the unit called the metre. We use units of length to describe the distance between two points, or the distance around the outside of a shape, called the perimeter.

R E V I S I ON

The Romans would have measured the perimeter of the Colosseum in paces.

Let’s start: Provide the perimeter In this diagram some of the lengths are given. Three students were asked to find the perimeter. • Will says that you cannot work out some lengths and so the perimeter cannot be found. • Sally says that there is enough information and the answer is 9 + 12 = 21 cm. • Greta says that there is enough information but the answer is 90 + 12 = 102 cm.

6 cm

45 mm

Who is correct? Discuss how each person arrived at their answer.

The common metric units of length include the kilometre (km), the metre (m), the centimetre (cm) and the millimetre (mm). ×1000 km

m ÷1000

■

×100

×10 cm

÷100

Key ideas

■

mm x

÷10

Perimeter is the distance around a closed shape. – All units must be of the same type when calculating the perimeter. – Sides with the same type of markings (dashes) are of equal length.

y

z P = 2x 2 +y+z

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122

Chapter 3 Measurement and Pythagoras’ theorem

Example 1 Converting length measurements Convert these lengths to the units shown in the brackets. a 5.2 cm (mm) b 85 000 cm (km) SOLUTION a

EXPLANATION

5.2 cm = 5.2 × 10 = 52 mm

1 cm = 10 mm so multiply by 10. ×10 cm

b 85 000 cm = 85 000 ÷ 100 ÷ 1000 = 0.85 km

mm

1 m = 100 cm and 1 km = 1000 m so divide by 100 and 1000.

Example 2 Finding perimeters Find the perimeter of this shape.

4 cm 3 cm SOLUTION

EXPLANATION

P = 2 × (3 + 3) + 2 × 4 = 12 + 8 = 20 cm

6 cm 4 cm

4 cm

3 cm 3 cm

Example 3 Finding an unknown length Find the unknown value x in this triangle if the perimeter is 19 cm. x cm P = 19 cm 5 cm SOLUTION

EXPLANATION

2 + 5 = 19 2x 2 = 14 2x x=7

2 + 5 makes up the perimeter. 2x 2x is the difference between 19 and 5. 2x If 22x = 14 then x = 7 since 2 × 7 = 14.

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123

Measurement and Geometry

WO

U

b 100 × 1000

3 Find the value of x in these diagrams. a b

xm

10 cm

HE

T

2 Evaluate the following. a 10 × 100

R

MA

1 Write the missing number in these sentences. a There are ___ mm in 1 cm. b There are ___ cm in 1 m. c There are ___ m in 1 km. d There are ___ cm in 1 km. e There are ___ mm in 1 m. f There are ___ mm in 1 km.

R K I N G

C

F PS

Y

REVISION

LL

Exercise 3A

M AT I C A

c 10 × 100 × 1000

xm

c xm 14 m

7m

12 m

3m 4 Find the perimeter of these quadrilaterals. a Square with side length 3 m b Rectangle with side lengths 4 cm and 7 cm c Rhombus with side length 2.5 mm d Parallelogram with side lengths 10 km and 12 km e Kite with side lengths 0.4 cm and 0.3 cm f Trapezium with side lengths 1.5 m, 1.1 m, 0.4 m and 0.6 m

6 Find the perimeter of these shapes. a b 5m

c 7m

6m

3 cm 5 cm

15 m 8m d

f 1 cm

e

10 cm 3 cm

5 km

4 cm 1 km

8 km © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

2 cm Cambridge University Press

R K I N G

C

F PS

Y

R

HE

3 m (mm) 38 000 cm (km) 342 000 cm (km) 994 000 mm (km)

T

Example 2

5 Convert these measurements to the units shown in the brackets. a 3 cm (mm) b 6.1 m (cm) c 8.93 km (m) d e 0.0021 km (m) f 320 mm (cm) g 9620 m (km) h i 0.0043 m (mm) j 0.0204 km (cm) k 23 098 mm (m) l m 194 300 mm (m) n 10 000 mm (km) o 0.02403 m (mm) p

MA

Example 1

U

LL

WO

M AT I C A

3A

g

4.3 cm

h

WO

U

in MA

R

T

HE

5.1 m

7.2 mm

R K I N G

C

F PS

Y

Chapter 3 Measurement and Pythagoras’ theorem

LL

124

M AT I C A

9.6 m 2.8 mm 7 Find the unknown value x in these shapes with the given perimeter (P). a b c 7 cm xm 3m 4m 4m P = 12 m d

P = 22 cm

xm P = 10 m e

10 mm

x cm

f

xm

x km 7m

x mm

P = 26 km

P = 46 mm

WO

U

T

1.1 cm 20 mm

15 mm d

e

7m

3m

f

10 cm

12 m

9 cm 44 m 20 m

7 cm 9 Find the unknown value x in these diagrams. a b

c 5 cm

x cm 5 cm P = 24 cm

x cm P = 34 cm

R

HE

4 cm

30 mm

10 cm

MA

8 Find the perimeter of these shapes. Give your answers in cm. a b c

12 m

xm P = 60 m

Cambridge University Press

R K I N G

C

F PS

Y

13 m P = 39 m

LL

Example 3

M AT I C A

125

Measurement and Geometry

R K I N G

MA

R

T

HE

C

F PS

Y

U

LL

WO

10 Jennifer needs to fence her country house block to keep her dog in. The block is a rectangle with length 50 m and breadth 42 m. Fencing costs $13 per metre. What will be the total cost of fencing?

M AT I C A

11 Gillian can jog 100 metres in 24 seconds. How long will it take her to jog 2 km? Assume Gillian can keep jogging at the same rate and give your answer in minutes. 12 A rectangular picture of length 65 cm and breadth 35 cm is surrounded by a frame of breadth 5 cm. What is the perimeter of the framed picture?

T

a

b

b

c

a b

a

b

a d

e

a

b

R

HE

f

a

b

a

b

Cambridge University Press

R K I N G

C

F PS

Y

MA

13 Write down rules using the given letters for the perimeter of these shapes, e.g. P = a + 2b.

U

LL

WO

M AT I C A

3A

WO

14 These shapes have perimeter P. Write a rule for x in terms of P, e.g. x = P – 10. a b c 3 x 4

U

MA

R

T

HE

2

R K I N G

C

x

x 4

e

f x

x

x

Enrichment: Disappearing squares 15 A square is drawn with a particular side length. A second square is drawn inside the square so that its side length is one-third that of the original square. Then a third square is drawn, with side length of one-third that of the second square and so on. a What is the minimum number of squares that would need to be drawn in this pattern (including the starting square), if the innermost square has a perimeter of less than 1 hundredth the perimeter of the outermost square? b Imagine now if the situation is reversed and each square’s perimeter is 3 times larger than the next smallest square. What is the minimum number of squares that would be drawn in total if the perimeter of the outermost square is to be at least 1000 times the perimeter of the innermost square?

Cambridge University Press

PS

M AT I C A

7

d

F Y

Chapter 3 Measurement and Pythagoras’ theorem

LL

126

127

Measurement and Geometry

3B Circumference of circles

R E V I S I ON

Since the ancient times, people have known about a special number that links a circle’s diameter to its circumference. We know this number as pi (π). π is a mathematical constant that appears in formulas relating to circles, but it is also important in many other areas of mathematics. The actual value of π has been studied and approximated by ancient and more modern civilisations over thousands of years. Some historians believe that the Egyptians knew π was slightly more than 3 and approximated it to be 256 25 339 ≈ 3.16. The Babylonians used ≈ 3.125 and the ancient Indians used ≈ 3.139. 81 8 108 It is believed that Archimedes of Syracus (287–212 bc) was the first person to use a mathematical technique to evaluate π. He was able to prove that π was greater 223 22 than and less than . In 480 ad, the Chinese mathematician Zu Chongzhi 71 7 335 showed that π was close to ≈ 3.1415929, which is accurate to six decimal places. 113 22 Before the use of calculators, the fraction was commonly used as a good and 7 simple approximation to π. Interestingly, mathematicians have been able to prove that π is an irrational number, which means that there is no fraction that can be found that is exactly equal to π. If the exact value of π was written down as a decimal, the decimal places would continue forever with no repeated pattern.

Let’s start: Discovering pi Here are the diameters and circumferences for three circles correct to 2 decimal places. Use a calculator to work out the value of Circumference ÷ Diameter and put your results in the third column. Add your own circle measurements by measuring the diameter and circumference of circular objects such as a can or a wheel.

Diameter d (mm)

Circumference C (mm)

4.46 11.88 40.99 Add your own

14.01 37.32 128.76 Add your own

C÷d

■

■

■

Features of a circle fe rcum rence Ci – Diameter (d) is the distance across the centre of a circle. – Radius (r) is the distance from the centre to the circle. Note d = 2r. eter Diam Circumference (C) is the distance around a circle. Radius – C = 2πr or C = πd Pi (π) ≈ 3.14159 (correct to 5 decimal places) 22 – Common approximations include 3.14 and . 7 – A more precise estimate for pi can be found on most calculators or on the internet. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

Cambridge University Press

Key ideas

• What do you notice about the numbers C ÷ d in the third column? • Why might the numbers in the third column vary slightly from one set of measurements to another? • What rule can you write down which links C with d?

128

Chapter 3 Measurement and Pythagoras’ theorem

Example 4 Finding the circumference with a calculator Find the circumference of these circles correct to 2 decimal places. Use a calculator for the value of pi. a b

4 cm 3.5 m SOLUTION

EXPLANATION

a C = 2πr = 2 × π × 3.5 = 7π = 21.99 m (to 2 decimal places)

Since r is given, you can use C = 2πr. Alternatively use C = πd with d = 7.

b C = πd =π×4 = 4π = 12.57 cm

Substitute d = 4 into the rule C = πd or use C = 2πr with r = 2. (to 2 decimal places)

Example 5 Finding circumference without a calculator Calculate the circumference of these circles using the given approximation of π. a b 10 m 14 cm π = 3.14

π=

22 7

SOLUTION

EXPLANATION

a C = πd = 3.14 × 10 = 31.4 m

Use π = 3.14 and multiply mentally. Move the decimal point one place to the right. Alternatively use C = 2πr with r = 5.

b C = 2πr 22 =2× × 14 7 = 88 cm

22 and cancel the 14 with the 7 before 7 calculating the final answer. 22 2× × 14 = 2 × 22 × 2 7

Use π =

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4 A circle has circumference (C) 81.7 m and diameter (d) 26.0 m correct to 1 decimal place. Calculate C ÷ d. What do you notice? c

2 mm d

39 cm

18 m

e

f 5 cm 7 km

4m

Example 5a

6 Calculate the circumference of these circles using π = 3.14. a b c 100 cm

20 m 3 km

Example 5b

22 7 Calculate the circumference of these circles using π = . 7 a b c 21 cm

70 m

7 mm

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9 An athlete trains on a circular track of radius 40 m and jogs 10 laps each day, 5 days a week. How far does he jog each week? Round the answer to the nearest whole number of metres.

10 These shapes are semicircles. Find the perimeter of these shapes including the straight edge and round the answer to 2 decimal places. a b c 25 cm 4.8 m 12 mm 11 Calculate the perimeter of these diagrams correct to 2 decimal places. a b c

45°

14 m

2 cm

8 cm 12 Calculate the perimeter of these shapes correct to 2 decimal places. a b c 5m

4 cm

9m 10m

13 Here are some student’s approximate circle measurements. Which students have incorrect measurements? r

C

Mick

4 cm

25.1 cm

Svenya

3.5 m

44 m

Andre

1.1 m

13.8 m

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17 We know that C = 2πr or C = πd. a Rearrange these rules to write a rule for: i r in terms of C ii d in terms of C b Use the rules you found in part a to find the following correct to 2 decimal places. i The radius of a circle with circumference 14 m ii The diameter of a circle with circumference 20 cm

Enrichment: Memorising pi 18 The box shows π correct to 100 decimal places. The Guinness World record for the most number of digits of π recited from memory is held by Lu Chao, a Chinese student. He recited 67 890 digits nonstop over a 24-hour period.

Challenge your friends to see who can remember the most number of digits in the decimal representation of π. Number of digits memorised

Report

10+

A good show

20+

Great effort

35+

Superb

50+

Amazing memory

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16 Find the exact answers for Question 12 in terms of π.

3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679

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3C Area

R E V I S I ON

Area is a measure of surface and is often referred to as the amount of space contained inside a twodimensional space. Area is measured in square units and the common metric units are square millimetres (mm2), square centimetres (cm2), square metres (m2), square kilometres (km2) and hectares (ha). The hectare is often used to describe area of land, since the square kilometre for such areas is considered to be too large a unit and the square metre too small. A school football oval might be about 1 hectare, for example, and a small forest might be about 100 hectares.

Let’s start: Squares of squares Consider this enlarged drawing of one square centimetre divided into square millimetres. • How many square millimetres are there on one edge of the square centimetre? • How many square millimetres are there in total in 1 square centimetre? • What would you do to convert between mm2 and cm2 or cm2 and mm2 and why? • Can you describe how you could calculate the number of square centimetres in 1 square metre and how many square metres in one square kilometre? What diagrams would you use to explain your answer?

1 cm = 10 mm

1 cm = 10 mm

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The common metric units for area include: ×10002 ×1002 ×102 – square millimetres (mm2) = 1 000 000 = 10 000 = 100 – square centimetres (cm2) – square metres (m2) km2 m2 cm2 mm2 2 – square kilometres (km ) ÷10002 ÷1002 ÷102 – hectares (ha) ×10 000 = 1 000 000

= 10 000

= 100

m2

ha ÷10 000 ■

Area of squares, rectangles and triangles – Square A = × = 2

– Rectangle A = × b = b b – Triangle

A=

1 1 × b × h = bbh 2 2

The dashed line which gives the height is perpendicular (at right angles) to the base. ■

h

Areas of composite shapes can be found by adding subtracting the area of more basic shapes.

b

1

or 2

Example 6 Converting units of area Convert these area measurements to the units shown in the brackets. a 0.248 m2 (cm2) b 3100 mm2 (cm2) SOLUTION

EXPLANATION

a 0.248 m2 = 0.248 × 10 000 = 2480 cm2

1 m2 = 1002 cm2 = 10 000 cm2 ×1002 Multiply since you are changing to a smaller unit. m2 cm2

b 3100 mm2 = 3100 ÷ 100 = 31 cm2

1 cm2 = 102 mm2 = 100 mm2 Divide since you are changing to a larger unit.

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cm2

mm2 ÷ 102

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Example 7 Finding areas of rectangles and triangles Find the area of these shapes. a 2 cm 6 cm

b 7m 13 m

SOLUTION

EXPLANATION

a A = b =6×2 = 12 cm2

Write the formula for the area of a rectangle and substitute = 6 and b = 2.

b

1 bbh 2 1 = x 13 x 7 2 = 45.5 m2

A=

Remember that the height is measured using a line that is perpendicular to the base.

Example 8 Finding areas of composite shapes Find the area of these composite shapes using addition or subtraction. a b 4m 6m 1 mm 3 mm 1.2 mm 10 m SOLUTION 1 bh 2 1 = 10 × 6 – × 10 × 4 2 = 60 – 20 = 40 m2

a A = b –

2

b A = + b = 32 + 1.2 × 1 = 9 + 1.2 = 10.2 mm2

EXPLANATION The calculation is done by subtracting the area of a triangle from the area of a rectangle. Rectangle – triangle 6m

10 m

4m 10 m The calculation is done by adding the area of a rectangle to the area of a square. A1

A2

Area = A1 + A2

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1 cm = 10 mm

b i How many cm2 in 1 m2? ii How many cm2 in 7 m2? iii How many m2 in 40 000 cm2?

1 m = 100 cm

c i How many m2 in 1 km2? ii How many m2 in 5 km2? iii How many km2 in 2 500 000 m2?

1 km = 1000 m

1 m2

1 km2

1 m = 100 cm

1 km = 1000 m

100 m

d i How many m2 in 1 ha? ii How many m2 in 3 ha? iii How many ha in 75 000 m2?

1 ha

100 m

2 Which length measurements would be used for the base and the height (in that order) to find the area of these triangles? a b c 10 cm 7m 3m

1.7 mm

6 cm 5m

8 cm

2.4 mm

2 mm

3 How many square metres are in one hectare?

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b

c

3m

7m

13 cm

3 cm

6 cm

d

e

f

12 mm

11 m

9 cm 4 cm 3m g

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3m

7m

3 km

2m

4 km 10 km

18 m Example 8

6 Find the area of these composite shapes by using addition or subtraction. a b c 9m 4m

14 cm

5m 8 cm

5m 10 m d

3m e

16 cm 7 cm

f

10 km 6 km

7 km

2 km

3 cm

6 mm 4 mm

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4 Convert these area measurements to the units shown in the brackets. a 2 cm2 (mm2) b 7 m2 (cm2) c 0.5 km2 (m2) d 3 ha (m2) e 0.34 cm2 (mm2) f 700 cm2 (m2) 2 2 2 2 g 3090 mm (cm ) h 0.004 km (m ) i 2000 cm2 (m2) j 450 000 m2 (km2) k 4000 m2 (ha) l 3210 mm2 (cm2) 2 2 2 m 320 000 m (ha) n 0.0051 m (cm ) o 0.043 cm2 (mm2) p 4802 cm2 (m2) q 19 040 m2 (ha) r 2933 m2 (ha) 2 2 s 0.0049 ha (m ) t 0.77 ha (m ) u 2.4 ha (m2)

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3m 12km 3m

3m

20 m 15 km

3m 9 Find the side length of a square if its area is: a 36 m2 b 2.25 cm2 10 a b c d

Find the area of a square if its perimeter is 20 m. Find the area of a square if its perimeter is 18 cm. Find the perimeter of a square if its area is 49 cm2. Find the perimeter of a square if its area is 169 m2.

11 A triangle has area 20 cm2 and base 4 cm. Find its height. 12 Paint costs $12 per litre and can only be purchased in a full number of litres. One litre of paint covers an area of 10 m2. A rectangular wall is 6.5 m long and 3 m high and needs two coats of paint. What will be the cost of paint for the wall?

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8 Find the area of these composite shapes. You may need to determine some side lengths first. a b c 14 m 9 cm 6 cm

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13 Write down expressions for the area of these shapes in simplest form using the letters a and b (e.g. A = 2ab + a2). a b c 2b a

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14 Using only whole numbers for length and breadth, answer the following questions. a How many distinct (different) rectangles have an area of 24 square units? b How many distinct squares have an area of 16 square units? 15 Write down rules for: a the breadth of a rectangle (b) with area A and length b the side length of a square () with area A c the height of a triangle (h) with area A and base b

Enrichment: The acre 16 Two of the more important imperial units of length and area that are still used today are the mile and the acre. Many of our country and city roads, farms and house blocks were divided up using these units. Here are some conversions 1 square mile = 640 acres 1 mile ≈ 1.609344 km 1 hectare = 10 000 m2 a Use the given conversions to find: i the number of square kilometres in 1 square mile (round to 2 decimal places) ii the number of square metres in 1 square mile (round to the nearest whole number) iii the number of hectares in 1 square mile (round to the nearest whole number) iv the number of square metres in 1 acre (round to the nearest whole number) v the number of hectares in 1 acre (round to 1 decimal place) vi the number of acres in 1 hectare (round to 1 decimal place) b A dairy farmer has 200 acres of land. How many hectares is this? (Round your answer to the nearest whole number.) c A house block is 2500 m2. What fraction of an acre is this? (Give your answer as a percentage rounded to the nearest whole number.)

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3D Area of special quadrilaterals The formulas for the area of a rectangle and a triangle can be used to develop the area of other special quadrilaterals. These quadrilaterals include the parallelogram, the rhombus, the kite and the trapezium. Knowing the formulas for the area of these shapes can save a lot of time dividing shapes into rectangles and triangles.

The area of each quadrilateral needs to be calculated to work out how many pavers are needed.

Let’s start: Developing formulas These diagrams contain clues as to how you might find the area of the shape using only what you know about rectangles and triangles. Can you explain what each diagram is trying to tell you? • Parallelogram

• Rhombus

• Kite

• Trapezium 1 2

h

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Area of a parallelogram Area = base × perpendicular height or A = bh Area of a rhombus and the area of a kite 1 Area = × diagonal x × diagonal y 2 1 x or A = xy 2 Area of a trapezium 1 Area = × sum of parallel sides × perpendicular height 2 1 or A = h(a + b) 2

h b y x

y

b h a

Example 9 Finding areas of special quadrilaterals Find the area of these shapes. a b 3m

8m

c 10 cm

20 cm

3 mm 5 mm 11 mm

SOLUTION

EXPLANATION

a A = bh =8×3 = 24 m2

The height is measured at right angles to the base.

1 xy 2 1 = × 10 × 20 2 = 100 cm2

b A=

1 h(a + b) 2 1 = × 5 × (11 + 3) 2 1 = × 5 × 14 2 = 35 mm2

c A=

1 xy since both diagonals are 2 given. This formula can also be used for a rhombus. Use the formula A =

The two parallel sides are 11 mm and 3 mm in length. The perpendicular height is 5 mm.

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Exercise 3D

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C 1 Find the value of A using these formulas and given values. R PS HE 1 M AT I C A a A = bh (b = 2, h = 3) b A = xy (x = 5, y = 12) 2 1 1 c A = h(a + b) (a = 2, b = 7, h = 3) d A = h(a + b) (a = 7, b = 4, h = 6) 2 2 2 Complete these sentences. a A perpendicular angle is __________ degrees. b In a parallelogram, you find the area using a base and the __________________. c The two diagonals in a kite or a rhombus are __________________. 1 d To find the area of a trapezium you multiply by the sum of the two __________________ sides 2 and then by the __________________ height. e The two special quadrilaterals that have the same area formula using diagonal lengths x and y are the __________________ and the __________________.

e

5c

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11 km

3.1 m 3 cm

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6.2 m

22 km h

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2 cm

20 mm

4 cm

1 mm 1.8 mm

30 mm j

7 cm 8 cm 17 cm

k

l 9 m

20 mm 16 mm

5 m

50mm 4m

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4 These trapeziums have one side at right angles to the two parallel sides. Find the area of each. a b c 2 cm 13 cm 4m 3 cm 2 cm 10 m 10 cm 5m 4 cm

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30 cm

60 cm

6 A parallelogram has an area of 26 m2 and its base length is 13 m. What is its perpendicular height? 7 A landscape gardener charges $20 per square metre of lawn. A lawn area is in the shape of a rhombus and its diagonals are 8 m and 14.5 m. What would be the cost of laying this lawn? 8 The parallel sides of a trapezium are 2 cm apart and one of the sides is 3 times the length of the other. If the area of the trapezium is 12 cm2, what are the lengths of the parallel sides?

9 Consider this shape. a What type of shape is it? b Find its area if a = 5, b = 8 and h = 3. All measurements are in cm.

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10 Write an expression for the area of these shapes in simplest form (e.g. A = 2a + 3ab). b a

c

3a

b

x

and 2x

a 2b

1 11 Would you use the formula A = xy to find the area of 2 this rhombus? Explain. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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10 cm

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Enrichment: Proof 12 Complete these proofs to give the formula for the area of a rhombus and a trapezium. a Rhombus A = 4 triangle areas 1 1 2y = 4 × × base × height 2 1 1x =4× × × 2 2

=

b Trapezium 1 A = Area (triangle 1) + Area (triangle 2) 1 1 = × base1 × height1 + × base2 × height2 2 2 1 1 = × × _____ + × × 2 2 = + = c Trapezium 2 A = Area (rectangle) + Area (triangle) 1 = length × breadth + × base × height 2 1 = _____ × _____ + × _____ × _____ 2 = _____ + _____ – _____ = _____ + _____ = __________

a h

1

2 h b

a h b =

+

13 Design an A4 poster for one of the proofs in Question 12 to be displayed in your class.

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3E Area of circles We know that the link between the perimeter of a circle and its radius has challenged civilisations for thousands of years. Similarly people have studied the link between a circle’s radius and its area. Archimedes (287–212 bc) attempted to calculate the exact area of a circle using a particular technique involving limits. If a circle is approximated by a regular hexagon, then the approximate area would be the sum of the areas of 6 triangles with base b and height h. 1 So A ≈ 6 × bh 2 If the number of sides (n) on the polygon increases, the approximation would improve. If n approaches infinity, the error in estimating the area of the circle would diminish to zero. Proof 1 A = n × bh 2 =

1 × nb × h 2

=

1 × 2πr × r 2

b h

Hexagon (n = 6) A = 6 × 1 bh 2

b

h

Dodecagon (n = 12) A = 12 × 1 bh 2

(As n approaches ∞, nb limits to 2πr as nb is the perimeter of the polygon, and h limits to r.)

= πr2

Let’s start: Area as a rectangle Imagine a circle cut into small sectors and arranged as shown. Now try to imagine how the arrangement on the right would change if the number of sector divisions was not 16 (as shown) but a much higher number. • What would the shape on the right look like if the number of sector divisions was a very high number? What would the length and breadth relate to in the original circle? • Try to complete this proof. A = length × breadth 1 = × _____ × r 2 = __________

Breadth

Length

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The ratio of the area of a circle to the square of its radius is equal to π. A i.e. = π so A = πr2 2 r A half circle is called a semicircle. 1 A = πr2 2 r A quarter circle is called a quadrant. 1 A = πr2 4 r

r A = πr2

Example 10 Finding circle areas without technology Find the area of these circles using the given approximate value of π. a b π = 3.14 22 π= 7 10 cm 7m

SOLUTION

EXPLANATION

a = πr2 22 = × 72 7 = 154 m2

Always write the rule. 22 Use π = and r = 7. 7 22 × 7 × 7 = 22 × 7 7

b A = πr2 = 3.14 × 102 = 314 cm2

Use π = 3.14 and substitute r = 10. 3.14 × 102 is the same as 3.14 × 100

Example 11 Finding circle areas using a calculator Use a calculator to find the area of this circle correct to 2 decimal places.

2 cm SOLUTION

EXPLANATION

A = πrr2 = π × 22 = 12.57 cm2 (to 2 decimal places)

Use the π button on the calculator and enter π × 22 or π × 4.

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Chapter 3 Measurement and Pythagoras’ theorem

Example 12 Finding areas of semicircles and quadrants Find the area of this quadrant and semicircle correct to 2 decimal places. a b 5 km 3m SOLUTION

EXPLANATION

1 × πr2 4 1 = × π × 32 4 = 7.07 m2 (to 2 decimal places)

1 The area of a quadrant is the area of a circle with the 4 same radius.

a A=

5 = 2.5 2 1 A = × πr2 2 1 = × π × 2.52 2 = 9.82 km2 (to 2 decimal places)

b r=

The radius is half the diameter. 1 The area of a semicircle is the area of a circle with 2 the same radius.

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3 What is the length of the radius in these shapes? a b 10 m

d

22 × 72 7

d π × 9.82 c

2.3 mm

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7 km

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28 m

14 km

π=

22 7

d

22 π= 7

e

π=

22 7

f 200 m

2m 10 km

π = 3.14

π = 3.14 π = 3.14 Example 11

5 Use a calculator to find the area of these circles correct to 2 decimal places. a b c 1.5 mm

6 m 3 cm

d

f

e

1.7 m

3.4 cm 10 km

Example 12

6 Find the area of these quadrants and semicircles correct to 2 decimal places. a b c 16 cm 17 mm

2 cm d

e

3.6 mm

f

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10 cm

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9 A circular oil slick has a diameter of 1 km. The newspaper reported an area of more than 1 km2. Is the newspaper correct? 10 Two circular plates have radii 12 cm and 13 cm. Find the difference in their area correct to 2 decimal places. 11 Which has the largest area, a circle of radius 5 m, a semicircle of radius 7 m or a quadrant of radius 9 m? 12 A square of side length 10 cm has a hole in the middle. The diameter of the hole is 5 cm. What is the area remaining? Round the answer to the nearest whole number. U

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14 The area of a circle with radius 2 could be written exactly as A = π × 22 = 4π. Write the exact area of these shapes. a b c 9

7 24

1 d. 2 1 a Substitute r = d into the rule A = πrr2 to find a rule for the area of a circle in terms of d. 2 b Use your rule from part a to check that the area of a circle with diameter 10 m is 25π m2.

15 We know that the diameter d of a circle is twice the radius r, i.e. d = 2rr or r =

Enrichment: Reverse problems 16 Reverse the rule A = πrr2 to find the radius in these problems. a If A = 10, use your calculator to show that r ≈ 1.78. b Find the radius of circles with these areas. Round the answer to 2 decimal places. i 17 m2 ii 4.5 km2 iii 320 mm2 c Can you write a rule for r in terms of A? Check that it works for the circles defined in part b. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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3F Area of sectors and composite figures A slice of pizza or a portion of a round cake cut from the centre forms a shape called a sector. The area cleaned by a windscreen wiper could also be thought of as a difference of two sectors with the same angle but different radii. Clearly the area of a sector depends on its radius, but it also depends on the angle between the two straight edges.

i

Let’s start: The sector area formula Complete this table to develop the rule for finding the area of a sector. Angle

180°

90°

Fraction of area 180 ° 360 °

90 ° 360 °

=

1 2

= ___

Area rule

A=

1 × πr2 2

Diagram

180°

A = ___ × πr2 90°

45° 30°

i

A = ___ × πr2

i

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■

A sector is formed by dividing a circle with two radii. r

i

i

r ■

A sector’s area is determined by calculating a fraction of the area of a circle with the same radius. – Fraction is

i 360

– Sector area =

■

θ × πr2 360

r

The area of a composite shape can be found by adding or subtracting the areas of more basic shapes.

i r

r

b 1

A = b + 2 πr2

Example 13 Finding areas of sectors Find the area of these sectors correct to 2 decimal places. a b 120° 70°

2 cm SOLUTION

θ × πr2 360 120 = × π × 22 360 1 = ×π×4 3 = 4.19 cm2 (to 2 decimal places)

a A=

b θ = 360 – 70 = 290 θ A= × πr2 360 290 = × π × 52 360 = 63.27 m2 (to 2 decimal places)

5 m

EXPLANATION First write the rule for the area of a sector. Substitute θ = 120 and r = 2. Note that 1 to . 3

120 simplifies 360

First calculate the angle inside the sector and remember that a revolution is 360°. Then substitute θ = 290 and r = 5.

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Measurement and Geometry

Example 14 Finding areas of composite shapes. 20 mm

Find the area of this composite shape correct to 1 decimal place. 10 mm SOLUTION

The area can be found by subtracting the area of a quadrant from the area of a rectangle.

1 = 20 × 10 – × π × 102 4 = 200 – 25π = 121.5 mm2 (to 1 decimal place)

Exercise 3F

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1 Simplify these fractions.

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A = b –

EXPLANATION

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180 20 210 b c × π × 22 × π × 72 × π × 2.32 360 360 360 3 What fraction of a circle in simplest form is shown by these sectors? a b c a

60°

120°

3 cm

e

2.5 cm

f 240°

270° 5.1 m

11.2 cm

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18.9 m 5 Find the area of these sectors correct to 2 decimal places. b a

c 115° 14.3 km

7.5 m

6 Find the areas of these composite shapes using addition or subtraction. Round the answer to 2 decimal places. a b c 2m 10 cm 3m

20 cm

5 m

d

9 mm

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3m

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7 A simple bus wiper blade wipes an area over 100° as shown. Find the area wiped by the blade correct to two decimal places.

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1.2 m

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Example 14

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8 At Buy-by-the-sector Pizza they offer a sector of a 15 cm radius pizza with an angle of 45° or a sector of a 13 cm radius pizza with an angle of 60°. Which piece gives the bigger area and by how much? Round the answer to 2 decimal places.

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9 An archway is made up of an inside and outside semicircle as shown. Find the area of the arch correct to the nearest whole cm2. 10 What percentage of the total area is occupied by the shaded region in these diagrams? Round the answer to 1 decimal place. a

60 cm 60 cm

b

c 249°

7.2 cm 3 m

4 cm

c

b

40°

1 mm

5 m

2 cm 3 mm d

and 3cm

f 15 km

5 m

10 m

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11 An exact area measure in terms of π might look like π × 22 = 4π. Find the exact area of these shapes in terms of π. Simplify your answer.

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Enrichment: Sprinkler waste 13 A rectangular lawn area has a 180° sprinkler positioned in the middle of one side as shown.

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Lawn

2.5 m 30°

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A 4.33 m

5 m

a Find the area of the sector OAB correct to 2 decimal places. b Find the area watered by the sprinkler outside the lawn area correct to 2 decimal places. c Find the percentage of water wasted, giving the answer correct to 1 decimal place.

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12 Consider the percentage of the area occupied by a circle inside a square and touching all sides as shown. a If the radius of the circle is 4 cm, find the percentage of area occupied by the circle. Round the answer to 1 decimal place. b Repeat part a for a radius of 10 cm. What do you notice? c Can you prove that the percentage area is always the same for any radius r? Hint: Find the percentage area using the letter r for the radius.

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3G Surface area of prisms

E X T E N S I ON

Many problems in three dimensions can be solved by looking at the problem or parts of the problem in two dimensions. Finding the surface area of a solid is a good example of this, as each face can usually be looked at in two-dimensional space. The approximate surface area of the walls of an unpainted house, for example, could be calculated by looking at each wall separately and adding to get a total surface area.

Let’s start: Possible prisms Here are three nets that fold to form three different prisms. • Can you draw and name the prisms? • Try drawing other nets of these prisms that are a different shape to the nets given here.

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A prism is a polyhedron with a constant (uniform) cross-section. – The cross-section is parallel to the two identical (congruent) ends. – The other sides are parallelograms (or rectangles for right prisms). A net is a two-dimensional representation of all the surfaces of a solid. It can be folded to form the solid.

Key ideas

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The surface area (A) of a prism is the sum of the areas of all its faces. h s A = 6s2

b A = 2b + 2h + 2bh

Cambridge University Press

Chapter 3 Measurement and Pythagoras’ theorem

Example 15 Calculating surface areas Find the surface area of this prism. 10 cm 8 cm 6 cm

15 cm

SOLUTION

EXPLANATION

Area of 2 triangular ends 1 A = 2 × × bh 2 1 =2× ×6×8 2 = 48 cm2 Area of 3 rectangles A = (6 × 15) + (8 × 15) + (10 × 15) = 360 cm2 Surface area A = 48 + 360 = 408 cm2

One possible net is: 15 cm 6 cm 8 cm

8 cm

8 cm

10 cm Work out the area of each shape or group of shapes and find the sum of their areas to obtain the surface area.

The surface area of this chocolate can be estimated by a similar process. EXTENSION

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3 How many rectangular faces are on these solids? a Triangular prism b Rectangular prism c Hexagonal prism d Pentagonal prism

Example 15

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5 This prism has two end faces that are parallelograms. a Use A = bh to find the combined area of the 2 cm two ends. b Find the surface area of the prism.

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7 A book 20 cm long, 15 cm wide and 3 cm thick is covered in plastic. What area of plastic is needed to cover 1000 books? Convert your answer to m2. 8 Find the surface area of these solids. b a 5m

c

6m

3m

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8m

7m

15 m

6m

10 m 3m 3 cm

9 The floor, sides and roof of this tent are made from canvas at a cost of $5 per square metre. The tent’s dimensions are shown in the diagram. What is the cost of the canvas for the tent? 2.8 m 2m 1.5 m

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40 cm a Find its surface area both inside and out. b If the box was made with wood that is 1 cm thick, what would be the increase in surface area?

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11 A cube of side length 1 cm has a surface area of 6 cm2. a What is the effect on the surface area of the cube if: i its side length is doubled? ii its side length is tripled? iii its side length is quadrupled? b Do you notice a pattern from your answers to part a. What effect would multiplying the side length by a factor of n have on the surface area?

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3H Volume and capacity Volume is a measure of the space occupied by a three-dimensional object. It is measured in cubic units. Common metric units for volume given in abbreviated form include mm3, cm3, m3 and km3. We also use mL, L, kL and ML to describe volumes of fluids or gas. The volume of space occupied by a room in a house, for example, might be calculated in cubic metres (m3) or the capacity of fuel tanker might be measured in litres (L) or kilolitres (kL).

Let’s start: Volume

Key ideas

We all know that there are 100 cm in 1 m, but do you know how many cubic centimetres are in 1 cubic metre? • Try to visualise 1 cubic metre – 1 metre long, 1 metre wide and 1 metre high. Guess how many cubic centimetres would fit into this space. • Describe a method for working out the exact answer. Explain how your method works.

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Volume is measured in cubic units. The common metric units for volume include: – cubic millimetres (mm3) 1 mm – cubic centimetre (cm3)

1 mm 1 mm

1 cm 1 cm

1 cm

– cubic metre (m3) 1m (Not drawn to scale.) 1m ■

1 m

Conversions for volume ×10003 km3 ÷10003

x 1003 m3

×103 cm3

÷1003

mm3 ÷103

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Capacity is the volume of fluid or gas that a container can hold. Common metric units are: – millilitre (mL) – litre (L) – kilolitre (kL) – megalitre (ML) Some common conversions are: – 1 mL = 1 cm3 – 1 L = 1000 mL – 1 kL = 1000 L = 1 m3

×1000 ML

×1000 kL

÷1000

L ÷1000

mL ÷1000

M I L K

M I L K

1L (1000 mL)

1000 cm3

Volume of a rectangular prism – Volume = length × breadth × height V = bh Volume of a cube V = s3

×1000

h b

s

Example 16 Finding the volume of a rectangular prism Find the volume of this rectangular prism.

2m

6m 4m

SOLUTION

EXPLANATION

V = bh

First write the rule and then substitute for the length, breadth and height. Any order will do since 6 × 4 × 2 = 4 × 6 × 2 = 2 × 4 × 6 etc.

= 6×4×2 = 48 m3

Example 17 Calculating capacity Find the capacity, in litres, for a container that is a rectangular prism 20 cm long, 10 cm wide and 15 cm high. SOLUTION

EXPLANATION

V = bh

First calculate the volume of the container in cm3. Then convert to litres using 1 L = 1000 cm3.

= 20 x 10 x 15 = 3000 cm3 = 3000 ÷ 1000 =3L

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Measurement and Geometry

Chapter 3 Measurement and Pythagoras’ theorem

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___ × 5 × 20 = 600 20 × 2 × ___ = 200

b d

3 Write the missing number in the following unit conversions. a 1 L = ___ mL b ___ kL = 1000 L c 1000 kL = ___ ML d 1 mL = ___ cm3 e 1000 cm3 = ___ L f 1 m3 = ___ L

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5 Convert the measurements to the units shown in the brackets. a 2 L (mL) b 5 kL (L) c 0.5 ML (kL) 3 3 e 4 mL (cm ) f 50 cm (mL) g 2500 cm3 (L) Example 17

d 3000 mL (L) h 5.1 L (cm3)

6 Find the capacity of these containers, converting your answer to litres. a

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30 cm

70 cm 60 cm

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7 A oil tanker has a capacity of 60 000 m3. a What is the ship’s capacity in: i litres? ii kilolitres? iii megalitres? b If the ship leaks oil at a rate of 300 000 litres per day, how long will it take for all the oil to leak out? Assume the ship started with full capacity.

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8 Water is being poured into a fish tank at a rate of 2 L every 10 seconds. The tank is 1.2 m long by 1 m wide by 80 cm high. How long will it take to fill the tank? Give the answer in minutes. 9 A city skyscraper is a rectangular prism 50 m long, 40 m wide and 250 m high. a What is the total volume in m3? b What is the total volume in ML? 10 If 1 kg is the mass of 1 L of water, what is the mass of water in a full container that is a cube with side length 2 m?

c 47 cubic units

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13 How many cubes with side lengths that are a whole number of centimetres have a capacity of less than 1 litre?

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11 Using whole numbers only, give all the possible dimensions of rectangular prisms with the following volume. Assume the units are all the same. a 12 cubic units b 30 cubic units

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Enrichment: Halving rectangular prisms 15 This question looks at using half of a rectangular prism to find the volume of a triangular prism. a Consider this triangular prism. i Explain why this solid could be thought of as half a rectangular prism. ii Find its volume. b Using a similar idea, find the volume of these prisms. i

2 8 cm

5m 8m

7m 4cm

10 cm

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4 cm 9 cm 5 cm

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3I Volume of prisms and cylinders We know that for a rectangular prism its volume V is given by the rule V = bh. Length × breadth (b) gives the number of cubes on the base, but it h also tells us the area of the base A. So V = bh could also be written as V = Ah. A The rule V = Ah can also be applied to prisms that have different shapes b as their bases. One condition, however, is that the area of the base must represent the area of the cross-section of the solid. The height h is measured perpendicular to the cross-section. Note that a cylinder is not a prism as it does not have sides that are parallelograms (or rectangles); however, it can be treated like a prism when finding its volume because it has a constant cross-section, a circle. Here are some examples of two prisms and a cylinder with A and h marked. h h h A

A

A

Let’s start: Drawing prisms Try to draw prisms (or cylinders) that have the following shapes as their cross-sections. • Circle

• Triangle

• Trapezium

• Pentagon

• Parallelogram

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A prism is a polyhedron with a constant (uniform) cross-section. – Its sides between the two congruent ends are parallelograms. – A right prism has rectangular sides between the congruent ends.

A

h

Volume of a prism = Area of cross-section × perpendicular height or V = Ah. Volume of a cylinder = Ah = πr2 × h = πr2h So V = πr2h

r

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Key ideas

The cross-section of a prism should be the same size and shape along the entire length of the prism. Check this property on your drawings.

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Chapter 3 Measurement and Pythagoras’ theorem

Example 18 Finding the volumes of prisms Find the volumes of these prisms. a A = 10 cm2

b

2m

3 cm

8m

4m SOLUTION

EXPLANATION

and V = Ah = 10 × 3 = 30 cm3

Write the rule and substitute the given values of A and h, where A is the area of the cross-section.

b V = Ah 1 = × 4 × 2 × 8 2 3 = 32 m

1 The cross-section is a triangle, so use A = bh with 2 base 4 m and height 2 m.

Example 19 Finding the volume of a cylinder Find the volumes of these cylinders, rounding to 2 decimal places. a b 2 cm 14 m 20 m

10 cm

SOLUTION a V = πr2h = π × 22 × 10 = 125.66 cm3 b V = πr2h = π × 72 × 20 = 3078.76 m3

EXPLANATION

(to 2 decimal places)

Write the rule and then substitute the given values for π, r and h. Round as required. The diameter is 14 m so the radius is 7 m. Round as required.

(to 2 decimal places)

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2 For these prisms and cylinder state the value of A and the value of h that could be used in the rule V = Ah to find the volume of the solid. b

a

A = 6 m2

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10 mm

2 cm A = 8 cm2

A = 12 mm2

1.5 m

3 Evaluate the following. 1 ×3×4×6 2

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1 (2 + 3) × 4 2

d 3.14 × 102 × 20 WO

4 Find the volume of these solids using V = Ah. A = 4 m2

b A = 20 cm2

8 cm

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11 mm

11 m A = 32 mm2

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Example 18a

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It is 5m

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2m 5m 6 Find the volume of these cylinders. Round the answer to 2 decimal places. a b 40 mm 10 m c

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8 These solids are made up of more than one rectangular prism. Use addition or subtraction to find the volume of the composite solid. a b c 4m 8 cm

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5 cm 2 cm

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Example 19

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9 Susan pours water from a full 4 L container into a number of water bottles for a camp hike. Each water bottle is a cylinder with radius 4 cm and height 20 cm. How many bottles can be filled completely?

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10 There are 80 liquorice cubes stacked in a cylindrical glass jar. The liquorice cubes have a side length of 2 cm and the glass jar has a radius of 5 cm and a height of 12 cm. How much air space remains in the jar of liquorice cubes? Give the answer correct to 2 decimal places.

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Enrichment: Complex composites 15 Use your knowledge of volumes of prisms and cylinders to find the volume of these composite solids. Round the answer to 2 decimal places where necessary. b

5 mm

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16 cm 15 cm

2 cm

5cm and

d 108 cm

f 8m

12 m

4m 10m 2m

8m

8 cm 24 cm

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13 A cylinder has a volume of 100 cm3. Give three different combinations of radius and height measurements that give this volume. Give these lengths correct to 2 decimal places.

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11 A swimming pool is a prism with a cross-section that is a trapezium. The pool is being filled at a rate of 1000 litres per hour. a Find the capacity of the pool in litres. b How long will it take to fill the pool?

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3J Time

R E V I S I ON

Time in minutes and seconds is based on the number 60. Other units of time, including the day and year, are defined by the rate at which the Earth spins on its axis and the time that the Earth takes to orbit the Sun. The origin of the units seconds and minutes dates back to the ancient Babylonians, who used a base 60 number system. The 24-hour day dates back to the ancient Egyptians, who described the day as 12 hours of day and 12 hours of night. Today, we use a.m. (ante meridiem, which is Latin for ‘before noon’) and p.m. (post meridiem, which is Latin for ‘after noon’) to represent the hours before and after noon (midday). During the rule of Julius Caesar, the ancient Romans introduced the solar calendar, which recognised that the The Earth takes 1 year to orbit the Sun. 1 Earth takes about 365 days to orbit the Sun. This gave 4 rise to the leap year, which includes one extra day (in February) every 4 years.

Let’s start: Knowledge of time

Key ideas

Do you know the answers to these questions about time and the calendar? • When is the next leap year? • Why do we have a leap year? • Which months have 31 days? • Why are there different times in different countries or parts of a country? • What do bce (or bc) and ce (or ad) mean on time scales?

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The standard unit of time is the second (s). Units of time include: – 1 minute (min) = 60 seconds (s) × 24 × 60 × 60 – 1 hour (h) = 60 minutes (min) day hour minute second – 1 day = 24 hours (h) ÷ 24 ÷ 60 ÷ 60 – 1 week = 7 days – 1 year = 12 months a.m. or p.m. is used to describe the 12 hours before and after noon (midday). 24-hour time shows the number of hours and minutes after midnight. – 0330 is 3:30 a.m. – 1530 is 3:30 p.m.

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The ‘degrees, minutes and seconds’ button on a calculator can be used to covert a particular time into hours, minutes and seconds. e.g. 4.42 hours = 4°25′12″ meaning 4 hours, 25 minutes and 12 seconds. The Earth is divided into 24 time zones (one for each hour). – Twenty-four 15° lines of longitude divide the Earth into its time zones. Time zones also depend on a country’s borders and its proximity to other countries. (See map on pages 172–173 for details.) – Time is based on the time in a place called Greenwich, United Kingdom, and this is called Coordinated Universal Time (UTC) or Greenwich Mean Time (GMT). – Places east of Greenwich are ahead in time. – Places west of Greenwich are behind in time. Australia has three time zones: – Eastern Standard Time (EST), which is UTC plus 10 hours. – Central Standard Time (CST), which is UTC plus 9.5 hours. – Western Standard Time (WST), which is UTC plus 8 hours.

Example 20 Converting units of time Convert these times to the units shown in brackets. a 3 days (minutes) b 30 months (years) SOLUTION

EXPLANATION

a 3 days = 3 × 24 h = 3 × 24 × 60 min = 4320 min

1 day = 24 hours 1 hour = 60 minutes

b 30 months = 30 ÷ 12 years 1 = 2 years 2

There are 12 months in 1 year.

Example 21 Using 24-hour time Write these times using the system given in brackets. a 4:30 p.m. (24-hour time) b 1945 (a.m./p.m.) SOLUTION

EXPLANATION

a 4:30 p.m. = 1200 + 0430 = 1630 hours

Since the time is p.m., add 12 hours to 0430 hours.

b 1945 hours = 7:45 p.m.

Since the time is after 1200 hours, subtract 12 hours.

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Measurement and Geometry

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11

10

9

8

7

5

6

4

6

3

4

2

1

GREENLAND 1

3

ALASKA 9

1

6

ICELAND

SWEDEN NORWAY K UNITED KINGDOM N GERMANY POLAND L

CANADA 4 Q

IRELAND 3½

FRANCE 8

P

UNITED STATES 5

6

7

R

1

ITALY

PORTUGAL SPAIN

GREECE

MOROCCO LIBYA

ALGERIA MEXICO

CUBA

MAURITANIA

MALI

NIGER

the NIGERIA

VENEZUELA

CHAD 1

COLUMBIA

DEM. RE OF THE CO

5 PERU

4 BRAZIL

J K L M N P Q R

ANGOLA

BOLIVIA

World cities key

NAMIBIA

Auckland Edinburgh Greenwich Johannesburg London New York Vancouver Washington, DC

3 ARGENTINA

CHILE

Sun 1:00

2:00

3:00

4:00

5:00

6:00

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Example 22 Using time zones Coordinated Universal Time (UTC) and is based on the time in Greenwich, United Kingdom. Use the world time zone map (on pages 172–173) to answer the following. a When it is 2 p.m. UTC, find the time in these places. i France ii China iii Queensland iv Alaska b When it is 9:35 a.m. in New South Wales, Australia, find the time in these places. i Alice Springs ii Perth iii London iv central Greenland SOLUTION

EXPLANATION

a i 2 p.m. + 1 hour = 3 p.m.

Use the time zone map to see that France is to the east of Greenwich and is in a zone that is 1 hour ahead.

ii 2 p.m. + 8 hours = 10 p.m.

From the time zone map, China is 8 hours ahead of Greenwich.

iii 2 p.m. + 10 hours = 12 a.m.

Queensland uses Eastern Standard Time, which is 10 hours ahead of Greenwich.

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2 p.m. – 9 hours = 5 a.m.

b i 9:35 a.m. –

1 hour = 9:05 a.m. 2

ii 9:35 a.m. – 2 hours = 7:35 a.m.

Alaska is to the west of Greenwich, in a time zone that is 9 hours behind. 1 Alice Springs uses Central Standard Time, which is hour 2 behind Eastern Standard Time. Perth uses Western Standard Time, which is 2 hours behind Eastern Standard Time.

iii 9:35 a.m. – 10 hours = 11:35 p.m. UTC (time in Greenwich, United Kingdom) is 10 hours (the day before) behind EST.

Exercise 3J

Central Greenland is 3 hours behind UTC in Greenwich, so is 13 hours behind EST.

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6 Write these times using the system shown in brackets. a 1:30 p.m. (24-hour) b 8:15 p.m. (24-hour) d 11:59 p.m. (24-hour) e 0630 hours (a.m./p.m.) g 1429 hours (a.m./p.m.) h 1938 hours (a.m./p.m.)

c 10:23 a.m. (24-hour) f 1300 hours (a.m./p.m.) i 2351 hours (a.m./p.m.)

7 Round these times to the nearest hour. a 1:32 p.m. b 5:28 a.m.

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c 2:37 p.m. and 5:21 p.m. f 1940 and 0629 hours

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9 Use the time zone map on pages 172–173 to find the time in the following places, when it is 10 a.m. UTC. a Spain b Turkey c Tasmania d Darwin e Argentina f Peru g Alaska h Portugal

Example 22b

10 Use the time zone map on pages 172–173 to find the time in these places, when it is 3:30 p.m. in Victoria. a United Kingdom b Libya c Sweden d Perth e Japan f central Greenland g Alice Springs h New Zealand

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13 Three essays are marked by a teacher. The first takes 4 minutes and 32 seconds to mark, the second takes 7 minutes and 19 seconds, and the third takes 5 minutes and 37 seconds. What is the total time taken to complete marking the essays? 14 Adrian arrives at school at 8:09 a.m. and leaves at 3:37 p.m. How many hours and minutes is Adrian at school? 15 On a flight to Europe, Janelle spends 8 hours and 36 minutes on a flight from Melbourne to Kuala Lumpur, Malaysia, 2 hours and 20 minutes at the airport at Kuala Lumpur, and then 12 hours and 19 minutes on a flight to Geneva, Switzerland. What is Janelle’s total travel time? 16 A phone plan charges 11 cents per 30 seconds. The 11 cents are added to the bill at the beginning of every 30-second block of time. a What is the cost of a 70-second call? b What is the cost of a call that lasts 6 minutes and 20 seconds? 17 A doctor earns $180 000 working 40 weeks per year, 5 days per week, 10 hours per day. What does the doctor earn in each of these time periods? a per day b per hour c per minute d per second (in cents) 18 A 2-hour football match starts at 2:30 p.m. Eastern Standard Time (EST) in Newcastle, NSW. What time will it be in the United Kingdom when the match finishes? 19 If the date is 29 March and it is 3 p.m. in Perth, what is the time and date in these places? a Italy b Alaska c Chile

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Enrichment: Daylight saving 26 Use the internet to investigate how daylight saving affects the time in some places. Write a brief report discussing the following points. a i Name the States in Australia that use daylight saving. ii Name five other countries that use daylight saving. b Describe how daylight saving works, why it is used and what changes have to be made to our clocks. c Describe how daylight saving in Australia affects the time difference between time zones. Use New South Wales and Greece as an example.

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20 Monty departs on a 20-hour flight from Brisbane to London, United Kingdom, at 5 p.m. on 20 April. Give the time and date of his arrival in London.

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3K Introducing Pythagoras’ theorem Pythagoras was a philosopher in ancient Greece who lived in the sixth century bc. He studied astronomy, mathematics, music and religion, but is most well known for his famous theorem. Pythagoras was known to provide a proof for the theorem that bears his name, and methods to find Pythagorean triads, which are sets of three whole numbers that make up the sides of right-angled triangles. The ancient Babylonians, 1000 years before Pythagoras’ time, and the Egyptians also knew that there was a relationship between the sides of a right-angled triangle. Pythagoras, however, was able to clearly explain and prove the theorem using mathematical symbols. The ancient theorem is still one of the most commonly used theorems today. Pythagoras’ theorem states that the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. An illustration of the theorem includes squares drawn on the sides of the right-angled triangle. The area of the larger square (c2) is equal to the sum of the two smaller squares (a2 + b2).

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The hypotenuse – It is the longest side of a right-angled triangle. Hyp c – It is opposite the right angle. a oten u Pythagoras’ theorem se – The square of the hypotenuse is the sum of the squares b of the other two shorter sides. 2 2 2 2 2 2 – a +b =c or c =a +b A Pythagorean triad is a set of three integers which satisfy Pythagoras’ theorem.

Example 23 Checking Pythagorean triads Decide if the following are Pythagorean triads a 6, 8, 10 b 4, 5, 9 SOLUTION

EXPLANATION

a a2 + b2 = 62 + 82 = 36 + 64 = 100 (= 102)

Let a = 6, b = 8 and c = 10 and check that a2 + b2 = c2.

b a2 + b2 = 42 + 52 = 16 + 25 = 41 ≠ 92

a2 + b2 + 41 and c2 = 81 so the set of numbers are not a Pythagorean triad.

Example 24 Deciding if a triangle has a right angle Decide if this triangle has a right angle.

7m 4m 9m

SOLUTION

EXPLANATION

a2 + b2 = 42 + 72 = 16 + 49 = 65 ≠ 92

Check to see if a2 + b2 = c2. In this case a2 + b2 = 65 and c2 = 81 so the triangle is not right angled.

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3L Using Pythagoras’ theorem From our understanding of equations, it may be possible to solve the equation to find the unknown. This is also the case for equations derived from Pythagoras’ theorem where if two of the side lengths of a rightangled triangle are known, then the third can be found. So if c2 = 32 + 42 then c2 = 25 and c = 5. We also notice that if c2 = 25 then c = 25 = 5 (if c > 0). This application of Pythagoras’ theorem has wide range of applications wherever right-angled triangles can be drawn.

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Note that a number using a sign may not always result in a whole number. For example, 3 and 24 are not whole numbers and cannot be written as a fraction. These types of numbers are called surds and are a special group of numbers (irrational numbers) that are often approximated using rounded decimals.

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Let’s start: Correct layout Three students who are trying to find the value of c in this triangle using Pythagoras’ theorem write their solutions on a board. There are only very minor differences between each solution and the answer is written rounded to 2 decimal places. Which student has all the steps correct? Give reasons why the other two solutions are not laid out correctly.

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Surds are numbers that have a sign when written in simplest form. – They are not a whole number and cannot be written as a fraction. – Written as a decimal, the decimal places would continue forever with no repeated pattern (just like the number pi), so surds are irrational numbers. c 2, 5, 2 3, – 3 90 and 7 30 are all examples of surds. a Using Pythagoras’ theorem 2 2 If c2 = a2 + b2 then c = a + b .

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Example 25 Finding the length of the hypotenuse Find the length of the hypotenuse for these right-angled triangles. Round the answer for part b to 2 decimal places. a b 9 c 6 c 8

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EXPLANATION

a c2 = a2 + b2 = 62 + 82 = 100 ∴ c = 100 = 10

Write the equation for Pythagoras’ theorem and substitute the values for the shorter sides.

b c2 = a2 + b2 = 72 + 92 = 130 ∴ c = 130 = 11.40 (to 2 decimal places)

First calculate the value of 72 + 92.

Find c by taking the square root.

130 is a surd (the exact answer), so round the answer as required.

Example 26 Applying Pythagoras’ theorem A rectangular wall is to be strengthened by a diagonal brace. The wall is 6 m wide and 3 m high. Find the length of brace required correct to the nearest cm.

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c2 = a2 + b2

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3M Calculating the length of a shorter side We know that if we are given the two shorter sides of a right-angled triangle we can use Pythagoras’ theorem to find the length of the hypotenuse. Generalising further, we can say that if given any two sides of a right-angled triangle we can use Pythagoras’ theorem to find the length of the third side.

Let’s start: What’s the setting out? The triangle shown has a hypotenuse length of 15 and one of the shorter sides is of length 12. Here is the setting out to find the length of the unknown side a. Can you fill in the missing gaps and explain what is happening at each step. a a2 + b2 = c2 2 2 2 a + ___ = ___ a2 + ___ = ___ 12 a2 = ___ (Subtract ___ from both sides) 15 (Hypotenuse) ______ ∴a=

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Pythagoras’ theorem can be used to find the length of the shorter sides of a right-angled triangle if the hypotenuse and another side are known. Use subtraction to make the unknown the subject of the equation. a2 + b2 = c2 24 a2 + 242 = 252 a2 + 576 = 625 a a2 = 49 (Subtract 576 from both sides.) 25 ∴ a = 49 =7

Example 27 Finding the length of a shorter side Find the value of a in this right-angled triangle.

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a2 + b2 = c2 a2 + 42 = 52 a2 + 16 = 25 a2 = 9 ∴a = 9

Write the equation for Pythagoras’ theorem and substitute the known values. Subtract 16 from both sides.

=3

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Example 28 Applying Pythagoras to find a shorter side A 10 m steel brace holds up a concrete wall. The bottom of the brace is 5 m from the base of the wall. Find the height of the concrete wall correct to 2 decimal places.

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Let a metres be the height of the wall. a2 + b2 = c2 a2 + 52 = 102 a2 + 25 = 100 a2 = 75 ∴ a = 75 = 8.66 (to 2 decimal places) The height of the wall is 8.66 metres.

Choose a letter (pronumeral) for the unknown height. Substitute into Pythagoras’ theorem. Subtract 25 from both sides. 75 is the exact answer. Round as required. Answer a worded problem using a full sentence.

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5 A yacht’s mast is supported by a 12 m cable attached to its top. On the deck of the yacht, the cable is 8 m from the base of the mast. How tall is the mast? Round the answer to two decimal places.

MA

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6 A circle’s diameter AC is 15 cm and the chord AB is 9 cm. Angle ABC is 90°. Find the length of the chord BC.

M AT I C A

9 cm B

14

cm

7 A 14 cm drinking straw just fits into a can as shown. The diameter of the can is 7 cm. Find the height of the can correct to 2 decimal places.

7 cm

8 Find the length AB is this diagram. Round to 2 decimal places.

11 A

B

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9 Describe what is wrong with the second line of working in each step. a a2 + 10 = 24 b a2 = 25 c a2 + 25 = 36 2 a = 34 =5 a+5=6

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10 The number 11 is an example of a surd that is written as an exact value. Find the surd that describes the exact lengths of the unknown sides of these triangles. a b c 100 5 7

F Y

24

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25

3M

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11 Show how Pythagoras’ theorem can be used to find the unknown length in these isosceles triangles. Complete the solution for part a and then try the others. Round to 2 decimal places.

HE

T

a2 + b2 = c2 x2 + x2 = 52 2x2 = 25 x2 = ___

a 5

x

∴x=

x b

10

______

c

x

x

x

d

34

61

x

Enrichment: Pythagorean families 12 (3, 4, 5) is called a Pythagorean triad because the numbers 3, 4 and 5 satisfy Pythagoras’ theorem (32 + 42 = 52). a Explain why (6, 8, 10) is also a Pythagorean triad. b Explain why (6, 8, 10) is considered to be in the same family as (3, 4, 5). c List three other Pythagorean triads in the same family as (3, 4, 5) and (6, 8, 10). d Find another triad not in the same family as (3, 4, 5), but which has all three numbers less than 20. e List five triads that are each the smallest triad of five different families.

3, 4, 5 is the best known of an infinite number of Pythagorean triads.

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GMT and travel As discussed in Section 3J, the world is divided into 24 time zones, which are determined loosely by each 15° meridian of longitude. World time is based on the time at a place called Greenwich near London, United Kingdom. This time is called Coordinated Universal Time (UTC) or Greenwich Mean Time (GMT). Places east of Greenwich are ahead in time and places west of Greenwich are behind. In Australia, the Western Standard Time is 2 hours behind Eastern Standard Time and Central 1 Standard Time is hour behind Eastern Standard Time. Use the world time zone map on pages 172–173 2 to answer these questions and to investigate how the time zones affect the time when we travel. 1

East and west Name five countries that are: a ahead of GMT

2

b

behind GMT

Noon in Greenwich When it is noon in Greenwich, what is the time in these places? a Sydney b Perth c Darwin d Washington, DC e Auckland f France g Johannesburg h Japan

3

2 p.m. EST When it is 2 p.m. Eastern Standard Time (EST) on Wednesday, find the time and day in these places. a Perth b Adelaide c London d western Canada e China f United Kingdom g Alaska h South America

4

Adjusting your watch a Do you adjust your watch forwards or backwards when you are travelling to these places? i India ii New Zealand b In what direction should you adjust your watch if you are flying over the Pacific Ocean?

5

Flight travel a You fly from Perth to Brisbane on a 4-hour flight that departed at noon. What is the time in Brisbane when you arrive? b You fly from Melbourne to Edinburgh on a 22-hour flight that departed at 6 a.m. What is the time in Edinburgh when you arrive? c You fly from Sydney to Los Angeles on a 13-hour flight that departed at 7:30 p.m. What is the time in Los Angeles when you arrive?

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Investigation

Measurement and Geometry

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d Copy and complete the following table. Departing

Arriving

Departure time

Flight time (hours)

Brisbane

Broome

7 a.m.

3.5

Melbourne

London

1 p.m.

23

Hobart

Adelaide

1.5

4 p.m.

Arrival time

London

Tokyo

12

11 p.m.

New York

Sydney

15

3 a.m.

Beijing

Vancouver

3:45 p.m.

7:15 p.m.

e Investigate how daylight saving alters the time in some time zones and why. How does this affect flight travel? Give examples.

Pythagorean triads and spreadsheets Pythagorean triads (or triples) can be grouped into families. The triad (3, 4, 5) is the base triad for the family of triads (3k, 4k, 5k). Here are some triads in this same family. k

1

2

3

Triad

(3, 4, 5) (base triad)

(6, 8, 10)

(9, 12, 15)

1 Write down three more triads in the family (3k, 4k, 5k). 2 Write down three triads in the family (7k, 24k, 25k). 3 If (3k, 4k, 5k) and (7k, 24k, 25k) are two triad families, can you find three more families that have whole numbers less than 100? 4 Pythagoras discovered that if the smaller number in a base triad is a then the other two numbers in the triad are given by the rules 1 2 1 (a + 1) and (a2 – 1) 2 2 Set up a spreadsheet to search for all the families of triads of whole numbers less than 200. Here is how a spreadsheet might be set up.

Fill down far enough so that c is a maximum of 200. 5 List all the base triads that have whole numbers (less than 200). How many are there?

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1 A cube has capacity 1 L. What are its dimensions in cm correct to 1 decimal place? 2 A fish tank is 60 cm long, 30 cm wide, 40 cm high and contains 70 L of water. Rocks with a volume of 3000 cm3 are placed into the tank. Will the tank overflow? 3 What proportion (fraction or percentage) of the semicircle does the full circle occupy? 4 What is the distance AB in this cube? (Pythagoras’ theorem is required.) B

1 m

A

5 By what number do you multiply the radius of a circle to double its area? 6 Find the exact value (as a surd) of a in this diagram. (Pythagoras’ theorem is required.) a

1

1

1 1

1 1

1

7 A cube of side length 3 cm has its core removed in all directions as shown. Find its surface area both inside and out.

3 cm 1 cm 1 cm 3 cm 8 A square just fits inside a circle. What percentage of the circle is occupied by the square?

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Puzzles and challenges

Measurement and Geometry

Chapter summary

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Chapter 3 Measurement and Pythagoras’ theorem

Perimeter

Triangle 4 cm

Quadrilaterals – Square A = 2 – Rectangle A = b – Parallelogram A = bh – Rhombus A = 12 xy – Kite A = 12 xy – Trapezium A = 12 h(a + b)

3 cm

10 cm

6 cm

P = 2 × 10 + 2 × 4 = 28 cm

A=

1 bh 2 1 ×6 2

= ×3 = 9 cm2 Units 1 km = 1000 m 1 m = 100 cm 1 cm = 10 mm

2

2

2

2

m

cm

2

2

3m

2

Sector θ 360 280 360

× πr 2

× π × 22 = 9.77 m2

=

Area

2m 280°

Measurement and Pythagoras’ theorem

Volume

Time

units cm3,

mm3

1 min = 60 s 1 h = 60 min 0311 is 03:11 a.m. 2049 is 08:49 p.m.

Pythagoras' theorem

×1000 ×1000 ×1000 ML

7 cm

÷10

A=

Length

m3,

2

mm

1 ha = 10,000 m2

Circumference

km3,

×10 2

÷1000 ÷100

A = πr 2 = π × 72 = 49π cm2

2

×1000 ×100 km

C = 2πr or πd =2×π×3 = 6π m2

Circle

Units

kL

L

mL

Theorem

÷1000 ÷1000 ÷1000 1 mL = 1 cm3 1 m3 = 1000 L

a

b a2 + b2 = c2

Rectangular prism V = bh = 10 × 20 × 30 = 6000 cm3 =6L

c

Prism and cylinders V = Ah = 12 × 3 × 1 × 2

30 cm

= 3 m3

10 cm

1 m

3m

c 2 = 52 + 72 = 74 — ∴ c = √ 74

7 5

c

A shorter side

V = πr2h = π × 22 × 6 = 75.40 cm3 2 cm

2m 20 cm

Finding c

6 cm

a a2 + 12 = 22 1 a2 + 1 = 4 a2 = 3 – a = √3

2

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Measurement and Geometry

Multiple-choice questions 1 The perimeter of this rectangle is 20 cm. The unknown value x is: A 4 B 16 C 5 D 10 E 6

4 cm

2 A wheel has a diameter of 2 m. Its circumference and area (in that order) are given by: A π, π2 B 2π, π C 4π, 4π D 2, 1 E 4, 4 3 The area of this triangle is: B 55 m A 27.5 m2 D 110 m E 16 m2

C

x cm

5 m

55 m2 11 m

4 Using π = 3.14, the area of a circular oil slick with radius 100 m is: A 7850 m2 B 314 m2 C 31 400 m2 D 78.5 m2

628 m2

5 A sector of a circle is removed as shown. The fraction of the circle remaining is: 29 7 C 36 36 7 3 D E 180 4 6 A cube has a volume of 64 cm3. The length of each side is: 64 A 32 cm B cm C 16 cm 3 D 8 cm E 4 cm A 290

B

x

7 The rule for the area of the trapezium shown is:

1 1 h (x + y) C xy y 2 2 1 D πxy2 E h (x + y) 2 8 The volume of a rectangular prism is 48 cm3. If its breadth is 4 cm and height 3 cm, its length would be: A 3 cm B 4 cm C 2 cm D 12 cm E 96 cm A

1 xh 2

70°

B

22 , its volume would be: 7 D 1540 cm3 E 220 cm3

9 A cylinder has radius 7 cm and height 10 cm. Using π = A 1540 cm2

B 440 cm3

C 440 L

10 The rule for Pythagoras’ theorem for this triangle would be: A a2 – b2 = c2 B x2 + y2 = z2 C z2 + y2 = x2 D x2 + z2 = y2

E

y = x2 − z2

x y

z

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Short-answer questions 1 Convert these measurements to the units given in the brackets. a 2 m (mm) b 50 000 cm (km) c 3 cm2 (mm2) d 4000 cm2 (m2) e 0.01 km2 (m2) f 350 mm2 (cm2) 3 g 400 cm (L) h 0.2 m3 (L) 2 Find the perimeter/circumference of these shapes. Round the answer to 2 decimal places where necessary. a b c 5m 6 cm 8m

3m 8 cm d

e

f

12 m

20 mm 8m

3 cm 2 cm

3 Find the area of these shapes. Round the answer to 2 decimal places where necessary. a b c 5 cm 18 m 2 cm 7m 6 cm 11 cm d

20 km

e

f

8 km

10 cm

16 m

4 cm

14 km 8m g

h

3 cm

i

110° 4m

2 cm

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Measurement and Geometry

4 Find the area of these composite shapes. b a 10 cm

c 5 cm

5 cm

8 cm 4 cm

9 cm

14 cm 6 cm 5 Find the volume of each prism, giving your answer in litres. Remember 1 L = 1000 cm3 and 1 m3 = 1000 L. a b c 3 cm 8 cm 12 cm

40 cm 20 cm

1 m

10 cm 6 Find the volume of these cylinders, rounding the answer to 2 decimal places. a b c 3 mm 10 m 14 cm 8m

7.5 mm

20 cm

7 An oven is heated from 23°C to 310°C in 18 minutes and 37 seconds. It then cools by 239°C in 1 hour, 20 minutes and 41 seconds. a Give the temperature: i increase ii decrease b What is the total time taken to heat and cool the oven? c How much longer does it take for the oven to cool down than heat up? 8 a What is the time difference between 4:20 a.m. and 2:37 p.m. of the same day? b Write 2145 hours in a.m./p.m. time. c Write 11:31 p.m. in 24-hour time. 9 When it is 4:30 p.m. in Western Australia, state the time in each of these places. a New South Wales b Adelaide c United Kingdom d China e Finland f South Korea g Russia (eastern tip) h New Zealand

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10 Use Pythagoras’ theorem to find the length of the hypotenuse in these right-angled triangles. Round the answer to 2 decimal places in part c. a b c 8 7 c

c

6

c 24 3

11 Use Pythagoras’ theorem to find the unknown length in these right-angled triangles. Round the answer to 2 decimal places in parts b and c. a b c 20 8 5

8 17

23

Extended-response questions 1 A company makes square nuts for bolts to use in building construction and steel structures. Each nut starts out as a solid steel square prism. A cylinder of diameter 2 cm is bored through its centre to make a hole. The nut and its top view are shown here. 2 cm 2 cm 4 cm

4 cm

4 cm 4 cm

The company is interested in how much paint is required to paint the nuts. The inside surface of the hole is not to be painted. Round all answers to 2 decimal places where necessary. a Find the area of the top face of the nut. b Find the total outside surface area of the nut. c If the company makes 10 000 nuts, how many square metres of surface needs to be painted? The company is also interested in the volume of steel used to make the nuts. d Find the volume of steel removed from each nut to make the hole. e Find the volume of steel in each nut. f Assuming that the steel removed to make the hole can be melted and reused, how many nuts can be made from 1 L of steel?

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Measurement and Geometry

2 A simple triangular shelter has a base width of 2 m, a height of 2 m and a length of 3 m. a Use Pythagoras’ theorem to find the hypotenuse length of one of the ends of the tent. Round the answer to 1 decimal place. b Assuming that all the faces of the shelter including the floor are covered with canvas material, what area of canvas is needed to make the shelter. 2 m Round the answer to the nearest whole number of square metres. 2m c Every edge of the shelter is to be sealed with a special tape. What length

3m

of tape is required? Round to the nearest whole number of metres. d The shelter tag says that is occupies 10 000 L of space. Show working to decide if this is true or false. What is the difference?

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Chapter

4

Fractions, decimals, percentages and financial mathematics

What you will learn

4A 4B 4C 4D 4E 4F 4G 4H 4I 4J 4K

Equivalent fractions REVISION Computation with fractions REVISION Decimal place value and fraction/decimal conversions REVISION Computation with decimals REVISION Terminating decimals, recurring decimals and rounding REVISION Converting fractions, decimals and percentages REVISION Finding a percentage and expressing as a percentage Decreasing and increasing by a percentage The Goods and Services Tax (GST) Calculating percentage change, proﬁ t and loss Solving percentage problems with the unitary method and equations

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NSW Syllabus

for the Australian Curriculum Strand: Number and Algebra

Substrands: FRACTioNS, DECiMAlS AND pERCENTAGES FiNANCiAl MATHEMATiCS

Outcomes A student operates with fractions, decimals and percentages. (MA4–5NA) A student solves ﬁ nancial problems involving purchasing goods. (MA4–6NA)

phi and golden rectangles Phi is a unique and mysterious decimal number approximately equal to 1.618. The value of phi has now been calculated to one trillion (1000 000 000 000) decimal places (2010 record). Phi’s decimal places continue forever; no pattern has been found and it cannot be written as a fraction. Golden rectangles are rectangles that have the proportion of length to width equal to phi : 1. This ratio is thought to be the most visually appealing proportion to the human eye. Proportions using the decimal phi are found in an astounding variety of places including ancient Egyptian pyramids, Greek architecture and sculptures, art, the nautilus shell, an ant, a dolphin, a beautiful human face, an ear, a tooth, the body, the graphic design of credit cards, websites and company logos, DNA and even the shape of the Milky Way galaxy.

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Chapter 4 Fractions, decimals, percentages and financial mathematics

pre-test

204

1 Match the following words to the types of fractions: whole number, improper fraction, proper fraction, mixed numeral 2 3 10 7 a 1 b c d 5 7 2 4 2 How many quarters are in 5 wholes? 3 Fill in the blanks. 3 75 = a 4

3 = 6 2

b

4 What fraction is shaded? a

c

3 1 = 5 5

d

3

=

16 12

b

5 Match the fractions on the left-hand side to their decimal form on the right. 1 a A 3.75 2 1 b B 0.625 100 3 c C 0.01 20 3 d 3 D 0.5 4 5 e E 0.15 8 6 Find: 1 1 1 1 a b 0.5 + c 3–1 + 2 4 2 3 2.4 d 0.3 + 0.2 + 0.1 e f 0.5 × 6 2 7 Write as i simple fractions and ii decimals. a 10% b 25% c 50% d 75% 8 Find 10% of: a $50

b $66

9 Find: a 25% of 40

c

d 6900 m

8 km

b 75% of 24

c 90% of $1

10 Copy and complete the following table. Fraction Decimal percentage

3

2

4

5 0.2

2 0.99

15%

1.6 100%

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Number and Algebra

4A Equivalent fractions

R E V I S I ON

Fractions are extremely useful in all practical situations whenever a proportion is required. Fractions are used by a chef measuring the ingredients for a cake, a builder measuring the ingredients for concrete and a musician using computer software to create music. A fraction is formed when a whole number or amount is divided into equal parts. The bottom number is referred to as the denominator (down) and tells you how many parts the whole is divided up into. The top number is referred to as the numerator (up) and tells you how many of the parts you have selected.

4 parts selected Numerator Denominator

205

Aerial view of farmland. The paddocks show the farmer’s land divided into parts. Each paddock is a particular fraction of the farmer’s land.

4 7 The w whole is divided into 7 parts.

Equivalent fractions are fractions that represent equal portions of a whole amount and so are equal in value. The skill of generating equivalent fractions is needed whenever you add or subtract fractions with different denominators.

let’s start: Know your terminology It is important to know and understand key terms associated with the study of fractions. Working with a partner and using your previous knowledge of fractions, write a one-sentence definition or explanation for each of the following key terms. Also give an example of each. ● Ascending ● • Numerator Highest common factor ● ● • Equivalent fraction Composite number Descending ● Denominator ● • Improper fraction Lowest common denominator ● ● • Multiples Proper fraction Rational numbers ● Mixed numeral ● • Reciprocal Irrational numbers ● • Lowest common multiple Factors

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Key ideas

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■

Equivalent fractions are equal in value. They mark the same place on a number line. For example, 3 and 6 are equivalent fractions. 5 10 0 5 0

■

1 5 1 10

3 10

4 10

3 5 5 10

6 10

4 5 7 10

8 10

5 5 9 10

1

Equivalent fractions are formed by multiplying or dividing a fraction by a number equal to 1, which can be written in the form a . a For example: 3 3 2 6 ×1 = × = 4 4 2 8

■

2 10

2 5

∴

3 6 and are equivalent fractions. 4 8

Equivalent fractions are therefore produced by multiplying the numerator and the denominator by the same whole number. For example: ×5

2 10 = 7 35

∴

2 10 and are equivalent fractions. 7 35

×5 ■

Equivalent fractions are also produced by dividing the numerator and the denominator by the same common factor. For example: ÷3

6 2 = 21 7

∴

6 2 and are equivalent fractions. 21 7

÷3 ■

The simplest form of a fraction is an equivalent fraction with the lowest possible whole numbers in the numerator and denominator. This is achieved by dividing the numerator and the denominator by their highest common factor (HCF). In the simplest form of a fraction, the HCF of the numerator and the denominator is 1. For example: 12 HCF of 12 and 18 is 6. 18 ÷6

12 2 = 18 3

∴

12 2 written in simplest form is . 18 3

÷6

■

This technique is also known as ‘cancelling’. It is important to understand that cancelling is division. The HCF is cancelled (divided) 12 2 × 6 2 6 = = ‘cancels’ to 1 from the numerator and the 18 3 × 6 3 6 denominator. because 6 ÷ 6 = 1 Two fractions are equivalent if they have the same simplest form.

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Number and Algebra

Example 1 Generating equivalent fractions Rewrite the following fractions with a denominator of 40. 3 5

a

b

1 2

7 4

c

3 = 5 1 = 2

b

36 120

ExplANATioN

SoluTioN a

d

Denominator has been multiplied by 8. Numerator must be multiplied by 8.

24 40 20 40

Multiply numerator and denominator by 20.

c

7 70 = 4 40

Multiply numerator and denominator by 10.

d

36 12 = 120 40

Divide numerator and denominator by 3.

Example 2 Converting to simplest form Write the following fractions in simplest form. 8 20

b

25 15

SoluTioN

ExplANATioN The HCF of 8 and 20 is 4. Both the numerator and the denominator are divided by the HCF of 4.

1

8 2× 4 2 = = 20 5 × 4 1 5

b

25 5 × 5 5 = = 15 3 × 5 1 3

1

Exercise 4A

The HCF of 25 and 15 is 5. The 5 is ‘cancelled’ from the numerator and the denominator.

REVISION

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3 2 14 = = = 9 3

d

18 6 15 = = = 24 4

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1 Fill in the missing numbers to complete the following strings of equivalent fractions. 3 12 120 4 8 80 a b = = = = = = 5 10 7 70

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M AT I C A

4A

11 1 2 and and are all equivalent fractions. 99 9 18

b d f

U

h

d h

b

7 14 = 9

c

7 1 = 14

d

e

4 = 3 21

f

8 80 = 5

g

3 = 12 60

h

5 12 7 8 3 1 150 300 21 = 30 10 7 28 = 11

7 State the missing numerators for the following sets of equivalent fractions.

b

3 = = = = = = 2 4 8 20 30 100 200

c

2 = = = = = = 5 10 15 20 35 50 75

8 State the missing denominators for the following sets of equivalent fractions. 1 2 4 8 10 25 100 = = = = = = a 3 b

1 2 3 4 5 10 12 = = = = = = 5

c

5 10 15 35 55 100 500 = = = = = = 4

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d

2 = 5 15

1 = = = = = = 2 4 6 10 20 32 50

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PS

4 2 can be simplified to . 5 5

a

a

F

3 1 and are equivalent fractions. 6 2 14 2 can be simplified to . 21 3

MA

4 Rewrite the following fractions with a denominator of 24. 1 2 1 a b c 3 8 2 5 5 3 e f g 6 1 4 5 Rewrite the following fractions with a denominator of 30. 1 2 5 a b c 5 6 10 2 22 5 e f g 3 60 2 6 Find the missing value to make the equation true.

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Example 1

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2 4 4 20 1 4 10 20 30 9 6 30 2 3 15 36 3 Are the following statements true or false? 1 1 a and are equivalent fractions. 2 4 8 c The fraction is written in its simplest form. 9

2 ? 3

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e

24 96 565 452

b f

36 108 637 273

c g

165 195 225 165

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9 Write the following fractions in simplest form. 3 4 10 15 a b c d 9 8 12 18 11 12 16 25 e f g h 44 20 18 35 15 22 120 64 i j k l 9 20 100 48 10 Using your calculator express the following fractions in simplest form.

M AT I C A

196 476 742 h 224 d

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C 11 Three of the following eight fractions are not written in simplest form. Write down these three R PS HE fractions and simplify them. M AT I C A 17 14 5 51 23 15 1 13 31 42 11 68 93 95 15 31 12 Group the following 12 fractions into six pairs of equivalent fractions. 5 3 7 8 2 20 6 9 15 1 16 6 11 5 21 22 7 50 21 15 33 3 44 15 13 A 24-hour swim-a-thon was organised to raise funds for a local charity. The goal was to swim 1500 laps during the 24-hour event. After 18 hours a total of 1000 laps had been swum. a On the basis of time, what fraction, in simplest form, of the swim-a-thon had been completed? b On the basis of laps, what fraction, in simplest form, of the swim-a-thon had been completed? c Were the swimmers on target to achieve their goal? Explain your answer by using equivalent fractions.

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14 a A particular fraction has a prime number in the numerator and a different prime number in the denominator. Using some examples, justify whether or not this fraction can be simplified. b A fraction has a composite number in the numerator and a different composite number in the denominator. Using some examples, justify whether or not this fraction can be simplified.

U

1 16 Are there a finite or infinite number of equivalent fractions of ? Write a one paragraph response to 2 justify your answer.

x 3a 17 Algebraic fractions contain pronumerals (letters). and are both examples of algebraic y 5 fractions. a Find the value of the 2a 4 a = i 3b

to produce equivalent algebraic fractions. x 3b 12b = = ii iii y 5y 20

4 of = p 3p x viii y = 2 y iv

a ac 3k 4a = vi vii = = b 2t 2ttm m 5b 20 20 bc b Simplify the following algebraic fractions. 4 5y 15b 3x i ii iii iv 8y 8 xy 20 5x 10 p 120 y mnop 15 x v vi vii viii 15qp 12 xy mnpq 5x 2 c Can an algebraic fraction be simplified to a non-algebraic fraction? Justify your answer by using examples. d Can a non-algebraic fraction have an equivalent algebraic fraction? Justify your answer by using examples. v

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15 a A pizza is cut into eight equal pieces. Can it be shared by four people in such a way that no-one receives an equivalent fraction of the pizza? b A pizza is cut into 12 equal pieces. Can it be shared by four people in such a way that no-one receives the same amount of the pizza? c What is the least number of pieces into which a pizza can be cut such that four people can share it and each receive a different amount?

Enrichment: Equivalent algebraic fractions

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Number and Algebra

4B Computation with fractions

211

R E V I S I ON

This section reviews the different techniques involved in adding, subtracting, multiplying and dividing fractions. Proper fractions, improper fractions and mixed numerals will be considered for each of the four mathematical operations.

let’s start: You write the question 1 1 1 3 1 , , , , 2, − . 8 4 2 4 4 Your challenge is to write six different questions which Here are six different answers:

■

Adding and subtracting fractions – When we add or subtract fractions, we count how many we have of a certain ‘type’ of fraction. For example, we could count eighths: 3 eighths plus 5 eighths minus 1 eighth equals 7 eighths. When we count ‘how many’ eighths, the answer must be in eighths. + 3 8

+

− 5 8

−

= 1 8

=

7 8

Examples: 1 2 + Different denominators 2 3 3 4 = + Same denominators 6 6 7 Add numerators = Denominator stays the same 6 – Denominators must be the same before you can proceed with adding or subtracting fractions. – If the denominators are different, use the lowest common multiple (LCM) of the denominators to find equivalent fractions. In this example we are counting sixths: 3 sixths plus 4 sixths equals 7 sixths. – When the denominators are the same, simply add or subtract the numerators as required. The denominator remains the same. Equivalent fractions

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Key ideas

will produce each of the above six answers. Each 1 question must use only the two fractions 2 1 and and one operation (+, – , ×, ÷). 4

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Common multiple (CM) and lowest common multiple (LCM) – A common multiple is found by multiplying whole numbers together. – The LCM is found by multiplying the whole numbers together and then dividing by the HCF of the whole numbers. For example, consider the whole numbers 4 and 6. 4 × 6 . A common multiple is 24 (4 × 6). The LCM is 12 2 – The lowest common denominator (LCD) is the lowest common multiple (LCM) of the denominators. Multiplying fractions 3 4 shaded Example: 1 3 of 1 1 1 2 4 4 4 4 1 3 = × 2 4 3 3 = 4 8 – Denominators do not need to be the same before you can proceed with fraction multiplication. – Simply multiply the numerators together and multiply the denominators together. – Mixed numerals must be converted to improper fractions before you can proceed. – If possible, simplify or ‘cancel’ fractions before multiplying.

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1 8

1 8

1 8

1 8

1 8

1 8

3 is one half of 3 8 4 Example:

Multiply numerators together. 2 5 2 × 5 10 × = = 3 9 3 × 9 27 Multiply denominators together.

Dividing fractions Example:

3 4

3 shaded 4 1 4

1 4

1 4

1 8

1 8

1 8

1 8

1 8

1 8

3 is one half of 3 4 8 3 1 is the same as ‘how many ÷ 4 4 1 3 quarters 4 are in ?’ 4 3 1 3 4 ÷ = × 4 4 4 1 =3

3 3 ÷ 2 is the same as ‘one half of .’ 4 4 3 3 1 ÷2= × 4 4 2 3 = 8

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Denominators do not need to be the same before you can proceed. Mixed numerals must be converted to improper fractions before you can proceed. To divide by a fraction multiply by its reciprocal. The reciprocal of a fraction is found by swapping the numerator and the denominator. This is known as inverting the fraction. Proceed as for multiplying fractions.

Example: 3 5 3 4 12 3 ÷ = × = = 8 4 8 5 40 10 Instead of ÷, × by the reciprocal fraction. Checking your answer – Final answers should be written in simplest form. – It is common to write answers involving improper fractions as mixed numerals.

Summary ■

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Adding and subtracting fractions When the denominators are the same, simply add or subtract the numerators. The denominator remains the same. Multiplying fractions Cancel where possible, then multiply the numerators together and multiply the denominators together. Dividing fractions To divide by a fraction multiply by its reciprocal.

Example 3 Adding and subtracting fractions Evaluate: 3 4 a + 5 5 SoluTioN a

b

b

5 3 − 3 4

ExplANATioN

3 4 7 + = 5 5 5 2 =1 5

The denominators are the same, therefore count the number of fifths by simply add the numerators. The final answer is written as a mixed numeral.

5 3 20 9 − = − 3 4 12 12 11 = 12

LCM of 3 and 4 is 12. Write equivalent fractions with a LCD of 12. The denominators are the same, so subtract the numerators.

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Example 4 Adding and subtracting mixed numerals Evaluate: a

5 3 3 +2 8 4

b

SoluTioN a

2

1 5 −1 2 6

ExplANATioN Convert mixed numerals to improper fractions. The LCM of 8 and 4 is 8. Write equivalent fractions with LCD. Add numerators together, denominator remains the same. Convert the answer back to a mixed numeral.

5 3 29 11 3 +2 = + 8 4 8 4 29 22 = + 8 8 51 3 = =6 8 8 Alternative method 5 3 5 3 3 + 2 = 3+ 2+ + 8 4 8 4 5 6 = 5+ + 8 8 11 3 = 5+ = 6 8 8

b

2

Add the whole number parts together. The LCM of 8 and 4 is 8. Write equivalent fractions with LCD. Add fraction parts together and simplify the answer.

Convert mixed numerals to improper fractions. The LCD of 2 and 6 is 6. Write equivalent fractions with LCD. Subtract numerators and simplify the answer.

1 5 5 11 −1 = − 2 6 2 6 15 11 = − 6 6 4 2 = = 6 3

Alternative method Subtract the whole numbers. The LCM of 2 and 6 is 6. Subtract numerators. Subtract fraction from whole number. Simplify the answer.

1 5 1 5 2 −1 =1 − 2 6 2 6 3 5 =1 − 6 6 2 = 1− 6 4 2 = = 6 3

Example 5 Multiplying fractions Evaluate: a

2 3 × 5 7

b

8 7 × 5 4

c

1 2 3 ×2 3 5

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SoluTioN a

ExplANATioN

2 3 2×3 × = 5 7 5×7 6 = 35

Multiply the numerators together. Multiply the denominators together. The answer is in simplest form.

2

Cancel first. Then multiply numerators together and denominators together. Write the answer as an improper fraction.

8 7 2×7 × = 5 5 ×1 14

b

=

c

14 4 =2 5 5

1 2 2 10 4 12 3 ×2 = × 3 5 15 13

Convert to improper fractions first. Simplify fractions by cancelling. Multiply ‘cancelled’ numerators and ‘cancelled’ denominators together. Write the answer in simplest form.

2×4 1×1 8 = =8 1 =

Example 6 Dividing fractions Evaluate: a

2 3 ÷ 5 7

b

SoluTioN a

c

2

1 1 ÷1 4 3

ExplANATioN

2 3 2 7 ÷ = × 5 7 5 3 14 = 15

Change ÷ sign to a × sign and invert the divisor (the second fraction). Proceed as for multiplication. Multiply numerators together and multiply denominators together. 2

b

5 15 ÷ 8 16

5 15 1 5 16 ÷ = × 8 16 1 8 15 3

Change ÷ sign to a × sign and invert the divisor (the second fraction). Proceed as for multiplication.

1 2 = × 1 3 2 = 3 c

2

1 1 9 4 ÷1 = ÷ 4 3 4 3 9 3 = × 4 4 27 11 = =1 16 16

Convert mixed numerals to improper fractions. Change ÷ sign to × sign and invert the divisor (the second fraction). Multiply and simplify.

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1 a Which two operations require the denominators to be the same before proceeding? b Which two operations do not require the denominators to be the same before proceeding?

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2 a For which two operations must you first convert mixed numerals to improper fractions? b For which two operations can you choose whether or not to convert mixed numerals to improper fractions before proceeding? 3 State the LCD for the following pairs of fractions. a

1 3 + 5 4

b

2 5 + 9 3

c

11 7 + 25 10

d

5 13 + 12 8

4 Rewrite the following equations and fill in the empty boxes. a

2 1 + 3 4 = =

b

8 + 12 12 11

7 9 − 8 16 = =

16

−

c 9 16

4 3 1 × 7 5 = =

16

7

×

d 3 5

5 2 ÷ 7 3 5 3 = 7 2 =

35

15

=

14

5 State the reciprocal of the following fractions. a

5 8

b

3 2

c

3

1 4

d

1

1 11

6 Write the ‘first step’ for the following division questions (no need to find the answer). b

1 1 ÷ 2 4

c

17 3 ÷ 10 1

d

1 3 2 ÷1 3 4 WO

7 Evaluate:

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a

1 2 + 5 5

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3 2 + 4 5

b

7 2 − 9 9

c

5 7 + 8 8

d

24 11 − 7 7

f

3 4 + 10 5

g

5 2 − 7 3

h

11 1 − 18 6

8 Evaluate: a

1 3 3 +1 7 7

b

2 1 7 +2 5 5

c

5 2 3 −1 8 8

d

8

e

1 1 5 +4 3 6

f

5 1 17 + 4 7 2

g

1 3 6 −2 2 4

h

2 5 4 −2 5 6

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5 3 −7 11 11

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b

2 5 × 9 7

c

7 6 × 5 5

d

5 8 × 3 9

e

4 3 × 9 8

f

12 5 × 10 16

g

12 2 × 9 5

h

24 5 × 8 3

b

2 1 3 × 7 3

c

1 3 4 ×3 6 5

d

1 1 10 × 3 2 3

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10 Evaluate: a

2

3 1 ×1 4 3

11 Evaluate: Example 6a, b

2 3 ÷ 9 5

b

1 2 ÷ 3 5

c

8 11 ÷ 7 2

d

11 5 ÷ 3 2

e

3 6 ÷ 4 7

f

10 1 ÷ 15 3

g

6 9 ÷ 5 10

h

22 11 ÷ 35 63

b

1 1 3 ÷8 5 3

c

1 2 3 ÷2 5 7

d

6

2 1 ÷2 4 6

b

3 9 14 ÷ × 7 2 16

c

1 1 3 2 ÷1 ÷1 3 4 5

d

4

1 1 × 3 ÷ 10 2 3

12 Evaluate: a

4 2 1 ÷1 7 3

13 Evaluate: a

1 3 2 × ÷ 2 4 5

14 For each fluency question above (questions 7–13), redo the first part of the question and the last part of the question on your calculator to check you answer is correct.

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3 15 Max and Tanya are painting two adjacent walls of equal area. Max has painted of his wall 7 2 and Tanya has painted of her wall. 5 a What fraction of the two walls have Max and Tanya painted in total? b What fraction of the two walls remains to be painted?

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16 Eilish was required to write a 600-word, literature essay. After working away for one hour, the word count on her computer showed that she had typed 240 words. What fraction of the essay does Eilish still need to complete? Write your answer in simplest form.

18 For Mary’s party, she asked her dad to buy 18 bottles of soft drink. Each bottle contained 1

1 4

1 litres. The glasses they had for the party could hold of a litre. How many glasses could be filled 5 from the 18 bottles of soft drink?

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1 3 kilograms of Granny Smith apples. Unfortunately of the apples were bruised 4 7 and unusable. How many kilograms of good apples did Vernald have to make his apple pies?

17 Vernald ordered 12

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c

3 1

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3

=1 =

13 15

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11 5 − = 7 3 21

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19 Fill in the empty boxes to make the following fraction equations true. More than one answer may be possible.

c

4 2

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5 = 6 12

b

2 5

d

5

=

21

÷

3 4

3 =2 7

1 1 × 4

=4

1 12

21 Write what you think are the four key bullet points when computing with fractions. Commit these to memory and then prepare a one minute oral presentation to give to your partner explaining your four key bullet points.

Enrichment: How small, how close, how large? 1 1 1 1 1 , , , , and four different operations +, –, ×, ÷ at your 2 3 4 5 6 disposal. You must use each fraction and each operation once and only once. You may use as many brackets as you need. Here is your challenge: a Produce an expression with the smallest possible positive answer. b Produce an expression with an answer of 1, or as close to 1 as possible. c Produce an expression with the largest possible answer. An example of an expression using each of the five fractions and four operations is 7 1 1 1 1 1 . + − × ÷ . This has an answer of 2 3 4 5 6 10

22 You have five different fractions

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20 Fill in the empty boxes to make the following fraction equations true. More than one answer may be possible. a

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4C Decimal place value and fraction/decimal

conversions

R E V I S I ON

Decimals are another way of representing ‘parts of a whole’. They are an extension of our base 10 number system. The term decimal is derived from the Latin word decem meaning ‘ten’. A decimal point is used to separate the whole number and the fraction part. Decimals have been studied extensively in Year 7. In this section we revisit the concepts of comparing decimals and converting between decimals and fractions.

let’s start: Order 10 The following 10 numbers all contain a whole number part and a Why do we use base 10? fraction part. Some are decimals and some are mixed numerals. Work with a partner. Your challenge is to place the 10 numbers in ascending order.

Key ideas

4

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1 11

3

5 6

3.3

3

72 100

3

1 3

2.885

3.09

2

3 4

3

2 5

3.9

When dealing with decimal numbers, the place value table is extended to involve tenths, hundredths, thousandths, etc. The number 517.364 means: Hundreds

Tens

units

Decimal point

Tenths

Hundredths

Thousandths

5

1

7

.

3

6

4

5 × 100

1 × 10

7 × 1

.

3×

500

10

7

.

1 10

6×

3 10

1 100

4×

6 100

1 1000

4 1000

Comparing and ordering decimals To compare two decimal numbers with digits in the same place value columns, you must compare the left-most digits first. Continue comparing digits as you move from left to right until you find two digits that are different. For example, compare 362.581 and 362.549. Hundreds

Tens

units

Decimal point

Tenths

Hundredths

Thousandths

3

6

2

.

5

8

1

3

6

2

.

5

4

9

Both numbers have identical digits in the hundreds, tens, units and tenths columns. Only when we get to the hundredths column is there a difference. The 8 is bigger than the 4 and therefore 362.581 > 362.549. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Converting decimals to fractions – Count the number of decimal places used. – This is the number of zeroes that you must place in the denominator. – Simplify the fraction if required. For example: 0.664 =

■

64 16 = 100 25

Converting fractions to decimals – If the denominator is a power of 10, simply change the fraction directly to a decimal from your knowledge of its place value. For example:

239 = 0.239 1000

– If the denominator is not a power of 10, try to find an equivalent fraction for which the denominator is a power of 10 and then convert to a decimal. For example:

3 3×5 15 = = = 0.115 20 20 × 5 100

– A method that will always work for converting fractions to decimals is to divide the numerator by the denominator. This can result in terminating and recurring decimals and is covered in Section 4E.

A fraction is equivalent to a division operation. In spreadsheets and some calculators fractions and division operations are typed the same way, using a slash (/) between the numbers.

Example 7 Comparing decimals Compare the following decimals and place the correct inequality sign between them. 57.89342 and 57.89631 SoluTioN

ExplANATioN

57.89342 < 57.89631

Digits are the same in the tens, units, tenths and hundredths columns. Digits are first different in the thousandths column. 3 6 < 1000 1000

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Example 8 Converting decimals to fractions Convert the following decimals to fractions in their simplest form. a 0.725 b 5.12 SoluTioN a

725 29 = 1000 40

b

5

ExplANATioN Three decimal places, therefore three zeroes in denominator. 0.725 = 725 thousandths. Two decimal places, therefore two zeroes in denominator. 0.12 = 12 hundredths.

12 3 =5 100 25

Example 9 Converting fractions to decimals Convert the following fractions to decimals. 239 100

b

ExplANATioN

a

239 39 =2 = 2.339 100 100

Convert improper fraction to a mixed numeral. Denominator is a power of 10.

b

9 36 = = 0.336 25 100

9 9×4 36 = = 25 25 × 4 100

Exercise 4C

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8

17 100

E

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75 1000

3 Compare the following decimals and place the correct inequality sign (< or >) between them. a 36.485 37.123 b 21.953 21.864 c 0.0372 0.0375 d 4.21753 4.21809 e 65.4112 64.8774 f 9.5281352 9.5281347

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2 Which of the following is the mixed numeral equivalent of 5.75? 75 25 3 15 A 5 B 5 C 5 D 5 10 50 4 20

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1 Which of the following is the mixed numeral equivalent of 8.17? 17 1 17 1 A 8 B 8 C 8 D 8 10 17 1000 7

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9 25

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7 Convert the following fractions to decimals.

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6 Convert the following decimals to fractions in their simplest form. a 0.31 b 0.537 c 0.815 d 0.96 e 5.35 f 8.22 g 26.8 h 8.512

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5 Arrange the following sets of decimals in descending order. a 3.625, 3.256, 2.653, 3.229, 2.814, 3.6521 b 11.907, 11.891, 11.875, 11.908, 11.898, 11.799 c 0.043, 1.305, 0.802, 0.765, 0.039, 1.326

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11 = = 0.5 20 100

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3 6 = = 0. 5

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17 301 405 76 b c d 100 1000 100 10 3 7 5 7 e f g h 25 20 2 4 8 Using a calculator convert the following decimals to fractions and fractions to decimals. a 0.052 b 6.125 c 317.06 d 0.424 11 3 17 29 f g h e 40 8 25 125 a

9 Convert the following mixed numerals to decimals and then place them in ascending order. Consider whether to use a calculator or not. Possibly a combination of mental arithmetic and some simple calculator keystrokes is the most efficient strategy. 2 1 3 7 9 3 2 , 2 , 2 , 2 , 2 , 2 5 4 8 40 50 10

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List Nam’s six classrooms in order of distance of his locker from the closest classroom to the one furthest away. 11 The prime minister’s approval rating is 0.35, while the opposition leader’s approval rating is 3 . 8 Which leader is ahead in the popularity polls and by how much?

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10 The distances from Nam’s locker to his six different classrooms are listed below: • Locker to room B5 (0.186 km) • Locker to gym (0.316 km) • Locker to room A1 (0.119 km) • Locker to room C07 (0.198 km) • Locker to room P9 (0.254 km) • Locker to BW Theatre (0.257 km)

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Write a decimal that lies midway between 2.65 and 2.66. Write a fraction that lies midway between 0.89 and 0.90. Write a decimal that lies midway between 4.6153 and 4.6152. Write a fraction that lies midway between 2.555 and 2.554.

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12 Jerome dug six different holes for planting six different types of fruit bushes and trees. He measured the dimensions of the holes and found them to be: A depth 1.31 m, width 0.47 m B depth 1.15 m, width 0.39 m C depth 0.85 m, width 0.51 m D depth 0.79 m, width 0.48 m E depth 1.08 m, width 0.405 m F depth 1.13 m, width 0.4 m a List the holes in increasing order of depth. b List the holes in decreasing order of width.

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Enrichment: Exchange rates 15 The table below shows a set of exchange rates between the US dollar (US$), the Great Britain pound (£), the Canadian dollar (C$), the euro (€) and the Australian dollar (A$). £

uS$

C$

A$

€

uS$

1

1.58781

0.914085

1.46499

0.866558

£

0.629795

1

0.575686

0.922649

0.545754

C$

1.09399

1.73705

1

1.60269

0.948006

€

0.682594

1.08383

0.623949

1

0.591507

A$

1.15399

1.83232

1.05484

1.69059

1

The following two examples are provided to help you to interpret the table. • A$1 will buy US$0.866558. • You will need A$1.15399 to buy US$1. Study the table and answer the following questions. a How many euros will A$100 buy? b How many A$ would buy £100? c Which country has the most similar currency rate to Australia? d Would you prefer to have £35 or €35? e C$1 has the same value as how many US cents? f If the cost of living was the same in each country in terms of each country’s own currency, list the five money denominations in descending order of value for money. g A particular new car costs £30 000 in Great Britain and $70 000 in Australia. If it costs A$4500 to freight a car from Great Britain to Australia, which car is cheaper to buy? Justify your answer by using the exchange rates in the table. h Research the current exchange rates and see how they compare to those listed in the table.

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4D Computation with decimals

R E V I S I ON

This section reviews the different techniques involved in adding, subtracting, multiplying and dividing decimals.

Precision electronic measuring instruments give a decimal read-out.

let’s start: Match the phrases

Key ideas

There are eight different sentence beginnings and eight different sentence endings below. Your task is to match each sentence beginning with its correct ending. When you have done this, write the eight correct sentences in your work book. Sentence beginnings

Sentence endings

When adding or subtracting decimals

the decimal point moves two places to the right.

When multiplying decimals

the decimal point in the quotient goes directly above the decimal point in the dividend.

When multiplying decimals by 100

make sure you line up the decimal points.

When dividing decimals by decimals

the number of decimal places in the question must equal the number of decimal places in the answer.

When multiplying decimals

the decimal point moves two places to the left.

When dividing by 100

start by ignoring the decimal points.

When dividing decimals by a whole number

we start by changing the question so that the divisor is a whole number.

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Adding and subtracting decimals – Ensure digits are correctly aligned in similar place value columns. – Ensure the decimal points are lined up directly under one another. 37.56 + 5.231

37.560 + 5 .231

✓

3 7.56 5.2231

✕

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Multiplying and dividing decimals by powers of 10 – When multiplying, the decimal point appears to move to the right the same number of places as there are zeroes in the multiplier. 13.7 5 3 13.753 × 100 = 1375.3 Multiply by 10 twice. – When dividing, the decimal point appears to move to the left the same number of places as there are zeroes in the divisor. 586.92 586.92 ÷ 10 = 58.692 Divide by 10 once.

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Multiplying decimals – Initially ignore the decimal points and carry out routine multiplication. – The decimal place is correctly positioned in the answer according to the following rule: ‘The number of decimal places in the answer must equal the total number of decimal places in the question.’ 573 × 86 49278

5.73 73 × 8.6

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5.73 73 × 8.6 = 4499.2278 (3 decimal places in question, 3 decimal places in answer)

Dividing decimals The decimal point in the quotient goes directly above the decimal point in the dividend. 56.34 ÷ 3

Divisor

Quotient (answer)

18.78 3 56.334

)

Dividend

We avoid dividing decimals by other decimals. Instead we change the divisor into a whole number. Of course, whatever change we make to the divisor we must also make to the dividend, so it is equivalent to multiplying by 1 and the value of the question is not changed. We avoid 27.354 ÷ 0.02 27.354 ÷ 0.02 Preferring to do 2735.4 ÷ 2

Example 10 Adding and subtracting decimals Calculate: a 23.07 + 103.659 + 9.9

b

9.7 – 2.86

SoluTioN

ExplANATioN

a

Make sure all decimal points and places are correctly aligned directly under one another. Fill in missing decimal places with zeroes. Carry out the addition of each column, working from right to left.

23.070 103.659 9.900 136.629

b

8

−

1

9 .16 7 0 2. 8 6 6. 8 4

Align decimal points directly under one another and fill in missing decimal places with zeroes. Carry out subtraction following the same procedure as for subtraction of whole numbers.

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Example 11 Multiplying and dividing by powers of 10 Calculate: a 9.753 ÷ 100

b

27.58 × 10 000

SoluTioN

ExplANATioN

a 9.753 ÷ 100 = 0.09753

Dividing by 100, therefore decimal point appears to move two places to the left. Additional zeroes are inserted as necessary. .09.753

b 27.58 × 10 000 = 275 800

Multiplying by 10 000, therefore decimal point appears to move four places to the right. Additional zeroes are inserted as necessary. 27.5800.

Example 12 Multiplying decimals Calculate: a 2.57 × 3 SoluTioN a

b

b

4.13 × 9.6 ExplANATioN

2 57 × 3 7 71

Perform multiplication ignoring decimal point. There are two decimal places in the question, so two decimal places in the answer.

2.57 × 3 = 7.71

Estimation is less than 10 (≈ 3 × 3 = 9).

1 2

413 96 2478 37170 39648

×

4.13 × 9.6 = 39.648

Ignore both decimal points. Perform routine multiplication. There is a total of three decimal places in the question, so there must be three decimal places in the answer. Estimation is about 40 (≈ 4 × 10 = 40).

In everyday life decimal numbers are most frequently encountered when dealing with money.

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Example 13 Dividing decimals Calculate: a 35.756 ÷ 4

b

64.137 ÷ 0.03

SoluTioN

ExplANATioN

a 8.939 8. 9 3 9

Carry out division, remembering that the decimal point in the answer is placed directly above the decimal point in the dividend.

)

4 35.3 7 1536 b 64.137 ÷ 0.03 = 6413.7 ÷ 3 = 2137.9

Instead of dividing by 0.03, multiply both the divisor and the dividend by 100. Move each decimal point two places to the right. Carry out the division question 6413.7 ÷ 3.

)

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3 If 56 × 37 = 2072, the correct answer to the problem 5.6 × 3.7 is: A 207.2 B 2072 C 20.72 D 2.072

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1 Which of the following is correctly set-up for the following addition problem? 5.386 + 53.86 + 538.6 5.386 5.386 A B C 5.386 D 538 + 53 + 5 53.86 53.860 53.86 + 0.386 + 0.8866 + 0.6 + 538.6 + 538.600 + 5388.6

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4 Which of the following divisions would provide the same answer as the division question 62.5314 ÷ 0.03? A 625.314 ÷ 3 B 6253.14 ÷ 3 C 0.625314 ÷ 3 D 625314 ÷ 3

6 Calculate: a 13.546 + 35.2 + 9.27 + 121.7 c 923.8 + 92.38 + 9.238 + 0.238

b d

5.623 + 18.34 123.8 – 39.21

d 92.3 + 1.872 h 14.57 – 9.8

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45.983 + 3.41 + 0.032 + 0.8942 4.572 + 0.0329 + 2.0035 + 11.7003

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b 0.08155 × 1000 c f 1.0615 ÷ 1000 g

7.5 ÷ 10 0.003 × 10 000

d 3.812 ÷ 100 h 0.452 ÷ 1000

9 Calculate: a 12.45 × 8 e 0.0023 × 8.1

b 4.135 × 3 f 300.4 × 2.2

c g

26.2 × 4.1 7.123 × 12.5

d 5.71 × 0.32 h 81.4 × 3.59

10 Calculate: a 24.54 ÷ 2 e 133.44 ÷ 12

b 17.64 ÷ 3 f 4912.6 ÷ 11

c g

0.0485 ÷ 5 2.58124 ÷ 8

d 347.55 ÷ 7 h 17.31 ÷ 5

11 Calculate: a 6.114 ÷ 0.03 e 0.02345 ÷ 0.07

b 0.152 ÷ 0.4 f 16.428 ÷ 1.2

c g

4023 ÷ 0.002 0.5045 ÷ 0.8

d 5.815 ÷ 0.5 h 541.31 ÷ 0.4

12 Calculate: a 13.7 + 2.59 e 13.72 × 0.97

b 35.23 – 19.71 f 6.7 – 3.083

c g

15.4 × 4.3 0.582 ÷ 0.006

d 9.815 ÷ 5 h 7.9023 + 34.81

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14 If the rental skis at Mt Buller were lined up end to end, they would reach from the summit of Mt Buller all the way down to the entry gate at Mirimbah. The average length of a downhill ski is 1.5 m and the distance from the Mt Buller summit to Mirimbah is 18.3 km. How many rental skis are there? 15 Design your own question similar to question 14 and ask a partner to solve it. For example, you might use the number of cars placed bumper to bumper from Sydney to Brisbane. Carry out some research so that your decimal lengths are approximately accurate.

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13 The heights of Mrs Buchanan’s five grandchildren are 1.34 m, 1.92 m, 0.7 m, 1.5 m, and 1.66 m. What is the combined height of Mrs Buchanan’s grandchildren?

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16 Joliet is a keen walker. She has a pedometer that shows she has walked 1 428 350 paces so far this year. Her average pace length is 0.84 metres. How many kilometres has Joliet walked so far this year? (Give your answer correct to the nearest kilometre.)

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17 A steel pipe of length 7.234 m must be divided into four equal lengths. The saw blade is 2 mm thick. How long will each of the four lengths be?

a×b×c

d c÷a–b

19 If a = 0.1, b = 2.1 and c = 3.1, without evaluating, which of the following alternatives would provide the biggest answer? A a+b+c B a×b×c C b÷a+c D c÷a×b

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18 If a = 0.12, b = 2.3 and c = 3.42, find: a a+b+c b c – (a + b)

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Enrichment: Target practice 20 In each of the following problems, you must come up with a starting decimal number/s that will provide an answer within the target range provided when the nominated operation is performed. For example: Find a decimal number that when multiplied by 53.24 will give an answer between 2.05 and 2.1. You may like to use trial and error, or you may like to work out the question backwards. Confirm these results on your calculator: 0.03 × 53.24 = 1.5972 (answer outside target range – too small) 0.04 × 53.24 = 2.1296 (answer outside target range – too large) 0.039 × 53.24 = 2.07636 (answer within target range of 2.05 to 2.1) Therefore a possible answer is 0.039. Try the following target problems. (Aim to use as few decimal places as possible.) Question

Starting number

operation (instruction)

Target range

1

0.039

× 53.24

2.05–2.1

2

× 0.034

100–101

3

÷ 1.2374

75.7–75.8

4

× by itself (square)

0.32–0.33

5

÷ (−5.004)

9.9–9.99

Try the following target problems. (Each starting number must have at least two decimal places.) Question

Two starting numbers

operation (instruction)

Target range

6

0.05, 3.12

×

0.1–0.2

7

÷

4.1–4.2

8

×

99.95–100.05

9

+

0.001–0.002

10

–

45.27

You may like to make up some of your own target problems.

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4E Terminating decimals, recurring decimals

and rounding

R E V I S I ON

Not all fractions convert to the same type of decimal. For example: 1 = 1 ÷ 2 = 0.5 2 1 = 1 ÷ 3 = 0.33333 … 3 1 = 1 ÷ 7 = 0.142857 … 7 Decimals that stop (or terminate) are known as terminating decimals, whereas decimals that continue on forever with some form of pattern are known as repeating or recurring decimals.

The fraction of space occupied by each solar panel could be written as a terminating or recurring decimal.

let’s start: Decimal patterns Carry out the following short divisions without using a calculator and observe the pattern of digits which occur. 0. 0 9 0 9... 1 For example: = 1 ÷ 1111 = 11 11 1. 10 10 0 10 10 0 10 = 0.0909090909... 0909090909... 11

)

Try the following:

1 2 4 5 8 25 , , , , , 3 7 9 11 13 99

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A terminating decimal has a finite number of decimal places (i.e. it terminates). 0. 6 2 5 Terminating decimal 5 For example: = 5 ÷ 8 = 0.625 8 5. 5 0 2 0 4 0 8

)

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A recurring decimal (or repeating decimal) has an infinite number of decimal places with a finite sequence of digits that are repeated indefinitely. 0. 3 3 3... Recurring decimal 1 1 1 1 1 For example: = 1 ÷ 3 = 0.333… 3 1. 0 0 0 0 3

)

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All terminating and recurring decimals can be expressed as fractions. All numbers that can a be expressed as a fraction (i.e. written in the form where a and b are integers and b ≠ 0) are b known as ‘rational’ numbers.

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Remember to keep adding zeroes to the dividend until you have a repetitive pattern. Which fraction has the longest repetitive pattern?

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A convention is to use dots placed above the digits to show the start and finish of a repeating cycle of digits. For example: 0.55555… = 0.5

and

0.3412412412… = 0.3412

Another convention is to use a horizontal line placed above the digits to show the repeating cycle of digits. For example: 0.55555… = 0.5 ■

and

0.3412412412… = 0.33412

Rounding decimals involves approximating a decimal number to fewer decimal places. When rounding, the ‘critical digit’ is the digit immediately after the rounding digit. – If the critical digit is less than five, the rounding digit is not changed. – If the critical digit is five or more, the rounding digit is increased by one. For example: 51.34721 rounded to 2 decimal places is 51.35. The critical digit is 7, which is greater than 5, hence the rounding digit is increased by 1. An illustration of rounding to 1 decimal place: The decimals 7.41 to 7.44 are closer in value to 7.4 and will all round to 7.4. The values 7.46 to 7.5 are closer in value to 7.5 and will all round to 7.5. 7.45 also rounds to 7.5. 7.4 7.41 7.42 7.43 7.44 7.45 7.46 7.47 7.48 7.49 7.5 These values are closer to 7.4 and round to 7.4.

These values are closer to 7.5 and round to 7.5.

Example 14 W r i t i n g t erminating decimals Convert the following fractions to decimals. 1 a b 4 SoluTioN a

1 = 0.225 4

b

7 = 0.875 8

7 8 ExplANATioN

)

0. 2 5

4 1. 10 2 0

)

0. 8 7 5

8 7. 7 0 6 0 4 0

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Number and Algebra

Example 15 Writing recurring decimals Express the following fractions as recurring decimals. a

2 3

b

SoluTioN a

3

5 7

ExplANATioN

2 = 0.6 3

)

0. 6 6...

3 2. 2 0 2 0 2 0

5 or 3.714285 b 3 = 3.714285 7

)

0. 7 1 4 2 8 5 7...

7 5. 5 0 1 0 3 0 2 0 6 0 4 0 5 0 1 0

Example 16 Rounding decimals Round each of the following to the specified number of decimal places. a 15.35729 (3 decimal places) b 4.86195082 (4 decimal places) SoluTioN

ExplANATioN

a 15.35729 ≈ 15.357

Critical digit is 2 which is less than 5. Rounding digit remains the same.

b 4.86195082 ≈ 4.8620

Critical digit is 5, therefore increase rounding digit by 1. There is a carry-on effect, as the rounding digit was a 9.

Example 17 Rounding recurring decimals Write

3 as a decimal correct to 2 decimal places. 7

SoluTioN 3 = 0.443 7

ExplANATioN (correct to 2 decimal places)

)

0. 4 2 8

7 3. 3 0 2 0 6 0 4 0

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REVISION

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1 State whether the following are terminating decimals (T) or recurring decimals (R). a 5.47 b 3.15415 … c 8.6 d 7.1834 e 0.333 f 0.534 g 0.5615 h 0.32727…

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2 Express the following recurring decimals using the convention of dots or a bar to indicate the start and finish of the repeating cycle. a 0.33333… b 6.21212121… c 8.5764444… d 2.135635635… e 11.2857328573… f 0.003523523… 3 Write down the ‘critical’ digit (the digit immediately after the rounding digit) for each of the following. a 3.5724 (rounding to 3 decimal places) b 15.89154 (rounding to 1 decimal place) c 0.004571 (rounding to 4 decimal places) d 5432.726 (rounding to 2 decimal places) 4 State how you round decimal numbers on your calculator.

6 Express the following fractions as recurring decimals. 1 5 5 a b c 3 9 6

d

11 20

d

8 11

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3 5 2 6 f g 3 h 4 7 13 15 7 7 Round each of the following to the specified number of decimal places, which is the number in the bracket. a 0.76581 (3) b 9.4582 (1) c 6.9701 (1) d 21.513426 (4) e 0.9457 (2) f 17.26 (0) g 8.5974 (2) h 8.10552 (3) e

Example 16

8 Write each of the following decimals correct to 2 decimal places. a 17.0071 b 0.9192 c 4.4444 d 14.5172 e 5.1952 f 78.9963 g 0.0015 h 2.6808 9 Round each of the following to the nearest whole number. a 65.3197 b 8.581 c 29.631 Example 17

d 4563.18

10 Write each of the following fractions as decimals correct to 2 decimal places. 6 2 4 5 a b c d 7 9 11 12

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5 Convert the following fractions to decimals. 3 3 a b c 5 4

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11 Round each of the following to the number of decimal places specified in the brackets. a 7.699951 (4) b 4.95953 (1) c 0.0069996 (5)

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12 Simone and Greer are two elite junior sprinters. At the Queensland State championships, Simone recorded her personal best 100 m time of 12.77 seconds, while Greer came a close second with a time of 12.83 seconds. a If the electronic timing equipment could only display times to the nearest second, what would be the time difference between the two sprinters? b If the electronic timing equipment could only record times to the nearest tenth of a second, what would be the time difference between the two sprinters? c What was the time difference between the two girls, correct to 2 decimal places? d If the electronic timing system could measure accurately to 3 decimal places, what would be the quickest time that Simone could have recorded? e Assume that Simone and Greer ran at a consistent speed throughout the 100 m race. Predict the winning margin (correct to the nearest centimetre). 3 is expressed in decimal form, what is the digit in the 19th decimal place? 7 1 14 Express as a recurring decimal. 17

13 When

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16 Two students gave the following answers in a short test on rounding. Both students have one particular misunderstanding. Study their answers carefully and write a comment to help correct each student’s misunderstanding Rounding question

Student A

Student B

0.543

(2)

0.54

✓

0.50

✗

6.7215

(3)

6.721

✗

6.722

✓

5.493

(1)

5.5

✓

5.5

✓

8.2143

(3)

8.214

✓

8.210

✗

11.54582

(2)

11.54

✗

11.55

✓

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15 Frieda stated that she knew an infinite non-recurring decimal. Andrew said that was impossible. He was confident that all decimals either terminated or repeated and that there was no such thing as an infinite non-recurring decimal. Who is correct?

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Enrichment: Will it terminate or recur? 17 Can you find a way of determining if a fraction will result in a terminating (T) decimal or a recurring (R) decimal? a Predict the type of decimal answer for the following fractions, and then convert them to see if you were correct. 1 1 1 i ii iii 8 12 14 1 1 1 iv v vi 15 20 60 A key to recognising whether a fraction will result in a terminating or recurring decimal lies in the factors of the denominator. b Write down the factors of the denominators listed above. c From your observations, write down a rule that assists the recognition of when a particular fraction will result in a terminating or recurring decimal. d Without evaluating, state whether the following fractions will result in terminating or recurring decimals. 1 1 1 1 ii iii iv i 9 42 50 16 v

1 75

vi

1 99

vii

1 200

viii

1 625

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Number and Algebra

4F Converting fractions, decimals and percentages

239

R E V I S I ON

Per cent is Latin for ‘out of 100’. One dollar equals 100 cents and one century equals 100 years. We come across percentages in many everyday situations. Interest rates, discounts, test results and statistics are just some of the common ways we deal with percentages. Percentages are closely related to fractions. A percentage is another way of writing a fraction with a denominator of 100. Therefore, 63% means that if something was broken into 100 parts you would have 63 63 of them (i.e. 63% = ). 100

let’s start: Estimating percentages

Creamy soda

Lime

Orange

Cola

Lemon

Milkshake

Raspberry Chocolate

• Estimate the percentage of drink remaining in each of the glasses shown above. • Discuss your estimations with a partner. • Estimate what percentage of the rectangle below has been shaded in.

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The symbol % means ‘per cent’. It comes from the Latin words per centum which translates to ‘out of 100’. 23 For example: 23% means 23 out of 100 or = 0.223. 100 To convert a percentage to a fraction – Change the % sign to a denominator of 100. – Simplify the fraction if required. 35 7 For example: 35% = = 100 20 To convert a percentage to a decimal – Divide by 100%. Therefore move the decimal point two places to the left. For example: 46% = 46% ÷ 100% = 46 ÷ 100 = 0.46

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Key ideas

• Use a ruler to draw several 10 cm × 2 cm rectangles. Work out an amount you would like to shade in and using your ruler measure precisely the amount to shade. Shade in this amount. • Ask your partner to guess the percentage you shaded. • Have several attempts with your partner and see if your estimation skills improve.

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To convert a fraction to a percentage – Multiply by 100%. For example:

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1 1 1 100 25 1 = × 100% = × %= % = 12 % 8 8 8 1 2 2

To convert a decimal to a percentage – Multiply by 100%. Therefore move the decimal point two places to the right. For example: 0.812 = 0.812 × 100% = 81.2% – Common percentages and their equivalent fractions and decimals are shown in the table below. It is helpful to know these.

Words

Diagram

Fraction

Decimal

Percentage

one whole

1

1

100%

one-half

1 2

0.5

50%

one-third

1 3

0.333... or 0.3

one-quarter

1 4

0.25

25%

one-fifth

1 5

0.2

20%

one-tenth

1 10

0.1

10%

one-hundredth

1 100

0.01

1%

33

1 % 3

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Number and Algebra

Example 18 Converting percentages to fractions Convert the following percentages to fractions or mixed numerals in their simplest form. a 160% b 12.5% SoluTioN a

b

ExplANATioN

160 100 8×2 20 = 5 × 220 8 3 = =1 5 5

160% =

12.5 100 25 = 200 1 = 8

12.5% =

Change % sign to a denominator of 100. Cancel the HCF. Convert answer to a mixed numeral.

or

12.5 100 125 = 1000 1 = 8 =

Change % sign to a denominator of 100. Multiply numerator and denominator by 2 or by 10 to make whole numbers. Simplify fraction by cancelling the HCF.

Example 19 Converting percentages to decimals Convert the following percentages to decimals. a 723% b 13.45% SoluTioN

ExplANATioN

a 723% = 7.23

723 ÷ 100

b 13.45% = 0.1345

13.45 ÷ 100 Decimal point appears to move two places to the left.

723.

Example 20 Converting fractions to percentages Convert the following fractions and mixed numerals into percentages. 3 7 1 a b c 2 5 40 4 SoluTioN a

d

2 3

ExplANATioN

3 3 × 100% = × 5 15 = 60%

20

100 % 1

Multiply by 100%. Simplify by cancelling the HCF. 3 6 60 Alternatively, = = = 60%. 5 10 100

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b

7 × 100% = 40 =

c

2

7 × 2 40

5

100 % 1

35 1 % = 17 % 2 2

1 9 × 100% = × 4 14

25

100 % 1

= 225%

d

2 2 100 × 100% = × % 3 3 1 200 2 = % = 66 % 3 3

Multiply by 100%. Simplify by cancelling the HCF. Write the answer as a mixed numeral. Alternatively, multiply numerator and denominator by 2.5.

Convert mixed numeral to improper fraction. Cancel and simplify. 1 1 Alternatively, 2 = 200% and = 25% so 2 = 225%. 4 4 Multiply by 100%. Multiply numerators and denominators. Write answer as a mixed numeral. 2 2 1 1 Alternatively, = 33 % so = 66 %. 3 3 3 3

1 This slice can be represented by , 25% or 0.25 of the whole. 4

Example 21 Converting decimals to percentages Convert the following decimals to percentages. a 0.458 b 17.5 SoluTioN

ExplANATioN

a 0.458 = 45.8%

0.458 × 100

b 17.5 = 1750%

17.5 × 100 The decimal point appears to move two places to the right.

0.4 5 8

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13.5 10

2700

D

C

3.7

D 37.00

47 is: 100 B 4.7%

C

47%

D 470%

4 The percentage equivalent of 0.57 is: A 57% B 5.7%

C

570%

D 0.57%

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3 The percentage equivalent of

Example 18b

6 Convert the following percentages to fractions in their simplest form. 1 1 2 a 37 % b 15.5% c 33 % d 66 % 2 3 3 e 2.25% f 4.5% g 10 1 % h 87.5% 5

Example 19

7 Convert the following percentages to decimals. a 65% b 37% c 158% e 6.35% f 0.12% g 4051%

d 319% h 100.05%

8 Convert the following fractions to percentages. Example 20a, b

to e

2 5 9 40

b f

1 4 17 25

c g

11 20 150 200

d h

13 50 83 200

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5 Convert the following percentages to fractions or mixed numerals in their simplest form. a 39% b 11% c 20% d 75% e 125% f 70% g 205% h 620%

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12 25

f

1

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10 Without using a calculator, convert the following fractions to percentages. Check your answers using a calculator. 1 1 1 1 a b c d 3 8 12 15 3 2 3 27 f g h 8 7 16 36 11 Without using a calculator, convert the following decimals to percentages. Check your answers using a calculator. e

at 0.42 and 0.0035

b 0.17 f 0.0417

c g

d 11.22 h 1.01

3.541 0.01

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12 Which value is larger? A 0.8 of a large pizza

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Fraction

Decimal

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b

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Fraction

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a

Decimal

%

0.3

1 4

0.6

2 4

0.9

3 4 4 4

c

Fraction

Decimal

% 20% 40% 60% 80% 100%

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B 70% of a large pizza

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Decimal

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11 5 0.003

3 8

6.5% 119%

5 40

4.2 0.7 62%

5 6

15 A cake is cut into equal 8 pieces. What fraction, what percentage and what decimal does one slice of the cake represent? 16 The Sharks team has won 13 out of 17 games for the season to date. The team still has three games to play. What is the smallest and the largest percentage of games the Sharks could win for the season?

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17 a Explain why multiplying by 100% is the same as multiplying by 1. b Explain why dividing by 100% is the same as dividing by 1.

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4F

Enrichment: Tangram percentages 19 You are most likely familiar with the Ancient Chinese puzzle known as a tangram. A tangram consists of seven geometric shapes (tans) as shown. The tangram puzzle is precisely constructed using vertices, midpoints and straight edges. a Express each of the separate tan pieces as a percentage, a fraction and a decimal amount of the entire puzzle. b Check your seven tans add up to a total of 100%. c Starting with a square, design a new version of a ‘modern’ tangram puzzle. You must have at least six pieces in your puzzle. An example of a modern puzzle is shown. d Express each of the separate pieces of your new puzzle as a percentage, a fraction and a decimal amount of the entire puzzle. e Separate pieces of tangrams can be arranged to make more than 300 creative shapes and designs, some of which are shown. You may like to research tangrams and attempt to make some of the images.

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Number and Algebra

4G Finding a percentage and expressing

as a percentage Showing values in percentages ‘standardises’ the value and makes it easier for comparisons to be made. For example, Huen’s report card could be written as marks out of each total or in percentages: French test

German test

14 20 54 75

French test 70%

German test 72%

It is clear that Huen’s German score was the stronger result. Comparison is easier when proportions or fractions are written as percentages (equivalent fractions with denominator of 100). Expressing one number as a percentage of another number is the technique covered in this section. Another common application of percentages is to find a certain percentage of a given quantity. Throughout your life you will come across many examples in which you need to calculate percentages of a quantity. Examples include retail discounts, interest rates, personal improvements, salary increases, commission rates and more.

let’s start: What percentage has passed? Answer the following questions. • What percentage of your day has passed? • What percentage of the current month has passed? • What percentage of the current season has passed? • What percentage of your school year has passed? • What percentage of your school education has passed? • If you live to an average age, what percentage of your life has passed? • When you turned 5, what percentage of your life was 1 year? • When you are 40, what percentage of your life will 1 year be? © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Chapter 4 Fractions, decimals, percentages and financial mathematics

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To express one quantity as a percentage of another 1 Write a fraction with the ‘part amount’ as the numerator and the ‘whole amount’ as the denominator. 2 Convert the fraction to a percentage by multiplying by 100%. For example: Express a test score of 14 out of 20 as a percentage. 14 is the ‘part amount’ that you want to express as a percentage out of 20, which is the ‘whole amount’. ×5 5 14 14 100 × 100% = × % = 70% 20 20 1 1

Alternatively,

14 70 = = 70%. 20 100 ×5

■

To ﬁnd a certain percentage of a quantity 1 Express the required percentage as a fraction. 2 Change the ‘of’ to a multiplication sign. 3 Express the number as a fraction. 4 Follow the rules for multiplication of fractions. For example: Find 20% of 80. 20% of 80 =

20 80 × = 100 1

4

1

20 × 5 100

4

80 = 16 1

1 × 80 5 = 80 ÷ 5 = 16.

Alternatively, 20% of 80 =

Example 22 Expressing one quantity as a percentage of another Express each of the following as a percentage. a 34 out of 40 b 13 out of 30 SoluTioN a

b

34 100 17 100 × %= × % 40 1 20 1 17 5 = × % 1 1 = 85% 13 100 13 10 × %= × % 30 1 3 1 130 = % 3 1 = 43 % or 43.3 % 3

ExplANATioN Write as a fraction, with the first quantity as the numerator and second quantity as the denominator. Multiply by 100%. Cancel and simplify.

Write quantities as a fraction and multiply by 100%. Cancel and simplify. Express the percentage as a mixed numeral or a recurring decimal.

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Example 23 Converting units before expressing as a percentage Express: a 60 cents as a percentage of $5

b

SoluTioN a

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2 km as a percentage of 800 m ExplANATioN

60 100 60 × %= % 500 1 5 = 12%

Convert $5 to 500 cents. Write quantities as a fraction and multiply by 100%. Cancel and simplify. 60 6 12 Alternatively, = = = 12%. 500 50 100

2000 100 2000 × %= % 800 1 8 = 250%

Convert 2 km to 2000 m. Write quantities as a fraction and multiply by 100%. Cancel and simplify. 2000 200 50 25 250 Alternatively, = = = = = 250%. 800 8 2 10 100

Example 24 Finding a certain percentage of a quantity Find: a 25% of 128

b

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155% of 60 ExplANATioN

25 128 × 100 1 1 128 = × = 32 4 1

a 25% of 128 =

Write the percentage as a fraction over 100. Cancel and simplify. Alternatively, 25% of 128 =

155 60 × 100 1 155 3 = × 5 1 31 3 = × = 93 1 1

b 155% of 60 =

1 of 128 = 128 ÷ 4 = 32. 4

Write the percentage as a fraction over 100. Cancel and simplify.

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42 × 100 65

3 Copy and complete the following sentences. a Finding 1% of a quantity is the same as dividing the quantity by ______. b Finding 10% of a quantity is the same as dividing the quantity by ______. c Finding 20% of a quantity is the same as dividing the quantity by ______. d Finding 50% of a quantity is the same as dividing the quantity by ______.

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5 Express the first quantity as a percentage of the second quantity, giving answers in fractional form where appropriate. a 3, 10 b 9, 20 c 25, 80 d 15, 18 e 64, 40 f 82, 12 g 72, 54 h 200, 75 6 Express the first quantity as a percentage of the second quantity, giving answers in decimal form, correct to 2 decimal places where appropriate. a 2, 24 b 10, 15 c 3, 7 d 18, 48 e 56, 35 f 15, 12 g 9, 8 h 70, 30 Example 23

7 Express: a 40 cents as a percentage of $8 c 3 mm as a percentage of 6 cm e 200 g as a percentage of 5 kg g 1.44 m as a percentage of 48 cm

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50 cents as a percentage of $2 400 m as a percentage of 1.6 km 8 km as a percentage of 200 m $5.10 as a percentage of 85 cents

8 Express each quantity as a percentage of the total. a 28 laps of a 50-lap race completed b Saved $450 towards a $600 guitar c 172 fans in a train carriage of 200 people d Level 7 completed of a 28-level video game e 36 students absent out of 90 total f 14 km mark of a 42 km marathon Example 24a

9 Find: a 50% of 36 e 5% of 60 i 15% of 880

b 20% of 45 f 2% of 150 j 45% of 88

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25% of 68 14% of 40 80% of 56

d 32% of 50 h 70% of 250 l 92% of 40

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39 out of 50 44 out of 55 11 out of 30 34 out of 36

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4 Express each of the following as a percentage. a 20 out of 25 b 13 out of 20 d 12 out of 40 e 26 out of 130 g 24 out of 60 h 6 out of 16 j 85 out of 120 k 17 out of 24

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c 30% of 150 kg f 75% of 4.4 km WO

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12 Find: 1 a 33 % of 16 litres of orange juice 3

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1 c 12 % of a $64 pair of jeans d 37.5% of 120 donuts 2 13 In a survey, 35% of respondents said they felt a penalty was too lenient and 20% felt it was too harsh. If there were 1200 respondents, how many felt the penalty was appropriate? 14 Four students completed four different tests. Their results were: Maeheala: 33 out of 38 marks Wasim: 16 out of 21 marks Francesca: 70 out of 85 marks Murray: 92 out of 100 marks Rank the students in decreasing order of test percentage. 15 Jasper scored 22 of his team’s 36 points in the basketball final. What percentage of the team’s score did Jasper shoot? Express your answer as a fraction and as a recurring decimal. 16 Due to illness, Vanessa missed 15 days of the 48 school days in Term 1. What percentage of classes did Vanessa attend in Term 1?

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18 Calculate 40% of $60 and 60% of $40. What do you notice? Can you explain your findings? 19 Which of the following alternatives would calculate a% of b? 100 ab a 100a A B C D ab 100 100 b b 20 Which of the following alternatives would express x as a percentage of y? 100 xy x 100x A B C D xy 100 100 y y © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Enrichment: Two-dimensional percentage increases

+25%

12 cm

20 cm

+25%

21 A rectangular photo has dimensions of 12 cm by 20 cm. a What is the area of the photo in cm2? The dimensions of the photo are increased by 25%. b What effect will increasing the dimensions of the photo by 25% have on its area? c What are the new dimensions of the photo? d What is the new area of the photo in cm2? e What is the increase in the area of the photo? f What is the percentage increase in the area of the photo? g What effect did a 25% increase in dimensions have on the area of the photo? h Can you think of a quick way of calculating the percentage increase in the area of a rectangle for a given increase in each dimension? i What would be the percentage increase in the area of a rectangle if the dimensions were: i increased by 10%? ii increased by 20%? 1 iii increased by 33 % ? iv increased by 50%? 3 You might like to draw some rectangles of particular dimensions to assist your understanding of the increase in percentage area. j What percentage increase in each dimension would you need to exactly double the area of the rectangle? k You might like to explore the percentage increase in the volume of a three-dimensional shape when each of the dimensions is increased by a certain percentage.

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4H Decreasing and increasing by a percentage Percentages are regularly used when dealing with money. Here are some examples. • Decreasing an amount by a percentage (Discount) All items in the store were reduced by 30% for the three-day sale. The value of the car was depreciating at a rate of 18% per annum. • Increasing an amount by a percentage (Mark-up) A retail shop marks up items by 25% of the Column 1 of this stock market display of share prices wholesale price. shows the current share price, column 2 shows the The professional footballer’s new contract was opening price, column three shows the change (in increased by 40%. dollars and cents), column 4 shows the percentage change. Minus signs indicate a decrease.

When dealing with questions involving money, you generally round your answers to the nearest cent. Do this by rounding correct to 2 decimal places. For example: $356.4781 rounds to $356.48 (suitable if paying by credit card). As our smallest coin is the five-cent coin, on many occasions it will be appropriate to round your answer to the nearest five cents. For example: $71.12 rounds to $71.10 (suitable if paying by cash).

let’s start: Original value ± % change = new value The table below consists of three columns: the original value of an item, the percentage change that occurs and the new value of the item. However, the data in each of the columns have been mixed up. Your challenge is to rearrange the data in the three columns so that each row is correct. This is an example of a correct row. original value

percentage change

New value

$65.00

Increase by 10%

$71.50

Rearrange the values in each column in this table so that each row is correct. original value

percentage change

New value

$102.00

Increase by 10%

$73.50

$80.00

Increase by 5%

$70.40

$58.00

Decreased by 2%

$76.50

$64.00

Decrease by 25%

$78.40

$70.00

Increase by 30%

$73.80

$82.00

Decrease by 10%

$75.40

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Common words used to represent a decrease in price include reduction, discount, sale, loss or depreciation. In each case the new value equals the original value minus the decrease. Common words used to represent an increase in price include mark-up and proﬁt. In each case the new value equals the original value plus the increase. Increasing or decreasing an item by a set percentage involves the technique of finding a percentage of a quantity. For example: A 15% mark-up on $200 involves finding 15% of $200 and then adding this amount to $200. In retail terms: Selling price = retail price – discount Selling price = wholesale price + mark-up

Example 25 Finding new values Find the new value when: a $160 is increased by 40% SoluTioN a 40% of $160 =

b

$63 is decreased by 20% ExplANATioN

40 160 × = $64 100 1

New price = $160 + $64 = $224 Alternative method: 100% + 40% = 140% = 1.4 $160 × 1.4 = $224 20 63 b 20% of $63 = × = $12.60 100 1 New price = $63 – $12.60 = $50.40

Calculate 40% of $160. Cancel and simplify. New price = original price + increase

The new value is 140% of the old value. Calculate 20% of $63. Cancel and simplify. New price = original price – decrease

Alternative method: 100% − 20% = 80% = 0.8 $63 × 0.8 = $50.40

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Example 26 Calculating discounts or mark-ups a Find the cost of a $860 television that has been discounted by 25%. b Find the cost of a $250 microwave oven that has been marked up by 12%. SoluTioN

ExplANATioN

a Discount = 25% of $860

Calculate 25% discount. Cancel and simplify.

=

25 860 × = $215 100 1

Selling price = $860 – $215 = $645

Selling price = cost price – discount

Alternative method: 100% − 25% = 75% = 0.75 $860 × 0.75 = $645 b Mark-up = 12% of $250 12 250 = × = $30 100 1 Selling price = $250 + $30 = $280

Calculate 12% of $250. Cancel and simplify. Selling price = cost price + mark-up

Alternative method: 100% + 12% = 112% = 1.12 $250 × 1.12 = $280

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2 Calculate the new price when: a an item marked at $80 is discounted by 50% b an item marked at $30 is marked up by 20% c an item marked at $45 is reduced by 10% d an item marked at $5 is increased by 200% 3 A toy store is having a sale in which everything is discounted by 10% of the recommended retail price (RRP). A remote-control car is on sale and has a RRP of $120. a Calculate the discount on the car (i.e. 10% of $120). b Calculate the selling price of the car (i.e. RRP – discount).

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1 Calculate the new price when: a an item marked at $15 is discounted by $3 b an item marked at $25.99 is marked up by $8 c an item marked at $17 is reduced by $2.50 d an item marked at $180 is increased by $45

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6 Find the cost of the following. a A $600 television that has been discounted by 20% b A $150 ripstick that has been reduced by 15% c A $52 jumper that has depreciated by 25% d A $80 framed Pink poster that has been marked up by 30% e A $14 meal that has been increased by 10% f A $420 stereo that has been marked up by 50% 7 Calculate the selling prices of the following items if they are to be reduced by 25%. a $16 thongs b $32 sunhat c $50 sunglasses d $85 bathers e $130 boogie board f $6.60 surfboard wax 8 Using a calculator, calculate the selling prices of the following items: a A $450 Blu-ray player reduced by 12% b A $12 990 second-hand car reduced by 15% c A $675 surfboard marked up by 24% d A $725 000 home with an additional 3% selling fee WO

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9 Shop C and shop D purchase Extreme Game packages at a cost price of $60. Shop C has a mark-up of $20 for retailers and shop D has a mark-up of 25%. Calculate the selling price for the Extreme Game package at each shop.

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10 A retail rug store marks up all items by 25% of the wholesale price. The wholesale price of a premier rug is $200 and for a luxury rug is $300. What is the customer price for the two different types of rugs? 11 A bookstore is offering a discount of 10%. Jim wants to buy a book with a RRP of $49.90. How much will it cost him if he pays by cash? How much will it cost him if he pays by credit card?

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5 Find the new value when: a $400 is increased by 10% c $80 is decreased by 20% e $5000 is increased by 8% g $15 is decreased by 10%

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12 Shipshape stores were having a sale and reducing all items by 20%. Gerry purchased the following items, which still had their original price tags: jeans $75, long-sleeved shirt $94, T-shirt $38 and shoes $125. What was Gerry’s total bill at the sale?

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13 At the end of the winter season an outdoors store had a 20% discount on all items in the store. Two weeks later they were still heavily overstocked with ski gear and so they advertised a further 40% off already discounted items. Calculate the new selling price of a pair of ski goggles with a RRP of $175.00.

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15 Patrick wants to purchase a trail bike and has visited three different stores. For the same model he has received a different deal at each store. Pete’s Trail Bikes has the bike priced at $2400 and will not offer any discount. Eastern Bikers has the bike priced at $2900 but will offer a 20% discount. City Trail Bikes has the bike priced at $2750 but will offer a 15% discount. What is the cheapest price for which Patrick can purchase his trail bike and which store is offering the best deal?

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14 Georgie desperately wants to buy a new mountain bike that has a RRP of $350. She only has $220. Fortunately, her local bike store regularly has sales in which the discounts are multiples of 5%. What is the smallest discount that will enable Georgie to purchase the bike?

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Enrichment: Commission 16 Many sales representatives are paid by commission. This means their wage is calculated as a certain percentage of their sales. For example, Ben, a car salesman is paid 5% commission. In one week Ben sold three cars for a total value of $42 000. His wage for this week was 5% of $42 000 = $2100. If in the next week Ben sells no cars, he would receive no wage. a Calculate the following amounts. i 10% commission on sales of $850 ii 3% commission on sales of $21 000 iii 2.5% commission on sales of $11 000 iv 0.05% commission on sales of $700 000 Generally sales representatives can be paid by commission only, by an hourly rate only, or by a combination of an hourly rate and a percentage commission. The combination is common, as it provides workers with the security of a wage regardless of sales, but also the added incentive to boost wages through increased sales. Solve the following three problems involving commission. b Stuart sells NRL records. He earns $8.50 per hour and receives 5% commission on his sales. Stuart worked for five hours at the Brisbane Broncos vs Canberra Raiders match and sold 320 records at $4 each. How much did Stuart earn? c Sam, Jack and Justin were all on different pay structures at work. Sam was paid an hourly rate of $18 per hour. Jack was paid an hourly rate of $15 per hour and 4% commission. Justin was paid by commission only at a rate of 35%. Calculate their weekly wages if they each worked for 40 hours and each sold $2000 worth of goods. d Clara earns an hourly rate of $20 per hour and 5% commission. Rose earns an hourly rate of $16 per hour and 10% commission. They each work a 40-hour week. In one particular week, Clara and Rose both sold the same value of goods and both received the same wage. How much did they sell and what was their wage? e Do you know anyone who receives a commission as part of their salary? Interview someone and understand the structure of their salary. Would you like to be paid a set salary or like the added incentive of commission? Summarise your findings and write a brief report.

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4I The Goods and Services Tax (GST) The Goods and Services Tax or GST is a broad–based tax on most goods and services sold or consumed in Australia. The advertised price of the goods in shops, restaurants and other businesses must include the GST. At present in Australia the GST is set at 10%. Not all goods and services are taxed under the GST. Items that are exempt from the goods and services tax include: most basic foods, some education courses and some medical and health care products and services.

let’s start: GST Look at the prices before and after GST was included. Before GST $180

After GST $198

How much GST was paid? What percentage is the GST? What number could be placed in the box? If the GST was 12%, what number would go in the box?

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The GST is 10% of the sale price. It is paid by the consumer and passed onto the government by the supplier. The final advertised price, inclusive of the GST represents 110% of the value of the product: the cost (100%) plus the GST of 10% gives 110%. The unitary method can be used to find the GST included in the price of an item or the pre-tax price. The unitary method involves finding the value of ‘one unit’, usually 1%, then using this information to answer the question.

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Example 27 Calculating the GST Calculate the GST payable on: a a table that a manufacturer values at $289 b a bill from a landscape gardener of $2190. Solut ion a b

Expl anati on

10 × 289 = 28.9 100 The GST is $28.90.

GST is 10% of the value. Find 10% of $289.

10 × 2190 = 219 100 The GST is $219.

Find 10% of $2190.

Example 28 Using the unitary method to find the full amount A bike has GST of $38 added to its price. What is the price of the bike: a before the GST? b after the GST? Solut ion a

Expl anati on

10% = 38 1% = 3.8 100% = 380 Before the GST was added, the cost of the bike was $380.

b After the GST, the price is $418.

Divide by 10 to find the value of 1%. Multiply by 100 to find 100%.

Add the GST onto the pre-GST price $380 + $38 = $418

Example 29 Using the unitary method to find the pre-GST price The final price at a café including the 10% GST was $137.50, what was the pre-GST price of the meal? Solut ion

Expl anatio n

110% = 137.50 1% = 1.25 100% = 125 The pre-GST cost is $125. Alternative method $137.50 ÷ 1.1 = $125

The GST adds 10% to the cost of the meal 100% + 10% = 110% \ final price is 110%. Divide by 110 to find the value of 1%. Multiply by 100 to find 100%. Dividing by 1.1 gives the original price excluding GST.

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1 Without using a calculator, evaluate the following. a 10% of $50 b 10% of $160 c 10% of $250 d 10% of $700 e 10% of $15 f 10% of $88 g 10% of $5 h 10% of $2.50

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a

$100

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$50

c

$150

d

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e

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f

$120

10% GST

price (inc. GST)

3 If the GST is $8, what was the original price? 4 The final advertised price of a pair of shoes is $99. The GST included in this price is: A $9.90 B $90 C $9 D $89.10

c $550 f $5.67

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6 Calculate the final price, including the GST, on items priced at: a $700 b $3000 c $450 d $34 e $56 700 f $4.90

Example 29

7 The final price to the consumer includes the 10% GST. Calculate the pre-GST price if the final price was: a $220 b $66 c $8800 d $121 e $110 f $0.99 8 Calculate the pre-GST price if the final cost to the consumer was: a $352 b $1064.80 c $506 d $52.25 e $10791 f $6.16

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Final cost including the 10% GST

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$599 $68 $70 $660 $789 $89.20 $709.50 $95.98

GYMEA FRUIT MARKET H AV E A N I C E D AY

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SUPERBARN SUPERBARN GYMEA Description

To t a l $

O / E PA S O TA C O K I T S 2 9 0 G M TO M ATO E S L A R G E K I L O 0.270kg @$4.99/kg WAT E R M E L O N S E E D L E S S W H O L E K I L O 1.675kg @$2.99/kg LETTUCE ICEBERG EACH * PA S M / M A L L O W S 2 5 0 G M

0.090 KG @ $14.99/kg BANANA SUGAR $1.35 SL MUSHROOM

$2.49

P I S TA C C H I O 1 1

$6.00

ROUND

$0.01

T O TA L

$9.85

CASH TA X 1

$9.85 $0.55

6.09 1.35 5.01 2.49 1.89

S u b To t a l Rounding

$16.83 $0.02

T OTA L ( I n c G S T )

$16.85

C a s h Te n d e re d Change Due GST Amount

$20.00 $3.15 $0.17

Xmart a How many toys were purchased? b What was the cost of the most expensive item? c Which of the toys attracted GST? d How much GST was paid in total? e What percentage of the total bill was the GST?

* Signifies items(s) with GST Thank you for shopping at Superbarn

XMART

C U S T O M E R R E C E I P T TA X I N V O I C E 07/13/11 15:1

*JUNGLE JUMP BALL *CR COLOUR SET CARDS

6.00 10.00

* C R G L O W S TAT I O N

10.00

* S TA R O T T O M A N P I N K

12.00

* J U N G L E H I D E AWAY

12.00

* L P A I R P O RT

29.00

*MY OWN LEAPTOP 2 @ 35.00 T OTA L CASH TENDER CHANGE

70.00 149.00 150.00 1.00

* TA X A B L E I T E M S P L E A S E R E TA I N T H I S R E C E I P T / TA X INVOICE AS PROOF OF PURCHASE WE NOW TRADE 2 4 H O U R S A D AY, 7 D AY S A W E E K

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10 Here are three real-life receipts (with the names of shops changed). GST rate is 10%. Answer the following based on each one. Superbarn a How much was spent at Superbarn? b How many kilograms of tomatoes were bought? c Which item included the GST, and how do you tell by looking at the receipt? d What was the cost of the item if the GST is not included?

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12 Use the technique outlined in question 11 above to find the GST already paid on goods and services costing: a $616 b $1067 c $8679 d $108.57

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13 The cost of a lounge suite is $990 and includes $90 in GST. Find 10% of $990 and explain why it is more than the GST included in the price.

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Enrichment: GST and the manufacturing process The final consumer of the product pays GST. Consider the following: A fabric merchant sells fabric at $66 (including the 10% GST) to a dressmaker. The merchant makes $60 on the sale and passes the $6 GST onto the Australian Tax Office (ATO). A dressmaker uses the fabric and sells the product onto a retail store for $143 (this includes the $13 GST). As the dressmaker has already paid $6 in GST when they bought the fabric they have a GST tax credit of $6 and they pass on the $7 to the ATO. The retailer sells the dress for $220, including $20 GST. As the retailer has already paid $13 in GST to the dressmaker, they are required to forward the extra $7 to the ATO. The consumer who buys the dress bears the $20 GST included in the price. 16 Copy and complete the following. Raw material $110 (includes the 10% GST) GST on sale = ________ GST credit = $0 Net GST to pay = ________ Production stage $440 (includes the 10% GST) GST on sale = ________ GST credit = _________ Net GST to pay = ________ Distribution stage $572 (includes the 10% GST) GST on sale = ________ GST credit = _________ Net GST to pay = ________ Retail stage $943.80 (includes the 10% GST) GST on sale = ________ GST credit = _________ Net GST to pay = ________ GST paid by the final consumer = ______________

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4J Calculating percentage change, proﬁt and loss If you sell something for more than you paid for it, you have made a profit. On the other hand, if you sell something for less than you paid for it, you have made a loss. Retailers, business people and customers are all interested to know the percentage profit or the percentage loss that has occurred. In other words, they are interested to know about the percentage change. The percentage change provides valuable information over and above the simple value of the change. For example:

Hat was $40 Discount $5 Now $35

Cap was $8 Discount $5 Now $3

The change for each situation is exactly the same, a $5 discount. However, the percentage change is very different. Hat was $40. Discount 12.5%. Now $35. Cap was $8. Discount 62.5%. Now $3. For the sunhat, there is a 12.5% change, and for the cap there is a 62.5% change. In this case, the percentage change would be known as a percentage discount.

let’s start: Name the acronym What is an acronym? How many of the following nine business and finance-related acronyms can you name? • RRP • GST • ATM • CBD • COD • EFTPOS • GDP • IOU • ASX Can you think of any others? How about the names of two of the big four Australian banks: NAB and ANZ? How do these acronyms relate to percentages? The following three acronyms are provided for fun and interest only. Do you know what they stand for? • SCUBA diving • LASER gun • BASE jumping

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■ ■ ■

Proﬁt = selling price – cost price Loss = cost price – selling price Calculating a percentage change involves the technique of expressing one quantity as a percentage of another (see Section 4G). Percentage change = Percentage proﬁt = Percentage loss =

change × 100% or original value

profit of × 100% or original value

loss × 100% or original value

Example 30 Calculating percentage change Calculate the percentage change (profit/loss) when: a $25 becomes $32 b $60 becomes $48 SoluTioN

ExplANATioN

a Profit = $7

Profit = $32 – $25

7 100 % Profit = × % 25 1 = 28%

b Loss = $12 12 100 % Loss = × % 60 1 = 20%

Percentage profit = Alternatively,

7 28 = = 28%. 25 100

Loss = $60 – $48 Percentage loss = Alternatively,

profit of × 100% or original value

loss × 100% or original value

12 1 20 = = = 20%. 60 5 100

Example 31 Solving worded problems Ross buys a ticket to a concert for $125, but is later unable to go. He sells it to his friend for $75. Calculate the percentage loss Ross made. SoluTioN

ExplANATioN

Loss = $125 – $75 = $50 50 100 % Loss = × % 125 1 = 40% Ross made a 40% loss on the concert ticket.

Loss = cost price – selling price loss Percentage loss = × 100% cost pric pr e

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1 Calculate the profit made in each of the following situations. a Cost price = $14, Sale price = $21 b Cost price = $75, Sale price = $103 c Cost price = $25.50, Sale price = $28.95 d Cost price = $499, Sale price = $935

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2 Calculate the loss made in each of the following situations. a Cost price = $22, Sale price = $9 b Cost price = $92, Sale price = $47 c Cost price = $71.10, Sale price = $45.20 d Cost price = $1121, Sale price = $874 3 Which of the following is the correct formula for working out percentage change? A % change =

change or original value

C % change = change × 100%

B

% change =

original value × 100% change

D

% change =

change × 100% or original value

loss × 100% cost pric pr e Which of the following would therefore be the formula for calculating the percentage discount?

4 The formula used for calculating percentage loss is % loss =

C % discount = disc discount di scount ount × 100%

B

% discount =

cost pr price × 100% discount

D

% discount =

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6 Find the percentage change (increase or decrease) when: a 15ºC becomes 18ºC b 18ºC becomes 15ºC c 4ºC becomes 24ºC d 12ºC becomes 30ºC 7 Find the percentage change in population when: a a town of 4000 becomes a town of 5000 b a city of 750 000 becomes a city of 900 000 c a country of 5 000 000 becomes a country of 12 000 000

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5 Find the percentage change (profit or loss) when: a $20 becomes $36 b $10 becomes $13 c $40 becomes $30 d $25 becomes $21 e $12 becomes $20 f $8 becomes $11 g $6 becomes $4 h $150 becomes $117

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Example 30

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discount × 100% cost pric pr e

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A % discount =

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8 Gari buys a ticket to a concert for $90, but is unable to go. She sells it to her friend for $72. Calculate the percentage loss Gari made.

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9 Estelle purchased a piece of sporting memorabilia for $120. Twenty years later she sold it for $900. Calculate the percentage profit Estelle made. 10 Xavier purchased materials for $48 and made a dog kennel. He later sold the dog kennel for $84. a Calculate the profit Xavier made. b Calculate the percentage profit Xavier made. 11 Gemma purchased a $400 foal, which she later sold for $750. a Calculate the profit Gemma made. b Calculate the percentage profit Gemma made. 12 Lee-Sen purchased a $5000 car, which she later sold for $2800. a Calculate Lee-Sen’s loss. b Calculate Lee-Sen’s percentage loss. WO

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14 Which of the following stores is offering a larger percentage discount? Store A: Jeans originally selling for $60, now on sale for $51 Store B: Jeans originally selling for $84, now on sale for $73.50 15 The circulation of a student newspaper at Burrough High School was 60 copies in Term 1, 120 copies in Term 2, 200 copies in Term 3 and 360 copies in Term 4. a Calculate the percentage growth in circulation between: i Term 1 and Term 2 ii Term 2 and Term 3 iii Term 3 and Term 4 b In which term was there the greatest percentage growth in circulation? c What was the overall percentage growth in circulation over the course of the year?

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13 Explain why an increase of 10% followed by a decrease of 10% on the new price gives an answer that is less than the original amount.

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Enrichment: Australia’s population 16 Australia’s population in early 2010 was 22 003 926 and the percentage population growth was 1.8% per year. a Given this population size and percentage growth, how many more Australians would there be after one year? b How many would there be after: i 2 years? ii 5 years? iii 10 years?

Annual gr growth rrate (%)

The Australian Bureau of Statistics carries out comprehensive population projections. One such projection assumes the following growth rates: • One birth every 1 minute and 44 seconds • One death every 3 minutes and 39 seconds • A net gain of one international migrant every 1 minute and 53 seconds • An overall total population increase of one person every 1 minute and 12 seconds c Calculate the percentage growth per annum, using the 2010 population of 22 003 926 and the projected total population increase of one person every 1 minute and 12 seconds. Give your answer correct to 1 decimal place. d Find out the current population of Australia and the population 10 years ago, and work out the percentage growth over the past decade. e How does Australia’s percentage growth compare with that of other countries? f Find out the current life expectancy of an Australian male and female. Find out their life expectancy 50 years ago. Calculate the percentage increase in life expectancy over the past 50 years. g A key factor in population growth is the total fertility rate (TFR) (i.e. the number of babies per woman). For Australia the TFR in 2010 was at 1.74 expected births per woman. The peak of Australia’s TFR over the past 100 years was 3.6 children per woman in 1961. What is the percentage decrease in TFR from 1961 to 2010? h Carry out some of your own research on Australia’s population. Population growth in Australia, 1950-2000 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 Year

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4K Solving percentage problems with the unitary

method and equations If you know a percentage of an amount, but do not actually know the amount, you can use a technique known as the unitary method to find the whole amount. The unitary method involves finding the value of a unit and then using this value to calculate the value of a whole. In this section, the value of a unit will be the value of one per cent (1%). Generally, the first step in all problems involving the unitary method is to divide the information given to find the value of a unit. The second step is to then multiply the value of a unit to find the value of the number of units required in the question.

let’s start: Using the unitary method

Key ideas

By first finding the value of ‘1 unit’, answer the following questions using mental arithmetic only. 1 Four tickets to a concert cost $100. How much will 3 tickets cost? 2 Ten workers can dig 40 holes in an hour. How many holes can 7 workers dig in an hour? 3 Six small pizzas cost $54. How much would 10 small pizzas cost? 4 If 8 pairs of socks cost $64, how much would 11 pairs of socks cost? 5 Five passionfruit cost $2.00. How much will 9 passionfruit cost? 6 If a worker travels 55 km in 5 trips from home to the worksite, how far will they travel in 7 trips?

■

■ ■

The unitary method involves finding the value of ‘one unit’ and then using this information to answer the question. When dealing with percentages, finding ‘one unit’ corresponds to finding one per cent (1%). Once the value of 1% of an amount is known, it can be multiplied to find the value of any desired percentage.

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Number and Algebra

Example 32 using the unitary method to find the full amount If 8% of an amount of money is $48, what is the full amount of money? SoluTioN

ExplANATioN

8% of amount is $48 ÷8 1% of amount is $6 × 100 100% of amount is $600 × 100 Full amount of money is $600.

Divide by 8 to find the value of 1%. Multiply by 100 to find the value of 100%.

÷8

Alternative solution 8% of A = $48 0.08 × A = 48 ÷ 0.08 A = 48 ÷ 0.08 ÷ 0.08 A = 600

Form an equation. Divide both sides by 0.08.

Example 33 using the unitary method to find a new percentage If 11% of the food bill was $77, how much is 25% of the food bill? SoluTioN 11% of food bill is $77 \ 1% of food bill is $7 \ 25% of food bill is $175 Alternative solution 11% of B = $77 0.11 × B = 77 B = 77 ÷ 0.11 B = 700 0.25 × B = 175

ExplANATioN Divide by 11 to find the value of 1%. Multiply by 25 to find the value of 25%.

Form an equation. Solve the equation. We need 25% of B.

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Example 34 using the unitary method to find the original price A pair of shoes has been discounted by 20%. If the sale price was $120, what was the original price of the shoes? SoluTioN

ExplANATioN

Only paying 80% of original price: \ 80% of original price is $120 \ 1% of original price is $1.50 \ 100% of original price is $150 The original price of the shoes was $150.

20% discount, so paying (100 – 20)%. Divide by 80 to find the value of 1%.

Form an equation Solve the equation

Exercise 4K

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1 If 10% of an amount of money is $75. How much is 1% of that amount? A $1 B $7.50 C $75 D $750

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2 If 10% of an amount of money is $22. How much is 100% of that amount? A $100 B $2.20 C $22 D $220 3 Which alternative best describes the unitary method when dealing with percentages? A Work out the factor required to multiply the given percentage to make 100%. B Find the value of 1% and then multiply to find the value of percentage required. C Once you find the full amount, or 100% of the amount, this is known as ‘the unit’. D Find what percentage equals $1 and then find the given percentage.

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Alternative solution: 80% of P = $120 0.8 × P = 120 P = 120 ÷ 0.8 P =

Multiply by 100 to find the value of 100%.

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5 Calculate the full amount of money for each of the following. a 3% of an amount of money is $27 b 5% of an amount of money is $40 c 12% of an amount of money is $132 d 60% of an amount of money is $300 e 8% of an amount of money is $44 f 6% of an amount of money is $15 Example 33

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4 If 6% of an amount of money is $30, how much is the full amount of money?

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6 If 4% of the total bill is $12, how much is 30% of the bill? 7 Calculate: a 20% of the bill, if 6% of the total bill is $36 b 80% of the bill, if 15% of the total bill is $45 c 3% of the bill, if 40% of the total bill is $200 d 7% of the bill, if 25% of the total bill is $75 8 What is the total volume if 13% of the volume is 143 litres? 9 What is the total mass if 120% of the mass is 720 kg?

11 Find the original price of the following items. a A pair of jeans discounted by 40% has a sale price of $30. b A hockey stick discounted by 30% has a sale price of $105. c A second-hand computer discounted by 85% has a sale price of $90. d A second-hand textbook discounted by 80% has a sale price of $6. e A standard rose bush discounted by 15% has a sale price of $8.50. f A motorbike discounted by 25% has a sale price of $1500.

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10 A necklace in a jewellery store has been discounted by 20%. If the sale price is $240, what was the original price of the necklace?

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12 Forty per cent of workers for a large construction company prefer to have an early lunch and 25% of workers prefer to work through lunch and leave an hour earlier at the end of the day. All other workers prefer a late lunch. If 70 workers prefer a late lunch, how many workers are employed by the construction company?

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13 Daryl receives an amount of money from his grandparents for his birthday. He spends 70% of the money buying a new music CD. He then spends 50% of the remaining money buying more credit for his mobile phone. After these two purchases, Daryl has $6 remaining. How much money did Daryl’s grandparents give him for his birthday?

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15 If y% of an amount of money is $8, how much is the full amount of money? 16 If C% of an amount of money is $D, how much is F% of the amount of money?

Enrichment: lots of ups and downs 17 A shirt is currently selling for a price of $120. a Increase the price by 10%, then increase that price by 10%. Is this the same as increasing the price by 20%? b Increase the price of $120 by 10%, then decrease that price by 10%. Did the price go back to $120? c The price of $120 is increased by 10%. By what percentage must you decrease the price for it to go back to $120? d If the price of $120 goes up by 10% at the end of the year, how many years does it take for the price to double?

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14 If 22% of an amount is $8540, which of the following would give the value of 1% of the amount? A $8540 × 100 B $8540 ÷ 100 C $8540 × 22 D $8540 ÷ 22

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phi, the golden number Constructing a golden rectangle You will need a ruler, a pencil, a pair of compasses and a sheet of A4 paper. 1 Rule a square of side length about 15 cm on the left side of your paper. 2 Rule a line from the midpoint (M) of one side of the square to the opposite corner (C) as shown. 3 With your compass point on M and the compass pencil on C (radius MC), draw an arc from C as shown. 4 Extend the base of the square to meet this arc. This new length is the base length of a golden rectangle. 5 Complete the golden rectangle and erase the vertical line from C and the arc. 6 You now have two golden rectangles, one inside the other. C

M

Constructing a golden spiral 1 In your smaller golden rectangle, rule the largest square that can be drawn next to the first square. 2 Continue adding squares in each smaller golden rectangle until there are at least 7 adjacent squares arranged as shown below. 3 In each square, mark the corners that are closest to the smallest square. 4 Using a compass with the point on these corners draw arcs equal to the side of each square. 5 Colour your golden spiral pattern.

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investigation

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Calculating phi Phi is the ratio of the length to the width (height) of a golden spiral or a golden rectangle. In the design you have drawn there are several golden rectangles. Measure the length and width of each golden rectangle and calculate the value of phi (length divided by width) for each. Work out the average (mean) of all your calculations of phi. Compare your average with the 1+ 5 = 1.61803… actual value of 2

Golden rectangles in the human body Investigate some of your own measurements and see how close they are to the golden ratio of phi : 1 ≈ 1.6:1. Golden ratios in the human body include: • total height : head to fingertips • total height : navel (belly button) to floor • height of head : width of head • shoulder to fingertips : elbow to fingertips • length of forearm : length of hand • hip to floor : knee to floor • length of longest bone of finger : length of middle bone of same finger • eyes to the bottom of the chin : width of ‘relaxed’ smile • width of two centre teeth : height of centre tooth

Research and class presentation Research the internet to discover examples of golden rectangles in nature, human anatomy, architecture, art or graphic design. Present your findings to your class using a PowerPoint presentation or a poster.

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1 The side lengths of a cube are all increased by 20%. As a percentage, by how much does its volume change?

2 The cost of an item in a shop increased by 20% and was later decreased by 20% at a sale. Express the final price as a percentage of the original price. 3 Evaluate

1 2 3 4 999 × × × ×… × without using a calculator. 2 3 4 5 1000

4 Complete these magic squares. All rows, columns and the two main diagonals sum to the same total. b a 4 5 3 3 1

1 2

2 3

11 6

2 3

2

1 6

2

5 Find the value of x if: a

x×x 1 = x+x 2

b

6 On line segment AD, AC is

x×x 1 = x+x 5

2 3 of AD and BD is of AD. What fraction of AD is BC? 3 4

? A

B

C

D

7 Fifty people said they liked apples or bananas or both. Forty per cent of the people like apples and 70% of the people like bananas. What percentage of people like only apples? 8 A bank balance of $100 has 10% of its value added to it at the end of every year, so that at the end of the second year the new balance is $110 + $11 = $121. How many full years will it take for the balance to be more than $10 000?

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Chapter summary

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2 3

=

5 –4 + 3 – 18 + 12

Numerator 2 parts selected

1 ×3 + 5 ×2 8 × 3 12 × 2 3 + 10 = –1 – 24 24 7 = –1 + 24 +7 = – 24 24 24 = – 17 24

1 3

0 6

1 6

Simplest form

2 3

2 6

3 6

4 6

×2 ×5 2 3

4 6

=

=

3 3 5 6 ÷2

20 30

×2 ×5

= ÷2

÷5 10 15

=

=

2 3

=

÷5

=

Equivalent fractions have the same simplest form and are equal in value.

3 + 1 HCF of 35 35 40 8 3 1 + × × 77 8 5×7 5×8 24 7 + 280 280 31 280

× 3 41 × 2 + = – 33 8 × 3 12 × 2

+ 82 = – 99 24 24 = – 17 24

Adding and subtracting fractions • Equivalent fractions • ‘Unit’ is the denominator • Add/subtract numerators • Keep denominator the same

Lowest common denominator (LCD) is lowest common multiple (LCM)

6 6

= –4 8 + 3 12

OR

Equivalent fractions represent equal proportions of one whole. 0 3

and 40 is 5

Negative fractions 5 6

1

3 25

=

+

3 5

=

Multiplying fractions

Improper OR

• Mixed numerals Improper fractions • Denominators do not need to be the same • Cancel: Divide common factors in numerator and denominator

Mixed 5×2

40 15

= 2 15 = 2 5 × 3 = 2 3

10

40 15

=

5×8 5×3

=

8 3

=

Negative fractions

2 2 3

1

× (– 4 )

–

5 8

2

× (– 5 )

Different signs

Same signs

Negative answer

Positive answer

= =

2 1 – 12 6 1 – 6

= =

– 5 × 4 15 14 5 1 21 3 = – 14 × 5 2 1

= – 32

1

= –1 2

10 1 40 4 1 4

= 56 +

1 6

Negative fractions

2

= 23

1 6

Dividing fractions • Mixed numerals Improper fractions • Denominators do not need to be the same • Multiply by reciprocal of divisor • Cancel: Divide common factors in numerator and denominator

Fractions

13 5

1

– (– 6 )

Add opposite

Subtract opposite

Improper 10 5

5 6

+ (– 6 )

= 56 – Mixed

5

1

= –1 –

3 parts in one whole Denominator

5

1

–4 8 – (–3 12 )

2 3 2 3

5 8 5 =–8

1

÷ (– 6 )

–

1

× (– 6 )

3

÷ (– 8 ) 8

× (– 3 )

Different signs

Same signs

Negative answer

Positive answer 5 3

= –12 3

=

= –4

=13

2

2 7 – 12 ÷ (– 4 3 )

Note =5÷5=1

5 5

21 14

÷7

=

3 2

14 7 = – 12 ÷ (– 3 ) 1

3 1

4

2

7 = – 12 × – 14

=

1 8

Reciprocal 3 is – 14 of – 14 3

÷7

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Number and Algebra

Fractions to decimals

Decimals as fractions 7 1 2 3.712 = 3 + 10 + 100 + 1000 3712 = 464 1000 125 28 7 = 25 100

3.712 = 0.28 =

× 25 25 1 = 100 4

Addition and subtraction Keep decimals places in line.

= 0.25

× 25 3 8

= 0.375

0.3 7 5 8 3. 0 0 0

2 9

= 0.2222...

0.2 2 2... 9 2.0 0 0...

= 0.2 Multiplying or dividing by powers of 10 9.807 + 26.350 11 36.157

3.672 × 100 = 367.2 47.81 ÷ 100 = 0.4781

Terminating decimals 0.24, 0.0381, 2.6 Recurring decimals 0.55555... = 0.5 0.4713 713 713... = 0.4 713 = 0.4 713

Rounding decimals When the critical digit is five or more, increase the rounding digit by one.

Decimals

Division Multiply by powers of ten to make divisior on integer.

6 2 5 9 9

2.35 2.34 2.35 5.90 6.00

–3.826 × 100 –0.02 × 100 –382.6 –2

= 191.3

or

216.04 - 38.57 177.47

Multiplication • Multiply without decimals. • Count decimal places in question. • Answer has this many decimal places.

681 × 52 1362 34050 35412

Rounding to two decimal places

2.34 2.34 2.34 5.89 5.99

20 15 9

681 × 5.2 = 3541.2 68.1 × 5.2 = 354.12 6.81 × 5.2 = 35.412 6.81 × 0.52 = 3.5412 0.681 × 0.52 = 0.35412 0.0681 × 0.52 = 0.035412

–3.826 ÷ –0.02 = –382.6 ÷ –2 =

191.3 2 382.6

= 191.3

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Goods and Services Tax (GST) Current GST rate in Australia is 10% Fraction to percentage 5 × 100% = 62.5% 8 Decimal to percentage 0.362 × 100% = 36.2%

Percentage of a quantity

× 1.1 A

B

Price (excluding GST)

Price (including GST)

7 7% of $40 = 100 × 40 = $2.80

÷ 1.1 GST = 10% of A or B – A or B ÷ 11

Percentage to fraction 56% =

56 100

1

=

14 × 4 25 × 4

=

‘A’ as a percentage of ‘B’ A × 100% B

14 25

25 g as a percentage of 1 kg Same 25 units × 100% = 2.5% 1000

Memorise 1 2 = 12 = 14 = 15 1 = 10 = 18 2 =3 = 34

50% = 33 13 % 25% 20% 10% 12 12 % 66 23 % 75%

= 0.5

Percentages

= 0.3 = 0.25 = 0.2

means ‘out of 100’

Percentage profit Profit = × 100% Cost price

= 0.1

Percentage loss Loss = × 100% Cost price

= 0.125 = 0.6 = 0.75

Percentage increase • mark-up • profit Selling price = Cost price + Mark-up 1 (Profit) 12 % mark-up on $300 2

• Increase =

1

12 2 100

× 300 = $37.50

• New price = 300 + 37.50 = $337.50

Percentage decrease • Discount • Reduce • Depreciate New price = Original price – Decrease $1700 TV discounted by 10% Discount = 1700

10 × 100

Unitary method 1 unit is 1% If ÷6 ×80

6% is $420 1% is $70 80% is $5600

find 80% ÷6 ×80

= $170

Sale price = 1700 – 170 = $1530

alternatively, 100% + 12.5% = 112.5% = 1.125 $300 × 1.125 = $337.50

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Multiple-choice questions 1 0.36 expressed as a fraction is: 36 36 A B 10 100 2

C

3 6

D

9 20

C

71 88

D

31 44

2 5 + is equal to: 11 8 A

7 19

B

10 19

3 When 21.63 is multiplied by 13.006, the number of decimal places in the answer is: A 2 B 3 C 4 D 5 4

124 is the same as: 36 A 88 B

34 9

C

3

4 9

D 3.49

1 is written as an improper fraction, its reciprocal is: 3 1 16 A B 53 C 53 3 6 Which decimal has the largest value? A 6.0061 B 6.06 C 6.016

D 6.0006

7 9.46 ×105 is the same as: A 94 600 000 B 946 000

5 When 5

3 16

D

C

94 605

D 0.0000946

C

84 × 100 ÷ 75

D

9 590% is the same as: A 59.0 B 0.59

C

5.9

D 0.059

10 $790 increased by 15% gives: A $118.50 B $908.50

C

$671.50

D $805

8 75% of 84 is the same as: 84 A 84 × 3 B ×4 3 4

(0.75 × 884) 100

11 A camera has $87 of GST included in its price. What is the price of the camera including the GST? A $870 B $957 C $790 D $1052.70

Short-answer questions 1 Copy and complete: a

7 = 20 60

b

25 5 = 40

c

350 = 210 6

b

36 12

c

102 12

2 Simplify: a

25 45

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3 Evaluate each of the following: 5 2 7 3 a b + − 11 11 8 4 1 2 3 −1 4 3 4 Evaluate: d

e

2 b × 12 3 1 d 3÷ e 2 5 Evaluate each of the following: a

c 3−1

1 4

2 1 2 +3 5 4

f

1 2 3 1 +2 − 2 3 5

3 1 ×1 7 12 2 ÷ 12 3

c

1 ×6 3 1 3 1 ÷ 2 4

a

1 2 − 5 3

b

−

d

3 1 − − 4 5

e

5 1 ÷ − 3 3

f

2

3 c − 5

3 1 × 4 5

f

2

1 1 −6 + −1 4 3

6 Insert >, < or = to make each of the statements below true. 11 0.5 55 20 7 Evaluate: a 12.31 + 2.34 + 15.73 c 569.74 × 100 e 7.4 × 104 a

8 Calculate: a 2.67 × 4 d 1.02 ÷ 4

b

2 3

c 0.763

0.7 b d f

0.7603

14.203 – 1.4 25.14 × 2000 5 – 2.0963

b 2.67 × 0.04 e 1.8 ÷ 0.5

c 1.2 × 12 f 9.856 ÷ 0.05

9 Round these decimals to 3 decimal places. a 0.6 b 3.57964

c 0.00549631

10 Copy and complete this table of conversions. 0.1

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1 4 5%

1 3

1 8

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11 Express each of the following as a percentage. a $35 out of $40 b 6 out of 24 1 d 16 cm out of 4 m e 15 g out of kg 4 12 Find: a 30% of 80 b 15% of $70

c $1.50 out of $1

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13 a Increase $560 by 10%. b Decrease $4000 by 18%. c Increase $980 by 5% and then decrease the result by 5%. Calculate the overall percentage loss. 14 A new plasma television was valued at $3999. During the end-of-year sale, it was discounted by 9%. What was the discount and the sale price? 15 Jenni works at her local pizza takeaway restaurant and is paid $7.76 per hour. When she turns 16, her pay will increase to $10 an hour. What will be the percentage increase in her pay (to the nearest per cent)? 16 An investment of $22 000 earns $2640 in interest over 2 years. What is the flat interest rate per year? 17 Johan saved 15% of his weekly wage. He saved $5304 during the year. Calculate Johan’s weekly wage. 18 A dress cost $89.10 after a 10% discount. How much was saved on the original price of the dress? 19 If 5% of an amount equals 56, what is 100% of the amount? 20 A shopping receipt quotes an A4 folder as costing $3.64 including 10% GST. What is the cost of the folder pre-GST, correct to the nearest cent? 21 Copy and complete the table. pre-GST price $250

10% GST

Final cost including the 10% GST

$45 $8 $737

Extended-response question indian rupee (iNR) 42 1 The following table shows the value of A$1 (one Australian Singapore dollar (SGD) 1.25 dollar) in foreign currency. Thai baht (THB) 30 Genevieve is planning an extended holiday to Asia, where 7 she plans on visiting India, Singapore, Phuket and Hong Kong. Hong Kong dollar (HKD) a She has decided to convert some Australian dollars to each of the above currencies before she flies out. How much of each currency will she receive if she converts A$500 to each currency? b During the exchange she needs to pay a fee of 1.5% for each transaction. How much has she paid in fees (in A$)? c i The day after she leaves, the exchange rate for Indian rupees is A$1 = 43.6 INR. How much more would she have received if she had waited until arriving in India to convert her Australian dollars? (Ignore transaction fees.) ii Express this as a percentage (to 1 decimal place). d i On her return to Australia, Genevieve has in her wallet 1000 INR, 70 SGD and 500 THB. Using the same exchange rate, how much is this in A$? ii She also needs to pay the 1.5% transaction fee. How much does she receive in A$, and is it enough to buy a new perfume at the airport for $96?

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What you will learn

5A 5B 5C 5D 5E 5F 5G 5H 5I

Introducing ratios Simplifying ratios Dividing a quantity in a given ratio Scale drawings Introducing rates Ratios and rates and the unitary method Solving rate problems Speed Distance/time graphs

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for the Australian Curriculum Strand: Number and Algebra

Substrands: RATIOS AND RATES FINANCIAL MATHEMATICS

Outcomes A student operates with ratios and rates, and explores their graphical representation. (MA4–7NA) A student solves ﬁ nancial problems involving purchasing goods. (MA4–6NA)

Bicycle gear ratios The bicycle is renowned as the most energy-efﬁ cient means of human transport. A bicycle with a range of gears is an extremely versatile and enjoyable machine. The ‘gear ratio’ of a bicycle is the ratio of pedal rotations compared to the rate at which the wheels rotate. Manufacturers apply the mathematics of gear ratios to provide cyclists with the range of gears that is most useful for the style of bike. Have you ever tried to ride up a steep hill in too high a gear or reached the bottom of a hill while still in a very low gear? A bicycle rider has a comfortable range of pressures and rotation rates that they can apply to the pedals. Gears make it easier to stay in this comfortable range over a wide range of riding conditions such as BMX, downhill racing, a fast time trial for racing bikes on

ﬂ at ground or climbing steep, rough trails on a mountain bike. Unicycles have only one wheel and so have a ratio of 1:1 for pedal turns to wheel turns. Unicycles can’t reach high speeds because it is very difﬁ cult to rotate pedals more than about 100 rpm; however, riding a unicycle is fun and it requires strength and agility to remain stable, especially when playing unicycle hockey!

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1 Express each fraction in its simplest form. 2 6 a b c 4 12 8 16 e f g 12 32 2 Copy and complete: 2 4 a = 5

b

8 24 40 60

15 20 200 h 350 d

4 = 7 28

3 Convert: a 5 m = ____ cm c 500 cm = ____ m e 120 cm = ____ m

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b d f

3 12 = 2

6 km = ____ m 80 mm = ____ cm 15 000 m = ____ km

4 Express the first quantity as a fraction of the second. a 50 cm, 1 m b 15 minutes, 1 hour c 200 g, 1 kg d 40 minutes, 2 hours 1 1 f 50 cm, m e 40 m, km 2 2 5 Find: 2 1 2 a of $40 b of 60 m c of 24 cm 5 3 3 3 1 1 d of 800 m e 1 lots of $5 f of $100 4 2 8 6 The cost of 1 kg of bananas is $4.99. Find the cost of: 1 a 2 kg b 5 kg c 10 kg d kg 2 7 Kevin walks 3 km in one hour. How far did he walk in 30 minutes? 8 Tao earns $240 for working 8 hours. How much did Tao earn each hour? 9 A car travels at an average speed of 60 km per hour. a How far does it travel in the following times? i

2 hours

ii 5 hours

iii

1 hour 2

b How long would it take to travel the following distances? i 180 km ii 90 km iii 20 km c If the car’s speed was 70 km per hour, how many minutes would it take to travel 7 km?

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5A Introducing ratios Ratios are regularly used in everyday life. They are used to show the relationship between two (or more) related quantities. Here are five common uses of ratios: • Ingredients – the ratio of different ingredients in a recipe (cooking, industrial) • Maps – most maps include a scale which is written as a ratio • Sporting success – showing a team’s win to loss ratio, or the ratio of kicking goals to points The ratio of eggs to butter to sugar will affect the ﬂ avour and texture of a cake. • Comparing size – the ratio of length, area or volume of different shapes • Legal requirements – minimum standards of supervision, staff to student ratios. When dealing with ratios, the order in which the ratio is written is very important. For example a team’s win:loss ratio of 5 : 2 is very different to a team’s win:loss ratio of 2 : 5. Ratios compare quantities of the same type and given in the same unit. Therefore a ratio is not generally written with a unit of measurement.

Let’s start: Estimating ratios

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Ratios show the relationship between two (or more) related quantities. The quantities must be of the same type and in the same unit. For example, a drink was made with the ratio of cordial to water of 1 : 3. The colon: is the mathematical symbol used to represent ratios. The written ratio of a : b is read as the ratio of ‘a to b’ or ‘a is to b’. The order in which the quantities are written in a ratio is important. Before ratios can be written the quantities must be expressed in the same unit. If each number in a ratio is multiplied or divided by the same amount, an equivalent ratio is formed. For example, 1 : 3 and 2 : 6 and 10 : 30 are equivalent ratios.

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Key ideas

Estimate the following ratios. • Ratio of boys to girls in your school. • Ratio of Year 8 students to Year 7 students in your school. • Ratio of your teacher’s age to your age! • Ratio of hours you spend outside to hours you spend inside. • Ratio of hours you are awake to hours you are asleep. • Ratio of parents to children in an average Australian family. • Ratio of the length to the width of an A4 sheet of paper. Discuss your answers as a class.

Chapter 5 Ratios and rates

Example 1 Writing ratios A sample of mixed nuts contains 5 cashews, 12 peanuts and 2 macadamia nuts. Write down: a the ratio of cashews to peanuts to macadamias b the ratio of cashews to the total number of nuts c the ratio of peanuts to other nuts SOLUTION

EXPLANATION

a 5 : 12 : 2

cashews : peanuts : macadamias

b 5 : 19

5 cashews, total nuts 19

c 12 : 7

12 peanuts, other nuts 7

Example 2 Producing equivalent ratios Complete each pair of equivalent ratios. a 4 : 9 = 16 : SOLUTION

b

30 : 15 =

:5

EXPLANATION

×4 a

4 : 9 = 16 : 36

Both numbers are multiplied by 4.

×4 ÷3 b

30 :15 = 10 : 5

Both numbers are divided by 3.

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3 Express each of the following pairs of quantities as a ratio. a 9 goals to 4 goals b 52 litres to 17 litres c 7 potatoes to 12 carrots d 3 blue marbles to 5 green marbles

5 Over the past fortnight, it has rained on eight days and it has snowed on three days. Write down the ratio of: a rainy days to snowy days b snowy days to total days c fine days to rainy and snowy days d rainy days to non-rainy days

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4 A bag contains 5 green marbles, 7 blue marbles and 3 yellow marbles. Write down the ratio of: a green marbles to yellow marbles b blue marbles to total marbles c yellow to blue to green marbles d green to yellow to blue marbles

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7 Complete each pair of equivalent ratios. a 1:3 = 4: b 1:7 = 2: c e 5 : 10 = 1 : f 12 : 16 = 3 : g i 4 : 7 = 44 : j 14 : 17 = 42 : k

2:5 = : 10 12 : 18 = :3 27 : 6 = :2

8 Complete each pair of equivalent ratios. a 2:3:5 = 4: : b c 1:7:9 = : : 63 d

4 : 12 :16 = :6: 22 : 110 : 66 = 11:

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d 3:7 = : 21 h 20 : 50 = : 25 l 121 : 66 = :6

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11 During a recent dry spell, it rained on only 3 days during the month of September. a What was the ratio of wet days to total days for the month of September? b Write equivalent ratios for a total of 10 days and 100 days. 12 On their way to work, Andrew passes 15 sets of traffic lights and Pauline passes 10 sets of traffic lights. One morning Andrew was stopped by 12 red traffic lights. How many green traffic lights would Pauline need to pass through to have the equivalent ratio of red to green traffic lights as Andrew? 13 Write the ratio of vowels to consonants for each of the following words. a Queensland b Canberra c Wagga Wagga d Australia 14 Name any Australian states that have a vowel to consonant ratio of 1 : 1.

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c

y : 2x = ___ : 8x

d 12xy : 6y = ___ : 1

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6 In a box of 40 ﬂavoured icy poles there were 13 green, 9 lemonade, 11 raspberry and 7 orange icy poles. Write down the ratio of: a green icy poles to orange icy poles b raspberry icy poles to lemonade icy poles c the four different ﬂavours of icy poles d green and orange icy poles to raspberry and lemonade icy poles

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Enrichment: Area ratios 19 Using the dimensions provided, find the ratio of the shaded area to the unshaded area for each of the following diagrams. a b 20 cm 10 cm 4 cm 2 cm

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20 Estimate the ratio of the shaded ﬂoor to the unshaded ﬂoor in this photo.

21 Design your own diagram for which the ratio of shaded area to unshaded area is: a 1:3 b 1:7

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5B Simplifying ratios In a similar way to fractions, ratios are simplified by dividing each term by a common factor. A ratio is said to be in its simplest form when it contains whole numbers only and the highest common factor (HCF) between the terms in the ratio is 1. If a ratio contains fractions or decimals, it can be simplified by multiplying rather than dividing or cancelling.

Let’s start: Class ratios

Key ideas

Look around your classroom and write down the following ratios: a Ratio of girls to boys b Ratio of teachers to students c Ratio of wearing a watch to not wearing a watch d Ratio of white socks to black socks e Ratio of textbooks open to textbooks closed f Ratio of not having a pencil case to having a pencil case g Ratio of blonde hair to brown hair to black hair h Ratio of blue eyes to brown eyes to other colour eyes Design your own ratio question for your class or classroom. Can any of your ratio answers be simplified?

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Simplifying ratios A ratio is simplified by dividing both numbers in the ratio by their highest common factor (HCF). For example, the ratio 15 : 25 can be simplified to 3 : 5. ÷5 15 : 25 = 3 : 5

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÷5 It is convention to express ratios in their simplest form. Ratios in simplest form use whole numbers only. If a ratio is expressed with fractions, it is simplified by converting the quantities to whole numbers. This is generally done by multiplying by the lowest common denominator (LCD). Before ratios are simplified the quantities must be expressed in the same unit.

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Example 3 Simplifying ratios Simplify the following ratios. a 7 : 21

b

SOLUTION

450 : 200 EXPLANATION

÷7 a

7 : 21 = 1 : 3

HCF of 7 and 21 is 7. Divide both quantities by 7.

÷7 ÷ 50 b 450 : 200 = 9 : 4

HCF of 450 and 200 is 50. Divide both quantities by 50.

÷ 50 450 ÷ 200 = 2.25 9 2.25 Frac = 4 Simplified ratio is 9 : 4

Calculators can assist with simplifying ratios. Convert 2.25 to a fraction. The fraction indicates the simplified ratio is 9 : 4.

Example 4 Simplifying ratios involving fractions Simplify the following ratios. 3 1 a : 5 2 SOLUTION

b

1 1 2 :1 3 4 EXPLANATION

× 10 a

3 1 : = 6:5 5 2 × 10

b

1 1 7 5 2 :1 = : 3 4 3 4 × 12 7 5 : = 28 : 15 3 4 × 12

LCD of 5 and 2 is 10. Multiply both quantities by 10.

Convert mixed numerals to improper fractions. LCD of 3 and 4 is 12. Multiply both quantities by 12. Using a calculator to assist: 1 1 28 2 ÷ 1 = . Therefore simplifie lif d ratio is 28:15. lifie 2 4 15

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Example 5 Changing quantities to the same units First change the quantities to the same unit, and then express each pair of quantities as a ratio in simplest form. a 4 mm to 2 cm b 25 minutes to 2 hours EXPLANATION

a 4 mm to 2 cm = 4 mm to 20 mm = 4 : 20 = 1:5

2 cm = 20 mm Once in same unit, write as a ratio. Simplify ratio by dividing by HCF of 4.

b 25 minutes to 2 hours = 25 minutes to 120 minutes = 25 : 120 = 5 : 24

2 hours = 120 minutes Once in same unit, write as a ratio. Simplify ratio by dividing by HCF of 5.

Exercise 5B

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2 To express the ratio 4 : 16 in simplest form, you would: A multiply both quantities by 2 B subtract 4 from both quantities C divide both quantities by 2 D divide both quantities by 4 3 Decide which of the following ratios is not written in simplest form. A 1: 5 B 3: 9 C 2:5 D 11 : 17 4 Decide which of the following ratios is written in simplest form. A 2 : 28 B 15 : 75 C 14 : 45 D 13 : 39

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4 : 24 21 : 28 45 : 35 300 : 550 200 : 125

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c g

42 : 60 : 12 270 : 420 : 60

d 85 : 35 : 15 h 24 : 48 : 84

6 : 18 24 : 80 81 : 27 150 : 75 90 : 75

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5 Simplify the following ratios. a 2:8 b 10 : 50 e 8 : 10 f 25 : 40 i 18 : 14 j 26 : 13 m 51 : 17 n 20 : 180 q 1200 : 100 r 70 : 420

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8 Simplify the following ratios. Use a calculator to check your answers. 1 3 1 3 1 3 1 2 a 1 : b 2 : c 3 :1 d 4 :3 3 4 2 4 5 5 3 5

Example 5

9 First change the quantities to the same unit, and then express each pair of quantities as a ratio in simplest form. a 12 mm to 3 cm b 7 cm to 5 mm c 120 m to 1 km d 60 mm to 2.1 m e 3 kg to 450 g f 200 g to 2.5 kg g 2 tonnes to 440 kg h 1.25 L to 250 mL i 400 mL to 1 L j 20 minutes to 2 hours k 3 hours to 15 minutes l 3 days to 8 hours m 180 minutes to 2 days n 8 months to 3 years o 4 days to 4 weeks p 8 weeks to 12 days q 50 cents to $4 r $7.50 to 25 cents U

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11 Kwok was absent from school on 8 days in Term 1. There were 44 school days in Term 1. Write the following ratios in simplest form. a Ratio of days absent to total days b Ratio of days present to total days c Ratio of days absent to days present 12 Over the past four weeks, the Paske family had eaten takeaway food for dinner on eight occasions. They had hamburgers twice, fish and chips three times and pizza three times. Every other night they had home-cooked dinners. Write the following ratios in simplest form. a Ratio of hamburgers to fish and chips to pizza b Ratio of fish and chips to pizza c Ratio of takeaway dinners to home-cooked dinners d Ratio of home-cooked dinners to total dinners 13 When Lisa makes fruit salad for her family, she uses 5 bananas, 5 apples, 2 passionfruit, 4 oranges, 3 pears, 1 lemon (for juice) and 20 strawberries. a Write the ratio of the fruits in Lisa’s fruit salad. b Lisa wanted to make four times the amount of fruit salad to take to a party. Write an equivalent ratio that shows how many of each fruit Lisa would need. c Write these ratios in simplest form: i bananas to strawberries ii strawberries to other fruits

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16 a Write two quantities of time, in different units, which have a ratio of 2 : 5. b Write two quantities of distance, in different units, which have a ratio of 4 : 3. c a : a2 f 24xyz : 60yz

Enrichment: Aspect ratios 18 Aspect ratio is the relationship between the width and height of the image as displayed on a screen.

A high-deﬁ nition television screen (left) compared to the same image on an analogue television screen (right).

a A standard analogue television has an aspect ratio of 1.3 : 1 . Write this ratio in simplest form. b A high definition digital television has an aspect ratio of 1.7 : 1 . Write this ratio in simplest form. c Although these two ratios are clearly not equivalent, there is a connection between the two. What is it? d The digital television aspect ratio of 1.7 : 1 was chosen as the best compromise to show widescreen movies on television. Many major films are presented in Panavision, which has an aspect ratio of 2.35 : 1. Write this ratio in simplest form. e Investigate the history of aspect ratio in films and television. f Research how formats of unequal ratios are converted through the techniques of cropping, zooming, letterboxing, pillarboxing or distorting. g Investigate aspect ratios of other common media. i Find the aspect ratio for several different newspapers. ii Find the aspect ratio for several common sizes of photographs. iii Find the aspect ratio for a piece of A4 paper (a surd is involved!) © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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15 Mariah has $4 and Rogan has 50 cents. To write a ratio in simplest form, values must be written in the same unit. a First convert both units to dollars, and then express the ratio of Mariah’s money to Rogan’s money in simplest form. b As an alternative, convert both units to cents and express the ratio in simplest form. Do you arrive at the same simplified ratio?

17 Simplify the following ratios. a 2a : 4b b 15x : 3y d 5f : 24f e hk : 3k

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5C Dividing a quantity in a given ratio Ratios are closely connected to fractions. To help solve problems involving ratios, we can think of each quantity in terms of a ‘number of parts’. A drink made in the ratio of cordial to water of 1 : 4 means that there is ‘1 part’ of cordial and ‘4 parts’ of water. There is a total of ‘5 parts’ in the drink. In terms of ratios, it does not matter whether the drink is a 250 mL glass, a 2 L bottle or a 50 L urn. Changing the size of the drink will not change the ratio of ‘parts’; it will simply change the size of each part. The fraction of cordial in the drink is The fraction of water in the drink is

number of cordial parts 1 = . 5 total number of parts

number of water parts 4 = . total number of parts 5

Thinking about ratios in terms of ‘parts’ helps us to divide quantities into particular ratios.

Let’s start: Sharing $120

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Think of a ratio in terms of ‘parts’. A ratio of 2 : 3 has 2 parts of one quantity for every 3 parts of another quantity and a total of 5 parts. Using the unitary method to divide a quantity in a given ratio: – Find the total number of parts in the ratio. – Find the value of one part. – Find the value of the number of parts required in the ratio. For example: Share $20 in ratio of 2 : 3. Think of sharing $20 into 2 parts and 3 parts. ÷ 5 $$20 = 5 parts ÷ 5 Total number of parts = 2 + 3 = 5. $$4 = 1 part × 2 ×2 Value of one part = $20 ÷ 5 = $4. $$8 = 2 parts Therefore 2 parts = $8, and 3 parts = $12.

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Key ideas

For each situation below you have $120 to share out in the given ratios. With a partner, work out how much each person receives. Situation 1: Two people, Jack and Pippa, ratio of 1 : 1 Situation 2: Two people, Andrew and Alex, ratio of 2 : 2 Situation 3: Two people, Jess and Teresa, ratio of 2 : 3 Situation 4: Two people, Kyle and Kara, ratio of 3 : 5 Situation 5: Three people, Matt, Leos and Djarrin, ratio of 1 : 1 : 1 Situation 6: Three people, Christine, Prue and Carolyn, ratio of 3 : 5 : 7 Discuss the strategy that you used to share the $120. Does your strategy work for every situation?

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Using fractions to divide a quantity in a given ratio: number in ratio – Fraction of the amount required = total number of parts – Calculate the fraction of the amount for each share of the ratio. For example, share $20 in ratio of 2 : 3. 2 3 Fractions of the amount required are and . 5 5 2 3 Therefore of $20 = $8 and of $20 = $12. 5 5 To find a total quantity from a given ratio: – Use the concept of ‘parts’ and the unitary method to find the value of one part and therefore the value of the total parts can be calculated. Or – Use equivalent ratios to find the value of each quantity in the ratio and then add the numbers together to find the total.

Example 6 Dividing a quantity in a particular ratio Divide 54 m in a ratio of 4 : 5. SOLUTION Unitary method Total number of parts = 9 ÷ 9 9 parts = 54 m ÷ 9 1 part = 6 m 1 part = 6 m ×4 ×4 ×5 4 parts = 24 m 5 parts = 30 m 54 m divided in a ratio of 4 : 5 is 24 m and 30 m. Fractions method 4 4 54 of 54 = × = 24 9 9 1 5 5 54 of 54 = × = 30 9 9 1 54 m divided in a ratio of 4 : 5 is 24 m and 30 m.

EXPLANATION Total number of parts = 4 + 5 = 9 Value of 1 part = 54 m ÷ 9 = 6 m ×5

Check numbers add to total: 24 + 30 = 54 Fraction =

number in ratio total number of parts

Check numbers add to total: 24 + 30 = 54

An example of dividing a quantity (a length of steel) in a particular ratio.

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Example 7 Dividing a quantity in a ratio with three terms Divide $300 in the ratio of 2 : 1 : 3. SOLUTION

EXPLANATION

Unitary method Total number of parts = 6 ÷ 6 6 parts = $300 ÷ 6 1 part = $50 1 part = $50 ×2 ×2 ×3 × 3 2 parts = $100 3 parts = $150 $300 divided in a ratio of 2 : 1 : 3 is $100, $50 and $150. Fractions method 2 2 300 of 300 = × = 100 6 6 1 1 1 300 of 300 = × = 50 6 6 1 3 3 300 of 300 = × = 150 6 6 1 $300 divided in a ratio of 2 : 1 : 3 is $100, $50 and $150.

Total number of parts = 2 + 1 + 3 = 6 Value of 1 part = $300 ÷ 6 = $50

Check numbers add to total: $100 + $50 + $150 = $300 Fraction =

number in ratio total number of parts

Check numbers add to total: $100 + $50 + $150 = $300

Example 8 Finding a total quantity from a given ratio The ratio of boys to girls at Birdsville College is 2 : 3. If there are 246 boys at the school, how many students attend Birdsville College? SOLUTION

EXPLANATION

Unitary method ÷ 2 2 parts = 246 ÷ 2 1 part = 123 ×5 ×5 5 parts = 615 615 students attend Birdsville College.

Ratio of boys: girls is 2 : 3 Boys have ‘2 parts’ = 246 Value of 1 part = 246 ÷ 2 = 123 Total parts = 2 + 3 = 5 parts 5 parts = 5 × 123 = 615

Equivalent ratios method boys : girls × 123 × 123 = 2 : 3 = 246 : 369 615 students attend Birdsville College.

Use equivalent ratios. Multiply each quantity by 123. Total number of students = 246 boys + 369 girls = 615

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4 What fraction of each of the drinks above is cordial?

c $1000 in the ratio of 3 : 17 f 360 kg in the ratio of 5 : 7 i 155 m in the ratio of 4 : 1

6 Share $400 in the ratio: a 1: 3 b 2:3

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7 Share $6 600 in the ratio: a 4: 7 b 2: 3

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8 Divide: a $200 in the ratio of 1 : 2 : 2 c 12 kg in the ratio of 1 : 2 : 3 e 320 kg in the ratio of 12 : 13 : 15

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9 Share 600 lollies in the ratio: a 1: 9 b 2 :1: 3

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b $110 in the ratio of 7 : 4 e 14 kg in the ratio of 4 : 3 h 40 m in the ratio of 3 : 5

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11 The angles of a triangle are in the ratio of 2 : 3 : 4. Find the size of each angle. 12 Three friends, Cam, Molly and Seb, share a prize of $750 in a ratio of 3 : 4 : 8. How much more money does Seb receive than Cam? 13 Trudy and Bella made 80 biscuits. If Trudy made 3 biscuits in the time that Bella made 2 biscuits, how many biscuits did Trudy make? 14 In Year 8, the ratio of boys to girls is 5 : 7. If there are 140 girls in Year 8, what is the total number of students in Year 8? 15 A light rye bread requires a ratio of wholemeal ﬂour to plain ﬂour of 4 : 3. A baker making a large quantity of loaves uses 126 cups of plain ﬂour. What is the total amount of ﬂour used by the baker? 16 A textbook contains three chapters and the ratio of pages in these chapters is 3 : 2 : 5. If there are 24 pages in the smallest chapter, how many pages are in the textbook?

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Enrichment: A fair share 19 Three students, Ramshid, Tony and Maria, entered a group Geography competition. • Ramshid spent 5 hours preparing the PowerPoint presentation. • Tony spent 3 hours researching the topic. 1 • Maria spent 2 hours designing the poster. 2 Their group won second prize in the competition and received a prize of $250. Ramshid, Tony and Maria decide to share the prize in a ratio of 3 : 2 : 1. a How much money did each student receive? Round the answer to nearest cent. b If the prize was divided up according to the time spent on the project, what would be the new ratio? Write the ratio in whole numbers. c How much money did each student receive with the new ratio? Round the answer to the nearest cent. d Although she spent the least time on the project, Maria would prefer to divide the prize according to time spent. How much more money did Maria receive with the new ratio? e Tony preferred that the original ratio remained. How much was Tony better off using the original ratio? f Which ratio would Ramshid prefer and why? g The group ended up going with a ratio based on time spent but then rounded amounts to the nearest $10. How much prize money did each student receive?

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5D Scale drawings Scale drawings are a special application of ratios. The purpose of a scale drawing is to provide accurate information about an object which has dimensions that are either too large or too small to be shown on a page. If a scale drawing has a ratio of 1 : 1, then the drawing would be exactly the same size as the actual (real-life) object. For example, it is not practical to have a map that is the same size as the actual distance you need to travel, so a much smaller map (a scaled drawing) is used. The map shows a scale to indicate the relationship of the map distance to the actual distance. The scale is displayed as a ratio. Three common travel maps with their scales are shown below. Wyong

Liverpool

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North Harbour

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Bouddi NP Broken Bay Pacific Palm Ocean Ku-ring-gai Beach Chase NP Mona Vale Hornsby

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Let’s start: Scaling down and scaling up

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• Brainstorm specific examples where you have come across scale drawings. Think of examples where a scale drawing has been used to show very large objects or distances, and also think of examples where a scale drawing has been used to show very small objects. • Share your list of examples with your partner. • As a class, make a list on the board of specific examples of scale drawings. • What are some examples of common scales that are used?

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A scale drawing has exactly the same shape as the original object, but a different size. All scale drawings should indicate the scale ratio. The scale on a drawing is written as a scale ratio: Drawing length : Actual length Drawing length represents the length on the diagram, map or model. Actual length represents the real length of the object. A scale ratio of 1 : 100 means the actual or real lengths are 100 times greater than the lengths measured on the diagram. A scale ratio of 20 : 1 means the scaled lengths on the model are 20 times greater than the actual or real lengths. It is helpful if scales begin with a 1. Then the second number in the ratio can be referred to as the scale factor. Scale ratios that do not start with a 1 can be converted using equivalent ratios. To convert a scaled distance to an actual distance you multiply by the scale factor. To convert an actual (real) distance to a scaled distance you divide by the scale factor. Multiply

Scale distance

Actual distance

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Conversion of units. It is important to remember how to correctly convert measurements of length when × 1000 dealing with scales. 1 km = 1000 m 1 m = 100 cm × 100 1 cm = 10 mm

km

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Example 9 Converting scale distance to actual distance A map has a scale of 1 : 20 000. Find the actual distance for each scaled distance. a 2 cm b 5 mm

c

14.3 cm

SOLUTION

EXPLANATION

a Actual distance = 2 cm × 20 000 = 40 000 cm = 400 m (or 0.4 km)

Scale factor = 20 000 Multiply scaled distance by scale factor. Convert answer into sensible units.

b Actual distance = 5 mm × 20 000 = 100 000 mm = 100 m

Multiply scaled distance by scale factor. Convert answer into sensible units.

c Actual distance = 14.3 cm × 20 000 = 286 000 cm = 2.86 km

Multiply scaled distance by scale factor. Shortcut: × 2, then × 10 000 Convert answer into sensible units.

Example 10 Converting actual distance to scaled distance A model boat has a scale of 1:500. Find the scaled length for each of these actual lengths. a 50 m b 75 cm

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4550 mm

SOLUTION

EXPLANATION

a Scaled distance = 50 m ÷ 500 = 5000 cm ÷ 500 = 10 cm (0.1 m)

Scale factor = 500 Convert metres to centimetres. Divide actual distance by scale factor. Convert answer into sensible units.

b Scaled distance = 75 cm ÷ 500 = 750 mm ÷ 500 = 1.5 mm

Convert centimetres to millimetres. Divide actual distance by scale factor.

c Scaled distance = 4550 mm ÷ 500 = 45.5 mm ÷ 5 = 9.1 mm

Divide actual distance by scale factor. Shortcut: ÷ 100, then ÷ 5 (or vice versa)

If the model is 90 cm long and the scale is 1:500, how long is the actual boat?

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Example 11 Determining the scale factor State the scale factor in the following situations. a 4 mm on a scale drawing represents an actual distance of 50 cm. b An actual length of 0.1 mm is represented by 3 cm on a scaled drawing. SOLUTION

EXPLANATION

a Scale ratio = 4 mm : 50 cm = 4 mm : 500 mm Scale ratio = 4 : 500 = 1 : 125 Scale factor = 125

Write the ratio drawing length : actual length. Convert to ‘like’ units. Write the scale ratio without units. Divide both numbers by 4. Ratio is now in the form 1 : scale factor. The actual size is 125 times larger than the scaled drawing.

b Scale ratio = 3 cm : 0.1 mm = 30 mm : 0.1 mm Scale ratio = 30 : 0.1 = 300 : 1 1 = 1: 300 1 Scale factor = 300

Write the ratio drawing length : actual length. Convert to ‘like’ units. Write the scale ratio without units. Multiply both numbers by 10. Divide both numbers by 300. Ratio is now in the form 1: scale factor. The actual size is 300 times smaller than the scaled drawing.

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b m

c km

4 Convert 560 m to: a km

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60 cm 75 cm 736.5 m 400 mm 1250 m 6.15 m 0.2 mm 8.2 mm

7 Determine the scale ratio for each of the following. a 2 mm on a scale drawing represents an actual distance of 50 cm. b 4 cm on a scale drawing represents an actual distance of 2 km. c 1.2 cm on a scale drawing represents an actual distance of 0.6 km. d 5 cm on a scale drawing represents an actual distance of 900 m. e An actual length of 7 mm is represented by 4.9 cm on a scaled drawing. f An actual length of 0.2 mm is represented by 12 cm on a scaled drawing. 8 Convert the two measurements provided in each scale into the same unit and then write the scale as a ratio of two numbers in simplest form. a 2 cm : 200 m b 5 mm : 500 cm c 12 mm : 360 cm d 4 mm : 600 m e 4 cm : 5 m f 1 cm : 2 km g 28 mm : 2800 m h 3 cm : 0.6 mm i 1.1 m : 0.11 mm

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10 Blackbottle and Toowoola are 17 cm apart on a map with a scale of 1 : 50 000. How far apart are the towns in real life?

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9 A model city has a scale ratio of 1 : 1000. a Find the actual height of a skyscraper that has a scaled height of 8 cm. b Find the scaled length of a train platform that is 45 m long in real life.

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6 Find the scaled length for each of these actual lengths. a Scale 1 : 200 i 200 m ii 4 km b Scale 1 : 500 i 10 000 m ii 1 km c Scale 1 : 10 000 i 1350 m ii 45 km d Scale 1 : 20 000 i 12 km ii 1800 m e Scale 1 : 250 000 i 5000 km ii 750 000 m f Scale 1 : 10 i 23 m ii 165 cm g Scale 1 : 0.1 i 30 cm ii 5 mm h Scale 200 : 1 i 1 mm ii 7.5 cm

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5 Find the actual distance for each of the following scaled distances. Give your answer in an appropriate unit. a Scale 1 : 10 000 i 2 cm ii 4 mm iii 7.3 cm b Scale 1 : 20 000 i 80 cm ii 18.5 mm iii 1.25 m c Scale 1 : 400 i 16 mm ii 72 cm iii 0.03 m d Scale 1 : 300 i 5 mm ii 8.2 cm iii 7.1 m e Scale 1 : 2 i 44 m ii 310 cm iii 2.5 mm f Scale 1 : 5 i 4m ii 24 cm iii 155 mm g Scale 1 : 0.5 i 12 cm ii 3.2 mm iii 400 m h Scale 100 : 1 i 3 cm ii 11.5 km iii 81.5 cm

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11 Using the house plans shown, state the actual dimensions of the following rooms. a Bedroom 1 b Family room c Patio

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12 From the scaled drawing below, calculate the actual length and height of the truck, giving your answer to the nearest metre.

13 The photograph on the right shows Jackson enjoying a walk with his dog. Jackson is 1.8 m tall in real life. a What is the scale of the photograph? b How tall is Jackson’s dog in real life? 1 14 A scale length of 5 cm is equal to an actual distance of 44 km. 2 a How many kilometres does a scale length of 3 cm equal? b How many kilometres does a scale length of 20 cm equal? c How many centimetres does an actual distance of 100 km equal?

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17 A map maker wants to make a map of a mountanious area that is 64 km2 (side length of 8 km). She decides to use a scale of 1 : 1000. Describe the problem she is going to have. 18 A scientist says that the image of a 1 mm long insect is magnified by a scale of 1:1000 through his magnifying glass. Explain what might be wrong with the scientist’s ratio. 19 Obtain a map of Australia. a What is the scale of your map? b Which two capital cities in Australia are the furthest apart? State the distance. c Which two capital cities in Australia are closest together? State the distance. d Which capital city is furthest away from Sydney? State the distance. e Which capital city is closest to Adelaide? State the distance. f Check the accuracy of your answers on the internet.

Enrichment: Scaled drawing 20 Design a scaled drawing of one of the following: • Your classroom and the associated furniture • Your bedroom and the associated furniture • A room in your house (bathroom, family room, garage ...) Your scaled drawing should fit comfortably onto an A4 page. a Measure the dimensions of your chosen room and the dimensions of an A4 page, and determine the most appropriate scale for your diagram. b Show size and location of doors and windows where appropriate. c Produce a second scale diagram of your room, but show a different arrangement of the furniture. Can you produce the scale diagram on a computer software package?

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16 Which of the following scales would be most appropriate if you wanted to make a scale model of: a a car? b your school grounds? c Mt Kosciuszko? A 1 : 10 000 B 1 : 1000 C 1 : 100 D 1 : 10

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5E Introducing rates If you were to monitor what you said each day, you might well find that you speak about rates many times per day! A ratio shows the relationship between the same type of quantities with the same units, but a rate shows the relationship between two different types of quantities with different units. The following are all examples of rates: • Cost of petrol was $1.45 per litre. • Rump steak was on special for $18/kg. • Dad drove to school at an average speed of 52 km/h. • After the match, your heart rate was 140 beats/minute.

The top readout on this heart monitor shows heart rate in beats per minute.

Unlike ratios, a rate compares different types of quantities, and so units must be shown. For example: • The ratio of boys to girls in a school was 4 : 5. • The average rate of growth of an adolescent boy is 12 cm/year.

Let’s start: State the rate For each of the following statements, write down a corresponding rate. • The Lodges travelled 400 km in 5 hours. • Gary was paid $98 for a 4-hour shift at work. • Felicity spent $600 on a two-day shopping spree. • Max had grown 9 cm in the last three months. • Vuong paid $37 for half a cubic metre of crushed rock. • Paul cycled a total distance of 350 km for the week. What was the rate (in questions/minute) at which you answered these questions?

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Rates compare quantities measured in different units. All rates must include units for each quantity. The two different units are separated by a slash ‘/’, which is the mathematical symbol for ‘per’. For example: 20 km/h = 20 km per hour = 20 km for each hour It is convention to write rates in their simplest form. This involves writing the rate for only one unit of the second quantity. For example: spent $45 in 3 hours = $45 in 3 hours ← Non simplified rate ÷3

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÷3

= $15 in 1 hour

← Simplified rate ($15/h)

The average rate is calculated by dividing the total change in one quantity by the total change in the second quantity. For example: reading a 400-page book in 4 days Average reading rate = 400 pages in 4 days ÷4

= 100 pages in 1 day

÷4

Therefore average reading rate = 100 pages/day.

Example 12 Writing simplified rates Express each of the following as a simplified rate. a 12 students for two teachers b $28 for 4 kilograms SOLUTION

EXPLANATION

a 6 students/teacher

12 students for 2 teachers ÷2

÷2 6 students for 1 teacher

b $7/kg

$28 for 4 kg ÷4

÷4 $7 for 1 kg

Example 13 Finding average rates Tom was 120 cm tall when he turned 10 years old. He was 185 cm tall when he turned 20 years old. Find Tom’s average rate of growth per year between 10 and 20 years of age. SOLUTION

EXPLANATION

Average rate = 65 cm/10 years = 6.5 cm/1 year Average rate of growth = 6.5 cm/year.

Growth = 185 – 120 = 65 cm Divide both numbers by 10.

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1 Which of the following are examples of rates? A $5.50 B 180 mL/min C $60/h

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90 people/day

Speed of a car

$2100/m2

Cost of building a new home

68 km/h

Population growth

64 beats/min

Resting heart rate

$15/h

3 Write typical units for each of the following rates. a Price of sausages b Petrol costs c Typing speed d Goal conversion rate e Energy nutrition information f Water usage in the shower g Pain relief medication h Cricket team’s run rate

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5 Find the average rate of change for each situation. a Relma drove 6000 kilometres in 20 days. b Holly saved $420 over three years. c A cricket team scored 78 runs in 12 overs. d Saskia grew 120 centimetres in 16 years. e Russell gained 6 kilograms in 4 years. f The temperature dropped 5ºC in 2 hours.

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4 Write each of the following as a simplified rate. a 12 days in 4 years b 15 goals in 3 games c $180 in 6 hours d $17.50 for 5 kilograms e $126 000 to purchase 9 acres f 36 000 cans in 8 hours g 12 000 revolutions in 10 minutes h 80 mm rainfall in 5 days i 60 minutes to run 15 kilometres j 15 kilometres run in 60 minutes

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6 A dripping tap filled a 9-litre bucket in 3 hours. a What was the dripping rate of the tap in litres/hour? b How long would it take the tap to fill a 21-litre bucket?

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8 If 30 salad rolls were bought to feed 20 people at a picnic, and the total cost was $120, find the following rates. a Salad rolls/person b Cost/person c Cost/roll 9 The number of hours of sunshine was recorded each day for one week in April. The results are listed. Monday 6 hours, Tuesday 8 hours, Wednesday 3 hours, Thursday 5 hours Friday 7 hours, Saturday 6 hours, Sunday 7 hours a Find the average number of hours of sunshine: i per weekday ii per weekend day iii per week iv per day b Given the above rates, how many hours of sunshine would you expect in April?

10 Harvey finished a 10 kilometre race in 37 minutes and 30 seconds. Jacques finished a 16 kilometre race in 53 minutes and 20 seconds. Calculate the running rate of each runner. Which runner had a faster running pace? 11 The Tungamah Football Club had 12 000 members. After five successful years and two premierships, they now have 18 000 members. a What has been the average rate of membership growth per year for the past 5 years? b If this membership growth rate continues, how many more years will it take for the club to have 32 000 members?

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7 Martine grew at an average rate of 6 cm/year for the first 18 years of her life. If Martine was 50 cm long when she was born, how tall was Martine when she turned 18?

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12 a A car uses 24 L of petrol to travel 216 km. Express these quantities as a simplified rate in: i km/L ii L/km b How can you convert km/L to L/km?

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Connectplus, a rival telecommunications company, charges a constant call rate of 60¢/minute. c If you normally made calls that were 15 minutes long, which company has the better deal for you? d If you normally made calls that were 25 minutes long, which company has the better deal for you? e What is the length of phone call for which both companies would charge the same amount?

Enrichment: Target 155 14 In Victoria, due to drought conditions, the state government urged all residents to save water. The goal was set for each Victorian to use no more than 155 litres of water per day. a How many people live in your household? b According to the Victorian government, how many litres of water can your household use per day? c Perform some experiments and calculate the following rates of water ﬂow: i Shower rate (L/min) ii Washing machine (L/load) iii Hose (L/min) iv Toilet (L/ﬂush, L/half ﬂush) v Running tap (L/min) vi Drinking water (L/day) vii Dishwasher (L/wash) viii Water for food preparation (L/day) d Estimate the average daily rate of water usage for your household. e Ask your parents for a recent water bill and find out what your family household water usage rate was for the past three months. Before the initiative, Victorians were using an average of 164 litres/day/person. Twelve months after the initiative, Victorians were using 151 litres/day/person. f How much water per year for the state of Victoria does this saving of 13 litres/day/person represent? g What is the rate at which your family is charged for its water?

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13 The Teleconnect telecommunications company has a variable call charge rate for phone calls of up to 30 minutes. The charges are 50c/min for first 10 minutes, 75c/min for the second 10 minutes and $1/min for the third 10 minutes. a Find the cost of phone calls of these given lengths. i 8 minutes ii 13 minutes iii 24 minutes iv 30 minutes b What is the average charge rate per minute for a 30 minute call?

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5F Ratios and rates and the unitary method The concept of solving problems using the unitary method was introduced in Chapter 4. The unitary method involves finding the value of ‘one unit’. This is done by dividing the amount by the given quantity. Once the value of ‘one unit’ is known, multiplying will find the value of the number of units required.

Let’s start: Finding the value of ‘ 1 unit ’ For each of the following, find the value of 1 unit or 1 item. • 8 basketballs cost $200. • 4 cricket bats cost $316. • 5 kg of watermelon cost $7.50. For each of the following, find the rate per 1 unit. • Car travelled 140 km in 2 hours. • 1000 L of water leaked out of the tank in 8 hours. • $51 was the price paid for 3 kg of John Dory fish.

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The unitary method involves finding the value of ‘one unit’ and then using this information to solve the problem. When dealing with ratios, find the value of 1 ‘part’ of the ratio. For example: Ratio of cars to motorbikes is 35:2 and there are 6 motorbikes. 2 parts = 6 motorbikes ÷2 ÷2 1 part = 3 motorbikes When dealing with rates, find the value of the rate per 1 ‘unit’. For example: Pedro earned $64 for a 4-hour shift at work. Therefore, wage rate = $64 per 4 hours = $16 per hour = $16/h. Once the value of one ‘part’ or the rate per one ‘unit’ is known, the value of any number of parts or units can be found by multiplying. The technique of dividing and/or multiplying values in successive one-step calculations can be applied to the concept of converting rates from a set of given units to a different set of units.

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Key ideas

For each of the following, find the value of 1 ‘part’. • Ratio of books to magazines read was 2 : 5. Milli had read 14 books. • Ratio of pink to red ﬂowers is 7 : 11. A total of 330 red ﬂowers are in bloom. • Ratio of girls to boys is 8 : 5. There are 40 girls in a group.

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Example 14 Reviewing the unitary method Andy travels 105 km in 7 identical car trips from home to school. How far would she travel in 11 such car trips? SOLUTION

EXPLANATION

7 car trips = 105 km 1 car trip = 15 km × 11 11 car trips = 165 km

÷7

÷7

× 11

Find the value of 1 unit by dividing both quantities by 7. Solve the problem by multiplying both quantities by 11.

Andy travels 165 km.

Example 15 Solving ratio problems using the unitary method The ratio of apples to oranges is 3:5. If there are 18 apples, how many oranges are there? SOLUTION 3 parts = 18 apples 1 part = 6 pieces × 5 5 parts = 30 oranges There are 30 oranges. ÷3

EXPLANATION Apples = 3 ‘parts’, Oranges = 5 ‘parts’ Need to find the value of 1 ‘part’. To find 5 ‘parts’ multiply the value of 1 ‘part’ by 5.

÷3 ×5

Example 16 Solving rate problems using the unitary method A truck uses 4 L of petrol to travel 36 km. How far will it travel if it uses 70 L of petrol? SOLUTION

EXPLANATION

Rate of petrol consumption ÷ 4 36 km for 4 L ÷4 9 km for 1 L × 70 × 70 630 km for 70 L Truck will travel 630 km on 70 L.

Find the petrol consumption rate of 1 unit by dividing both quantities by 4. Solve the problem by multiplying both quantities by 70.

Example 17 Converting units using the unitary method Melissa works at the local supermarket and earns $57.60 for a 4-hour shift. How much does she earn in c/min? SOLUTION Wage = $57.60 for 4 hours ÷ 4 $14.40 for 1 hour ÷ 60

1440c for 60 minutes 24c for 1 minute

EXPLANATION ÷4 ÷ 60

Write down Melissa’s wage rate. Find the rate of $ per 1 hour. Convert $ to cents and hours to minutes. Divide rate by 60 to find rate of cents per minute.

Melissa earns 24c/min.

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6 Solve the following rate problems. a A tap is dripping at a rate of 200 mL every 5 minutes. How much water drips in 13 minutes? b A professional footballer scores an average of 3 goals every 6 games. How many goals is he likely to score in a full season of 22 games? c A snail travelling at a constant speed travels 400 mm in 8 minutes. How far does it travel in 7 minutes? d A computer processor can process 500 000 kilobytes of information in 4 seconds. How much information can it process in 15 seconds? 7 Leonie, Spencer and Mackenzie have just won a prize. They decide to share it in the ratio of 4 : 3 : 2. If Spencer receives $450, how much do Leonie and Mackenzie receive, and what was the total value of the prize?

Example 17

8 Convert the following rates into the units given in the brackets. a $15/h (c/min) b $144/h (c/s) c 3.5 L/min (L/h) d 20 mL/min (L/h) e 0.5 kg/month (kg/year) f 120 g/day (kg/week) g 60 g/c (kg/$) h $38/m (c/mm) i 108 km/h (m/s) j 14 m/s (km/h)

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5 Solve the following ratio problems. a The required staff to student ratio for an excursion is 2 : 15. If 10 teachers attend the excursion, what is the maximum number of students who can attend? b The ratio of commercials to actual show time for a particular TV channel is 2 : 3. How many minutes of actual show were there in 1 hour? c A rectangle has length and breadth dimensions in a ratio of 3 : 1. If a particular rectangle has a length of 21 m, what is its breadth? d Walter and William have a height ratio of 7 : 8. If William has a height of 152 cm, how tall is Walter?

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9 The Mighty Oats breakfast cereal is sold in boxes of three different sizes: small (400 g) for $5.00, medium (600 g) for $7.20, large (750 g) for $8.25 a Find the value of each box in $/100 g. b What is the cheapest way to buy a minimum of 4 kg of the cereal?

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10 Gemma runs 70 metres in 8.4 seconds. If she maintains the same average speed, in what time will she run 100 metres? 11 Zana’s hair grew 6 cm in 5 months. a Find Zana’s average rate of hair growth in cm/month and in m/year. b How long would it take for Zana’s hair to grow 30 cm?

12 Maria can paint 15 m2 in 20 minutes. a What is the rate at which Maria paints in m2/h? b What area can Maria paint in 20 hours? c Maria must paint 1000 m2 in 20 hours. Find the rate at which she will need to paint in m2/min.

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13 If x donuts cost $y: a how much would 1 donut cost? b how much would one dozen donuts cost? c how much would z donuts cost?

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A triangle has side lengths in a ratio of 3:5:4. If the longest side is 35 cm, find the lengths of the other two sides and the perimeter of the triangle. b A triangle has side lengths in a ratio of x:(x + 3):(x – 2). If the shortest side is 17 cm, find the lengths of the other two sides and the perimeter of the triangle.

15 In a faraway galaxy, a thriving alien colony uses the following units: For money they have puks and paks: 1 puk (pu) = 1000 pak (pa) For length they have doits and minidoits: 1 doit (D) = 80 minidoits (mD) Polynaute rope is priced at 4 pu/D. Find the cost of the rope in terms of pa/mD.

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Enrichment: Where will we meet? 16 Phil lives in Perth and his friend Werner lives in Sydney. The distance, by road, between their two houses is 4200 km (rounded to the nearest 100 km). Phil decides to drive to Sydney and Werner decides to drive to Perth. They leave home at the same time and travel the same route, but in opposite directions. Phil drives at a constant speed of 75 km/h and Werner drives at a constant speed of 105 km/h. a Will they meet on the road at a spot closer to Sydney or closer to Perth? b How long will it take Phil to travel to Sydney? c How long will it take Werner to travel to Perth? d State the location of each friend after they have been driving for 15 hours. e At what location (distance from Sydney and/or Perth) will they meet? When they meet, Phil decides to change into Werner’s car and they drive back to Werner’s home at an average speed of 105 km/h. f How long did it take Phil to travel to Sydney? g Design a similar problem for two friends travelling at different constant speeds between two different capital cities in Australia.

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5G Solving rate problems ‘One thing that is certain in life is that change is inevitable.’ We are constantly interested in rates of change and how things change over a period of time. We are also regularly faced with problems involving specific rates. Strong arithmetic skills and knowing whether to multiply or divide by the given rate allows many rate problems to be solved quickly and accurately. Over the next few days, keep a record of any rates you observe or experience. Look out for the slash ‘/’ sign and listen for the ‘per’ word.

Let’s start: Estimate the rate

Key ideas

For each of the following statements, estimate a corresponding rate. • Commercial rate: The number of commercials on TV per hour • Typing rate: Your typing speed in words per minute • Laughing rate: The number of times a teenager laughs per hour • Growth rate: The average growth rate of a child from 0 to15 years of age • Running rate: Your running pace in metres per second • Homework rate: The number of subjects with homework per night • Clapping rate: The standard rate of audience clapping in claps per minute • Thank you rate: The number of opportunities to say thank you per day Compare your rates. Which rate is the ‘highest’? Which rate is the ‘lowest’? Discuss your answers.

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When a rate is provided, a change in one quantity implies that an equivalent change must occur in the other quantity. For example: Patrick earns $20/hour. How much will he earn in 6 hours? ×6

$20 for 1 hour $120 for 6 hours

×6

For example: Patrick earns $20/hour. How long will it take him to earn $70? $20 for 1 hour 1 ×3 2

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$70 for

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Carefully consider the units involved in each question and answer.

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Example 18 Solving rate problems a Rachael can touch type at 74 words/minute. How many words can she type in 15 minutes? b Leanne works in a donut van and sells on average 60 donuts every 15 minutes. How long is it likely to take her to sell 800 donuts? SOLUTION a

× 15

EXPLANATION 74 words in 1 minute 1110 words in 15 minutes

× 15

Rachael can type 1110 words in 15 minutes.

74 ×15 370 740 1110

b

60 donuts in 15 minutes ÷ 15 4 donuts in 1 minute × 200 × 200 800 donuts in 200 minutes ÷ 15

Selling rate = 60 donuts/15 minutes Divide both quantities by HCF of 15.

Leanne is likely to take 3 hours and 20 minutes Multiply both quantities by 200. Convert answer to hours and minutes. to sell 800 donuts.

Example 19 Solving combination rate problems Three water hoses from three different taps are used to fill a large swimming pool. The first hose alone takes 200 hours to fill the pool. The second hose alone takes 120 hours to fill the pool and the third hose alone takes only 50 hours to fill the pool. How long would it take to fill the pool if all three hoses were used? SOLUTION

EXPLANATION

In 600 hours: hose 1 would fill 3 pools hose 2 would fill 5 pools hose 3 would fill 12 pools Therefore in 600 hours the three hoses together would fill 20 pools. Filling rate = 600 h/20 pools ÷ 20 ÷ 20 30 h/pool

LCM of 200, 120 and 50 is 600. Hose 1 = 600 h ÷ 200 h/pool = 3 pools Hose 2 = 600 h ÷ 120 h/pool = 5 pools Hose 3 = 600 h ÷ 50 h/pool = 12 pools Together = 3 + 5 + 12 = 20 pools filled. Simplify rate by dividing by HCF.

It would take 30 hours to fill the pool if all three hoses were used.

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5 A truck travels 7 km per litre of fuel. How many litres are needed for the truck to travel 280 km? 6 Daniel practises his guitar for 40 minutes every day. a How many days will it take him to log up 100 hours of practice? b How many days will it take him to log up 10 000 hours of practice? c If Daniel practises six days per week for 50 weeks per year, how many years will it take him to log up 10 000 hours of practice? 7 A ﬂywheel rotates at a rate of 1500 revolutions per minute. a How many revolutions does the ﬂywheel make in 15 minutes? b How many revolutions does the ﬂywheel make in 15 seconds? c How long does it take for the ﬂywheel to complete 15 000 revolutions? d How long does it take for the ﬂywheel to complete 150 revolutions? 8 Putra is an elite rower. When training, he has a steady working heart rate of 125 beats per minute (bpm). Putra’s resting heart rate is 46 bpm. a How many times does Putra’s heart beat during a 30 minute workout? b How many times does Putra’s heart beat during 30 minutes of ‘rest’? c If his coach says that he can stop his workout once his heart has beaten 10 000 times, for how long would Putra need to train?

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4 Mario is a professional home painter. When painting a new home, he uses an average of 2.5 litres of paint per hour. How many litres of paint would Mario use in a week if he paints for 40 hours? Example 18b

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1 Fill in the gaps. a 60 km in 1 hour ×3 180 km in ___

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9 What is the cost of paving a driveway that is 18 m long and 4 m wide, if the paving costs $35 per square metre?

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10 A saltwater swimming pool requires 2 kg of salt to be added for every 10 000 litres of water. Joan’s swimming pool is 1.5 metres deep, 5 metres wide and 15 metres long. How much salt will she need to add to her pool? 11 The Bionic Woman gives Batman a 12 second start in a 2 kilometre race. If the Bionic Woman runs at 5 km/min, and Batman runs at 3 km/ min, who will win the race and by how many seconds? 12 At a school camp there is enough food for 150 students for 5 days. a How long would the food last if there were only 100 students? b If the food ran out after only 4 days, how many students attended the camp?

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13 Michelle can complete a landscaping job in 6 days and Danielle can complete the same job in 4 days. Working together, in how many days could they complete the job?

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14 Three bricklayers, Maric, Hugh and Ethan, are cladding a new home. If Maric were to work alone, the job would take him 8 days to complete. If Hugh were to work alone, the job would take him 6 days to complete and if Ethan were to work by himself, the job would take him 12 days to complete. a If the three men work together, how long will it take them to complete the job? b What fraction of the house will each bricklayer complete?

16 If it takes 4 workers, 4 hours to dig 4 holes, how long would it take 2 workers to dig 6 holes? 17 State the units required for the answer to each of the following rate calculations. a $205/kg × 48 kg b 62 s × 12 m/s c 500 beats ÷ 65 beats/min (bpm) d 4000 revolutions ÷ 120 revs/min

Enrichment: Value for money 18 Soft drink can be purchased from supermarkets in a variety of sizes. Below are the costs for four different sizes of a certain brand of soft drink. 1.25 L bottles

2 L bottles

10 × 375 mL cans

$2.70 each

$1.60 each

$2.20 each

$6.00 per pack

a Find the economy rate (in $/L) for each size of soft drink. b Find and compare the cost of 30 litres of soft drink purchased entirely in each of the four different sizes. c If you only have $60 to spend on soft drink for a party, what is the difference between the greatest amount and the least amount of soft drink you could buy? Assume you have less than $1.60 left. Most supermarkets now include the economy rate of each item underneath the price tag to allow customers to compare value for money. d Carry out some research at your local supermarket on the economy rates of a particular food item with a range of available sizes (such as drinks, breakfast cereals, sugar, ﬂour). Write a report on your findings.

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15 Four cans of dog food will feed 3 dogs for 1 day. a How many cans are needed to feed 10 dogs for 6 days? b How many dogs can be fed for 9 days from 60 cans? c For how many days will 40 cans feed 2 dogs?

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5H Speed A rate that we come across almost every day is speed. Speed is the rate of distance travelled per unit of time. On most occasions, speed is not constant and therefore we are interested in the average speed of an object. Average speed is calculated by dividing the distance travelled by the time taken. Distance travelled Average speed = Time taken Given the average speed formula, we can tell that all units of speed must have a unit of length in the numerator, followed by a unit of time in the denominator. Therefore ‘mm/h’ is a unit of speed and could be an appropriate unit for the speed of a snail! The most common units of speed are m/s and km/h.

Let’s start: Which is faster?

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Speed is a measure of how fast an object is travelling. If the speed of an object does not change over time, the object is travelling at a constant speed. ‘Cruise control’ helps a car travel at a constant speed. When speed is not constant, due to acceleration or deceleration, we are often interested to know the average speed of the object. Average speed is calculated by the formula: d Distance travelled Average speed = or s= Time taken t Depending on the unknown value, the above formula can be rearranged to make d or t the subject. The three formulas involving s, d, and t are: s=

d t

d=s×t

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Care must be taken with units for speed, and sometimes units will need to be converted. The most common units of speed are m/s and km/h.

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With a partner, determine the faster of the two listed alternatives. a Car A travelling at 10 m/s Car B travelling at 40 km/h b Walker C travelling at 4 km/h Walker D travelling at 100 m/min This snail’s average speed won’t be high, c Jogger E running at 1450 m/h Jogger F running at 3 m/s but it can still be calculated. d Plane G ﬂying at 700 km/h Plane H ﬂying at 11 km/min

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Example 20 Finding average speed Find the average speed in km/h of: a a cyclist who travels 140 km in 5 hours SOLUTION d t 140 km = 5h = 28 km/h

a s=

Alternative unitary method 140 km in 5 hours ÷5 ÷5 28 km in 1 hour

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a runner who travels 3 km in 15 minutes EXPLANATION The unknown value is speed. Write the formula with s as the subject. Distance travelled = 140 km Time taken = 5 h Speed unit is km/h. Write down the rate provided in the question. Divide both quantities by 5.

Average speed = 28 km/h d t 3 km = 15 min 1 = 3÷ 4 = 3 × 4 = 12 km/h Alternative unitary method 3 km in 15 minutes ×4 ×4 12 km in 60 minutes Average speed = 12 km/h

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Distance travelled = 3 km Time taken = 15 min or

1 h 4

1 4 Dividing by , is the same as multiplying by . 4 1 Write down the rate provided in the question. Multiply both quantities by 4.

Example 21 Finding the distance travelled Find the distance travelled by a truck travelling for 15 hours at an average speed of 95 km/h. SOLUTION

EXPLANATION

d =s×t = 95 km/h × 15 h = 1425 km Alternative unitary method 95 km in 1 hour × 15 × 15 1425 km in 15 hours

The unknown value is distance. Write the formula with d as the subject. Distance unit is km. Write the rate provided in the question. Multiply both quantities by 15.

Truck travels 1425 km in 15 hours.

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Number and Algebra

Example 22 Finding the time taken Find the time taken for a hiker walking at 4 km/h to travel 15 km. SOLUTION

EXPLANATION

d with 15 km = 4 km/h

The unknown value is time. Write the formula with t as the subject. The time unit is h. Leave answer as a decimal or convert to hours and minutes. 0.75 h = 0.75 × 60 = 45 min

t=

= 3.75 h = 3 h 45 min Alternative unitary method 4 km in 1 hour ÷4 1 1 km in hour 4 × 15 15 15 km in hours 4

Express the rate as provided in the question. Divide both quantities by 4.

÷4 × 15

Multiply both quantities by 15.

It takes 3 h 45 min to travel 15 km.

Example 23 Converting units of speed a Convert 72 km/h to m/s.

b

Convert 8 m/s to km/h.

SOLUTION

EXPLANATION

a 72 km in 1 hour 72 000 m in 60 minutes ÷ 60 ÷ 60 1200 m in 1 minute 1200 m in 60 seconds ÷ 60 ÷ 60 20 m in 1 second ∴ 72 km/h = 20 m/s

Express rate in kilometres per hour. Convert km to m and hour to minutes. Divide both quantities by 60. Convert 1 minute to 60 seconds. Divide both quantities by 60. Shortcut for converting km/h → m/s ÷ 3.6.

b 8 m in 1 second × 60 × 60

8 m in 1 second 480 m in 1 minute 28 800 m in 1 hour

∴ 28.8 km in 1 hour

× 60

Express rate in metres per second. Multiply by 60 to find distance in 1 minute.

× 60

Multiply by 60 to find distance in 1 hour. Convert metres to kilometres. Shortcut: m/s × 3.6 → km/h. 8 m/s × 3.6 = 28.8 km/h

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Distance travelled then the Distance travelled must equal: Time taken Average speed Time taken A Average speed × Time taken B C Time taken Average speed Distance travelled 3 If Average speed = then Time taken must equal: Time taken Distance travelled Average speed A Distance travelled × Average speed B C Average speed Distance travelled

2 If Average speed =

4 If an object travels 800 metres in 10 seconds, the average speed of the object is: A 8000 m/s B 800 km/h C 80 km/h D 80 m/s

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6 Calculate the distance travelled by: a a cyclist travelling at 12 m/s for 90 seconds b an ant travelling at 2.5 m/s for 3 minutes c a bushwalker who has walked for 8 hours at an average speed of 4.5 km/h d a tractor ploughing fields for 2.5 hours at an average speed of 20 km/h

Example 22

7 Calculate the time taken by: a a sports car to travel 1200 km at an average speed of 150 km/h b a bus to travel 14 km at an average speed of 28 km/h c a plane to ﬂy 6900 km at a constant speed of 600 km/h d a ball moving through the air at a speed of 12 m/s to travel 84 m

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9 Convert the following speeds to km/h. a 15 m/s b 2 m/s

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d 1 km/s

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5 Calculate the average speed of: a a sprinter running 200 m in 20 seconds b a skateboarder travelling 840 m in 120 seconds c a car travelling 180 km in 3 hours d a truck travelling 400 km in 8 hours e a train travelling 60 km in 30 minutes f a tram travelling 15 km in 20 minutes

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d 210 km/h

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11 A plane is ﬂying at a cruising speed of 900 km/h. How far will the plane travel from 11:15 a.m. to 1:30 p.m. on the same day?

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13 The back end of a 160-metre-long train disappears into a 700-metre-long tunnel. Twenty seconds later the front of the train emerges from the tunnel. Determine the speed of the train in m/s. 14 Anna rode her bike to school one morning, a distance of 15 km, at an average speed of 20 km/h. It was raining in the afternoon, so Anna decided to take the bus home. The bus trip home took 30 minutes. What was Anna’s average speed for the return journey to and from school?

16 The Ghan train is an Australian icon. You can board The Ghan in Adelaide and 2979 km later, after travelling via Alice Springs, you arrive in Darwin. (Round the answers correct to 1 decimal place.) a If you board The Ghan in Adelaide on Sunday at 2:20 p.m. and arrive in Darwin on Tuesday at 5:30 p.m., what is the average speed of the train journey? b There are two major rest breaks. The train stops for 1 4 hours at Alice Springs and 4 hours at Katherine. 4 Taking these breaks into account, what is the average speed of the train when it is moving?

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12 The wheels on Charlie’s bike have a circumference of 1.5 m. When Charlie is riding fastest, the wheels rotate at a speed of five turns per second. a What is the fastest speed Charlie can ride his bike, in km/h? b How far would Charlie travel in 5 minutes at his fastest speed?

15 In Berlin 2009, Jamaican sprinter Usain Bolt set a new 100 m world record time of 9.58 seconds. Calculate Usain Bolt’s average speed in m/s and km/h for this world record. (Round the answers correct to 1 decimal place.)

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18 Nina, Shanti and Belle run a 1000 m race at a constant speed. When Nina crossed the finish line first, she was 200 m ahead of Shanti and 400 m ahead of Belle. When Shanti crossed the finish line, how far ahead of Belle was she?

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19 Julie and Jeanette enjoy finishing their 6 km morning run together. Julie runs at an average speed of 10 km/h and Jeanette runs at an average speed of 3 m/s. If Julie leaves at 8 a.m., at what time should Jeanette leave if they are to finish their run at the same time?

Enrichment: Speed research 20 Carry out research to find answers to the following questions. Light and sound a What is the speed of sound in m/s? b What is the speed of light in m/s? c How long would it take sound to travel 100 m? d How long would it take light to travel 100 km? e How many times quicker is the speed of light than the speed of sound? f What is a Mach number? Spacecraft g What is the escape velocity needed by a spacecraft to ‘break free’ of Earth’s gravitational pull? Give this answer in km/h and also km/s. h What is the orbital speed of planet Earth? Give your answer in km/h and km/s. i What is the average speed of a space shuttle on a journey from Earth to the International Space Station? Knots Wind speed and boat speed are often given in terms of knots (kt). j What does a knot stand for? k What is the link between nautical miles and a system of locating positions on Earth? l How do you convert a speed in knots to a speed in km/h?

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5I Distance/time graphs

Distance travelled

Distance/time graphs show the distance on the A train journey vertical axis and the time on the horizontal axis. When an object moves at a constant speed, then the graph of distance vs. time will be a straight line segment. The steepness of a line segment shows the speed of that part of the journey. A ﬂat line segment shows that there is no movement for a given time. Several different line segments joined together can make up a total journey. For example, a train journey could be graphed with a series of sloping line segments showing Time travel between stations and ﬂat line segments Note: the acceleration and deceleration phases are not showing when the train has stopped at stations. shown on this simplified distance/time graph.

Let’s start: Matching graphs and journeys Journey A

Journey B Back

Distance

Back

Distance

A teacher drew four distance/ time graphs on the whiteboard and some students volunteered to follow a graph for a walking journey between the front and back of the classroom. Work in pairs and match each graph with the student who correctly walked that journey.

Front

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• Ella walked from the back of the room, stopped for a short time and then turned around and walked to the back of the room again. • Jasmine walked from the front of the room to the back of the room where she stopped for a short time. She then turned and walked to the front of the room. • Lucas walked from the back to the front of the room where he stopped for a short time. He then turned and walked halfway to the back, brieﬂy stopped and then turned again and walked to the front. • Riley walked from the front of the room, stopped brieﬂy partway and then completed his walk to the back of the room where he stopped for a short time. He then turned and walked to the front. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Graphs of distance versus time usually consist of line segments. A distance-time graph Constant movement Distance

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Each segment shows whether the object is moving or at rest. The steepness of a line segment shows the speed. Steeper lines show greater speed than less steep lines. Horizontal lines show that the person or vehicle is stationary. Distance Speed = Time To draw a graph of a journey, use time on the horizontal axis and distance on the vertical axis.

Example 24 Reading information from a graph This distance/time graph shows the journey of a car from one town (A) to another (B). a How far did the car travel? b How long did it take the car to complete the journey? c What was the average speed of the car?

300 Distance (km)

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SOLUTION

EXPLANATION

a distance of 250 km

Draw an imaginary line from point B to the vertical axis; i.e. 250 km.

b 2.5 hours

Draw an imaginary line from point B to the horizontal axis; i.e. 2.5 hours.

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Distance Time

Distance = 250 km Time = 2.5 hours

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Example 25 Calculations from a distance/time graph

Distance from home ((km)

This distance/time graph shows Noah’s cycle journey from home one morning. a Find Noah’s speed in the first hour of the journey. 50 b How many minutes was Noah’s first rest break? c Find Noah’s speed between 7:15 a.m. and 8:30 a.m. 40 d At what time did Noah start and finish his second rest break? 30 e At approximately what time had Noah ridden a total 20 of 65 km? f Find Noah’s speed on the return journey to one 10 decimal place. g What was Noah’s average speed for the whole journey? 6 am 7 am 8 am 9 am 10 am 11 am Time

SOLUTION

EXPLANATION

at 25 km/h

The first hour is 6 a.m. to 7 a.m. Noah cycles 25 km in one hour.

b 15 minutes

The first horizontal line segment is at 7 a.m. to 7:15 a.m.

c

d t 20 km = 1.25 h = 16 km/h

s=

Distance travelled = 45 km – 25 km. Time taken = 7:15 a.m. to 8:30 a.m. = 1.25 hours

d 8:30 a.m. to 9:15 a.m.

A horizontal line segment shows that Noah is stopped.

e 10 a.m.

45 km plus 20 km of the return trip or 25 km from home.

d t 45 km = 1.75 h = 25.7 km/h

Distance travelled = 45 km. Time taken = 9:15 a.m. to 11 a.m. = 1.75 hours

d t 90 km = 5h = 18 km/h

Distance travelled Time takenn Distance travelled = 2 × 45 = 90 km. Time taken = 6 a.m. to 11 a.m. = 5 hours.

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2 This graph shows a train journey from one town (A) to another town (B). a How far did the train travel? b How long did it take to complete this journey? c What was the average speed of the train?

Distance (km)

Journey A: A man walks slowly away from home, stops for a short time, then continues walking away from home at a fast pace and finally stops again. Journey B: A girl walks slowly away from home and then stops for a short time. She then turns around and walks at a fast pace back to her home. Journey C: A boy walks quickly away from home, brieﬂy stops, then continues slowly walking away from home and finally stops again. Journey D: A woman walks quickly away from home and then stops brieﬂy. She then turns around and slowly walks back home.

3 The Wilson family drove from their home to a relative’s place for lunch and then back home again. This distance/time graph shows their journey. Distance from home (km)

Example 24

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5 This distance/time graph shows Levi’s walk from his home to school one morning.

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LL

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MA

For each description below, choose the line segment of the graph that matches it. 1 a A hour rest break is taken from 10 a.m. to 10:15 a.m. 4 b The Wilson’s drove 100 km in the first hour. 3 c The car is stopped for 1 hours. 4 d The car travels from 100 km to 200 km away from home in this section of the journey. e At the end of this segment, the Wilsons start their drive back to their home. f The Wilsons travel from 1 p.m. to 3 p.m. without stopping. g At the end of this segment, the Wilsons have arrived back home again.

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M AT I C A

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100 d

80

c b

60 40

it's at

20 20

a b c d e f g h i

40

60

80

100 120 140 Time (seconds)

160

180

200 220

Which line segments show that Ava was stopped? For how long was Ava not walking? What distance had Ava travelled by 20 seconds? How long did Ava take to walk 70 m? When did Ava turn around to start to walk back towards the lift? At what times was Ava 60 m from the lift? What was the total distance that Ava walked from the lift and back again? For which section of Ava’s walk does the line segment have the ﬂattest slope? What does this tell you about Ava’s speed for this section? For which section of Ava’s walk is the line segment steepest? What does this tell you about Ava’s speed for this section?

Cambridge University Press

R K I N G

C

F PS

Y

WO

MA

6 This distance/time graph shows Ava’s short walk in a shopping centre from the lift and back to the lift.

LL

Example 25

5I

Chapter 5 Ratios and rates

Distance from the lift (m)

336

M AT I C A

337

Number and Algebra

T

Distance from home (km)

20 16 12 8 4

11 a.m. noon 1 p.m. 2 p.m. 3 p.m. Time (hours) a At what time did Enzo arrive at the shop? b Find Enzo’s speed (to 1 decimal place) between his home and the shop. c Find Enzo’s speed when cycling between the shop and his friend’s place. d What time did Enzo arrive at and leave his friend’s place? e What was the total time that Enzo was stopped on this journey? f At what times was Enzo 12 km from home? g At what time had Enzo ridden a total of 30 km? h At what speed did Enzo travel when returning home from his friend’s place? i Calculate Enzo’s average speed (to 1 decimal place) over the whole journey. 8 Draw a separate distance/time travel graph for each of these journeys. Assume each section of a journey was at a constant speed. a A motor bike travelled 200 km in 2 hours. b A car travelled 100 km in one hour, stopped 1 for hour then travelled a further 50 km in 2 1 hour. 2 c In a 90 minute journey, a cyclist travelled 18 km in 45 minutes, stopped for 15 minutes then returned home.

R

HE

Cambridge University Press

R K I N G

C

F PS

Y

MA

7 Enzo cycles from home to a friend’s place. After catching up with his friend he then cycles back home. On the way to his friend’s place, Enzo stopped brieﬂy at a shop. His journey is shown on this distance/time graph.

U

LL

WO

M AT I C A

5I

R K I N G

R

U

MA

T

20 15 10 5 1

2

3

4 5 Time (hours)

R

HE

A bush hike

6

7

8

a What were the fastest and slowest speeds (in km/h) of the hikers? Suggest a feature of the hiking route that could have made these speeds so different. b Write a story of the journey shown by this graph. In your story, use sentences to describe the features of the hike shown by each straight line segment of the graph, including the time taken and distance travelled. Also include a sentence comparing the average speeds for the different sections of the hike. c Suppose the hikers turn back after 1 their hour rest at 10 km and return 2 to where they started from. Redraw the graph above showing the same sections with the same average speeds but with ‘distance from start’ vs. time.

Cambridge University Press

PS

M AT I C A

WO

10 A one day, 20 km bush hike is shown by this graph of ‘total distance travelled’ versus time.

F

R K I N G

C

F PS

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C

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9 A train travels 5 km in 8 minutes, stops at a station for 2 minutes, travels 12 km in 10 minutes, stops at another station for 2 minutes and then completes the journey by travelling 10 km in 15 minutes. a Calculate the speed of the train in km/h (to 1 decimal place) for each section of the journey. b Explain why the average speed over the whole journey is not the average of these three speeds. c Calculate the average speed of the train in km/h (to 1 decimal place) over the whole journey. d On graph paper, draw an accurate distance/time graph for the journey.

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Chapter 5 Ratios and rates

Total distance travelled (km)

338

M AT I C A

339

Number and Algebra

Distance from home

HE

E

F

Time

Distance from front of room

Distance from front of room

Journey B

Time

Time

Time

Journey D Distance from front of room

Distance from front of room

Journey C

Time

Adam: Walks from the front, stops, a fast walk, stops, slow walk to the back, stops, and walks to the front. Max: Walks from the back, stops, walks to the front, stops, walks to the back. Ruby: Walks from the back, stops halfway, turns, walks to the back, stops, walks to the front. Conner: Fast walk from front to back, stops, walks towards the front, stops, walks to the back, stops, slower walk to the front. Isla: Walks from the front, stops, walks to the back, stops, walks at a fast pace to the front.

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F PS

M AT I C A

12 Match each distance/time graph with the student below who correctly walked the journey shown by that graph. Explain the reasons for your choice. Journey A

C

Y

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R K I N G

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e Deanna was cycling faster at F than she was at A. f Deanna was further from home at F than at D. g Deanna had ridden further at F than at D.

WO

MA

11 This distance/time graph shows Deanna’s bike journey from home one morning. Choose True or False for each of these statements and give a reason for your answer. a Deanna was the same distance from home at B and F. b Deanna was not cycling at B. D c Deanna was cycling faster at A than she was B at D. C d Deanna was facing the same direction at A C and E.

5I

HE

Graph C

Graph D

Time

Distance from home

Time

Time

Distance from home

14 Jayden and Cooper cycled towards each other along the same track. Jayden was resting when Cooper caught up to him and stopped for a chat and then they both continued their ride. Which of these distance/time graphs show their journeys? Explain why. Graph B

Distance

Distance

Graph A

Time

Time Graph D

Distance

Graph C

Time

Time

Cambridge University Press

C

F PS

M AT I C A

Graph B

Time

R K I N G

Y

R

Distance from home

Distance from home

U

T

Graph A

WO

MA

13 One afternoon, Sienna cycled from school to a friend’s place, then to the library and finally she cycled home. Which of these distance/time graphs are appropriate to show Sienna’s journey? Explain why.

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Chapter 5 Ratios and rates

Distance

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Number and Algebra

Enrichment: More than one journey 15 Northbrook is 160 km north of Gurang. Archie leaves Gurang to drive north and at the same time Heidi leaves Northbrook to drive south. After 40 minutes, Archie is halfway to Northbrook when his car collides with a tree. Five minutes later Heidi sees Archie’s car and stops. Heidi immediately calls an ambulance which comes from 15 km away and arrives in 10 minutes. a Using graph paper, draw a distance/time graph to show the journey of the two cars and the ambulance. b A tow-truck and the police also arrived at this accident. Write the rest of the story and graph the completed journey for the all vehicles involved that day. Use colours and include a key. c Explain the journeys shown in your graph to a classmate. d Would any of the drivers have been charged for speeding if it was a 110 km/h speed limit?

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341

Investigation

342

Chapter 5 Ratios and rates

Fun run investigation Three maths teachers, Mrs M, Mr P and Mr A, trained very hard to compete in a Brisbane Fun Run. The 10.44 km route passed through the botanical gardens, along the Brisbane River to New Farm Park, and back again. Their times and average stride lengths were: • Mrs M: 41 minutes, 50 seconds: 8 strides per 9 m • Mr P: 45 minutes, 21 seconds: 7 strides per 9 m • Mr A: 47 minutes, 6 seconds: 7.5 strides per 9 m Copy and complete the following table and determine which rates are the most useful representations of fitness. Give the answers rounded to 1 decimal place. Justify your conclusions. Running rates Seconds per 100 m Seconds per km Metres per minute Km per hour Strides per 100 m Strides per minute Strides per hour

Mrs M

Mr. P

Mr. A

World record 10 km runners

Your family member or friend

Fitness investigation 1 Using a stopwatch, measure your resting heart rate in beats per minute. 2 Run on the spot for one minute and then measure your working heart rate in beats per minute. 3 Using a stopwatch, time yourself for a 100 m run and also count your strides. At the end, measure your heart rate in beats per minute. Also calculate the following rates. • Your running rate in m/s, m/min and km/h • Your running rate in time per 100 m and time per km. • Your rate of strides per minute, strides per km and seconds per stride 4 Run 100 m four times without stopping and, using a stopwatch, record your cumulative time after each 100 m. • Organise these results into a table. • Draw a graph of the distance ran (vertical) vs time taken (horizontal). • Calculate your running rate in m/min for each 100 m section. • Calculate your overall running rate in m/min for the 400 m. • Explain how and why your running rates changed over the 400 m. 5 Try sprinting really fast over a measured distance and record the time. Calculate your sprinting rate in each of the following units: • minutes per 100 m • metres per minute

• time per km • km per hour

6 Research the running rate of the fastest schoolboy and schoolgirl in Australia. How do their sprinting rates compare to the running rates of Australian Olympian athletes? © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

Cambridge University Press

1 This diagram is made up of 8 equal-sized squares.

How many squares need to be shaded if the ratio of shaded squares to unshaded squares is: a 1 : 3? b 2 : 3? c 1 : 2? 2 Bottle A has 1 L of cordial drink with a cordial to water ratio of 3 : 7. Bottle B has 1 L of cordial drink with a cordial to water ratio of 1 : 4. The drink from both bottles is combined to form a 2 L drink. What is the new cordial to water ratio? 3 Brothers Marco and Matthew start riding from home into town, which is 30 km away. Marco rode at 10 km/h and Matthew took 20 minutes longer to complete the trip. Assuming that they both rode at a constant speed, how fast was Matthew riding? 4 a If 1 person takes 1 hour to dig 1 post hole, how long will it take 2 people to dig 2 post holes? b If 3 people take 3 hours to make 3 wooden train sets, how many train sets can 6 people make in 6 hours? 5 At a market you can trade 2 cows for 3 goats or 3 goats for 8 sheep. How many sheep are 3 cows worth? 6 Two cars travel toward each other on a 100 km straight stretch of road. They leave at opposite ends of the road at the same time. The cars’ speeds are 100 km/h and 80 km/h. How long does it take for the cars to pass each other? Car 2 80 km/h

Car 1 100 km/h 100 km

7 A river is ﬂowing downstream at a rate of 1 km/h. In still water Michael can swim at a rate of 2 km/h. Michael dives into the river and swims downstream then turns around and swims back to the starting point, taking 0.5 hours in total. How far did he swim? 8 A fitness fanatic walks at 4 km/h for time t1, and then runs at 7 km/h for time t2. He travels a total of 26 km. If he had run for the same time that he had walked (t1) and walked for the same time that he had run (t2), then he would have travelled 29 km. What was the total time spent walking and running?

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343

Puzzles and challenges

Number and Algebra

Chapter 5 Ratios and rates

Ratios are writen in • simplest form • with whole numbers

Equivalent ratios 2:3 ×4 8 : 12 ÷2 ÷2 4:6

Dividing a quantity in a given ratio

×4

Divide $180 in the ratio 4 : 5

Simplest form

Eliminate decimals

×10

HCF 4

÷4

0.8 : 1.2 8 : 12 2:3

a:b a is to b a to b

×10 ÷4

Ratio

5.3 : 2.6 1 2 53 : 23

Recurring decimals to fractions

16 3

×3

25

:

8 3

16 : 8

÷8

Same units

Fraction method

2:1

1 minutes: 1 4

25 : 75

÷25

1:3

Total number of parts

4+5=9

Each fraction

4 9

180

and

5 9

× 180

Divide $180 in the ratio 4 : 5

÷8

Total number of parts Finding a quantity from a ratio

Sentence answer

Cows to horses in ratio 5 : 2 There are 8 horses, how many cows?

4+5=9 ÷9 9 parts = $180 ÷9 1 part = $20 ×5 ×4 1 part = $20 ×4 ×5 5 parts = $100 4 parts = $80

One part

hours ÷25

5 9

= $80 = $100 $180 divided in the ratio 4 : 5 is $80 and $100

Sentence answer

• same type and • same unit

×3

4 × 9

Each amount

Comparison of two quantities of the

and

Unitary method

$180 divided in the ratio 4 : 5 is $80 and $100

Actual size from model Equivalent ratio method C:H=5:2 5:2 C:8= ×4 ×4 20 : 8 = There are 20 cows.

Unitary method C:H 5:2 2 parts = 8 horses 1 part = 4 horses 5 parts = 20 cows Length units km ÷1000 m ×100 ÷100 cm ×10 ÷10 mm

Model car 17 cm long Scale 1 : 25 Scale factor = 25 Actual car = 25 × 17 cm = 425 cm = 4.25 m

Scale ratios

Drawing Actual : length length =

1 : scale factor

Map length from actual length

×1000

Scale ratio

Scale factor

Actual object

125

125 times larger than drawing

4 : 500 1 : 125 50 : 1 1 1 : 50 Average rates 720 km driven in 10 hours Average speed =

720 10

= 72 km/h

1 th 50

1 50

of the

drawing size

Actual length

= 4 cm

divide scale factor

Distance–time graph

Simplified rates • 25 km/h many km per one hour? • $12/kg many $ per one kg?

Rates s : speed d : distance t : time

d s t d s t d s t

s=

d t

Unitary method to change units 1

t = ds

• Steeper lines show greater speed than less steep lines. • Flat segment means the object is at rest. Reading the graph: • Start on given distance. move across to line then down to time scale (or in reverse).

1

1

d= s × t

Time

Comparing two quantities with different units

For a 18 km jog in 1 2 hours determine average rate in m/min 18 km in 1 2 hours

÷1 2

1

1 part

÷60 200 m in 1 min Average speed = 200 m/min

1 part

÷1 2

18 1

12

÷60

km in 1 hour

12 000 m in 60 min

Actual distance = 1 km Map scale = 1 : 25000 Scale factor = 25000 Map length = 1 km ÷ 25000 1 = 25000 km same units 1000 × 100 = 25000 cm

times scale factor Drawing (or map) length

same units

Distance

Chapter summary

344

Unitary method with rates 4 L paint to cover 80 m2 ? L to cover 165 m2 ÷80 ×165

4 L for 80 m2 4 L for 1 m2 80 4 × 165 for 165 m2 80 8.25 L for 165 m2

÷80

1 part

×165

Cambridge University Press

Number and Algebra

Multiple-choice questions 1 A school has 315 boys, 378 girls and 63 teachers. The ratio of students to teachers is; A 11 : 1 B 1 : 11 C 5:6 D 6:5 2 Find the ratio of the shaded area to the unshaded area in this triangle. A 3:5 B 8:5 C 5:3 D 5:8

3 The ratio 500 mm to A 50 : 2

1 m is the same as: 5 B 2500 : 1 C

1 3 4 The ratio 1 : simplifies to: 2 4 A 2:1 B 1:2

C

2:5

D 5:2

3:4

D 4:3

5 $750 is divided in the ratio 1 : 3 : 2. The smallest share is: A $250 B $125 C $375

D $750

6 The ratio of the areas of two triangles is 5 : 2. The area of the larger triangle is 60 cm2. What is the area of the smaller triangle? A 12 cm2 B 24 cm2 C 30 cm2 D 17 cm2 7 Callum fills his car with 28 litres of petrol at 142.7 cents per litre. His change from $50 cash is: A $10 B $39.95 C $10.05 D $40 8 45 km/h is the same as: A 0.25 m/s B 25 m/s

C

12.5 m/s

D 75 m/s

9 A ﬂag is created by enlarging the shaded rectangle as shown. What is the length of the original rectangle? 20 cm A 20 cm B 30 cm C 40 cm D 13 cm

80 cm

120 cm 10 On a map, Sydney and Melbourne are 143.2 mm apart. If the cities are 716 km apart, what scale has been used? A 1:5 B 1 : 5000 C 1 : 50 000 D 1 : 5 000 000

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345

Chapter 5 Ratios and rates

11 A toddler walked slowly away from his mum, had a short stop, then turned and walked quickly back to his mum again. In each graph below, the distance is measured from the toddler’s mum. Choose the correct distance/time graph for this journey. Graph A Distance

Distance

Graph B

Time

Graph C

Graph D Distance

Time

Distance

346

Time

Time

Short-answer questions 1 In Lao’s pencil case there are 6 coloured pencils, 2 black pens, 1 red pen, 5 textas, 3 lead pencils and a ruler. Find the ratio of: a lead pencils to coloured pencils b black pens to red pens c textas to all pencils 2 True or false? a 1:4 = 3:6 b The ratio 2 : 3 is the same as 3 : 2. c The ratio 3 : 5 is written in simplest form. d 40 cm : 1 m is written as 40 : 1 in simplest form. 1 e 1 :2 = 5:8 4 3 Copy and complete. a 4 : 50 = 2 : c

2 :4 = 1: 3

4 Simplify the following ratios. a 10 : 40 b 36 : 24 f 5 : 25 g 6:4 k

2 5 : 7 7

l

1

1 2 : 10 10

b

1.2 : 2 =

d

1:

: 20

: 5 = 5 : 15 : 25

c 75 : 100 h 52 : 26

d 8 : 64 i 6b : 9b

1 3 m 2 : 2 4

n 12 : 36 : 72

e 27 : 9 j 8a : 4

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Number and Algebra

5 Simplify the following ratios by first changing to the same units. a 2 cm : 8 mm b 5 mm : 1.5 cm c 3 L : 7500 mL d 30 min : 1 h 1 e 400 kg : 2 t f 6 h : 1 day g 120 m : 1 km h 45 min : 2 h 2 6 Divide: a $80 in the ratio 7 : 9 b 200 kg in the ratio 4 : 1 c 40 m in the ratio 6 : 2 d $1445 in the ratio 4 : 7: 6 e $1 in the ratio 3 : 1 : 1 7 a The ratio of the cost price of a TV to its retail price is 5 : 12. If its cost price is $480, calculate its retail price. b The ratio of Sally’s height to Ben’s height is 12 : 17. If the difference in their heights is 60 cm, how tall is Sally? c Orange juice, pineapple juice and guava juice are mixed in the ratio 4 : 3 : 2. If 250 mL of guava juice is used, how many litres of drink does this make? 8 Express each rate in simplest form. a 10 km in 2 hours b $650 for 13 hours

c 2800 km in 20 days

9 Copy and complete. a 400 km on 32 litres = _____ km/L = _____ L/100 km b 5 grams in 2 min = _____ g/min = _____ g/h 1 c $1200 in day = _____ $/day 2 = _____ $/h 10 For a scale of 1 : 1000, find the real length (in metres) if the scale length is given as: a 0.0002 m b 2.7 cm c 140 mm 11 Two cities are 50 km apart. How many millimetres apart are they on a map that has a scale of 1 : 100 000? 12 Find the scale factor for the following pairs of similar figures, and find the value of x. a 4 cm

6 cm

x cm

6 cm b

3 cm

5 cm 4 cm

9 cm

15 cm

x cm

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347

Chapter 5 Ratios and rates

13 Copy and complete. a $120/h = _____ c/min

b 6 m/s = _____ km/h

c 720 km/h = _____ m/s

14 a A truck uses 12 litres of petrol to travel 86 km. How far will it travel on 42 litres of petrol? b Samira earns $67.20 for a 12-hour shift. How much will she earn for a 7-hour shift? c Tap 1 fills the pool in 12 hours, while tap 2 fills the same pool in 15 hours. How long does it take to fill this pool if both taps are used? 15 a

Sandra drives to her mother’s house. It takes 45 minutes. Calculate Sandra’s average speed if her mother lives 48 km away. b How long does it take Ari to drive 180 km along the freeway to work if he manages to average 100 km/h for the trip? c How far does Siri ride his bike if he rides at 4.5 km/h for 90 minutes?

16 This distance/time graph shows Mason’s cycle journey from home one morning. Distance from home (km)

348

40 30

Q

20 10

R P

8 a.m. 9 a.m. 10 a.m. 11 a.m. noon 1 p.m. 2 p.m. Time a b c d e f g h

How far was Mason from home at 8:45 a.m.? What was Mason’s speed in the first hour of the journey? How many minutes was Mason’s first rest break? What was the furthest distance that Mason was from his home? Was Mason further away from home at point Q or point R? When had Mason ridden 60 km? How far did Mason ride between points P and R? Describe each section of Mason’s journey from 9:30 a.m. to the finish. Include the distances, times, speeds and any features.

Cambridge University Press

Number and Algebra

Extended-response question The Harrison family and the Nguyen family leave Wahroonga, Sydney, at 8 a.m. on Saturday morning for a holiday in Coffs Harbour which is 500 km north of Sydney. The Harrisons’ 17-year-old son drives for the first 2 hours at 90 km/h. They then stop for a rest of 1 1 hours. Mr Harrison drives the rest of the way arriving at 3 p.m. 2 The Nguyen family drives the first 300 km in 3 hours and then have a break for 1 hour. Then the Nguyens continue with no more stops and arrive in Coffs Harbour at 2:15 p.m. a Using graph paper, draw a distance/time graph showing both journeys. b How far apart were the two families at 10 a.m.? c At what times did the Nguyens and Harrisons arrive at the halfway point? d What was the Nguyen’s speed for the last section of their trip (correct to 1 decimal place)? e Calculate the average speed of the Nguyen’s journey. f At what time did the Harrisons resume their journey after their rest stop? g How many kilometres did the Harrisons still have to cover after their rest break before arriving in Coffs Harbour? h For how long did Mr Harrison drive and at what speed? i How far south of Coffs Harbour were the Harrisons when the Nguyen family arrived there? j Calculate the cost of the petrol for each family’s trip if petrol cost is 139 c/L and the Harrison car’s consumption is 9 L/100 km, while the Nguyen’s car uses 14 L/100 km.

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349

Semester review 1

350

Semester review 1

Chapter 1: Algebraic techniques 2 and indices Multiple-choice questions 83 × 84 is the same as: A 8−1 B 647

C

87

D 812

2 4x + 5 + 3xx is the same as: A 7x + 5 B 12x 12x

C

12 + x2

D 2x 2 + 12

3 12m + 18 factorises to: A 2(6m – 9) B −6(2m – 3)

C

6(3-2m)

D 6(2m + 3)

4 5a + 5 – 4a – 4 – a – 1 equals: A a–1 B 0

C

2–a

D a+1

1

5 Which answer is NOT equivalent to (m × n) ÷ (p ( × q)? A

mn pq

B

m×

n pq

C

m n × p q

b d

the product of p and 3 the sum of x and y, divided by 2

D

mnq p

Short-answer questions 1

Write an expression for: a the sum of p and q c half the square of m

2 If a = 6, b = 4 and c = −1, evaluate: a a+b+c b ab – c d 3a2 + 2b

c a(b2 – c)

e abc

f

ab c

3 Simplify each algebraic expression. a 4 × 6kk b a+a+a c e 3ab + 2 + 4ab f 7x + 9 – 6xx – 10 g

a×a×a 18xy ÷ 9xx

d 7p 7 ÷ 14 h m + n – 3m + n

4 Simplify: 5 xy a 5

w w + 5 2

d

5 Simplify: m 5 a × 5 6

b

3x 2 x − 7 7

b

c

ab 1 ÷ 7 7

c

3a +

a 2

m n mn × ÷ 3 2 4

6 Expand, and simplify where necessary. a 6(2m – 3) b 10 + 2(m – 3)

c 5(A 5( + 2) + 4(A 4( – 1)

7 Factorise: a 18a – 12

c −8m2 – 16mn

b 6m2 + 6m

Cambridge University Press

Semester review 1

8 Write an expression for the rectangle’s: a perimeter

b

area

3x x + 10 9 Simplify: a 57 × 52 d 410 ÷ 43

b 38 × 32 e 712 ÷ 76

c 67 × 6 f 67 ÷ 6

10 a (52)3

b (23)4

c 120

d 5 × 40

Extended-response question

4a cm

x cm

3a cm a b c d e

Write an expression for the perimeter of this triangle. Write an expression for the area of this triangle. Use Pythagoras’ theorem to find a relationship between x and a. Use your relationship to write an expression for the perimeter in terms of only a. If the perimeter equals 72 cm, what is the area of this triangle?

Chapter 2: Equations 2 Multiple-choice questions 1

The sum of a number and three is doubled. The result is 12. This can be written as: A x + 3 × 2 = 12 B 2x 2 + 6 = 12 C 2x 2 + 3 = 12 D x + 3 = 24

2 The solution to the equation 2m − 4 = 48 is: A m=8 B m = 22 C m = 20 D m = 26 3 The solution to the equation −5(m − 4) = 30 is: A m = −10 B m = −2 − 34 C m=2 D m= 5

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351

Semester review 1

352

4 Which equation has NO solution? A 2x = 9 C x2 = −9

B D

x2 = 4 2 =x 2x

5 Which equation has two solutions? A 2x 2 =9 C x2 = −9

B D

x2 = 4 2 =x 2x

Short-answer questions 1

Solve each of these equations. m a 3w = 27 b 12 = 6 d 4a + 2 = 10 e 2w + 6 = 0

2 Solve: a 6 = 4 − 4m x x + =1 d 5 3

b 3a + 4 = 7a + 8 e 2(5 − a) = 3(2a − 1)

c 4−x=3 x f −1=6 5 c 3(x + 5) = 15 a+7 f =3 2a

3 Double a number less three is the same as 9. What is the number? 4 A father is currently six times as old as his son. In 10 years’ time his son will be 20 years younger than his dad. How old is the son now? 5 Solve the following equations where possible. a x2 = 9 b x2 = 36 d y2 = 0 e b2 = −4

c a2 = 144 f c2 = −100

6 The formula F = ma relates force, mass and acceleration. a Find F if m = 10 and a = 3. b Find m if a = 5 and F = 20. Extended-response question EM Publishing has fixed costs of $1500 and production costs of $5 per book. Each book has a retail price of $17. a Write an equation for the cost (C) of producing n books. b Write an equation for the revenue (R) for selling n books. c A company ‘breaks even’ when the cost of production equals the revenue from sales. How many books must the company sell to break even? d Write an equation for the profit (P) of selling n books. e Calculate the profit if 200 books are sold. f What is the profit if 100 books are sold? Explain the significance of this answer.

Cambridge University Press

Semester review 1

Chapter 3: Measurement and Pythagoras’ theorem Multiple-choice questions 1

A cube has a volume of 8 cubic metres. The area of each face of the cube is: A 2 m2 B 4 m2 C 24 m2 D 384 m2

2 The area of this triangle is: A 48 m2 B 24 m2 C 30 m2 D 40 m2

8m

6m

3 The perimeter of this semicircle is closest to: A 38 cm B 30 cm C 19 cm D 31 cm 4 The value of x in this triangle is closest to: A 176 B 13 C 274 D 17

5 The area of this rectangle is: A 48 m2 B 48 000 cm2 C 480 cm2 D 0.48 m2

10 m

12 cm

15

x

7 1.2 m 40 cm

Cambridge University Press

353

Semester review 1

354

Short-answer questions 1

Complete these conversions. a 5 m = ______ cm

b

1.8 m = ______ cm

c 9 m2 = ______ cm2

d

e 4 L = ______ cm3

f

1800 mm = ______ m 1 km2 = ______ m2 100

2 Find the perimeter of these shapes. a

b

c

14 m

7m

6.2 cm

4m 18 m

3 Find correct to two decimal places: i the circumference a

ii

3m

the area

b 15 cm

4m

4 Find correct to two decimal places: i the perimeter a

ii

b

10 m

the area c

60°

5 cm

18 mm 5 Find the area of these shapes. a 6m

4m

5 m

b

4m

6m 4m 6m

9m

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Semester review 1

6 Find the volume of these solids. a b

c 2.5 m

4.2 m

3m

4m

4.2 m

4m

5 m

4m

4.2 m 7 Find the volume of each cylinder correct to 2 decimal places. a b c 200 cm 4m

40 cm

50 cm r=1m

r=7m

8 Find the value of x in these triangles. Round to 2 decimal places for part b. a b 5

x

x

8

12

12 9 Write these times using 24-hour time. a 3:30 p.m.

b

7:35 a.m.

Extended-response question A square sheet of metal 15 m by 15 m has equal squares of sides x m cut from each corner as shown. It is then folded to form an open tray. a What is the length of the base of the tray? Write an expression. b What is the height of the tray? c Write an expression for the volume of the tray. d If x = 1, find the volume of the tray. e What value of x do you think produces the maximum volume? x x 15 m x

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Semester review 1

356

Chapter 4: Fractions, decimals, percentages and financial mathematics Multiple-choice questions 1

150 simplifies to: 350 A

6 14

2 Sienna spends A $470

B

3 70

C

15 35

3 7

D

3 of $280 her income on clothes and saves the rest. She saves: 7 B $120 C $160 D $2613

3 0.008 × 0.07 is equal to: A 0.056 B 0.0056

C

0.00056

D 56

C

12 5

D

C

4y

D 10y

4 0.24 expressed as a fraction is: A

1 24

B

6 25

24 10

5 If 5% of x = y,, then 10% of 22xx equals: A

1 y 2

B2y

6 When paying for a meal at a local café the bill came to $97.90. The GST included in this cost is: A $89 B $9.79 C $8.90 D $88.11 Short-answer questions 1

Copy and complete these equivalent fractions. a

3 = 5 30

b

11

=

5 55

4 c 1 = 6 3

2 Evaluate each of the following. a

3 1 − 4 2

b

4 3 + 5 5

1 3 c 1 +1 2 4

d

4 2 − 7 3

e

4 3 × 9 4

f

1 3 1 × 2 5

c

4

3 Write the reciprocal of: 2 a 5 4 Evaluate: 1 4 a 2 ×1 2 5

b 8

b

1 1 ÷2 2

1 5

1 1 3 c 1 × ÷ 2 4 5

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Semester review 1

5 Calculate each of the following: a 3.84 + 3.09 b 10.85 – 3.27 d 6.59 – 0.2 × 0.4 e 96.37 × 40

c 12.09 ÷ 3 f 15.84 ÷ 0.02

6 Evaluate: a 5.3 × 103

c 61.4 ÷ 100

b 9.6 × 105

7 Copy and complete this table of decimals, fractions and percentages. Fraction

1 4

1 2

1 5

1 3

2 3

Decimal

0.99

Percentage

80%

8 Find: a 10% of 56

b 12% of 98 1 e 12 % of $840 2

d 99% of $2

0.005

95%

c 15% of 570 m f

58% of 8500 g

9 a Increase $560 by 25%. b Decrease $980 by 12%. c Increase $1 by 8% and then decrease the result by 8%. 10 A $348 dress sold for $261. This represents a saving of x%. What is the value of x? 11 Complete the table of pre- and post-GST prices; remember in Australia the GST is 10%. Pre-GST

GST

Post-GST

$550.00 $7.00 $47.50 $962.50

Extended-response question A laptop decreases in value by 15% a year. a Find the value of a $2099 laptop at the end of: i 1 year ii 2 years iii 3 years b After how many years is the laptop worth less than $800? c Is the laptop ever going to have a value of zero dollars? Explain.

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Semester review 1

358

Chapter 5: Ratios and rates Multiple-choice questions 1

The ratio of the length to the width of this rectangle is: A 12 : 80 B 3 : 20 C 3:2 D 20 : 3 3 km. 4 B 1 : 150

80 cm 1.2 m

2 Simplify the ratio 500 cm to A 2:3

C

D 150 : 1

3:2

3 $18 is divided in the ratio 2 : 3. The larger part is: A $3.60 B $7.20 C $10.80

D $12

4 Calvin spent $3 on his mobile phone card for every $4 he spent on his email account. Calvin spent $420 on his phone last year. How much did he spend on his email account for the same year? A $140 B $315 C $560 D $240 5 A boat sailed 30 kilometres in 90 minutes. What was the average speed of the boat? A 15 km/h B 45 km/h C 3 km/h D 20 km/h Short-answer questions 1

Simplify these ratios. a 24 to 36 d 15 cents to $2

b 15 : 30 : 45 3 e to 2 4

c 0.6 m to 70 cm f

60 cm to 2 m

2 a Divide 960 cm in the ratio of 3 : 2. b Divide $4000 in the ratio of 3 : 5. c Divide $8 in the ratio of 2 : 5 : 3. 3 A 20-metre length of wire is used to fence a rectangular field with dimensions in the ratio 3 : 1. Calculate the area of the field. 4 A business has a ratio of profit to costs of 5 : 8. If the costs were $12 400, how much profit was made? 5 Complete these rates. a 5 g/min = _____ g/h b $240 in 8 hours = $_____ /h 1 c 450 km in 4 h = _____ km/h 2 1 6 A shop sells 1 kg bags of apples for $3.40. Find the cost of a one kilogram at this rate. 2 7 A car travels the 1035 km from southern Sydney to Melbourne in 11.5 hours. Calculate its average speed.

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Semester review 1

Extended-response questions 1

A small car uses 30 litres of petrol to travel 495 km. a At this rate, what is the maximum distance a small car can travel on 45 litres of petrol? b What is the average distance travelled per litre? c Find the number of litres used to travel 100 km, correct to 1 decimal place. d Petrol costs 117.9 cents/litre. Find the cost of petrol for the 495 km trip. e A larger car uses 42 litres of petrol to travel 378 km. The smaller car holds 36 litres of petrol while the larger car holds 68 litres. How much further can the larger car travel on a full tank of petrol?

2 The training flight of a racing pigeon is shown in this distance/time graph.

Distance from home ((km) km)

50 40 30 20 10

6 am 7 am 8 am 9 am 10 am 11 am T Time a How long did the pigeon take to fly the first 20 km? b How long was each of the rest stops? c How far had the pigeon flown by 7:15 a.m.? d At what times was the pigeon 30 km from home? e Calculate the pigeon’s speed for each flight section of the journey, listing them in order. f What was the furthest distance the pigeon was from home? g How far did the pigeon fly between 7:30 a.m. and 9:15 a.m.? h At what time had the pigeon flown a total of 80 km? i Calculate the pigeon’s average speed for the flight (correct to 1 decimal place). j Describe the pigeon’s flight from 6 a.m. to 7:45 a.m. including the starting and finishing times, the distances flown and the speed of each section.

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360

Chapter 6 Angle relationships and properties of geometrical figures 1

Chapter

6

Angle relationships and properties of geometrical figures 1

What you will learn

6A 6B 6C 6D 6E 6F 6G

The language, notation and conventions of angles REVISION Transversal lines and parallel lines REVISION Triangles Quadrilaterals Polygons EXTENSION Line symmetry and rotational symmetry Euler’s formula for three-dimensional solids FRINGE

Cambridge University Press

361

NSW Syllabus

for the Australian Curriculum

Strand: Measurement and Geometry Substrands: ANGLE RELATIONSHIPS PROPERTIES OF GEOMETRICAL FIGURES

Outcomes A student identiﬁes and uses angle relationships, including those related to transversals on sets of parallel lines. (MA4–18MG) A student classiﬁes, describes and uses the properties of triangles and quadrilaterals, and determines congruent triangles to ﬁnd unknown side lengths and angles. (MA4–17MG)

The geometry of honey The cells in honeycomb made by bees are hexagonal in shape, but each cell is not exactly a hexagonal prism. Each cell is actually a dodecahedron (12-faced polyhedron) with 6 rectangular sides (giving the hexagonal appearance) and 3 faces at each end. Each of the end faces is a rhombus and forms angles of 120° with the other sides of the honeycomb cell. The shape of each cell is not just a random arrangement of quadrilateral faces. Instead, the geometry of the cells allows the cells to ﬁt neatly together to form a very efﬁcient geometrical construction. The structure allows for a minimum surface area for a given volume, maximising the use of space in which to store honey.

Cambridge University Press

Chapter 6 Angle relationships and properties of geometrical figures 1

Pre-test

362

1 Find the value of a in these diagrams. a b a° a°

c 50°

a°

30°

220°

2 Name these angles as acute, right, obtuse, straight, reflex or revolution. a 360° b 90° c 37° d 149° e 180° f 301° 3 This diagram includes a pair of parallel lines and a third line called a transversal. a° b° e° f° c° d° g°

50°

a What is the value of a? b Which pronumerals are equal in value to a? c Which pronumerals are equal in value to 50? 4 Name the triangle that best ﬁts the description. Choose from scalene, isosceles, equilateral, acute, right or obtuse. Draw an example of each triangle to help. a One obtuse angle b Two equal length sides c All angles acute d Three different side lengths e Three equal 60° angles f One right angle 5 Name the six special quadrilaterals. Draw an example of each. 6 Find the value of x in these shapes, using the given angle sum. a Angle sum = 180° b Angle sum = 360° x°

120° x°

20°

70°

7 Name each solid as a cube, a triangular prism or a pyramid. a b c

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Measurement and Geometry

6A The language, notation and conventions of angles For more than 2000 years school geometry has been based on the work of Euclid, the Greek mathematician who lived in Egypt in about 300 bc. Before this time, the ancient civilisations had demonstrated and documented an understanding of many aspects of geometry, but Euclid was able to produce a series of 13 books called the Elements, which contained a staggering 465 propositions. This great work is written in a well-organised and structured form, carefully building on solid mathematical foundations. The most basic of these foundations, called axioms, are fundamental geometric principles from which all other geometry can be built. There are ﬁve axioms described by Euclid: • Any two points can be joined by a straight line. • Any ﬁnite straight line (segment) can be extended in a straight line. • A circle can be drawn with any centre and any radius. • All right angles are equal to each other. • Given a line and a point not on the line, there is only one line through the given point and in the same plane that does not intersect the given line. These basic axioms are considered to be true without question and do not need to be proven. All other geometrical results can be derived from these axioms.

Let’s start: Create a sentence or definition The ﬁve pronumerals a, b, c, d and e represent the number of degrees in ﬁve angles in this diagram. Can you form a sentence using two or more of these pronumerals and one of the following words? Using simple language, what is the meaning of each of your sentences? • Supplementary • Revolution • Adjacent • Complementary • Vertically opposite • Right

c° d°

b° a° e°

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R E V I S I ON

Key ideas

364

Chapter 6 Angle relationships and properties of geometrical figures 1

■■

A point represents a position. –– It is shown using a dot and generally labelled with an upper case letter. C –– The diagram shows points A, B and C.

A

B ■■

This diagram shows intervals AC and CB. These are sometimes called line segments. –– AC and CB form two angles. One is acute and one is reflex. –– C is called the vertex. The plural is vertices.

A C B

■■

This diagram shows acute angle ACB. It can be written as: ˆ or BCA ˆ ∠C or ∠ACB or ∠BCA or ACB

A

–– CA and CB are sometimes called arms. –– The pronumeral x represents the number of degrees in the angle.

C

x° B

■■

This diagram shows reflex ∠ACB .

A x° C B

Type of angle

Size of angle

acute

greater than 0° but less than 90°

right

exactly 90°

obtuse

greater than 90° but less than 180°

straight

exactly 180°

reflex

greater than 180° but less than 360°

revolution

exactly 360°

Diagram

Cambridge University Press

■■

This diagram shows two angles sharing a vertex and an arm. They are called adjacent angles.

A C

B D

■■

This diagram shows two angles in a right angle. They are adjacent complementary angles. a° is the complement of b°. a + b = 90

■■

■■

■■

It is possible to have three or more angles in a right angle. They are not complementary. d + e + f = 90

This diagram shows two angles on a straight line. They are adjacent supplementary angles. g° is the supplement of h°. g + h = 180

a° b°

d° e° f°

g°

It is possible to have three or more angles on a straight line. They are not supplementary. i + j + k = 180 j°

h°

k°

i°

■■

This diagram shows angles at a point and angles in a revolution.

p + q + r + s = 360

p° q° s° r°

Cambridge University Press

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Key ideas

Measurement and Geometry

Key ideas

366

Chapter 6 Angle relationships and properties of geometrical figures 1

■■

When two straight lines meet they form two pairs of vertically opposite angles. Vertically opposite angles are equal. t=v u=w t° w° u° v°

■■

■■

If one of the four angles in vertically opposite angles is a right angle, then all four angles are right angles. C x = 90 A y = 90 z = 90 z° x° y° B D A – AB and CD are perpendicular lines. This is written as AB ⊥ CD. The markings in these diagrams indicate that: B D – AB = AD – BC = CD – ∠ABC = ∠ADC

C

Example 1 Finding the value of pronumerals in geometrical figures Determine the value of the pronumeral in these diagrams, giving reasons. a b b° 65°

a° 30°°

b°

a°

SOLUTION

EXPLANATION

a a + 30 = 90 (angles in a right angle) a = 60 b + 90 = 360 (angles in a revolution) b = 270

a° and 30° are adjacent complementary angles.

b a + 65 = 180 (angles on a straight line) a = 115 b = 65 (vertically opposite angles)

a° and 65° are adjacent supplementary angles.

b° and 90° make a revolution.

b° and 65° are vertically opposite angles.

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Measurement and Geometry

Architects and engineers use a variety of angles to create complex structures.

WO

U

d

d° 120°

A

130° 50° 20° 70°

D

B

C c c°

120°

and and°

3 Estimate the size of these angles, then measure with a protractor. a ∠AOB b ∠AOC c Reflex ∠AOE

C B D

O

A

E © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

R

HE

T

2 State the value of the pronumeral in these diagrams. b a a° 45° 45 b° 50°

E

MA

1 Complete these sentences for this diagram. a 20° and 70° are ______________________ angles. b 20° is the _________________________ of 70°. c 130° and 50° are ______________________ angles. d 130° is the __________________________ of 50°. e The ﬁve angles in the diagram form a complete _______________.

Cambridge University Press

R K I N G

C

F PS

Y

REVISION

LL

Exercise 6A

M AT I C A

6A

WO

4 Determine the value of the pronumerals, giving reasons. a b 65° a° b°

c 52°

b°

20°

a°

b°

d

e

f a° 35°

146°

a°

a°

30°

b°

50°

g

h

a°

120°

i

130° a°

a°

b° 130°

120° j

k

l a°

b° a°

a°

41° 41

20°

c°

32° b°

5 Consider the given diagram. a Name an angle that is: i vertically opposite to ∠AOB ii complementary to ∠BOC iii supplementary to ∠AOE iv supplementary to ∠AOC b Copy and complete: i AD ⊥ ____ ii ____ ⊥ OD

R

HE

T

a°

MA

Example 1

U

B

C

A

O D E

Cambridge University Press

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C

F PS

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Chapter 6 Angle relationships and properties of geometrical figures 1

LL

368

M AT I C A

369

Measurement and Geometry

d 17°

45°

R

T

HE

MA

T

b

c

a° a° a°

70°

a° a°

a° a°

e

f (a − 5)°

(2a)° a°

(3a)°

(a + 5)°

10 Write down an algebraic expression for the size of the ∠ABC ∠ABC . a b C A C

x° D

B c

x°

E

B

A

D d

D

A C

x° B

A C

x°

R

HE

B

Cambridge University Press

R K I N G

C

F PS

Y

U

9 Find the value of a in these diagrams.

(5a)°

PS

d 17° h 167°

45° 135°

8 A round birthday cake is cut into sectors for nine friends (including Jack) at Jack’s birthday party. After the cake is cut, there is no cake remaining. What will be the angle at the centre of the cake for Jack’s piece if: a everyone receives an equal share? b Jack receives twice as much as everyone else? (In parts b, c and d assume his friends have equal shares of the rest.) c Jack receives four times as much as everyone else? d Jack receives ten times as much as everyone else?

d

F

M AT I C A

WO

a

C

LL

7 What is the supplement of the following? a 30° b 80° c e 120° f 95° g

R K I N G

LL

U

Y

WO

MA

6 What is the complement of the following? a 30° b 80° c

M AT I C A

6A

WO

U

MA

11 Explain what is wrong with these diagrams. a b A B

HE

T

c

A

D 49°

141°

43°

C 260°

C

D

37°

A B C 12 Write down an equation (e.g. 2a + b = 90) for these diagrams. a b a° (2b)°

B

And c a°

a° (3b)°

b° 13 Consider this diagram (not drawn to scale). a Calculate the value of a. b Explain what is wrong with the way the diagram is drawn.

(a + 50)° (a − 90)°

Enrichment: Geometry with equations 14 Equations can be helpful in solving geometric problems in which more complex expressions are involved. Find the value of a in these diagrams. a b c (3a)° a°

a°

(2a)°

d

130°

(4a)°

e

(2a + 20)°

f a° a°

a°

(3a)°

(3a)° (2a − 12)°

(7a + 21)°

(4a)° (3a − 11)°

R

Cambridge University Press

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C

F PS

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Chapter 6 Angle relationships and properties of geometrical figures 1

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370

M AT I C A

Measurement and Geometry

6B Transversal lines and parallel lines

R E V I S I ON

Euclid’s ﬁfth axiom is: Given a line (shown in blue) and a point not on the line, there is only one line (shown in red) through the given point and in the same plane that does not intersect the given line. line

point

parallel line

In simple language, Euclid’s ﬁfth axiom says that parallel lines do not intersect. All sorts of shapes and solids both in the theoretical and practical worlds can be constructed using parallel lines. If two lines are parallel and are cut by a third line called a transversal, special pairs of angles are created.

Parallel lines never intersect.

Let’s start: Hidden transversals This diagram can often be found as part of a shape such as a parallelogram or other more complex diagram. To see the relationship between a and b more easily, you can extend the lines to form this second diagram. In this new diagram, you can now see the pair of parallel lines and the relationships between all the angles. • Copy the new diagram. • Label each of the eight angles formed with the pronumeral a or b, whichever is appropriate. • What is the relationship between a and b? Can you explain why?

a°

a°

b°

b°

Parallel lines and transversals are visible in many structures. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

371

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Key ideas

372

Chapter 6 Angle relationships and properties of geometrical figures 1

■■ ■■

A transversal is a line cutting two or more other lines. When a transversal crosses two or more lines, pairs of angles can be: – corresponding (in corresponding positions)

corresponding

– alternate (on opposite sides of the transversal and inside the other two lines)

alternate

– co-interior (on the same side of the transversal and inside the other two lines)

co-interior

– vertically opposite

– angles on a straight line

■■

Lines are parallel if they do not intersect. – Parallel lines are marked with the same number of arrows. or

– In the diagram below, it is acceptable to write AB || DC or BA || CD but not AB || CD.

B A

C

D ■■

If two parallel lines are cut by a transversal – the corresponding angles are equal Note: There are four pairs of corresponding angles.

b°

a=b

a°

Cambridge University Press

– the alternate angles are equal Note: There are two pairs of alternate angles.

a° b°

a°

b°

a=b

– the co-interior angles are supplementary (sum to 180°). Note: There are two pairs of co-interior angles.

y°

y° x°

x° ■■

x + y = 180

The eight angles can be grouped in the following way: In this diagram: a=c=e=g b=d=f=h

a° b° d° c° e° f ° h° g°

Example 2 Finding angles on parallel lines Find the value of the pronumerals in these diagrams, giving reasons. a b 72° b° a° 120° b°

c°

a°

SOLUTION

EXPLANATION

a a = 120 (corresponding angles on parallel lines)

Corresponding angles on parallel lines are equal.

b = 120 (vertically opposite angles)

Vertically opposite angles are equal.

c + 120 = 180 (co-interior angles on parallel lines) c = 60

b° and c° are co-interior angles and sum to 180°. Alternatively, look at a° and c°, which are adjacent supplementary angles.

b a + 72 = 180 (co-interior angles on parallel lines) a = 108 b + 72 = 180 (co-interior angles on parallel lines) b = 108

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Key ideas

Measurement and Geometry

The pairs of angles are co-interior, which are supplementary because the lines are parallel.

Cambridge University Press

Chapter 6 Angle relationships and properties of geometrical figures 1

WO

F

F PS

H D C

B A G

E

a°

f

a° b°

a°

R

HE

T

e

U

MA

3 State whether the following pairs of marked angles are corresponding, alternate or co-interior. For each diagram write down the relationship between a° and b°. a b c a° a° b° a° b° b°

b°

C

M AT I C A

WO

d

R K I N G

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HE

T

2 Name the angle that is: a corresponding to ∠ABF b corresponding to ∠BCG c alternate to ∠FBC d alternate to ∠CBE e co-interior to ∠HCB f co-interior to ∠EBC g vertically opposite to ∠ABE h vertically opposite to ∠HCB

R

MA

1 Two parallel lines are cut by a transversal. Write the missing word. (equal or supplementary) a Corresponding angles are _______________. b Co-interior angles are _______________. c Alternate angles are _______________.

U

Y

REVISION

b°

Cambridge University Press

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C

F PS

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Exercise 6B

LL

374

M AT I C A

375

Measurement and Geometry

c

110°

120°

b° b°

a° b°

d

e

a°

c°

a°

c°

74°

a°

f

b°

40°

a°

80° b°

b°

a°

95°

Example 2b

5 Find the value of the pronumerals in these diagrams, giving reasons. a b c a° 39° 122°

b°

80°

b°

d

and 118° to°

a°

a°

b°

b° 61°

75° a° b°

f 64°

R

HE

T

b

MA

a

U

a° 30° 25° b°

Cambridge University Press

R K I N G

C

F PS

Y

WO

4 Find the value of the pronumerals in these diagrams, giving reasons.

LL

Example 2a

M AT I C A

6b

WO

U

MA

R

T

HE

A

D 71° A

99° 81°

69°

C

F PS

M AT I C A

C

A

B

U

c 61°

a°

70° a°

d

e

a°

40° 40

67°

f 117°

65° a° 37° a° 31° 31

g

h a°

i

a°

R

HE

T

40°

MA

7 Find the value of a in these diagrams. a b a°

a° 82°

65°

120° 85°

Cambridge University Press

R K I N G

C

F PS

Y

WO

290°

C

97°

93° D

R K I N G

LL

6 For each of the following diagrams decide if AB and CD are parallel. Explain each answer. a b c C D B B

Y

Chapter 6 Angle relationships and properties of geometrical figures 1

LL

376

M AT I C A

377

Measurement and Geometry

70°

C

B Apply a similar technique to ﬁnd ∠AOB in these diagrams. a

b

A O 80°

A B

50°

B

c

d

O

O

A

20° 110°

116°

A

B

e

45° 45 O

50°

B

26°

147°

f

A

B

39° 39

B

155°

119°

A

O

O

9 Write an algebraic expression for the size of ∠AOB. a

B

O

c

O b°

A

A

a° A

b

a°

O a°

B

B

R

HE

T

O

40° 40° 70°

MA

A

Cambridge University Press

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C

F PS

Y

8 Sometimes parallel lines can be added to a diagram to help ﬁnd an unknown angle. For example, ∠AOB can be found in this diagram by ﬁrst drawing the dashed line and ﬁnding ∠AOC (40°) and ∠COB (70°). So ∠AOB = 40° + 70° = 110°.

U

LL

WO

M AT I C A

378

6b

Chapter 6 Angle relationships and properties of geometrical figures 1

Enrichment: Pipe networks 10 A plan for a natural gas plant includes many intersecting pipe lines some of which are parallel. Help the designers ﬁnish the plans by calculating the values of a, b etc.

146°

110°

130°

a°

e°

115° 170°

g°

145°

b° c° 50°

d°

f°

165°

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379

6C Triangles A triangle is a polygon with three sides. The triangle is a very rigid shape and this leads to its use in the construction of houses and bridges. It is one of the most commonly used shapes in design and construction. Knowing the properties of triangles can help to solve many geometrical problems and this knowledge can be extended to explore other more complex shapes.

Triangles are often used to striking effect in architecture, as shown by part of the National Gallery of Canada.

Let’s start: Illustrating the angle sum and the triangle inequality

■■

A triangle has: – 3 sides: AB, AC, BC – 3 vertices (singular ‘vertex’): A, B, C – 3 interior angles: ∠ABC, ∠BAC, ∠ACB

Key ideas

You can complete this task using a pencil and ruler or using dynamic geometry software. • Draw any triangle and measure each interior angle and each side. • Add all three angles to ﬁnd the angle sum of your triangle. • Compare your angle sum with the results of others. What do you notice? • Add the two shorter sides together and compare that sum to the longest side. • Is the longest side opposite the largest angle? If dynamic geometry is used, drag one of the vertices to alter the sides and the interior angles. Now check to see if your conclusions remain the same.

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Triangles classiﬁed by side lengths – Sides with the same number of dashes are of equal length. Scalene Isosceles

Equilateral 60°

60°

60°

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■■

Triangles classiﬁed by interior angles Acute-angled Right-angled (All angles acute) (one right angle)

Obtuse-angled (one obtuse angle)

The angle sum of a triangle is 180°.

a + b + c = 180

a° b°

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c°

The exterior angle theorem: The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

x° Exterior angle

y° z°

x+y=z

Example 3 Using the angle sum of a triangle Find the value of a in these triangles. a a°

b

a° 26°

38°

92°

SOLUTION

EXPLANATION

a a + 38 + 92 = 180 (angle sum of triangle) a + 130 = 180 a = 50

The angle sum of the three interior angles of a triangle is 180°. Also 38 + 92 = 130 and 180 – 130 = 50.

b 2a + 26 = 180 (angle sum of a triangle) 2a = 154 a = 77

The angles opposite the equal sides are equal.

a° 26° a°

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Example 4 Using the exterior angle theorem Find the value of a.

C

161°

a°

B EXPLANATION

a + 90 = 161 (exterior angle of ABC ) a = 71

Use the exterior angle theorem for a triangle. The exterior angle (161°) is equal to the sum of the two opposite interior angles.

or ∠ABC ∠ABC = 180 – 161 = 19 (angles on a straight line) so a = 180 – (19 + 90) (angle sum of a triangle) = 71

Alternatively ﬁnd ∠ABC ∠ABC (19°), then use the triangle angle sum to ﬁnd the value of a.

Exercise 6C

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d

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3 Classify these triangles as acute, right or obtuse. a b

f

c

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1 Give the common name of a triangle with these properties. a One right angle b Two equal side lengths c All angles acute d All angles 60° e One obtuse angle f Three equal side lengths g Two equal angles h Three different side lengths

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4 Use the angle sum of a triangle to help ﬁnd the value of the pronumeral in these triangles. a b c a° a° 116°° a° 30°

24° 32°

d

e

f

a°

127° a°

a° 54°

92°

17°

71°

5 These triangles are isosceles. Find the value of a. a b

c a° 80°

a° Example 3b

37°

68°

a°

6 Find the value of a in these isosceles triangles. a b 50° a° a°

c a°

100° 28°

Example 4

7 Find the value of a. a

b

c

a°

130°

a°

70° 80° 70°

50° d

e

a°

f a°

110°

100°

a° 60° 115°

a°

60°

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Example 3a

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8 Decide if it is possible to draw a triangle with the given description. Draw a diagram to support your answer. a Right and scalene b Obtuse and equilateral c Right and isosceles d Acute and isosceles e Acute and equilateral f Obtuse and isosceles

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9 Use your knowledge of parallel lines and triangles to ﬁnd the unknown angle a. a b c 74° 55° 35° 70° a°

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85° a°

10 Find the value of a in these diagrams, giving reasons. a b A a° B

C

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A

B 72°

22°

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a°

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29°

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B e A

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116° a°

A 155°

150°

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49°

a° B

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11 A triangle is constructed using two radii and a chord. a What type of triangle is ∆AOB ∆ ? Explain. b Name two angles that are equal. c Find ∠ABO ∠ if ∠BAO is 30°. d Find ∠AOB ∠ if ∠OAB is 36°. e Find ∠ABO ∠ if ∠AOB ∠ is 100°.

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13 Prove that a triangle cannot have two right angles. 14 Prove that an equilateral triangle must have 60° angles. 15 A different way of proving the angle sum of a triangle is to use this diagram. a Give a reason why ∠BCD = b. b What do you know about the two angles ∠BAC and ∠ACD and why? c What do parts a and b above say about the pronumerals a, b and c, and what does this say about the triangle ABC?

B D b°

A

a°

c° C

Steel girders arranged in triangular shapes combine strength and lightness in load-bearing structures.

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Enrichment: Angle in a semicircle 16 The angle sum of a triangle can be used to prove other theorems, one of which relates to the angle in a semicircle. This theorem says that ∠ACB in a semicircle is always 90° where AB is a diameter.

C B

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A C a Use your knowledge of isosceles triangles to ﬁnd the value of a, b and c in this circle. b What do you notice about the sum of a and c?

a°

c° b°

30°

c Repeat parts a and b above for this circle. a°

c°

b°

60°

d Repeat parts a and b above for this circle. a°

c° b°

16°

e What do you notice about the sum of a and c for all the circles above?

f

Prove this result generally by ﬁnding: i a, b and c in terms of x ii the value of a + c.

c° a° x°

b°

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6D Quadrilaterals Quadrilaterals are four-sided polygons. All quadrilaterals have the same angle sum, but other properties depend on such things as pairs of sides of equal length, parallel sides and lengths of diagonals. All quadrilaterals can be drawn as two triangles and, since the six angles inside the two triangles make up the four angles of the quadrilateral, the angle sum is 2 × 180° = 360°.

180° 180°

Some quadrilaterals possess special properties and can therefore be classiﬁed as one or more of the following: • kite • trapezium • parallelogram • rhombus • rectangle • square

Quadrilaterals are frequently used in tiling patterns.

Let’s start: Which quadrilateral? Name all the different quadrilaterals you can think of that have the properties listed below. There may be more than one quadrilateral for each property listed. Draw each quadrilateral to illustrate the shape and its features. • The opposite sides are parallel • The opposite sides are equal • The adjacent sides are perpendicular • The opposite angles are equal

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Every quadrilateral has two diagonals. – In some quadrilaterals the diagonals bisect each other (i.e. cut each other in half) Quadrilaterals can be convex or non-convex. – Convex quadrilaterals have all vertices pointing outwards. – Non-convex (or concave) quadrilaterals have one vertex pointing inwards. – Both diagonals of convex quadrilaterals lie inside the ﬁgure. Non-convex Convex

All interior angles less than 180°

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Special quadrilaterals Square

One reflex interior angle

Rectangle

Parallelogram

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Rhombus

Kite

Trapezium

The angle sum of any quadrilateral is 360°. b° a°

a° b° d°

c° c°

d° a + b + c + d = 360 ■■

Quadrilaterals with parallel sides contain two pairs of co-interior angles.

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a°

d°

c°

c + d = 180

b° a + b = 180

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Chapter 6 Angle relationships and properties of geometrical figures 1

Example 5 Using the angle sum of a quadrilateral Find the value of the pronumerals in these quadrilaterals. a b 100° b° a°

a°

115°

77° EXPLANATION Two angles inside parallel lines are co-interior and therefore add to 180°. Opposite angles in a parallelogram are equal.

b a + 100 + 90 + 115 = 360 (angle sum of a quadrilateral) a + 305 = 360 a = 55

The sum of angles in a quadrilateral is 360°.

Exercise 6D

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b

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2 Use a ruler and protractor to make a neat and accurate drawing of these special quadrilaterals. 3 cm rectangle

3 cm kite

3 cm

square 7 cm

7 cm

7 cm

rhombus 70° 5 cm

parallelogram 70° 7 cm

3 cm

m

5 cm

trapezium

70° 7 cm

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a a + 77 = 180 (co-interior angles in parallel lines) a = 103 b = 77 (opposite angles in parallelogram)

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SOLUTION

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trapezium

kite

parallelogram rectangle

rhombus

square

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Property

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Use your shapes to write YES in the cells in the table, for the statements that are deﬁnitely true.

M AT I C A

The opposite sides are parallel All sides are equal The adjacent sides are perpendicular The opposite sides are equal The diagonals are equal The diagonals bisect each other The diagonals bisect each other at right angles The diagonals bisect the angles of the quadrilateral

a°

c a° 52°

a°

76°

Example 5b

4 Use the quadrilateral angle sum to ﬁnd the value of a in these quadrilaterals. a b c a° 88° 110° 115° 70°

a° a°

96°

81°

84°

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Example 5a

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25°

75°

5 By considering the properties of the given quadrilaterals, give the values of the pronumerals. a b c b cm 2 cm 3 cm 70° c° 100°

b°

a

a cm

5

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cm b°

50°

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a°

85° 27°

97°

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and a°

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62°

31° 31

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106°

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a° 55°

7 Use your algebra skills to ﬁnd the value of x in these diagrams. a b c 95° 95° (5x)° (2x)° x°

(4x)° (3x)° x°

(2x)°

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8 The word ‘bisect’ means to cut in half. a Which quadrilaterals have diagonals that bisect each other? b Which quadrilaterals have diagonals that bisect all their interior angles?

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10 Is it possible to draw a non-convex quadrilateral with two or more interior reflex angles? Explain and illustrate.

11 Complete these proofs of two different angle properties of quadrilaterals. a Angle sum = a + b + c + ____ + ____ + ____ = ____ + ____ (angle sum of a triangle) = ____

b ∠ADC ∠ADC = 360 – (____ + ____ + ____) (angle sum of a quadrilateral) Reflex ∠ADC ∠ADC = 360 – ∠ADC ∠ = 360 – (360 – (____ + ____ + ____)) = ____ + ____ + ____

c°

dd°

e°

b° a°

f°

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A

a° b°

c°

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B

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9 By considering the properties of special quadrilaterals, decide if the following are always true. a A square is a type of rectangle. b A rectangle is a type of square. c A square is a type of rhombus. d A rectangle is a type of parallelogram. e A parallelogram is a type of square. f A rhombus is a type of parallelogram.

Enrichment: Proof

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Chapter 6 Angle relationships and properties of geometrical figures 1

6E Polygons

E X T E N S I ON

The word ‘polygon’ comes from the Greek words poly, meaning ‘many’, and gonia, meaning ‘angles’. The number of interior angles equals the number of sides and the angle sum of each type of polygon depends on this number. Also, there exists a general rule for the angle sum of a polygon with n sides, which we will explore in this section. The Pentagon is a famous government ofﬁce building in Washington, USA.

Let’s start: Developing the rule The following procedure uses the fact that the angle sum of a triangle is 180°, which was proved in an earlier section. Complete the table and try to write the general rule in the ﬁnal row. Shape

Number of sides

Number of triangles

Angle sum

Triangle

3

1

1 × 180° = 180°

Quadrilateral

4

2

____ × 180° = ____

Pentagon

5

Hexagon

6

Heptagon

7

Octagon

8

n-sided polygon

n

( ____ ) × 180°

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Polygons can be convex or non-convex. – Convex polygons have all vertices pointing outwards. – Non-convex (or concave) polygons have at least one vertex pointing inwards. Con x Conve

Non-conve Non-with x

This means all interior angles are less than 180°

This means there is at least one reflex interior angle

Polygons are named according to their number of sides. Number of sides

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Name

3

Triangle

4

Quadrilateral

5

Pentagon

6

Hexagon

7

Heptagon

8

Octagon

9

Nonagon

10

decagon

11

Undecagon

12

Dodecagon

The angle sum S of a polygon with n sides is given by the rule: S = (n – 2) × 180. A regular polygon has sides of equal length and equal interior angles.

A regular octagon

Example 6 Finding the angle sum Find the angle sum of a heptagon. SOLUTION

EXPLANATION

S = (n – 2) × 180 = (7 – 2) × 180 = 5 × 180 = 900

A heptagon has 7 sides so n = 7. Simplify (7 – 2) before multiplying by 180°.

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Chapter 6 Angle relationships and properties of geometrical figures 1

Example 7 Finding angles in polygons Find the value of a in this pentagon.

80°

95° 170° a°

SOLUTION

EXPLANATION

S = (n – 2) × 180 = (5 – 2) × 180 = 540

First calculate the angle sum of a pentagon using n = 5.

a + 170 + 80 + 90 + 95 = 540 a + 435 = 540 a = 105

Sum all the angles and set this equal to the angle sum of 540°. Then simplify and solve for a.

Example 8 Finding interior angles of regular polygons Find the size of an interior angle in a regular octagon. EXPLANATION

S = (n – 2) × 180 = (8 – 2) × 180 = 1080

First calculate the angle sum of a octagon using n = 8.

Angle size = 1080 ÷ 8 = 135

All 8 angles are equal in size so divide the angle sum by 8.

Exercise 6E

EXTENSION

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c n = 22

b n = 10

3 What is the common name given to these polygons? a Regular quadrilateral b Regular triangle 4 Regular polygons have equal interior angles. Find the size of an interior angle for these regular polygons with the given angle sum. a Pentagon (540°) b Decagon (1440°) c Octagon (1080°) © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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1 State the number of sides on these polygons. a Hexagon b Quadrilateral d Heptagon e Pentagon

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6 Find the value of a in these polygons. a b a° 110°

a°

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e 60°

f 30°

30°

320°

a°

40°

215°

a°

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265° 30°

30°

7 Find the size of an interior angle of these regular polygons. Round the answer to one decimal place where necessary. a Regular pentagon b Regular heptagon c Regular undecagon d Regular 32-sided polygon

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5 Find the angle sum of these polygons. a Hexagon b Nonagon d 15-sided polygon e 45-sided polygon

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40°

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16° x° 85°

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10 Consider a regular polygon with a very large number of sides (n). a What shape does this polygon look like? b Is there a limit to the size of a polygon angle sum or does it increase to inﬁnity as n increases? c What size does each interior angle approach as n increases?

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11 Let S be the angle sum of a regular polygon with n sides. a Write a rule for the size of an interior angle in terms of S and n. b Write a rule for the size of an interior angle in terms of n only. c Use your rule to ﬁnd the size of an interior angle of these polygons. Round to 2 decimal places where appropriate. i Regular dodecagon ii 82-sided regular polygon

Enrichment: Unknown challenges 12 Find the number of sides of a regular polygon if each interior angle is a 120° b 162° c 147.272727…° 13 With the limited information provided, ﬁnd the value of x in these diagrams. a b c 40° 60° 75° 50°

x°

100° x°

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120°

x°

150°

40° 40

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6F Line symmetry and rotational symmetry You see many symmetrical geometrical shapes in nature. The starﬁsh and sunflower are two examples. Shapes such as these may have two types of symmetry: line and rotational.

Let’s start: Working with symmetry On a piece of paper draw a square (with side lengths of about 10 cm) and a rectangle (with length of about 15 cm and width of about 10 cm), then cut them out. • How many ways can you fold each shape in half so that the two halves match exactly? The number of creases formed will be the number of lines of symmetry.

Starﬁsh and sunflowers are both symmetrical, but in different ways.

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An axis or line of symmetry divides a shape into two equal parts. It acts as a mirror line, with each half of the shape being a reflection of the other. This isosceles triangle has one line (axis) of symmetry. The plural of axis is axes. A rectangle has two lines (axes) of symmetry.

Cambridge University Press

Key ideas

• Now locate the centre of each shape and place a sharp pencil on this point. Rotate the shape 360°. How many times does the shape make an exact copy of itself in its original position? This number describes the rotational symmetry.

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The order of rotational symmetry refers to the number of times a ﬁgure coincides with its original position in turning through one full rotation. We say that there is no rotational symmetry if the order of rotational symmetry is 1.

2 1

3

Example 9 Finding the symmetry of shapes Give the number of lines of symmetry and the order of rotational symmetry for each of these shapes. a rectangle b regular pentagon

SOLUTION

EXPLANATION

a 2 lines of symmetry

rotational symmetry: order 2

b 5 lines of symmetry

rotational symmetry: order 5

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iii 3

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iii 3

iv 4

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7 Of the capital letters of the alphabet shown here, state which have: a 1 line of symmetry A B C D E F G H b 2 lines of symmetry N O P Q R S T U c rotational symmetry of order 2

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three and

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12 A trapezium has one pair of parallel sides. a State whether trapeziums always have: i line symmetry ii rotational symmetry b What type of trapezium will have one line of symmetry? Draw one.

Enrichment: Symmetry in 3D 13 Some solid objects also have symmetry. Rather than line symmetry, they have plane symmetry. This cube shows one plane of symmetry, but there are more that could be drawn.

State the number of planes of symmetry for each of these solids. a Cube b Rectangular prism

c Right square pyramid

d Right triangular prism

f

e Cylinder

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6G Euler’s formula for three-dimensional solids A solid occupies three-dimensional space. The outside surfaces could be flat or curved and the number of surfaces will vary depending on the properties of the solid. A solid with all flat surfaces is called a polyhedron, plural polyhedra or polyhedrons. The word ‘polyhedron’ comes from the Greek words poly, meaning ‘many’, and hedron, meaning ‘faces’.

The top of this Canary Wharf building in London (left) is a large, complex polyhedron. Polyhedra also occur in nature, particularly in rock or mineral crystals such as quartz (right).

Let’s start: Developing Euler’s formula Create a table with each polyhedron listed below in its own row. Include the name and a drawing of each polyhedron. Add columns to the table for faces (F), vertices (V), edges (E) and faces plus vertices added together (F + V). Count the faces, vertices and edges for each polyhedron and record the numbers in the table. Tetrahedron

Hexahedron

Pentagonal pyramid

• What do you notice about the numbers in the columns for E and F + V? • What does this suggest about the variables F, V and E? Can you write a formula? • Add rows to the table, draw your own polyhedra and test your formula by ﬁnding the values for F, V and E.

Cambridge University Press

F R I N GE

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A polyhedron (plural: polyhedra) is a closed solid with flat surfaces (faces), vertices and edges. – Polyhedra can be named by their number of faces: e.g. tetrahedron (4 faces), pentahedron (5 faces) and hexahedron (6 faces) Euler’s formula for polyhedra with F faces, V vertices and E edges is given by: E=F+V–2 Prisms are polyhedra with two identical (congruent) Hexagonal ends. The congruent ends deﬁne the cross-section prism of the prism and also its name. The other faces are parallelograms. If these faces are rectangles, as shown, then the solid is a right prism. Pyramids are polyhedra with a base face and all other faces meeting at the same vertex point called the apex. They are named by the shape of the base.

Some solids have curved surfaces. Common examples include: Cylinder Sphere

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A cube is a hexahedron with six square faces.

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A cuboid is a common name used for a rectangular prism.

apex

Square pyramid

Cone

Natural hexagonal prisms of rock at the Giant’s Causeway in Northern Ireland formed when basalt lava cooled quickly and cracked in quite regular patterns.

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Key ideas

Measurement and Geometry

Chapter 6 Angle relationships and properties of geometrical figures 1

Example 10 Classifying solids a Classify these solids by considering the number of faces. i ii

iii

b Name these solids as a type of prism or pyramid, e.g. hexagonal prism or hexagonal pyramid. SOLUTION

EXPLANATION

a i Hexahedron

The solid has 6 faces.

ii Heptahedron

The solid has 7 faces.

iii Pentahedron

The solid has 5 faces.

b i Rectangular prism

It has two rectangular ends with rectangular sides.

ii Pentagonal prism

It has two pentagonal ends with rectangular sides.

iii Square pyramid

It has a square base and four triangular faces meeting at an apex.

Example 11 Using Euler’s formula Use Euler’s formula to ﬁnd the number of faces on a polyhedron that has 10 edges and 6 vertices. EXPLANATION

E=F+V–2 10 = F + 6 – 2 10 = F + 4 F=6

Write down Euler’s formula and make the appropriate substitutions. Solve for F, which represents the number of faces.

Exercise 6G

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1 Write the missing word in these sentences. a A polyhedron has faces, ____________ and edges. b A heptahedron has ______ faces. c A prism has two ____________ ends. d A pentagonal prism has ____________ faces. e The base of a pyramid has 8 sides. The pyramid is called a ____________ pyramid.

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c 12 = 6 + V – 2

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3 Count the number of faces, vertices and edges (in that order) on these polyhedra. a b c

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2 Find the value of the pronumeral in these equations. a E = 10 + 16 – 2 b 12 = F + 7 – 2

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4 Which of these solids are polyhedra (i.e. have only flat surfaces)? A Cube B Pyramid C Cone D Sphere E Cylinder F Rectangular prism G Tetrahedron H Hexahedron

Example 10b

Example 11

6 How many faces do you think these polyhedra have? a Octahedron b Hexahedron c Tetrahedron e Heptahedron f Nonahedron g Decahedron

d Pentahedron h Undecahedron

7 Name these prisms. a

b

c

8 Name these pyramids. a

b

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Number of faces ((F )

Number of vertices (V )

Number of edges ((E )

F+V

Cube Square pyramid Tetrahedron Octahedron

b Compare the number of edges (E) with the value F + V for each polyhedron. What do you notice?

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9 a Copy and complete this table. Solid

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5 Name a polyhedron that has the given number of faces. a 6 b 4 c 5 e 9 f 10 g 11

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6 ___ 5 7 ___ 11

8 5 ___ ___ 4 11

___ 8 9 12 6 ___

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10 Use Euler’s formula to calculate the missing numbers in this table.

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11 a A polyhedron has 16 faces and 12 vertices. How many edges does it have? b A polyhedron has 18 edges and 9 vertices. How many faces does it have? c A polyhedron has 34 faces and 60 edges. How many vertices does it have?

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13 Decide if it is possible to cut the solid using a single straight cut, to form the new solid given in the brackets. a Cube (rectangular prism) b Square based pyramid (tetrahedron) c Cylinder (cone) d Octahedron (pentahedron) e Cube (heptahedron) 14 This solid is like a cube but is open at the top and bottom and there is a square hole in the middle forming a tunnel. Count the number of faces (F), F), vertices (V F V) and edges (E) then decide if Euler’s formula is true for such solids.

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12 Decide if the following statements are true or false. Make drawings to help. a A tetrahedron is a pyramid. b All solids with curved surfaces are cylinders. c A cube and a rectangular prism are both hexahedrons. d A hexahedron can be a pyramid. e There are no solids with 0 vertices. f There are no polyhedra with 3 surfaces. g All pyramids will have an odd number of faces.

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16 Rearrange Euler’s formula. a Write V in terms of F and E. b Write F in terms of V and E. 17 Show that Euler’s formula applies for these solids. a Heptagonal pyramid b Octagonal prism c Octahedron 18 Are the following statements true or false? a For all pyramids, the number of faces is equal to the number of vertices. b For all convex polyhedra, the sum E + V + F is even.

Enrichment: Convex solids 19 Earlier you learnt that a convex polygon will have all interior angles less than 180°. Notice also that all diagonals in a convex polygon are drawn inside the shape. Convex polygon Non-convex polygon

Solids can also be classiﬁed as convex or non-convex. Convex solid Non-convex solid

To test for a non-convex solid, join two vertices or two faces with a line segment that passes outside the solid. a Decide if these solids are convex or non-convex. i ii iii

b Draw your own non-convex solids and check by connecting any two vertices or faces with a line segment outside the solid.

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15 a A cuboid is a common name for a solid with six rectangular faces. Name the solid in two other different ways. b A pyramid has base with 10 sides. Name the solid in two ways.

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Chapter 6 Angle relationships and properties of geometrical figures 1

Constructions Geometric construction involves a precise set of mathematical and geometric operations that do not involve any approximate measurements or other guess work. The basic tools for geometric construction are a pair of compasses, a straight edge and a pencil or drawing pen. Computer geometry or drawing packages can also be used, and include digital equivalents of these tools.

For the following constructions use only a pair of compasses, a straight edge and a pencil. Alternatively, use computer geometry software and adapt the activities where required.

Perpendicular line Construct: a a segment AB b a circle with centre A c a circle with centre B d a line joining the intersection points of the two circles

A

B

Perpendicular bisector Repeat the construction for a perpendicular line, but ensure that the two circles have the same radius. If computer geometry is used, use the length of the segment AB for the radius of both circles.

A 60° angle Construct: a a ray AB b an arc with centre A c the intersection point C d an arc with centre C and radius AC e a point D at the intersection of the two arcs f a ray AD

D

60° A

C

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Measurement and Geometry

Equilateral triangle Repeat the construction for a 60° angle then construct the segment CD.

Angle bisector Construct: a any angle ∠BAC b an arc with centre A c the two intersection points D and E d two arcs of equal radius with centres at D and E e the intersection point F f the ray AF

B D F A C

E

Parallel line through a point Construct: a a line AB and point P b an arc with centre A and radius AP c the intersection point C d an arc with centre C and radius AP e an arc with centre P and radius AP f the intersection point D g the line PD

D P B C A

Rhombus Repeat the construction for a parallel line through a point and construct the segments AP and CD.

Construction challenges For a further challenge try to construct these objects. No measurement is allowed. a 45° angle b Square c Perpendicular line at the end of a segment d Parallelogram e Regular hexagon

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Chapter 6 Angle relationships and properties of geometrical figures 1

1 This shape includes 12 matchsticks. (To solve these puzzles all matches remaining must connect to other matches at both ends.) a Remove 2 matchsticks to form 2 squares. b Move 3 matchsticks to form 3 squares. 2 a Use 9 matchsticks to form 5 equilateral triangles. b Use 6 matchsticks to form 4 equilateral triangles. 3 Find the value of x in these diagrams. a

b

x°

x°

4 Find the size of ∠ABC in this quadrilateral. A 60°

B

110°

D

C 5 Is it possible to draw a net for these solids? If so, draw the net. b a

Sphere Cone 6 Find the value of a + b + c + d + e in this star. Give reasons for your answer.

b°

a° and°

c° d°

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Angles in a right angle a + 25 = 90 a = 65

a° 25°

Angles on a straight line 125° a°

Angles in a revolution

a + 125 = 180 a = 55

(2a)°

Vertically opposite angles a°

40°

Parallel lines • Corresponding angles are equal (a = b) • Alternate angles are equal (a = c) • Co-interior angles are supplementary (a + d = 180)

a = 40

Angle relationships and properties of geometrical figures 1

Triangles

b°

Angle sum = 180° Scalene 100° 35°

a°

c°

Isosceles a°

Quadrilaterals (Square, Rectangle, Rhombus, Parallelogram, Kite, Trapezium) Angle sum = 360° Non-convex Parallelogram

2a + 50 = 180 2a = 130 a = 65

Exterior angles c° c = a + b

40° 240°

a°

a° 70°

35°

a° 120°

d° a°

50°

a + 135 = 180 a = 45

b°

2a + 240 = 360 2a = 120 a = 60 240°

a + 315 = 360 a = 45

b + 300 = 360 b = 60 a = 120 b° a°

a + 70 = 180 a = 110

Symmetry

• 5 lines of symmetry • Rotational symmetry of order 5 regular pentagon

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Chapter summary

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Chapter 6 Angle relationships and properties of geometrical figures 1

Multiple-choice questions 1 What is the name given to two angles that add to 90°? A Right B Supplementary C Revolutionary D Complementary E Vertically opposite 2 The value of a in this diagram is: A b+c B c+d C b+d D 180 – a E d + 180

c° b°

3 The value of a in this diagram is: A 45 B 122 C 241 D 119 E 61

d°

a°

119° a°

4 Which quadrilateral has four equal sides and diagonals equal in length? A Kite B Rhombus C Square D Parallelogram E Rectangle 5 The size of an interior angle of an equilateral triangle is: A 30° B 45° C 60° D 120° E 180° 6 The size of an exterior angle on an equilateral triangle is: A 60° B 120° C 180° D 100°

E 360°

7 What is the complement of 60°? A 15° B 30°

C 120°

D 300°

E 360°

8 What is the supplement of 60°? A 15° B 30°

C 120°

D 300°

E 360°

D 300

E 360

9 What is the value of a? a°

A 15

B 30

60°

C 120

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Measurement and Geometry

10 What is the value of a? 150° a° A 15

B 30

C 120

D 300

E 360

Short-answer questions 1 Find the value of a in these diagrams. a a° 40°

b

c a° 115°

d

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f 29°

36° a°

a°

a°

2 These diagrams include parallel lines. Find the value of a. a b a°

120°

a°

42°

c a°

81°

103° a°

132° d

e

51°

f 116°

80°

150°

a° a°

a° 59°

3 Use the dashed construction line to help ﬁnd the size of ∠AOB in this diagram. A 45° O 50°

B

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4 Use the side lengths and angle sizes to give a name for each triangle (e.g. right-angled isosceles) and ﬁnd the value of a. a b c a°

25° 120° d

a°

75°

a° e

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71°

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19° a°

5 These triangles include exterior angles. Find the value of a. a b a° 85° 152° a° 70°

a°

c 145°

71° a°

140° 6 Find the value of a and b in these quadrilaterals. a b b° b° a°

a°

79°

29°

c 30° a°

95°

b°

47° 52°

82° 7 How many axes of symmetry do the following shapes possess? a Rectangle b Parallelogram c Square d Kite 8 What is the order of rotational symmetry for these shapes? a Rectangle b Parallelogram c Square 9 Find the value of a. a a°

b

d Equilateral triangle c

a°

38°

a°

40° a°

60°

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Measurement and Geometry

10 Find the value of x in this diagram. 30°

70°

x°

Extended-response questions 1 Find the value of a, giving reasons. D A E

a° 50°

70°

B

C

2 Find the size of ∠ABE ∠ , giving reasons. B A 130° C

E

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3 In the diagram, ABDE is a rectangle. AB = 5 cm, AE = 8 cm, EC = 13 cm. Find the value of a, giving reasons. A

B

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a° C

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Chapter

7

Linear relationships 1

What you will learn

7A 7B 7C 7D 7E 7F 7G 7H 7I

The Cartesian plane Using rules, tables and graphs to explore linear relationships Finding the rule using a table of values Gradient EXTENSION Gradient–intercept form EXTENSION The x-intercept EXTENSION Solving linear equations using graphical techniques Applying linear graphs EXTENSION Non-linear graphs

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NSW Syllabus

for the Australian Curriculum Strand: Number and Algebra

Substrand: LINEAR RELATIONSHIPS

Outcome A student creates and displays number patterns; graphs and analyses linear relationships; and performs transformations on the Cartesian plane. (MA4–11NA)

Mining optimisation The Australian mining companies of today spend millions of dollars planning and managing their mining operations. The viability of a mine depends on many factors including product price and quality and availability as well as environmental considerations. Through a process called linear programming, many of these factors are represented using linear equations and straight line graphs. From these equations and graphs an optimal solution can be found, which tells the company the most efﬁ cient and cost-effective way to manage all of the given factors. Such use of simple linear graphs can save companies many millions of dollars.

Cambridge University Press

Chapter 7 Linear relationships 1

1 This graph shows the relationship between age and height of two people. a Who is older? b Who is taller?

Height

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Amanda George Age

2 Write the x-coordinate for these points. a (1, 2) c (0, -1)

b d

(1, 5) (-3, 0)

3 Write the y-coordinate for these points. a (1, 6) b (-4, -1)

c

(-3, 0)

d (-4, 2)

4 The coordinates of the point A on this graph are (2, 3). What are the coordinates of these points? a B b C c D

y 3 2 1

A B

O 5 If y = x + 4, find the value of y when: a x=3 b x=0

c x = -2

6 If y = -2x - 3, find the value of y when: a x=2 b x=0

c x = -3

7 Complete the tables for the given rules. a y = 3x - 2 x

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7A The Cartesian plane On a number plane, a pair of coordinates gives the exact position of a point. The number plane is also called the Cartesian plane after its inventor, Rene Descartes, who lived in France in the seventeenth century. The number plane extends both a horizontal axis (x) and vertical axis (y) to include negative numbers. The point where these axes cross over is called the origin and it provides a reference point for all other points on the plane.

Let’s start: Make the shape In groups or as a class, see if you can remember how to plot points on a number plane. Then decide what type of shape is formed by each set. y

Coordinates on the number plane can be used to pinpoint locations on a map.

4 3 2 1 –4 –3 –2 –1–1O –2 –3 –4

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A number plane (or Cartesian plane) includes a vertical y-axis and a horizontal x-axis intersecting at right angles. – There are 4 quadrants labelled as shown. y A point on a number plane has coordinates (x, y). 4 – The x-coordinate is listed first followed by the 3 Quadrant 1 Quadrant 2 y-coordinate. 2 The point (0, 0) is called the origin. (0, 0) Origin 1 horizontal vertical x –4 –3 –2 –1–1O 1 2 3 4 ( x , y) = units from , units froom –2 origin origin Quadrant 3 –3 Quadrant 4 –4

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Key ideas

• A(0, 0), B(3, 1), C(0, 4) • A(-2, 3), B(-2, -1), C(-1, -1), D(-1, 3) • A(-3, -4), B(2, -4), C(0, -1), D(-1, -1) Discuss the basic rules for plotting points on a number plane.

Chapter 7 Linear relationships 1

Example 1 Plotting points Draw a number plane extending from -4 to 4 on both axes then plot and label these points. a A(2, 3) b B(0, 4) c C(-1, 2.5) d D(-3.5, 0) e E(-2, -2.5) f F(2, -4) SOLUTION

EXPLANATION The x-coordinate is listed first followed by the y-coordinate.

y

–4 –3 –2 –1–1O –2 E –3 –4

A

1 2 3 4

x

For each point start at the origin (0, 0) and move left or right or up and down to suit both x- and y-coordinates. For point C(-1, 2.5), for example, move 1 to the left and 2.5 up.

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1 Complete these sentences. a The coordinates of the origin are ______. b The vertical axis is called the ___-axis. c The quadrant that has positive coordinates for both x and y is the ______ quadrant. d The quadrant that has negative coordinates for both x and y is the ______ quadrant. e The point (-2, 3) has x-coordinate ___. f The point (1, -5) has y-coordinate ___. y 2 Write the missing number for the coordinates g(— , 3) a(3, —) of the points a–h. 4 3 2 f(–3, —) h(— , 2) 1

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4 Draw a number plane extending from -4 to 4 on both axes and then the plot and label these points. a A(4, 1) b B(2, 3) c C(0, 1) d D(-1, 3) e E(-3, 3) f F(-2, 0) g G(-3, -1) h H(-1, -4) i I(0, -2.5) j J(0, 0) k K(3.5, -1) l L(1.5, -4) m M(-3.5, -3.5) n N(-3.5, 0.5) o O(3.5, 3.5) p P(2.5, -3.5)

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5 Using a scale extending from -5 to 5 on both axes, plot and then join the points for each part. Describe the basic picture formed. 1 1 1 a (-2, -2), (2, -2), (2, 2), (1, 3), (1, 4), 2 , 4 , 2 , 3 2 , (0, 4), (-2, 2), (-2, -2) b (2, 1), (0, 3), (-1, 3), (-3, 1), (-4, 1), (-5, 2), (-5, -2), (-4, -1), (-3, -1), (-1, -3), (0, -3), (2, -1), (1, 0), (2, 1)

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7 Each set of points forms a basic shape. Describe the shape without drawing a graph if you can. a A(-2, 4), B(-1, -1), C(3, 0) b A(-3, 1), B(2, 1), C(2, -6), D(-3, -6) c A(-4, 2), B(3, 2), C(4, 0), D(-3, 0) d A(-1, 0), B(1, 3), C(3, 0), D(1, -9) 8 The midpoint of a line segment (or interval) is the point that cuts the segment in half. Find the midpoint of the line segment joining these pairs of points. a (1, 3) and (3, 5) b (-4, 1) and (-6, 3) c (-2, -3) and (0, -2) d (3, -5) and (6, -4) e (-2.5, 1) and (2.5, 1) f (-4, 1.5) and (0, 0.5) © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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6 One point in each set is not ‘in line’ with the other points. Name the point in each case. a A(1, 2), B(2, 4), C(3, 4), D(4, 5), E(5, 6) b A(-5, 3), B(-4, 1), C(-3, 0), D(-2, -3), E(-1, -5) c A(-4, -3), B(-2, -2), C(0, -1), D(2, 0), E(3, 1) d A(6, -4), B(0, -1), C(4, -3), D(3, -2), E(-2, 0)

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Example 1

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3 Write the coordinates of the points labelled A to M.

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9 List all the points, using only integer values of x and y that lie on the line segment joining these pairs of points. a (1, -3) and (1, 2) b (-2, 0) and (3, 0) c (-3, 4) and (2, -1) d (-3, -6) and (3, 12)

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10 If (a, b) is a point on a number plane, name the quadrant or quadrants that matches the given description. a a > 0 and b < 0 b a < 0 and b > 0 c a<0 d b<0 11 A set of points has coordinates (0, y) where y is any number. What does this set of points represent?

Enrichment: Distances between points 12 Find the distance between these pairs of points. a (0, 0) and (0, 10) b (0, 0) and (-4, 0) d (0, -4) and (0, 7) e (-1, 2) and (5, 2)

c (-2, 0) and (5, 0) f (4, -3) and (4, 1)

13 When two points are not aligned vertically or horizontally, Pythagoras’ theorem can be used to find the distance between them. Find the distance between these pairs of points. a (0, 0) and (3, 4) b (0, 0) and (5, 12) c (-3, -4) and (4, 20) d (1, 1) and (4, -1 ) e (-1, -2) and (2, 7) f (-3, 4) and (3, -1)

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7B Using rules, tables and graphs to explore

linear relationships From our earlier study of formulas we know that two (or more) variables that have a relationship can be linked by a rule. A rule with two variables can be represented on a graph to illustrate this relationship. The rule can be used to generate a table that shows coordinate pairs (x, y). The coordinates can be plotted to form the graph. Rules that give straight-line graphs are described as being linear. For example, the rule linking degrees Celsius (C) with degrees Fahrenheit (F) is 5 given by C = (F - 32) and gives a straight-line graph. 9

Let’s start: They’re not all straight Not all rules give a straight-line graph. Here are three rules which can be graphed to give lines or curves. 6 1 y= 2 y = x2 3 y = 2x + 1 x • In groups, discuss which rule(s) might give a straight line graph and which might give curves. • Use the rules to complete the given table of values. 1

2

y

3

y1 y2 y3

• Discuss how the table of values can help you decide which rule(s) give a straight line. • Plot the points to see if you are correct.

9 8 7 6 5 4 3 2 1 O

■■ ■■ ■■

■■

The rule for converting Celsius to Fahrenheit gives a straight-line graph.

1 2 3

x

A rule or formula is an equation connecting two or more variables. A straight-line graph will result from a rule that is linear. For two variables, a linear rule is often written with y as the subject. For example: y = 2x - 3 or y = -x + 7 One way to graph a linear relationship using a formula or rule is to follow these steps. 1 Construct a table of values finding a y-coordinate for each given x-coordinate. Substitute each x-coordinate into the rule. 2 Plot the points given in the table on a set of axes. 3 Draw a line through the points to complete the graph.

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Chapter 7 Linear relationships 1

Example 2 Plotting a graph from a rule For the rule y = 2x 2x - 1 construct a table and draw a graph. SOLUTION

EXPLANATION

x

-3

-2

-1

1

2

3

y

-7

-5

-3

-1

1

3

5

y 5 4 3 2 1

(3, 5)

Substitute each x-coordinate in the table into the rule to find the y-coordinate. Plot each point (-3, -7), (-2, -5) … and join them to form the straight-line graph.

(2, 3) (1, 1)

–3 –2 –1–1O 1 2 3 (0, –1) –2 (–1, –3) –3 –4 (–2, –5) –5 –6 (–3, –7) –7

x

Example 3 Checking if a point lies on a line Decide if the points (1, 3) and (-2, -4) lie on the graph of y = 3x. SOLUTION

EXPLANATION

Substitute (1, 3). y = 3x LHS = y RHS = 3x =3 =3×1 =3 LHS = RHS So (1, 3) is on the line.

Substitute (1, 3) into the rule for the line. The point satisfies the equation, so the point is on the line.

Substitute (-2, -4). y = 3x LHS = y RHS = 3x = -4 = 3 × -2 = -6 LHS ≠ RHS So (-2, -4) is not on the line.

Substitute (-2, -4) into the rule for the line. The point does not satisfy the equation, so the point is not on the line.

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1 For the rule y = 2x + 3 find the y-coordinate for these x-coordinates. a 1 b 2 c 0 d -1 e -5 f -7 g 11 h -12 2 Write the missing number in these tables for the given rules. a y = 2x b y=x-3

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Exercise 7B

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y = 5x + 2

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2

x

-1

y

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-8

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2

3 Complete the graph to form a straight line from the given rule and table. Two points have been plotted for you. y = 2x - 2 y -1

1

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-6

-4

-2

2

3 2 1 –3 –2 –1–1O –2 –3 –4 –5 –6

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4 For each rule construct a table like the one shown here.

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Example 2

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3

y

a y=x+1 and y = -2x + 3 Example 3

b y=x-2 f y = -3x - 1

c g

y = 2x - 3 y = -x

R

d y = 2x + 1 h y = -x + 4

5 Decide if the given points lie on the graph with the given rule. a Rule: y = 2x Points: i (2, 4) and ii (3, 5) b Rule: y = 3x - 1 Points: i (1, 1) and ii (2, 5) c Rule: y = 5x - 3 Points: i (-1, 0) and ii (2, 12) d Rule: y = -2x + 4 Points: i (1, 2) and ii (2, 0) e Rule: y = 3 - x Points: i (1, 2) and ii (4, 0) f Rule: y = 10 - 2x Points: i (3, 4) and ii (0, 10) g Rule: y = -1 - 2x Points: i (2, -3) and ii (-1, 1)

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Chapter 7 Linear relationships 1

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C 6 For x-coordinates from -3 to 3, construct a table and draw a graph for these rules. For R PS HE M AT I C A parts c and d remember that subtracting a negative number is the same as adding its opposite, for example that 3 - (-2) = 3 + 2 1 1 a y= x+1 b y= − x-2 c y=3-x d y = 1 - 3x 2 2 7 For the graphs of these rules, state the coordinates of the two points at which the line cuts the x- and y-axes. a y=x+1 b y=2-x c y = 2x + 4 d y = 10 - 5x e y = 2x - 3 f y = 7 - 3x

8 The rules for two lines are y = x + 2 and y = 5 - 2x. At what point do they intersect?

MA

9 a What is the minimum number of points needed to draw a graph of a straight line? b Draw the graph of these rules by plotting only two points. Use x = 0 and x = 1. 1 i y= x ii y = 2x - 1 2

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11 The y-coordinates of points on the graphs of the rules in Question 4 parts a to d increase as the x-coordinates increase. Also the y-coordinates of points on the graphs of the rules in Question 4 parts e to h decrease as the x-coordinates increase. a What do the rules in Question 4 parts a to d have in common? b What do the rules in Question 4 parts e to h have in common? c What feature of a rule tells you that a graph increases or decreases as the x-coordinate increases?

Enrichment: x- and y-intercepts

2

Cambridge University Press

PS

M AT I C A

10 a The graphs of y = x, y = 3x and y = -2x all pass through the origin (0, 0). Explain why. b The graphs of y = x - 1, y = 3x - 2 and y = 5 - 2x do not pass through the origin (0, 0). Explain why.

12 A sketch of a straight-line graph can be achieved by finding only the x- and y y-intercepts and labelling these on the graph. The y-intercept is the y value of the point where x = 0. The x-intercept is the x value of the point 4 where y = 0. For example: y = 4 - 2x y-intercept (x = 0) x-intercept (y = 0) y=4-2×0 0 = 4 - 2x y=4 -4 = -2x 2=x Sketch graphs of these rules using the method outlined above. a y=x+4 b y = 2x - 4 c y=5-x d y = -1 + 2x e y = 7x - 14 f y = 5 - 3x g y = 3 - 2x h 3y - 2x = 6

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7C Finding the rule using a table of values A mathematical rule is an efficient way of describing a relationship between two variables. While a table and a graph are limited by the number of points they show, a rule can be used to find any value of y for any given x value quickly. Finding such a rule from a collection of points on a graph or table is an important step in the development and application of mathematics.

Let’s start: What’s my rule?

Businesses attempt to ﬁ nd mathematical rules from the data they collect and graph.

Each of the tables here describe a linear relationship between y and x. x

y

x

y

x

y

4

-3

-5

-2

5

1

5

-2

-3

-1

4

2

6

-1

-1

3

3

7

1

1

2

4

8

1

3

2

1

■■

A rule must be true for every pair of coordinates (x, y) in a table or graph. coefficient of x y=

■■

×x+

constant Consider a linear rule of the form y = ×x+ . – The coefﬁcient of x will be the increase in y as x increases by 1. If there is a decrease in y, then the coefficient will be negative. x

-2

-1

1

2

x

-2

-1

1

2

y

-1

1

3

5

7

y

1

-1

-2

-3

+2

+2

+2

+2

y = 2x + 3

-1

-1

-1

-1

y = -x - 1

– The constant will be the value of y when x = 0.

Cambridge University Press

Key ideas

• For each table write a rule making y the subject. • Discuss your strategy for finding the three different rules. What patterns did you notice and how did these patterns help determine the rule?

Key ideas

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Chapter 7 Linear relationships 1

– If the value of y when x = 0 is not given in the table, substitute another pair of coordinates to find the value of the constant. x

2

3

4

5

y

5

7

9

11

+2 y = 2x + 5=2×2+ So =1

+2

+2 substituting (2, 5)

Alternatively, extend the table to the left so that 0 appears in the top row. x

1

2

3

4

5

y

1

3

5

7

9

11

-2

-2

Example 4 Finding rules from tables Find the rule for these tables of values. a x

-2

-1

1

2

y

-8

-5

-2

1

4

SOLUTION a Coefficient of x is 3. Constant is -2. y = 3xx - 2

b

x

3

4

5

y

-5

-7

-9

6

7

-11 -13

EXPLANATION x

-2

-1

1

2

y

-8

-5

-2

1

4

+3

+3

+3

+3

y = 3x - 2

b Coefficient of x is -2. y = -2x 2x + 2x Substitute (3, -5). -5 = -2 × 3 + -5 = -6 + So = 1. y = -2x 2x + 1

x

3

4

5

6

7

y

-5

-7

-9

-11

-13

-2

-2

-2

-2

y = -2x 2 + 2x To find the constant substitute a point and choose the constant so that the equation is true. This can be done mentally.

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Number and Algebra

Example 5 Finding a rule from a graph and write the rule in words Find the rule for this graph by first constructing a table of (x, y) values. Write the rule in words. y 3 2 1

(3, 1)

(2, 0) O 1 2 3 –3 –2 –1–1 (1, –1) –2 (0, –2) (–1, –3)

x

SOLUTION

EXPLANATION

1

2

3

y

-3

-2

-1

1

Construct a table using the points given on the graph. Change in y is 1 for each increase by 1 in x.

Coefficient of x is 1. When x = 0, y = -2 y=x-2 Rule in words: To find a y value, start with x then subtract 2.

x

-1

1

2

3

y

-3

-2

-1

1

+1

+1

+1

+1

y = 1x - 2 or y = x - 2

Exercise 7C

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-1

1

2

y

-1

1

3

5

7

y

3

d

y (1, 3) 3 2 1 (0, 1) –3 –2 –1 OR (–1, –1) –1 –2 (–2, –3) –3

4

1 2 3

2

1

y (–1, 3) 3 2 1

x

–3 –2 –1–1O –2 –3

1 2 3

x

(1, –3)

R

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c

x

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1 By how much does y increase for each increase by 1 in x? If y is decreasing, give a negative answer. a b x -3 -2 -1 0 1

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M AT I C A

7C

MA

R

3 Find the missing number in these equations. a 4=2+ b 3=5+ d 5=3×2+ e -2 = 2 × 4 + g -8 = -4 × 2 + h -12 = -5 × 3 +

T

HE

c -1 = 2 + f -10 = 2 × (-4) + i 16 = -4 × (-3) +

Example 4b

2

4

6

8

x

-3

-2

-1

1

y

4

3

2

1

x

1

2

3

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5

y

5

9

13

17

21

x y

Example 5

d

5 Find the rule for these tables of values. a

c

5

6

7

b

8

9

b

d

-12 -14 -16 -18 -20

x

-2

-1

1

2

y

-7

-4

-1

2

5

x

-1

1

2

3

y

8

6

4

2

-4

-3

-2

-1

y

-13 -11

-9

-7

-5

x

-6

-5

-4

-3

-2

y

10

9

8

7

6

x

-5

6 Find the rule for these graphs by first constructing a table of (x, y) values. Write the rule in words. a b y y

(–1, 0)

–3 –2 –1–1O (–2, –1) –2 –3

c

x

1 2 3

d

(–1, 5) 5 4 3 2 (0, 2) 1 1 2 3 (1, –1)

(2, 2) (1, 0)

–3 –2 –1–1O 1 2 3 –2 (0, –2)

y

–3 –2 –1–1O

(3, 4)

4 3 2 1

3 2 (1, 2) 1 (0, 1)

x

x

y 3 2 (–2, 2) (–1, 1) 1 –3 –2 –1–1O 1 2 3 (1, –1) –2 (2, –2) –3

R

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c

y

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4 Find the rule for these tables of values. a x -2 -1 0 1 2

x

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Example 4a

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2 For each of the tables and graphs in Question 1, state the value of y when x = 0.

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Number and Algebra

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8 Write a rule for these matchstick patterns. Write the rule in words. a x = number of squares y = number of matchsticks

Shape 1

Shape 2

Shape 3

Shape 1

y = number of matchsticks Shape 2

Shape 3

Shape 4

b x = number of hexagons

Shape 4

c x = number of matchsticks on top row

Shape 1

y = number of matchsticks

Shape 2

Shape 3

Shape 4

9 A graph passes through the two points (0, -2) and (1, 6). What is the rule of the graph? 10 A graph passes through the two points (-2, 3) and (4, -3). What is the rule of the graph?

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7 Find the rule for these set of points. Try to do it without drawing a graph or table. a (1, 3), (2, 4), (3, 5), (4, 6) b (-3, -7), (-2, -6), (-1, -5), (0, -4) c (-1, -3), (0, -1), (1, 1), (2, 3) d (-2, 3), (-1, 2), (0, 1), (1, 0)

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11 The rule y = -2x + 3 can be written as y = 3 - 2x. Write these rules in a similar form. a y = -2x + 5 b y = -3x + 7 c y = -x + 4 d y = -4x + 10

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13 A straight line has two points (0, a) and (1, b). a Write an expression for the coefficient of x in the rule linking y and x. b Write the rule for the graph in terms of a and b. 14 In Question 8a you can observe that 3 extra matchsticks are needed for each new shape and 1 matchstick is needed to complete the first square (so the rule is y = 3x + 1). In a similar way, describe how many matchsticks are needed for the shapes in: a Question 8b b Question 8c

15 Consider this table of values.

a b c d

x

-2

2

4

y

-4

-2

2

The increase in y for each unit increase in x is not 2. Explain why. If the pattern is linear, state the increase in y for each increase by 1 in x. Write the rule for the relationship. Find the rule for these tables. i

iii

x

-4

-2

2

4

y

-5

-1

3

7

11

x

-6

-3

3

6

y

15

9

3

-3

-9

ii

iv

x

-3

-1

1

3

5

y

-10

-4

2

8

14

x

-10

-8

-6

-4

-2

y

20

12

4

-4

-12

Cambridge University Press

PS

M AT I C A

12 A straight line has two points (0, 2) and (1, b). a Write an expression for the coefficient of x in the rule linking y and x. b Write the rule for the graph in terms of b.

Enrichment: Skipping x-values

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Number and Algebra

7D Gradient

433

E X T E N S I ON

The gradient of a line is a measure of how steep the line is. The steepness or slope of a line depends on how far it rises or falls over a given horizontal distance. This is why the gradient is calculated by dividing the vertical rise by the horizontal run between two points. Lines that rise (from left to right) have a positive gradient and lines that fall (from left to right) have a negative gradient.

i tiv

Pos

No

nt

die

gra e v i it

t

ien

rad

It’s more difﬁ cult for a plane to take off at a steep gradient than a gradual one.

Let’s start: Which is the steepest?

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The gradient is a measure of slope. – It is the increase in y as x increases by 1. – It is the ratio of the change in y over the change in x. rise – Gradient = run – Rise = change in y – Run = change in x – The run is always considered to be positive when moving from left to right.

Wall

Positive gradient rise run 4 = 2 2 = 1 =2

Gradient = 4 2 1 2

Negative gradient rise run −3 = 2

Gradient =

3 2

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Key ideas

At a children’s indoor climbing centre there are three types of sloping walls to climb. The blue wall rises 2 metres for each metre across. The red wall rises 3 metres for every 2 metres across and the yellow wall rises 7 metres for every 3 metres across. • Draw a diagram showing the slope of each wall. • Label your diagrams with the information given above. • Discuss which wall might be the steepest giving reasons. • Discuss how it might be possible to accurately compare the slope of each wall.

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Chapter 7 Linear relationships 1

■■ ■■ ■■

A gradient is negative if y decreases as x increases. The rise is considered to be negative. The gradient of a horizontal line is 0. The gradient of a vertical line is undefined. Gradient = 2 , 0 which is undefined.

Gradient = 0 2 =0

Example 6 Defining a type of gradient Decide if the lines labelled a, b, c and d on this graph have a positive, negative, zero or undefined gradient. y d c x a b SOLUTION

EXPLANATION

a Negative gradient

As x increases y decreases.

b Undefined gradient

The line is vertical.

c Positive gradient

y increases as x increases.

d Zero gradient

There is no increase or decrease in y as x increases.

Cambridge University Press

435

Number and Algebra

Example 7 Finding the gradient from a graph Find the gradient of these lines. a y 3 2 1

b

y 4 (0, 4) 3 2 1

(2, 3)

x

O1 2

–1–1O

SOLUTION

(4, 0) x 1 2 3 4

EXPLANATION rise run 3 = or 1.5 2

a Gradient =

The rise is 3 for every 2 across to the right.

rise = 3 run = 2

Exercise 7D

run = 4 rise = –4

EXTENSION

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1 Decide if the gradients of these lines are positive or negative. b

c

2 Simplify these fractions. 12 4 a b 4 2 e −6 3

f

−

20 4

d

c

6 4

g

−

14 7 15 h − 9 d

16 6

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a

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The y value falls 4 units while the x value increases by 4.

LL

rise run −4 = 4 = −1

b Gradient =

M AT I C A

7D

3 Write down the gradient using

d

HE

2

2

2

4

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5

3

1

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rise for each of these slopes. run c

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Chapter 7 Linear relationships 1

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436

1 e

f

g

1

3

h 2

2 4

6

2

2

y

T

HE

a

d x c b Example 7a

rise 5 Find the gradient of these lines. Use Gradient = . run b a y y

c

y

(1, 3) (2, 2) x

O

d

O

e

y

(2, 1)

x

O

f

y

x

y

(1, 4)

(0, 4) (0, 2)

(0, 1) O

x

(–3, 0) O

x

(–1, 0)

O

R

x

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4 Decide if these lines labelled a, b, c and d on this graph have a positive, negative, zero or undefined gradient.

MA

Example 6

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Number and Algebra

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b

y

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6 Find the gradient of these lines.

M AT I C A

x

O

(1, –2) (5, –3) d

y

y (0, 3)

x

O

(3, 0) O

(3, –4) e

f

y x

(–1, 0) O

x

y

(–2, 5) (0, 2)

(0, –3)

O

x

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8 A submarine falls 200 m for each 40 m across and a torpedo falls 420 m for each 80 m across in pursuit of the submarine. Which has the steeper gradient, the submarine or torpedo?

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7 Abdullah climbs a rocky slope that rises 12 m for each 6 metres across. His friend Jonathan climbs a nearby grassy slope that rises 25 m for each 12 m across. Which slope is steeper?

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Example 7b

M AT I C A

7D R K I OF

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M AT I C A

(3, 5)

(4, 5)

(–4, –1)

WO

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9 Find the gradient of these lines. You will need to first calculate the rise and run. b a y y

LL

Chapter 7 Linear relationships 1

x

O

x

O (–1, –5)

c

d

y (–3, 4)

y (–4, 4)

O

x (3, –5)

x

O (6, –3)

10 Find the gradient of the line joining these pairs of points. a (0, 2) and (2, 7) b (0, -1) and (3, 4) c (-3, 7) and (0, -1) d (-5, 6) and (1, 2) e (-2, -5) and (1, 3) f (-5, 2) and (5, -1) WO

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12 A line joins the point (0, 0) with the point (a, b) with a gradient of 2. a If a = 1 find b. b If a = 5 find b. c Write an expression for b in terms of a. d Write an expression for a in terms of b. 1 13 A line joins the point (0, 0) with the point (a, b) with a gradient of − . 2 a If a = 1 find b. b If a = 3 find b. c Write an expression for b in terms of a. d Write an expression for a in terms of b.

Cambridge University Press

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C 11 A line with gradient 3 joins (0, 0) with another point A. R PS HE a Write the coordinates of three different positions for A, using positive integers for M AT I C A both the x- and y-coordinates. b Write the coordinates of three different positions for A using negative integers for both the x- and y-coordinates.

LL

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439

Number and Algebra

Enrichment: Rise and run as differences 14 The run and rise between two points can be calculated by finding the difference between the pairs of x- and pairs of y-coordinates. For example: Rise = 4 - 1 = 3 Run = 4 - (-3) = 7

(4, 4) (–3, 1) O

3 7 Use this method to find the gradient of the line joining the given points.

y

Gradient =

a

y

b

y

(5, 4)

O

x

O

x (–4, –1)

(3, –3)

(–4, –2)

c

y

d

y

(4, 5) (0, 4) O

x

(–3, 0) O

x

(–5, –3) e (-3, 1) and (4, -2)

f

(-2, -5) and (1, 7)

g (5, -2) and (-4, 6)

h

1 − 2 ,

j

3 2 2 − 2 , 3 and 4, − 3

i

2 4 − 4, 3 and 3,− 3

7 k − , 4

2 and 9 , − 3 2

1 1 2 and , 3 2

Cambridge University Press

x

440

Chapter 7 Linear relationships 1

7E Gradient–intercept form

E X T E N S I ON

From previous sections in this chapter you may have noticed some connections between the numbers that make up a rule for a linear relationship and the numbers that are the gradient and y-coordinate with x = 0 (the y-intercept). This is no coincidence. Once the gradient and y-intercept of a graph are known, the rule can be written down without further analysis.

Let’s start: What’s the connection? To explore the connection between the rule for a linear relationship and the numbers that are the gradient and the y-intercept, complete the missing details for the graph and table below. y 5 (2, 5) 4 (1, 3) 3 2 1 (0, 1) –4 –3 –2 –1–1O (–1, –1) –2 –3

1 2 3

x

-1

y

-1

1

1

2

+2 y=

×x+

Gradient =

x

rise run

= ___ y -intercept = ___

• What do you notice about the numbers in the rule including the coefficient of x and the constant (below the table) and the numbers for the gradient and y-intercept? • Complete the details for this new example below to see if your observations are the same. y (–2, 3) (–1, 0)

x

-1

1

y

3 2 1

–5 –4 –3 –2 –1–1O 1 2 3 –2 –3 (0, –3) –4 –5 (1, –6) –6

-2

y= x

×x+ rise run = ___

Gradient =

y -intercept = ___

Cambridge University Press

■■

The rule for a straight line graph is given by y = mx + b where: –– m is the gradient –– b is the y-intercept y 2 1

(1, 1)

–2 –1–1O 1 2 (0, –1) –2

■■

m= 2 =2 1 b = –1

x

So y = mx + b becomes y = 2x − 1

A horizontal line (parallel to the x-axis) has the rule y = b. y y=5

(0, 5)

x (0, –4)

■■

y = –4

A vertical line (parallel to the y-axis) has the rule x = k. y

(–2, 0)

x = –2

(4, 0)

x

x=4

441

Key ideas

Number and Algebra

Cambridge University Press

442

Chapter 7 Linear relationships 1

Example 8 Stating the gradient and y-intercept from the rule Write down the gradient and y-intercept for the graphs of these rules. 1 a y = 2x 2x + 3 b y= x-4 3 SOLUTION a

EXPLANATION

y = 2x 2x + 3 m=2 b=3

The coefficient of x is 2 and this number is the gradient. The y-intercept is the constant.

1 b y= x-4 3 1 m= 3 b = -4

The gradient (m) is the coefficient of x. 1 1 Remember that y = x - 4 is the same as y = x + (-4) 3 3 so the constant is -4.

Example 9 Finding a rule from a graph Find the rule for these graphs by first finding the values of m and the y-intercept. a b y y 3 2 1

(1, 3)

–2 –1–1O 1 2 (0, –1) –2 SOLUTION a m=

rise run

4 =4 1 b = -1 y = 4xx - 1

x

4 3 2 1 –1–1O

(0, 4)

(2, 0) x 1 2

EXPLANATION Between (0, -1) and (1, 3) the rise is 4 and the run is 1.

=

b m=

rise run

−4 2 = -2 b=4 y = -2x 2x + 4 2x

The line cuts the y-axis at -1. Substitute the value of m and b into y = mx + b. Between (0, 4) and (2, 0) y falls by 4 units as x increases by 2.

=

The line cuts the y-axis at 4.

Cambridge University Press

Number and Algebra

Example 10 Finding rules for horizontal and vertical lines Write the rule for these horizontal and vertical lines. y a

b

y

x (7, 0)

x

(0, –3)

SOLUTION

EXPLANATION

to y = -3

All points on the line have a y value of -3 and the gradient is 0.

b x=7

Vertical lines take the form x = k. Every point on the line has an x value of 7.

Cambridge University Press

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Chapter 7 Linear relationships 1

WO

a

b

y 3 (1, 3) 2 1 (0, 1) –1–1O

d

3 2 1 x

1 2

e

3 2 1–1–1O

–1–1 O 1 (0, –1) (1, –2) –2 –3

x

f

1–1–1O –2 –3

(3, 0) x 1 2 3

M AT I C A

y

(2, 2)

y

(0, 3)

1 2

x

y 3 (0, 3) 2 (–1, 0) 1 x –1–1O 1

x

(1, –3)

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(0, 2)

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7 Sketch horizontal or vertical graphs for these rules. a y=2 b y = -1 c y = -4 e x = -3 f x=4 g x=1

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9 Find the rule for the graph of the lines connecting these pairs of points. a (0, 0) and (2, 6) b (-1, 5) and (0, 0) c (-2, 5) and (0, 3) d (0, -4) and (3, 1) 10 A line passes through the given points. Note that the y-intercept is not given. Find m and b and write the linear rule. A graph may be helpful. a (-1, 1) and (1, 5) b (-2, 6) and (2, 4) c (-2, 4) and (3, -1) d (-5, 0) and (2, 14) 11 Find the rectangular area enclosed by these sets of lines. a x = 4, x = 1, y = 2, y = 7 b x = 5, x = -3, y = 0, y = 5

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12 a Explain why the rule for the x-axis is given by y = 0. b Explain why the rule for the y-axis is given by x = 0.

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x (a, –a)

14 Some rules for straight lines may not be written in the form y = mx + b. The rule 2y + 4x = 6, for example, can be rearranged to 2y = -4x + 6 then to y = -2x + 3. So m = -2 and b = 3. Use this idea to find m and b for these rules. b 3y - 6x = 9 c 2y - 3x = 8 d x - 2y = -6 a 2y + 6x = 10

Enrichment: Sketching with m and b 15 The gradient and y-intercept can be used to sketch a graph without the need to plot more than two points. 2 For example, the graph of the rule y = 2x - 1 has m = 2 = and b = -1. By plotting the point 1 (0, -1) for the y-intercept and moving 1 to the right and 2 up for the gradient, a second point (1, 1) can be found. y 2 1 –1–1O

(1, 1) 1 2 (0, –1)

x

Use this idea to sketch the graphs of these rules. a y = 3x - 1 b y = 2x - 3 c y = -x + 2 1 e y = 4x f y = -5x g y = x - 2 2

d y = -3x - 1 3 h y = − x + 1 2

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7F The x-intercept

E X T E N S I ON

We know that the y-intercept marks the point where a line cuts the y-axis. This is also the value of y for the rule where x = 0. Similarly the x-intercept marks the point where y = 0. This can be viewed in a table of values or found using the algebraic method. y 4 3 2 x-intercept 1

x-intercept where y = 0 x

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Let’s start: Discover the method

Key ideas

These rules all give graphs that have x-intercepts at which y = 0. A y = 2x - 2 B y=x-5 C y = 3x - 9 D y = 4x + 3 • First try to guess the x-intercept by a trial and error (guess and check) method. Start by asking what value of x makes y = 0. • Discuss why the rule for D is more difficult to work with than the others. • Can you describe an algebraic method that will give the x-intercept for any rule? How would you show your working for such a method?

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The x-intercept is the point on a graph where y = 0. Find the x-intercept by substituting y = 0 into the rule. Solve using algebraic steps. y = 2x + 4 0 = 2x + 4 -4 -4 -4 = 2x ÷ 2 -2 = x ÷2 ∴ x-intercept is -2

y 4 y-intercept = 4 x-intercept = –2 –2

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Example 11 Finding the x-intercept For the graphs of these rules, find the x-intercept. a y = 3xx - 6 b y = -2x 2x + 1 2x SOLUTION

EXPLANATION

y = 3xx - 6 0 = 3xx - 6 +6 +6 6 = 3x ÷3 ÷3 2=x ∴ x-intercept is 2

a

Substitute y = 0 into the rule. Add 6 to both sides. Divide both sides by 2.

y = -2x 2x + 1 2x 0 = -2x 2x + 1 2x - 1 -1 = -2x -1 2 2x ÷2 1 ÷2 =x 2 1 ∴ x-intercept is 2 b

Substitute y = 0 into the rule. Subtract 1 from both sides. Divide both sides by -2.

Example 12 Sketching with intercepts Find the x- and y-intercepts and then sketch the graph of the rule y = 2x 2x - 8. SOLUTION

EXPLANATION

y = 2x 2x - 8 0 = 2x 2x - 8 +8 +8 8 = 2x 2 ÷2 ÷2 4=x x-intercept is 4. y-intercept is -8.

Substitute y = 0 into the rule. Add 8 to both sides. Divide both sides by 2. The y-intercept is the value of b in y = mx + b. Alternatively substitute x = 0 to get y = -8. Sketch by showing the x- and y-intercepts. There is no need to show a grid.

y

(4, 0) 0

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(0, –8)

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3 Solve each of these equations for x. a 0=x-4 b 0=x+2 d 0 = 2x + 10 e 0 = 3x - 12 g 0 = 3x + 1 h 0 = 5x + 2 j 0 = -5x - 20 k 0 = -x + 2

c f i l

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5 Find the x- and y-intercepts and then sketch the graphs of these rules. a y=x+1 b y=x-4 c y = 2x - 10 d y = 3x + 9 e y = -2x - 4 f y = -4x + 8 g y = -x + 3 h y = -x - 5 i y = -3x - 15

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4 For the graphs of these rules, find the x-intercept. a y=x-1 b y=x-6 d y = 2x - 8 e y = 4x - 12 g y = 2x + 20 h y = -2x + 4 j y = -2x - 4 k y = -x - 7

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7 Write the rule for a graph that matches the given information. a y-intercept 4, x-intercept -4 b y-intercept -1, x-intercept 2 c x-intercept -2, gradient 3 d x-intercept 5, gradient -5 8 The height of water (H cm) in a tub is given by H = -2t + 20 where t is the time in seconds. a Find the height of water initially (i.e. at t = 0). b How long will it take for the tub to empty?

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11 Decide if the x-intercept will be positive or negative for y = mx + b under these conditions. a m is positive and b is positive, e.g. y = 2x + 4 b m is positive and b is negative, e.g. y = 2x - 4 c m is negative and b is negative, e.g. y = -2x - 4 d m is negative and b is positive, e.g. y = -2x + 4

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Enrichment: Using ax + by = d 13 The x-intercept can be found if the rule for the graph is given in any form. Substituting y = 0 starts the process whatever the form of the rule. Find the x-intercept for the graphs of these rules. a x+y=6 b 3x - 2y = 12 c y - 2x = 4 d 2y - 3x = -9 e y - 3x = 2 f 3y + 4x = 6 g 5x - 4y = -10 h 2x + 3y = 3 i y - 3x = -1

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12 Write the rule for the x-intercept if y = mx + b. Your answer will include the pronumerals m and b.

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9 The amount of credit (C cents) on a phone card is given by the rule C = -t + 200 where t is time in seconds. a How much credit is on the card initially (t = 0)? b For how long can you use the phone card before the money runs out?

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7G Solving linear equations using graphical techniques The rule for a straight line shows the connection between the x- and y-coordinate of each point on the line. We can substitute a given x-coordinate into the rule to calculate the y-coordinate. When we substitute a y-coordinate into the rule it makes an equation that can be solved to give the x-coordinate. So, for every point on a straight line, the value of the x-coordinate is the solution to a particular equation. The point of intersection of two straight lines is the shared point where the lines cross over each other. This is the only point with coordinates that satisfy both equations; i.e. makes both equations true (LHS = RHS).

y 5 4 3 2 1

y = 2x – 1

(2, 3)

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1 2 3 4

x

For example, the point (2, 3) on the line y = 2x - 1 shows us that when 2x - 1 = 3 the solution is x = 2.

Let’s start: Matching equations and solutions When a value is substituted into an equation and it makes the equation true (LHS = RHS), then that value is a solution to that equation. a From the lists below, match each equation with a solution. Some equations have more than one solution. Equations 2x - 4 = 8 3x + 2 = 11 y = 10 - 3x

y=x+4 y = 2x - 5 5x - 3 = 2

Possible solutions x=1 (1, 5) x = -1 x=6 (-2, -9) (-2, 16)

x=2 (2, -1) x=3

(3, 1) (2, 6) (2, 4)

■■

The x-coordinate of each point on the graph of a y y = 2x – 1 straight line is a solution to a particular linear equation. 6 – A particular linear equation is formed by 2x – 1 = 4 Equation 5 substituting a chosen y-coordinate into a (2.5, 4) 4 linear relationship. 3 E.g. if y = 2x - 1 and y = 4, then the linear x = 2.5 Solution 2 equation is 2x - 1 = 4. 1 – The solution to this equation is the x-coordinate x of the point with the chosen y-coordinate. –1–1O 1 2 3 4 E.g. the point (2.5, 4) shows that x = 2.5 is the solution to 2x - 1 = 4.

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Key ideas

b Which two equations share the same solution and what is this solution? c List the equations that have only one solution. What is a common feature of these equations? d List the equations that have more than one solution. What is a common feature of these equations?

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A point (x, y) is a solution to the equation for a line if its coordinates make the equation true. – An equation is true when LHS = RHS after the coordinates are substituted. – Every point on a straight line is a solution to the equation for that line. – Every point that is not on a straight line is a not a solution to the equation for that line. The point of intersection of two straight lines is the only solution y that satisfies both equations. 5 – The point of intersection is the shared point where two 4 straight lines cross each other. (1, 3) 3 – This is the only point with coordinates that make both Point of 2 equations true. intersection 1 E.g. (1, 3) is the only point that makes both y = 6 - 3x x and y = 2x + 1 true. O 1 2 3 Substituting (1, 3) y = 6 - 3x y = 2x + 1 3=6-3×1 3=2×1+1 3 = 3 (True) 3 = 3 (True)

Example 13 Using a linear graph to solve an equation Use the graph of y = 2x 2x + 1 shown here to solve each of the following equations: a 2x 2x + 1 = 5 b 2x 2x + 1 = 0 c 2x 2x + 1 = -4

y

y = 2x + 1

5 4 3 2 1 –4 –3 –2 –1–1 O 1 2 3 4 –2 –3 –4 –5

x

SOLUTION

EXPLANATION

a x=2

Locate the point on the line with y-coordinate 5. The x-coordinate of this point is 2 so x = 2 is the solution to 22x + 1 = 5.

b x = 0.5

Locate the point on the line with y-coordinate 0. The x-coordinate of this point is −0.5 so x = -0.5 0.5 is the solution to 22x + 1 = 0.

c x = -2.5

Locate the point on the line with y-coordinate -4. The x-coordinate of this point is −2.5 so x = -2.5 2.5 is the solution to 22x + 1 = -4.

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Example 14 Using the point of intersection of two lines to solve an equation Use the graph of y = 4 - x and y = 2x 2 + 1, shown here, to answer these questions. a Write two equations that each have x = -2 as a solution. b Write four solutions (x, y) for the line with equation y = 4 - x. c Write four solutions (x, y) for the line with equation y = 2x 2 + 1. d Write the solution (x, y) that is true for both lines and show that it satisfies both line equations. e Solve the equation 4 - x = 2x 2 + 1.

y

y=4–x

6 5 4 3 2 1

y = 2x + 1

–3 –2 –1–1 O 1 2 3 4 5 –2 –3

x

SOLUTION

EXPLANATION

and 4-x=6 2 + 1 = -3 2x

(-2, 6) is on the line y = 4 - x so 4 - x = 6 has solution x = -2. (-2, -3) is on the line y = 2x 2 + 1 so 22x + 1 = -3 has solution x = -2.

b (-2, 6)(-1, 5)(1, 3)(4, 0)

Many correct answers. Each point on the line y = 4 - x is a solution to the equation for that line.

c (-2, -3)(0, 1)(1, 3)(2, 5)

Many correct answers. Each point on the line y = 2x 2 + 1 is a solution to the equation for that line.

d (1, 3) y=4-x 3=4-1 3 = 3 True

The point of intersection (1, 3) is the solution that satisfies both equations. Substitute (1, 3) into each equation and show that it makes a true equation (LHS = RHS).

(1, 3) y = 2x 2 +1 3=2×1+1 3 = 3 True

The solution to 4 - x = 2x 2 + 1 is the x-coordinate at the point of intersection. The value of both rules is equal for this x-coordinate.

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4 For each of these graphs write down the coordinates of the point of intersection (i.e. the point where the lines cross over each other). a b y y 5 4 3 2 1

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–5 –4 –3 –2 –1–1 O 1 2 3 4 5 –2 –3 7 Graph each pair of lines on the same set of axes and read off the point of intersection. a y = 2x - 1 b y = -x x

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8 Use digital technology to sketch a graph of y = 1.5x - 2.5 for x and y values between −7 and 7. Use the graph to solve each of the following equations. Round answers to 2 decimal places. a 1.5x - 2.5 = 3 b 1.5x - 2.5 = -4.8 c 1.5x - 2.5 = 5.446 9 Use digital technology to sketch a graph of each pair of lines and find the coordinates of the points of intersection. Round answers to 2 decimal places. a y = 0.25x + 0.58 and y = 1.5x - 5.4 b y = 2 - 1.06x and y = 1.2x + 5 © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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11 Jayden and Ruby are saving all their y money for the school ski trip. 110 • Jayden has saved $24 and earns 100 $6 per hour mowing lawns. 90 • Ruby has saved $10 and earns 80 $8 per hour babysitting. 70 This graph shows the total Amount (A) 60 in dollars of their savings for the number 50 (n) of hours worked. 40 a Here are two rules for calculating the 30 Amount (A) saved for working for 20 n hours. 10 A = 10 + 8n and A = 24 + 6n x Which rule applies to Ruby and 1 2 3 4 5 6 7 8 9 10 11 12 which to Jayden? Explain why. Number (n) of hours worked b Use the appropriate line on the above graph to find the solution to the following equations. i 10 + 8n = 42 ii 24 + 6n = 48 iii 10 + 8n = 66 iv 24 + 6n = 66 v 10 + 8n = 98 vi 24 + 6n = 90 c From the graph write three solutions (n, A) that satisfy A = 10 + 8n. d From the graph write three solutions (n, A) that satisfy A = 24 + 6n. e Write the solution (n, A) that is true for both Ruby’s and Jayden’s equations and show that it satisfies both equations. f From the graph find the solution to the equation: 10 + 8n = 24 + 6n (i.e. find the value of n that makes Ruby’s and Jayden’s savings equal to each other). g Explain how many hours have been worked and what their savings are at the point of intersection of the two lines. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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y=x+2

Amount (A) saved in $

Example 14

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a Copy and complete this table showing the distance run by each athlete. Time (t) in seconds

Max’s distance (d) in metres

Jessica’s distance (d) in metres

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b Plot these points on a distance-time graph and join to form two straight lines labelling them ‘Jessica’ and ‘Max’. c Find the rule linking distance d and time t for Max. d Using the rule for Max’s race, write an equation that has the solution: i t = 3 ii t = 5 iii t = 8 e Find the rule linking distance d and time t for Jessica. f Using the rule for Jessica’s race, write an equation that has the solution: i t = 3 ii t = 5 iii t = 8 g Write the solution (t, d) that is true for both distance equations and show that it satisfies both equations. h Explain what is happening in the race at the point of intersection and for each athlete state the distance from the starting line and time taken. 13 Some equations can be re-arranged so that a given graph can be used to find a solution. For example 4x - 1 = 2(x - 3) can be solved using the graph of y = 2x - 3 with this re-arrangement: 4x - 1 = 2(x - 3) 4x - 1 = 2x - 6 - 2x - 2x 2x - 1 = - 6 - 2 - 2 2x - 3 = - 8 y = - 8 on the graph gives the solution x = -2.5 Re-arrange the following equations so the left side of each is 2x - 3 and then use this graph of y = 2x - 3 to find each solution. a 4x - 1 = 2(x - 5) b 5x + 7 = 3(x + 4) c 3 = 6 - 2(x - 3) d 3 + 4x - 1 = x + 14 + x - 4 e 3(x - 3) - 2x = 4x + 3 - 5x f 4(x - 1) + x = 5x - 7 - 2x

y 12 10 8 6 4 2

y = 2x – 3

–5 –4 –3 –2 –1–2 O 1 2 3 4 5 6 7 8 –4 –6 –8 –10 –12

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12 Jessica and Max have a 10 second running race. • Max runs at 6 m/second. • Jessica is given a 10 m head-start and runs at 4 m/second.

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15 Here is a table and graph for the line y = 2x - 1. x

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Enrichment: More than one solution 16 a Use this graph of y = x2 to solve the following y equations. i x2 = 4 25 ii x2 = 9 2 iii x = 16 20 iv x2 = 25 b Explain why there are two solutions to each of 15 the equations in question a above. 2 c Use digital technology to graph y = x and 10 graphically solve the following equations, rounding answers to 2 decimal places. 5 i x2 = 5 2 ii x = 6.8 x iii x2 = 0.49 –5 –4 –3 –2 –1 O 1 2 3 4 5 2 iv x = 12.75 v x2 = 18.795 d Give one reason why the graph of y = x2 does not give a solution to the equation x2 = -9. e List three more equations of the form x2 = ‘a number’ that can’t be solved from the graph of y = x2. f List the categories of numbers that will give a solution to the equation: x2 = ‘a number’. g Graph y = x + 2 and y = x2 on the same screen and graphically solve x2 = x + 2 by finding the xvalues of the points of intersection. h Use digital technology to solve the following equations using graphical techniques. Round answers to 2 decimal places. i x2 = 3x + 16 ii x2 = 27 - 5x iii x2 = 2x - 10 iv x2 = 6x - 9

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7H Applying linear graphs

E X T E N S I ON

Rules and graphs can be used to help analyse many situations in which there is a relationship between two variables. If the rate of change of one variable with respect to another is constant, then the relationship will be linear and a graph will give a straight line. For example, if a pile of dirt being emptied out of a pit increases at a rate of 12 tonnes per hour, then the graph of the mass of dirt over time would be a straight line. For every hour, the mass of dirt increases by 12 tonnes.

Let’s start: Water storage

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If the rate of change of one variable with respect to another is constant, then the relationship between the two variables is linear. When applying straight line graphs, choose letters to replace x and y to suit the variables. For example, V for volume and t for time. y = mx + b can be used to help find the linear rule linking two variables.

A constant rate of excavation creates a linear relationship and a straight-line graph.

m = –25 b = 200 V = –25t + 200

200 V (litres)

Key ideas

The volume of water in a tank starts at 1000 litres and with constant rainfall the volume of water increases by 2000 litres per hour for 5 hours. • Describe the two related variables in this situation. • Discuss whether or not the relationship between the two variables is linear. • Use a table and a graph to illustrate the relationship. • Find a rule that links the two variables and discuss how your rule might be used to find the volume of water in the tank at a given time.

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Number and Algebra

Example 15 Linking distance with time A hiker walks at a constant rate of 4 kilometres per hour for 4 hours. a Draw a table of values using t for time in hours and d for distance in kilometres. Use t between 0 and 4. b Draw a graph by plotting the points given in the table in part a. c Write a rule linking d with t. d Use your rule to find the distance travelled for 2.5 hours of walking. e Use your rule to find the time taken to travel 8 km. SOLUTION a

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Plot the points on a graph using a scale that matches the numbers in the table.

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1 2 3 4 t (hours hour ) hours

4 =4 1 b=0 y = mx + b d = 4tt + 0 So d = 4t

c m=

Using y = mx + b find the gradient (m) and the y-intercept (b). Replace y with d and x with t.

d d = 4t = 4 × 2.5 = 10 The distance is 10 km after 2.5 hours of walking.

Substitute t = 2.5 into your rule and find the value for d.

e d = 4t 8 = 4t 2=t It takes 2 hours to travel 8 km.

Substitute d = 8 into your rule then divide both sides by 4.

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Example 16 Applying graphs when the rate is negative The initial volume of water in a dish in the sun is 300 mL. The water evaporates and the volume decreases by 50 mL per hour for 6 hours. a Draw a table of values using t for time in hours and V for volume in millilitres. b Draw a graph by plotting the points given in the table in part a. c Write a rule linking V with t. d Use your rule to find the volume of water in the dish after 4.2 hours in the sun. e Use your rule to find the time taken for the volume to reach 75 mL. SOLUTION a

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Use numbers from 0 to 300 on the V V-axis and 0 to 6 on the t-axis to accommodate all the numbers in the table.

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b = 300 y = mx + b V = -50tt + 300 d V = -50tt + 300 = -50 × 4.2 + 300 = 90 The volume of water in the dish is 90 millilitres after 4.2 hours. e

The volume starts at 300 millilitres and decreases by 50 millilitres every hour.

V = -50tt + 300 75 = -50tt + 300 -225 = -50t 4.5 = t It takes 4.5 hours for the volume to reach 75 mL.

− The gradient m in y = mx + b is given by rise = −50 . run 1 b is the y-intercept -intercept = 300

Substitute t = 4.2 into your rule to find V. V

Substitute V = 75 into your rule. Subtract 300 from both sides. Divide both sides by -50.

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1 A rule linking distance d and time t is given by d = 10t + 5. Use this rule to find the value of d for the given values of t. a t=1 b t=4 c t=0 d t = 12

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2 Write down the gradient (m) and the y-intercept (b) for these rules. a y = 3x + 2 b y = -2x - 1 c y = 4x + 2 d y = -10x - 3 e y = -20x + 100 f y = 5x - 2 3 The height (in cm) of fluid in a flask increases at a rate of 30 cm every minute starting at 0 cm. Find the height of fluid in the flask at these times. a 2 minutes b 5 minutes c 11 minutes 4 The volume of gas in a tank decreases from 30 L by 2 L every second. Find the volume of gas in the tank at these times. a 1 second b 3 seconds c 10 seconds

6 A paddle steamer moves up the Murray River at a constant rate of 5 kilometres per hour for 8 hours. a Draw a table of values using t for time in hours and d for distance in kilometres. Use t between 0 and 8. b Draw a graph by plotting the points given in the table in part a. c Write a rule linking d with t. d Use your rule to find the distance travelled after 4.5 hours. e Use your rule to find how long it takes to travel 20 km.

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5 A jogger runs at a constant rate of 6 kilometres per hour for 3 hours. a Draw a table of values using t for time in hours and d for distance in kilometres. Use t between 0 and 3. b Draw a graph by plotting the points given in the table in part a. c Write a rule linking d with t. d Use your rule to find the distance travelled for 1.5 hours of jogging. e Use your rule to find how long it takes to travel 12 km.

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7 The volume of water in a sink is 20 L. The plug is pulled out and the volume decreases by 4 L per second for 5 seconds. a Draw a table of values using t for time in seconds and V for volume in litres. b Draw a graph by plotting the points given in the table in part a. c Write a rule linking V with t. d Use your rule to find the volume of water in the sink 2.2 seconds after the plug is pulled. e Use your rule to find how long it takes for the volume to fall to 8 L.

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8 A weather balloon at a height of 500 m starts to descend at a rate of 125 m per minute for 4 minutes. a Draw a table of values using t for time in minutes and h for height in metres. b Draw a graph by plotting the points given in the table in part a. c Write a rule linking h with t. d Use your rule to find the height of the balloon after 1.8 minutes. e Use your rule to find how long it takes for the balloon to fall to a height of 125 m. R K I NG

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9 A BBQ gas bottle starts with 3.5 kg of gas. Gas is used at a rate of 0.5 kg per hour for a long lunch. a Write a rule for the mass of gas M in terms of time t. b How long will it take for the gas bottle to empty? c How long will it take for the mass of the gas in the bottle to reduce to 1.25 kg?

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10 A cyclist races 50 km at an average speed of 15 km per hour. a Write a rule for the distance travelled d in terms of time t. b How long will it take the cyclist to travel 45 km? c How long will the cyclist take to complete the 50 km race? Give your answer in hours and minutes. 11 An oil well starts to leak and the area of an oil slick increases by 8 km2 per day. How long will it take the slick to increase to 21 km2? Give your answer in days and hours. WO

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13 The cost of a phone call is 10 cents plus 0.5 cents per second. a Explain why the cost c cents of the phone call for t seconds is given by c = 0.5t + 10. b Explain why the cost C dollars of the phone call for t seconds is given by C = 0.005t + 0.1. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Enrichment: Danger zone 14 Two small planes take off and land at the same airfield. One plane takes off from the runway and gains altitude at a rate of 15 metres per second. At the same time, the second plane flies near the runway and reduces its altitude from 100 metres at rate of 10 metres per second. a Draw a table of values using t between 0 and 10 seconds and h for height in metres of both planes. t (s)

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On the one set of axes draw a graph of the height of each plane during the 10-second period. How long does it take for the second plane to touch the ground? Write a rule for the height of each plane. At what time are the planes at the same height? At what time is the first plane at a height of 37.5 m? At what time is the second plane at a height of 65 m? At the same time, a third plane at an altitude of 150 m descends at a rate of 25 m per second. Will all three planes ever be at the same height at the same time? What are the heights of the three planes at the 4-second mark?

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7I Non-linear graphs Not all relationships between two variables are linear. The amount of money invested in a compound interest account, for example, will not increase at a constant rate. Over time, the account balance will increase more rapidly, meaning that the graph of the relationship between Amount and Time will be a curve and not a straight line.

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Compound interest means your account balance increases more and more over time.

Let’s start: Drawing curves by hand Here are three common non-linear patterns shown as a set of points on a graph. y

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• Test your drawing skill by copying these graphs and drawing a smooth curve through the points on each graph. For the third graph you will need to draw two separate curves. • Discuss why it might not be correct to join each point with a straight line. 4 • The three graphs above are y = 2x, y = and y = x2 - 4. Can you match each rule with its graph? x Explain how you can check your answers.

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To plot non-linear curves given their rule: – Construct a table of values using the rule. – Plot the points on a set of axes. – Join the plotted points to form a smooth curve. The graph of y = x2 is an example of a non-linear graph called a parabola. The table shows five points that lie on the parabola. In each point, the y value is the square of the x value. x

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EXPLANATION

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Plot the points and join with a smooth curve. The curve is called a parabola.

y 7 6 5 4 3 2 1 –3 –2 –1–1O –2

Find the value of y by substituting each value of x into the rule.

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x This bridge support arch is an inverted parabola.

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4 Plot points to draw the graph of each of the given rules. Use the table and set of axes as a guide.

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6 What are the lowest and highest values of y on a graph of these rules if x ranges from -3 to 3? a y = x2 b y = x2 - 10 c y = 6 - x2 d y = x3 e y = -x3 –1 f y = (x - 1)3

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Parabola

Hyperbola

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Name the type of graph that is produced by each of these rules. 2 a y = x2 + 7 b y = -2x + 4 c y = d y = 7x x −3 e y = g y = 1 - x h y = (0.5)x + 1 f y = 4 - x2 x

Enrichment: Families of parabolas 8 For each family of parabolas, plot graphs by hand or use technology to draw each set on the same set of axes. Then describe the features of each family. Describe the effect on the graph when the number a changes. 1 a Family 1: y = ax2 Use a = , a = 1, a = 2 and a = 3. 2 1 b Family 2: y = -ax2 Use a = , a = 1, a = 2 and a = 3. 2 2 Use a = -3, a = -1, a = 0, a = 2 and a = 5. c Family 3: y = x + a d Family 4: y = (x - a)2 Use a = -2, a = 0, a = 1 and a = 4.

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Families of straight lines

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A set of lines are said to be in the same family if the lines have something in common. For example, four lines that pass through the point (1, 2) could be called a family.

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The parallel family 1 Complete the table for these rules. a y1 = 2x - 5 b y2 = 2x - 2 c y3 = 2x d y4 = 2x + 3

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2 Plot the points given in your table to draw graphs of the four rules in part a on the one set of axes. Label each line with its rule. 3 What do you notice about the graphs of the four rules? Describe how the numbers in the rule relate to its graph. 4 How would the graphs for the rules y = 2x + 10 and y = 2x - 7 compare with the graphs you have drawn above? Explain.

The point family x -3 -2 -1 0 1 2 1 Complete the table for these rules. y1 a y1 = x + 1 y2 b y2 = 2x + 1 y3 c y3 = 1 d y4 = -x + 1 y4 1 y5 e y=− x+1 2 2 Plot the points given in your table to draw graphs of the five rules in part a on one set of axes. Label each line with its rule.

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3 What do you notice about the graphs of the five rules? Describe how the numbers in the rule relate to its graph. 1 4 How would the graphs for the rules y = 3x + 1 and y = − x + 1 compare with the graphs you have 3 drawn above? Explain.

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Number and Algebra

Exploring families with technology Graphics or CAS calculators, graphing software and spreadsheets are useful tools to explore families of straight lines. Here are some screenshots showing the use of technology.

y y = 2x + 1 y = 2x + 3 y = 2x − 1 y = 2x − 3

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b

3 Make up your own design then use technology to produce it. Explain how your design is built and give the rules that make up the design. © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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1 A trekker hikes down a track at 3 km per hour. Two hours later, a second trekker sets off on the same track at 5 km per hour. How long is it before the second trekker to catches up with the first? 2 Find the rules for the non-linear relations with these tables. a b

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3 A line with a gradient of 3 intersects another line at right angles. Find the gradient of the other line. 4 Two cars travel toward each other on a 100 km stretch of road. One car travels at 80 km per hour and the other at 70 km per hour. If they set off at the same time, how long will it be before the cars meet? 5 Find the y-intercept of a line joining the two points (-1, 5) and (2, 4). 6 Find the rule of a line that passes through the two points (-3, -1) and (5, 3). 7 Find the number of matchsticks needed in the 100th diagram in the pattern given below. The first three diagrams in the pattern are given.

8 At a luxury car hire shop, a Ferrari costs $300 plus $40 per hour. A Porsche costs $205 plus $60 per hour. What hire time makes both cars the same cost? Give the answer in hours and minutes.

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–4 –3 –2 –1

Cartesian plane (number plane)

y = 2x – 1

y = –2x + 3

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(2, 3)

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Rules, tables and graphs

To graphically solve 2x – 1 = 3:

• locate point with y = 3 • x-coordinate is solution,

(–1, 3) 3 2 1 (–3, 0)

x –2 –1 0 y 7 5 3

(2, 3)

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O 1 2 3 –3 –2 –1–1 –2 origin (–2, –2) –3 (0, –3) (0, 0)

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x –2 –1 0 1 2 y 2 –1 –2 –1 2 y 3 2 1 0 –2 –1–1 –2

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A parabola

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Linear relationships 1 Finding the rule from tables

The point of intersection is the only solution that satisfies both linear equations.

Non-linear graphs

1 2 1 –1

7 6 5 4 3 2 1

i.e. x = 2 y Point of intersection

Chapter summary

y

x –2 –1 0 1 y –8 –5 –2 1

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+3 +3 +3 +3

y = mx + b Gradient

y= ×x+ = 3x + (–2) = 3x – 2

y-intercept

y 5 (2, 5) 4 3 2 1 (0, 1) x –1–1 O 1 2 3

m = 42 = 2 b=1 y = 2x + 1 Gradient

Applications • Distance increases by 20 km per hour.

t 0 1 2 3 d 0 20 40 60 d = 20t • Volume decreases from 1000 L by 200 L per minute. t 0 1 2 3 4 5 V 1000 800 600 400 200 0 V = –200t + 1000

rise = – 86 run = – 43

x-intercept y=0

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Chapter 7 Linear relationships 1

Multiple-choice questions 1 This graph shows the relationship between the height and age of three people. Who is the tallest person? A Ralph B Lucy C Kevin D Lucy and Ralph together E Kevin and Lucy together

Lucy Height

478

Kevin Ralph Age

2 The name of the point (0, 0) on a number (Cartesian) plane is: A y-intercept B gradient C origin D axis E x-intercept 3 Which point is not in line with the other points? A(-2, 3), B(-1, 2), C(0, 0), D(1, 0), E(2, -1) A A B B C C D D E E 4 Which of the points A(1, 2), B(2, -1) or C(3, -4) lie on the line y = -x + 1? A C B A and C C A D B E None 5 A rule gives this table of values. x

-2

-1

1

2

y

3

2

1

-1

Which points is on the y-axis? A (-2, 3) B (-1, 2) D (1, 0) E (2, -1)

C (0, 1)

6 Which line passes through the point (3, 2)? A y = 2x + 1 B y = 2x - 1 C y = 2x D y = 2x - 4 E y = 2x + 4 7 Which line does not pass through (3, 2)? A y=x+1 B y=x-1 C y=5-x D y = 3x - 7 E y = 8 - 2x 8 Which line passes through the y-axis at (0, 3)? A y = 3x B y=x-3 C y = 3x + 1 D y = -3x E y=x+3

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Number and Algebra

9 Consider the line y = 3x + 2. There is a point on the curve (a, 0). The value of a is: A 0 D -

B 2 2 3

E

C 3

2 3

10 Which point is the point of intersection of the two lines graphed here? A (-4, 0) B (2, 0) C (0, 4) D (-1, 3) E (0, 2)

y 5 4 3 2 1 –5 –4 –3 –2 –1–1O –2

1 2 3 4 5

x

Short-answer questions 1 Write the coordinates of all the points A–J in the graph below. y C

D

4 3 B 2 1

A

J O 1 2 3 4 –4 –3 –2 –1–1 H –2 E I –3 G –4 F

x

2 For each rule create a table using x values from -3 to 3 and plot to draw a straight-line graph. a y = 2x b y = 3x - 1 c y = 2x + 2 d y = -x + 1 e y = -2x + 3 f y=3-x

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3 Write the rule for these tables of values. a x 0 1 2 -2 -1 c

e

y

-3

-1

1

3

5

x

3

4

5

6

7

y

6

7

8

9

10

x

-1

1

2

3

y

3

-1

-5

-9

-13

b

d

f

x

-2

-1

1

2

y

-4

-1

2

5

8

x

-3

-2

-1

1

y

4

3

2

1

x

-6

-5

-4

-3

-2

y

8

7

6

5

4

4 Find the rule from these graphs. a

b

and 4 3 2 1 OR

3 2 1

(1, 3)

1 2

d

e

y 3 2 (–1, 1) 1 OR –2 –1–1

1

x

y (1, 0) OR 1 2 3 –1 –2 (0, –2) –3

(2, 1) O

x

c

y

1 2

x

y 4 (0, 4) 3 2 1 (1, 0) x O –1–1 1 2

f

x

y (–2, 6)

7 6 5 4 3 (0, 3) 2 1

–2 –1–1O

1 2 3

5 Find the equation of the line joining these pairs of points. a (0, 0) and (3, -12) b (-4, 2) and (0, 0) c (1, 1) and (4, 4) d (-5, 3) and (1, -9)

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Number and Algebra

6 Write the rule for these graphs. a

y

b

y

4 3 2 1

4 3 (1, 3) 2 1 (0, 1) –1–1O

x

1 2

7

x

f

x

y

1

–3 –2 –1–1O 1 –2 –3 –4 (0, –4) –5

x

1

(2, 0) O 1 2 3 –1–1 –2 (0, –2) –3

y (–2, 0)

–2 –1–1O

1

e

(–1, 4) 4 3 2 1

y

(1, 3)

–1–1 O 1 2

y

d

c

(–2, 2)

x

2 (0, 1) 1

–3 –2 –1–1O

x

1 2 3

a Consider this table: x

1

2

3

4

y

5

8

11

14

17

Complete this sentence: To find a value of y, __________________ x by then ________________ ________________. b Repeat part a for these tables. i

x

1

2

3

y

2

8

14

ii

x

1

2

3

20

y

6

8

10

12

8 A café arranges its tables like this: T = table C C = chairs

C

C

T

T

C

C

C

T

C

C

C

C

C

C

C

C

T

T

T

C

C

C

C

a Complete the table. T

1

2

3

4

5

C

b Complete the sentence: To find the number of chairs, multiply ________________________________________. c Find the rule that begins with C = d How many chairs would be needed for ten tables?

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9

a Use the appropriate line on the above graph to find the solution to the following linear equations. i 3-x=4 ii 2x - 3 = -1 iii 3 - x = 2 iv 2x - 3 = 4 v 3 - x = -1.5 b Write the coordinates of the point of intersection of these two lines. c Show whether (-1, 4) is a solution or not a solution to y = 3 - x and y = 2x - 3. d Show that the point of intersection is a solution to both equations y = 3 - x and y = 2x - 3.

y 5 4 3 2 1 –2 –1–1O –2 –3

y = 2x – 3

1 2 3 4 5

x

y=3–x

10 Using a table with x values between -2 and 2 draw a smooth curve for the non-linear graphs of these rules. 2 a y = x2 - 2 b y = x3 c y= x

Extended-response questions 1 A seed sprouts and the plant grows 3 millimetres per day in height for 6 days. a Construct a table of values using t for time in days and h for height in millimetres. b Draw a graph using the points from your table. Use t on the horizontal axis. c Find a rule linking h with t. d Use your rule to find the height of the plant after 3.5 days. e If the linear pattern continued, what would be the height of the plant after 10 days? f How long will it be before the plant grows to 15 mm in height?

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Number and Algebra

2 A speed boat at sea is initially 12 km from a distant buoy. The boat travels towards the buoy at a rate of 2 km per minute. The distance between the boat and the buoy will therefore decrease over time. a Construct a table showing t for time in minutes and d for distance to the buoy in kilometres. b Draw a graph using the points from your table. Use t on the horizontal axis. c How long does it take the speed boat to reach the buoy? d What is the gradient of the line drawn in part b? e Find a rule linking d with t. f Use your rule to find the distance from the buoy at the 2.5 minute mark. g How long does it take for the distance to reduce to 3.5 km?

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Chapter 8 Transformations and congruence

Chapter

8

Transformations and congruence

What you will learn

8A 8B 8C 8D 8E 8F 8G 8H

Reﬂection Translation Rotation Congruent ﬁgures Congruent triangles Similar ﬁgures EXTENSION Similar triangles EXTENSION Using congruent triangles to establish properties of quadrilaterals

Cambridge University Press

Measurement and Geometry

NSW Syllabus

for the Australian Curriculum Strands: N umber and Algebra

Measurement and Geometry Substrands: L INEAR RELATIONSHIPS PROPERTIES OF GEOMETRICAL FIGURES

Outcomes A student creates and displays number patterns; graphs and analyses linear relationships; and performs transformations on the Cartesian plane. (MA4–11NA) A student classiﬁes, describes and uses the properties of triangles and quadrilaterals, and determines congruent triangles to ﬁnd unknown side lengths and angles. (MA4–17MG)

Symmetrical architecture Geometry is at the foundation of design and architecture. In modern times, many public buildings as well as private residences have been designed and built with a strong sense of geometric expression. In the ancient Roman world, the famous architect Marcus Vitruvius said ‘For without symmetry and proportion no temple can have a regular plan’. The Pantheon is about 2000 years old and is one of the oldest and best kept buildings in Rome. Its rectangular portico is supported by eight monolithic cylindrical columns, and these combined with the triangular roof provide perfect line symmetry at the entrance of the building. Inside the main dome the symmetry changes. At the centre of the dome roof is a

large oculus (hole) that lets in the sunlight. This natural light allows visitors to see the perfect rotational symmetry within. The height of the dome is the same as its width, meaning that a sphere of the same diameter would ﬁt perfectly under the dome. It is no wonder that people travel to the Pantheon to pray.

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Pre-test

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1 Complete the simple transformations of the given shapes as instructed. a Reﬂect this shape over the mirror line.

b Shift this shape 2 units to the right and 1 unit down.

c Rotate this shape 180° around point C. C

D 2 Consider this pair of quadrilaterals. a Name the vertex, E, F, G or H, that matches (corresponds to) vertex: i B ii D b Name the angle, ∠A, ∠B, ∠C or ∠D, that matches (corresponds to) angle: A i ∠E ii ∠G E c Name the side, EF, FG, GH or HE, that matches (corresponds to) side: i CD ii BC

C

H

G

3 Find the unknown angle in these triangles. a b 40° a° 110° a° 80°

c

B F

d

a° a°

4 Which of the special quadrilaterals definitely fit the description? A Square B Rectangle C Rhombus D Parallelogram E Kite F Trapezium a Opposite sides are of equal length. b It has at least one pair of equal opposite angles. c Diagonals are of equal length. d Diagonals intersect at right angles.

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Measurement and Geometry

8A Reﬂ ection When an object is shifted from one position to another, rotated about a point, reﬂected over a line or enlarged by a scale factor, we say the object has been transformed. The names of these types of transformations are reﬂection, translation, rotation and dilation. The first three of these transformations are called isometric transformations because the object’s geometric properties are unchanged and the transformed object will be congruent to the original object. The word ‘isometric’ comes from the Greek words isos meaning ‘equal’ and metron meaning ‘measure’. Dilation (or enlargement) results in a change in the size of an object to produce a ‘similar’ figure and this will be studied later in this chapter. The first listed transformation, reﬂection, can be thought of as the creation Reﬂection creates an image reversed as in a of an image over a mirror line. mirror or in water.

Let’s start: Visualising the image This activity could be done individually by hand on a page, in a group using a whiteboard or using dynamic geometry projected onto a white board. • Draw any shape with straight sides. • Draw a vertical or horizontal mirror line outside the shape. • Try to draw the reﬂected image of the shape in the mirror line. • If dynamic geometry is used, reveal the precise image (the answer) using the Reﬂection tool to check your result. • For a further challenge, redraw or drag the mirror line so it is not horizontal or vertical. Then try to draw the image.

Dynamic geometry software provides a reﬂection tool.

Cambridge University Press

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Key ideas

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Chapter 8 Transformations and congruence

■

■ ■ ■

■

■

■

Reﬂection is an isometric transformation in which the size of the object is unchanged. The image of a point A is denoted A′. Each point is reﬂected at right angles to the mirror line. The distance from a point A to the mirror line is equal to the distance from the image point A′ to the mirror line. When a line of symmetry is used as a reﬂection line, the image looks the same as the original figure. When a point is reﬂected across the x-axis, – the x-coordinate remains unchanged – the y-coordinate undergoes a sign change, e.g. (5, 3) becomes (5, −3); that is, the y values are equal in magnitude but opposite in sign When the point is reﬂected across the y-axis, – the x-coordinate undergoes a sign change – the y-coordinate remains unchanged, e.g. (5, 3) becomes (−5, 3)

C

A

A′

B

B′

C C′

Example 1 Drawing reflected images Copy the diagram and draw the reﬂected image over the given mirror line. a A b C B C

E

A

D

SOLUTION

EXPLANATION

a A

B′

B C

E b

A′ A

Reﬂect each vertex point at right angles to the mirror line. Join the image points to form the final image.

C C′

D′

D

E′ E Reﬂect points A, B and C at right angles to the mirror line to form A′, B′ and C′. Note that A′ is in the same position as A as it is on the mirror line. Join the image points to form the image triangle.

C

B′ A A′ A

B

B

C C′

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Measurement and Geometry

Example 2 Using coordinates State the coordinates of the vertices A′, B′ and C′ after this triangle is reﬂected in the given axes. a x-axis b y-axis

y 4 3 2 1

B C

–4 –3 –2 –1–1O A1 2 3 4 –2 –3 –4 SOLUTION

x

EXPLANATION

a A′ = (1, 0) B′ = (2, −3) C′ = (4, −2)

y 4 B 3 C C′′ C 2 b 1 A′ A x –4 –3 –2 –1–1O 1 2 3 4 a –2 C′ C′ –3 B′ –4 B′

WO

e

f

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d

c

MA

1 Use the grid to precisely reﬂect each shape in the given mirror line. a b

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Exercise 8A

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b A′ = (−1, 0) B′ = (−2, 3) C′ = (−4, 2)

M AT I C A

U

R

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HE

d

e

U

MA

f

4 Copy the diagram and accurately locate and draw the mirror line. Alternatively, pencil in the line on this page. a

b

c

d

e

f

R

HE

T

e

F PS

f

3 Copy the diagram and draw the reﬂected image over the given mirror line. a b c

d

C

M AT I C A

WO

Example 1b

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2 Copy the diagram and draw the reﬂected image over the given mirror line. a b c

LL

Example 1a

8A

Chapter 8 Transformations and congruence

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Measurement and Geometry

C A 1 2 3 4

–4 –3 –2 –1–1O –2 –3 –4

x

U

MA

R

HE

T

c

Rhombus e

g

f

h

Isosceles triangle

Parallelogram

Trapezium i

Equilateral triangle

8 A point (2, 4) is reﬂected in the given horizontal or vertical line. State the coordinates of the image point. As an example, the graph of the mirror lines x = 1 and y = −2 are shown here. The mirror lines are: a x = 1 b x = 3 c x = 0 d x = −1 e x = −4 f x = −20 g y = 3 h y = −2 i y = 0 j y = 1 k y = −5 l y = −37

Regular octagon y (2, 4)

4 3 2 1 –4 –3 –2 –1–1O –2 –3 –4

1 2 3 4

x y = –2

x=1 © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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R K I N G

C

F PS

M AT I C A

Rectangle

Kite

PS

x

7 How many lines of symmetry do these shapes have? a b

d

F

M AT I C A

WO

Square

C

Y

R

HE

B

R K I N G

Y

–4 –3 –2 –1 O –11 2 3 4 A B –2 –3 –4 C D

4 3 2 1

T

4 3 2 1

U

MA

6 State the coordinates of the vertices A′, B′, C′ and D′ after this rectangle is reﬂected in the given axes. a x-axis b y-axis y

WO

y

LL

5 State the coordinates of the vertices A′, B′ and C′ after the triangle (right) is reﬂected in the given axes. a x-axis b y-axis

LL

Example 2

8A

Chapter 8 Transformations and congruence

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9 A shape with area 10 m2 is reﬂected in a line. What is the area of the image shape? Give a reason for your answer.

U

10 How many lines of symmetry does a regular polygon with n sides have? Write an expression. 11 A point is reﬂected in the x-axis then in the y-axis and finally in the x-axis again. What single reﬂection could replace all three reﬂections? 12 Two important lines of reﬂection on the coordinate plane are the line y = x and the line y = −x as shown. a Draw the coordinate plane shown here. Draw a y=x y triangle with vertices A(−1, 1), B(−1, 3) and C(0, 3). Then complete these reﬂections. i Reﬂect triangle ABC in the y-axis. ii Reﬂect triangle ABC in the x-axis. iii Reﬂect triangle ABC in the line y = x. x iv Reﬂect triangle ABC in the line y = −x. O b Draw a coordinate plane and a rectangle with vertices A(−2, 0), B(−1, 0), C(−1, −3) and D(−2, −3). Then complete these reﬂections. i Reﬂect rectangle ABCD in the y-axis. ii Reﬂect rectangle ABCD in the x-axis. y = –x – iii Reﬂect rectangle ABCD in the line y = x. iv Reﬂect rectangle ABCD in the line y = −x. 13 Points are reﬂected in a mirror line but do not change position. Describe the position of these points in relation to the mirror line. 14 Research to find pictures, diagrams, cultural designs and artwork which contain line symmetry. Make a poster or display electronically.

Enrichment: Reﬂ ection through a point 15 Rather than completing a reﬂection in a line, it is possible to reﬂect an object through a point. An example of a reﬂection through point P is shown here. A goes to A′, B goes to B′ and C goes to C′ through P. a Draw and reﬂect these shapes through the point P. i ii iii P

P

P

C

B′

A′

P

A

B

C C′

b Like line symmetry, shapes can have point symmetry if they can be reﬂected onto themselves through a point. Decide if these shapes have any point symmetry. i ii iii

c How many special quadrilaterals can you name that have point symmetry? © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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8B Translation Translation is a shift of every point on an object in a given direction and by the same distance. The direction and distance is best described by the use of a translation vector. This vector describes the overall direction using a horizontal component (for shifts left and right) and a vertical component (for shifts up and down). Negative numbers are used for translations to the left and down. Designers of animated movies translate images in many of their scenes. Computer software is used and translation vectors help to define the specific movement of the objects and characters on the screen.

Animated characters move through a series of translations.

Let’s start: Which is further? Consider this shape on the grid. The shape is translated by the vector 3 (3 to the right and 2 down). −2

( )

3 2

Now consider the shape being translated by these different vectors. a

( ) 3

b

( ) −1

c

() 5

d

( ) −2

■

■

Translation is an isometric transformation that involves a shift by a given distance in a given direction. a A vector can be used to describe the distance and direction of a translation. b Right 2 2 Vector Left 1 −3

()

( )

Down 3

■

Vector

( ) −1 4

Up 4

– If a is positive you shift to the right. – If a is negative you shift to the left. – If b is positive you shift up. – If b is negative you shift down. The image of a point A is denoted A′.

Cambridge University Press

Key ideas

−3 −4 0 4 • By drawing and looking at the image from each translation, which vector do you think takes the shape furthest from its original position? • Is there a way that you can calculate the exact distance? How?

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Chapter 8 Transformations and congruence

Example 3 Finding the translation vector State the translation vector that moves the point A(−1, 3) to A′(2, 0). y A

3 2 1

–1–1O

A′ 1 2 3

x

SOLUTION

EXPLANATION

( )

To shift A to A′ move 3 units to the right and 3 units down.

3

−3

Example 4 Drawing images using translation

( )

Draw the image of the triangle ABC after a translation by the vector − 3 . 2 y 3 2 1

A O –3 –2 –1–1 –2 B –3

C 3

x

SOLUTION

EXPLANATION

First translate each vertex, A, B and C, left 3 spaces, then up 2 spaces.

already'

C C′

3 2 1 A

B–2 ′ –1 O –1 –2 B –3

C 3

x

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Measurement and Geometry

WO

1 Use the words left, right, up or down, to complete these sentences. 2 a The vector means to move 2 units to the ______ and 4 units _____. 4 −5 b The vector means to move 5 units to the _____ and 6 units _____. 6 3 c The vector means to move 3 units to the _____ and 1 unit _____. −1

MA

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d The vector

() ( ) ( ) ( )

R

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Exercise 8B

M AT I C A

−10 means to move 10 units to the _____ and 12 units _____. −12

2 Write the vector that describes these transformations. b 2 units to the left and 6 units down d 9 units to the right and 17 units up

()

()

( )

( ) −6

WO

5 Copy the diagrams and draw the image of the shapes translated by the given vectors. a

() 2

b

3

y

4

c

−2

( ) −3

1

y

x

( )

y

x

x

R

HE

T

Example 4

4 Write the vector that takes each point to its image. Use a grid to help you. a A(2, 3) to A′(3, 2) b B(1, 4) to B′(4, 3) c C(−2, 4) to C′(0, 2) d D(−3, 1) to D′(−1, −3) e E(−2, −4) to E′(1, 3) f F(1, −3) to F′(−2, 2) g G(0, 3) to G′(2, 0) h H(−3, 5) to H′(0, 0) i I(5, 2) to I′(−15, 10) j J(−3, −4) to J′(−12, −29)

MA

Example 3

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3 Decide if these vectors describe vertical or horizontal translation. 2 0 0 d a b c 0 7 −4

LL

a 5 units to the right and 2 units down c 7 units to the left and 4 units up

M AT I C A

8B

y

( )

f

−1

( )

C

F PS

M AT I C A

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y

x

R

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( )

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x

x

6 Write the coordinates of the image of the point A(13, −1) after a translation by the given vectors. 2 8 0 −4 a b c d 3 0 7 3 −2 −10 −2 6 e f g h −1 −5 −8 −9 12 −26 −4 −21 i j k l −3 14 18 −38

() ( ) ( )

() ( ) ( )

() ( ) ( )

( ) ( ) ( )

WO

( )()() ()

2 8 A car makes its way around a city street grid. A vector represents travelling 200 m east and 3 300 m north. a Find how far the car travels if it follows these vectors: 2 −5 3 −2 , , and 3 1 −3 −4

()( )( ) ( )

b What vector takes the car back to the origin (0, 0), assuming it started at the origin? 9 A point undergoes the following multiple translations with these given vectors. State the value of x and y of the vector that would take the image back to its original position. 3 −1 x 2 −7 −1 x a , , b , , , 4 −2 y 5 2 −3 y 0 7 −4 x −4 12 −36 x c , , , d , , , 4 0 −6 y 20 0 40 y

()( )() ()()( )()

()( )( )() ( )( )( )()

R

HE

T

( )()()

MA

7 Which vector from each set takes an object the greatest distance from its original position? You may need to draw diagrams to help, but you should not need to calculate distances. −1 0 7 −1 4 0 a , , b , , 3 3 0 −4 0 3

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Measurement and Geometry

( )

( ) ( )( )

()

()( )( )

( )

()( )( )

Enrichment: How many options for the rabbit?

Hole

12 Hunters spot a rabbit on open ground and it has 1 second to find a hole before getting into big trouble with the hunter’s gun. It can run a xm 4m maximum of 5 metres in one second. a Use Pythagoras’ theorem to check that the distance x m in this diagram is less than 5 m. Rabbit b The rabbit runs a distance and direction described by the 2m −4 translation vector . Is the rabbit in trouble? 3 c The rabbit’s initial position is (0, 0) and there are rabbit holes at every point that has integers as its coordinates, e.g. (2, 3) and (−4, 1). How many rabbit holes can it choose from to avoid the hunters before its 1 second is up? Draw a diagram to help illustrate your working.

( )

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10 A reverse vector takes a point in the reverse direction by the same distance. Write the reverse vectors of these vectors. 3 −5 x −x a b c d −2 0 y −y 11 These translation vectors are performed on a shape in succession (one after the other). What is a single vector that would complete all transformations for each part in one go? 2 −3 0 6 6 −11 a c −a a , , b , , c , , 1 −4 3 4 −2 0 b −a a – c

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Chapter 8 Transformations and congruence

8C Rotation When the arm of a crane moves left, right, up or down, it undergoes a rotation about a fixed point. This type of movement is a transformation called a rotation. The pivot point on a crane would be called the centre of rotation and all other points on the crane’s arm move around this point by the same angle in a circular arc.

Let’s start: Parallelogram centre of rotation This activity will need a pencil, paper, ruler and scissors.

Every point on this crane’s arm rotates around the pivot point when it moves.

Key ideas

• Accurately draw a large parallelogram on a separate piece of paper and cut it out. • Place the tip of a pencil at any point on the parallelogram and spin the shape around the pencil. • At what position do you put the pencil tip to produce the largest circular arc? • At what position do you put the pencil tip to produce the smallest circular arc? • Can you rotate the shape by an angle of less than 360° so that it exactly covers the area of the shape in its original position? Where would you put the pencil to illustrate this?

■

■ ■

Rotation is an isometric transformation about a centre point and by a given angle. An object can be rotated clockwise or anticlockwise . Each point is rotated on a circular arc about the centre of rotation C. This diagram shows a 90° anticlockwise rotation about the point C.

Y′ Z

Z Z′

X′

X C

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A shape has rotational symmetry if it can be rotated about a centre point to produce an exact copy covering the entire area of the original shape. – The number of times the shape can make an exact copy in a 360° rotation is called the order of rotational symmetry. If the order of rotation is 1, then it is said that the shape has no rotational symmetry. – This equilateral triangle has rotational symmetry of order 3.

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Example 5 Finding the order of rotational symmetry Find the order of rotational symmetry for these shapes. a b

SOLUTION

EXPLANATION

a Order of rotational symmetry = 2

1

2 b Order of rotational symmetry = 3

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Example 6 Drawing a rotated image Rotate these shapes about the point C by the given angle and direction. a Clockwise by 90° b 180°

C

C

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Measurement and Geometry

Chapter 8 Transformations and congruence

SOLUTION

EXPLANATION

a

Take each vertex point and rotate about C by 90°, but it may be easier to visualise a rotation of some of the sides first. Horizontal sides will rotate to vertical sides in the image and vertical sides will rotate to horizontal sides in the image.

C

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You can draw a dashed line from each vertex through C to a point at an equal distance on the opposite side. C

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2 Complete these sentences. a A rotation clockwise by 90° is the same as a rotation anticlockwise by _____. b A rotation anticlockwise by 180° is the same as a rotation clockwise by _____. c A rotation anticlockwise by _____ is the same as a rotation clockwise by 58°. d A rotation clockwise by _____ is the same as a rotation anticlockwise by 296°.

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Example 6a

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Example 6b

5 Rotate these shapes about the point C by the given angle and direction. a Clockwise by 90° b Anticlockwise by 90° c Anticlockwise by 180° C

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9 Which capital letters of the alphabet (A, B, C, . . . , Z) have rotational symmetry of order 2 or more? 10 Draw an example of a shape that has these properties. a Rotational symmetry of order 2 with no line symmetry b Rotational symmetry of order 6 with 6 lines of symmetry c Rotational symmetry of order 4 with no line symmetry

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11 What value of x makes these sentences true? a Rotating x degrees clockwise has the same effect as rotating x degrees anticlockwise. b Rotating x degrees clockwise has the same effect as rotating 3x degrees anticlockwise.

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12 When working without a grid or without 90° angles, a protractor and a pair of compasses are needed to accurately draw images under rotation. This example shows a rotation of 120° about C. a Copy this triangle with centre of rotation C onto a sheet of paper. C Z Z′

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C b Construct three circles with centre C and passing through the vertices of the triangle. c Use a protractor to draw an image after these rotations. i 120° anticlockwise ii 100° clockwise

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13 Make a copy of this diagram and rotate the shape anticlockwise by 135° around point C. You will need to use compasses and a protractor as shown in Question 12. C

Enrichment: Finding the centre of rotation 14 Finding the centre of rotation if the angle is known involves the calculation of an angle inside an isosceles triangle. For the rotation shown, the angle of rotation is 50°. The steps are given as: i Calculate ∠CAA′ and ∠CA′A. (2x + 50 = 180, so x = 65) ii Draw the angles ∠AA′C and ∠A′AC at 65° using a protractor. iii Locate the centre of rotation C at the point of intersection of AC and A′C. a On a sheet of paper draw two points A and A′ about 4 cm apart. Follow the steps above to locate the centre of rotation if the angle of rotation is 40°. b Repeat part a using an angle of rotation of 100°. c When a shape is rotated and the angle is unknown, there is a special method for accurately pinpointing the centre of rotation. Research this method and illustrate the procedure using an example.

A′

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8D Congruent ﬁ gures In mathematics, if two objects are identical we say they are congruent. If you ordered 10 copies of a poster from a printer, you would expect that the image on one poster would be congruent to the image on the next. Even if one poster was ﬂipped over, shifted or rotated you would still say the images on the posters were congruent.

Let’s start: Are they congruent?

Each brochure being printed is identical, so the images on them are congruent.

Here are two shapes. To be congruent they need to be exactly the same shape and size.

Key ideas

• Do you think they look congruent? Give reasons. • What measurements could be taken to help establish whether or not they are congruent? • Can you just measure lengths or do you need to measure angles as well? Discuss.

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A ﬁgure is a shape, diagram or illustration. Congruent ﬁgures have the same size and shape. They also have the same area. They are identical in every way.

The image of a figure that is reﬂected, translated or rotated is congruent to the original figure. Corresponding (matching) parts of a figure have the same geometric properties. – Vertex C corresponds to vertex E. D E C – Side AB corresponds to side FD. – ∠B corresponds to ∠D. F A congruence statement can be written using the symbol ≡, e.g. ABC ≡ FDE. A – The symbol ≅ can also be used for congruence. B – In a congruence statement vertices are named in matching order, e.g. ABC ≡ FDE not ABC ≡ DEF because B matches D. Two circles are congruent if they have equal radii.

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Example 7 Naming corresponding pairs These two quadrilaterals are congruent. Name the objects in quadrilateral EFGH that correspond to these objects in quadrilateral ABCD. a C b AB c ∠C A G F

H B E

SOLUTION

EXPLANATION

a G

C sits opposite A and ∠A is the smallest angle. G sits opposite E and ∠E is also the smallest angle.

b EH

Sides AB and EH are both the longest sides of their respective shapes. A corresponds to E and B corresponds to H.

c ∠G

∠C and ∠G are both the largest angle in their corresponding quadrilateral.

Exercise 8D

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1 In this diagram ABC has been reﬂected to give the image triangle DEF. a Is DEF congruent to ABC; i.e. is DEF ≡ ABC? C b Name the vertex on DEF which corresponds to: i A ii B iii C A B c Name the side on DEF which corresponds to: i AB ii BC iii AC d Name the angle in DEF which corresponds to: i ∠B ii ∠C iii ∠A

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7 Using the lines in the diagram as the sides of triangles, how many triangles are there that have: 1 a area ? b area 1? c area 2? 2 d area 4? e area 8?

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4 8 List the pairs of the triangles below that look congruent. A

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9 Write the pairs of corresponding vertices for these congruent shapes, e.g. (A, E), (B, D). a C E A B Z W D b

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10 An isosceles triangle is cut as shown, using the midpoint of AB. a Name the two triangles formed. b Will the two triangles be congruent? Explain.

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11 A kite ABCD has diagonals AC and BD. a The kite is cut using the diagonal BD. i Name the two triangles formed. ii Will the two triangles be congruent? Explain. b The kite is cut using the diagonal AC. i Name the two triangles formed. ii Will the two triangles be congruent? Explain.

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12 If a parallelogram is cut by either diagonal will the two triangles be congruent?

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13 What is the condition for two circles to be congruent?

Enrichment: Combined congruent transformations 14 Describe the combination of transformations (reﬂections, translations and/or rotations) that map each shape to its image under the given conditions. The reﬂections that are allowed include only those in the x- and y-axes and rotations will use (0, 0) as its centre. a A to A′ with a reﬂection and then a translation b A to A′ with a rotation and then a translation c B to B′ with a rotation and then a translation d B to B′ with 2 reﬂections and then a translation e C to C′ with 2 reﬂections and then a translation f C to C′ with a rotation and then a translation y A

5 4 3 2 1

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8E Congruent triangles Imagine the sorts of design and engineering problems we would face if we could not guarantee that two objects such as window panes or roof truss frames were not the same size or shape. Further, it might not be possible or practical to measure every length and angle to test for congruence. In the case of triangles, it is possible to consider only a number of pairs of sides or angles to test whether or not they are congruent.

This wall at Federation Square in Melbourne includes many congruent triangles.

Let’s start: How much information is enough? Given one corresponding angle (say 30°) and one corresponding equal side length (say 3 cm), it is clearly not enough information to say two triangles are congruent. This is because more than one triangle can be drawn with the given information. 30° 3 cm

30° 3 cm 30° 3 cm

Decide if the following information is enough to determine if two triangles are congruent. If you can draw two non-identical triangles, then there is not enough information. • ABC with AC = 4 cm and ∠C = 40° • ABC with AB = 5 cm and AC = 4 cm • ABC with AB = 5 cm, AC = 4 cm and ∠A = 45° • ABC with AB = 5 cm, AC = 4 cm and BC = 3 cm • ABC with AB = 4 cm, ∠A = 40° and ∠B = 60° © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Key ideas

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Chapter 8 Transformations and congruence

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Two triangles are congruent if one of these four sets of conditions is satisfied. – If the three sides of a triangle are respectively equal to the three sides of another triangle, then the two triangles are congruent (SSS test) – If two sides and the included angle of a triangle are respectively equal to two sides and the included angle of another triangle, then the two triangles are congruent (SAS test).

– If two angles and one side of a triangle are respectively equal to two angles and the matching side of another triangle, then the two triangles are congruent (AAS test) – If the hypotenuse and a second side of a right-angled triangle are respectively equal to the hypotenuse and a second side of another right-angled triangle, then the two triangles are congruent (RHS test)

Example 8 Choosing a test for congruence Which of the tests (SSS, SAS, AAS or RHS) would you choose to test the congruence of these pairs of triangles? a b 12 m 12 m 70° 60° 5 cm

70° 60°

5 cm

15 m

15 m

SOLUTION

EXPLANATION

a AAS

Two angles on one triangle are equal to two angles on the other triangle. The side that is 5 cm is opposite to the 70° angle on both triangles.

b RHS

Both triangles are right-angled. The hypotenuse and a second side are the same on both triangles.

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50° 80°

3 cm

4 cm © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Example 8

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4 These pairs of triangles are congruent. Find the values of the pronumerals.

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5 Decide if these pairs of triangles are congruent. If they are, write the test (SSS, SAS, AAS or RHS). a

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7 Decide if each pair of triangles is congruent. You may first need to use the angle sum of a triangle to help calculate some of the angles.

a

b 35°

100°

75°

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5 cm 30° 100°

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40 40°

35°° 10 cm

10 cm 65° 65

5 cm 8 Two adjoining triangular areas of land have the dimensions shown. Use Pythagoras’ theorem to help decide if the two areas of land are congruent. 5 km 12 km

13 km

5 km 9 a E xplain why SSA is not sufficient to prove that two triangles are congruent. Draw diagrams to show your reasoning. b Explain why AAA is not sufficient to prove that two triangles are congruent. Draw diagrams to show your reasoning.

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∠B = 30°, ∠C = 20°, EF = 3 m, FD = 3 m, AB = 6 cm, AC = 6 cm, ∠C = 20°, ∠A = 20° a A

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Enrichment: Proof challenge 11 Write a logical set of reasons (proof) as to why the following are true. Refer to Question 10 for an example of how to set out a simple proof. A (Apex) a The segment joining the apex of an isosceles triangle to the midpoint M of the base BC is at right angles to the base, i.e. prove ∠AMB = ∠AMC = 90°. D b The segment AC = 2AB in this diagram. c A kite has one pair of equal opposite angles. D A

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8F Similar figures

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When you look at an object through a telescope or a pair of binoculars you expect the image to be much larger than the one you would see with the naked eye. Both images would be the same in every way except for their size. Such images in mathematics are called similar figures. Images that have resulted in a reduction in size (rather than an enlargement in size) are also considered to be similar figures.

Let’s start: Are they similar? Here are two quadrilaterals.

An item seen through a telescope is similar to how it appears to the naked eye.

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Similar ﬁgures have the same shape but can be of different size. All corresponding angles are equal. All corresponding sides are in the same ratio. The ratio of sides is often written as a fraction, shown here. It is often called the scale factor.

E

F

A

B C

G

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H Scale factor =

EF FG GH HE = = = AB BC CD DA

Cambridge University Press

Key ideas

• Do they look similar in shape? Why? • How could you use a protractor to help decide if they are similar in shape? What would you measure and check? Try it. • How could you use a ruler to help decide if they are similar in shape? What would you measure and check? Try it. • Describe the geometrical properties that similar shapes have. Are these types of properties present in all pairs of shapes that are ‘similar’?

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Example 9 Identifying corresponding features For the following pairs of similar figures complete these tasks. y cm B a° C 2 cm 1 cm

F

A

G x cm

D a b c d

6 cm

E

50° 6 cm

H List the pairs of corresponding sides. List the pairs of corresponding angles. Find the scale factor. Find the values of the pronumerals.

SOLUTION

EXPLANATION

a (AB, EF), (BC, FG), (CD, GH), (DA, HE)

Pair up each vertex, noticing that G corresponds with C, H corresponds with D and so on.

b (∠A, ∠E), (∠B, ∠F), (∠C, ∠G), (∠D, ∠H)

∠G and ∠C are clearly the largest angles in their respective shapes. Match the other angles in reference to these angles.

c

HE 6 = =3 DA 2

HE and DA are corresponding sides both with given measurements.

d a = 50

x = 3 × 1 = 3 cm

y = 6 ÷ 3 = 2 cm

∠B and ∠F are equal corresponding angles. HE GH The scale factor should also equal 3. = 3 so DA CD Alternatively, say that GH is 3 times the length CD. Similarly, EF should be 3 times the length AB (y cm).

Example 10 Deciding if shapes are similar Decide if these shapes are similar by considering corresponding angles and the ratio of sides. a D b F G C C F E 2 cm 7 cm

7 cm

D A

A

3 cm

B

E

14 cm

5 cm

B

H

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SOLUTION

EXPLANATION

a All corresponding angles are equal. EH 14 = =2 BC 7 GH 7 1 = =2 AB 3 3

All angles are 90° and so corresponding angles are equal.

The scale factor needs to be equal if the shapes are to be similar.

Match pairs of sides to find scale factors.

Scale factors are not equal. Shapes are not similar.

b All angles are 60°. DE 2 = = 0.4 AB 5 EF 2 = = 0.4 BC 5 FD 2 = = 0.4 CA 5 Triangles are similar, because the corresponding sides are in the same ratio.

All scale factors are equal so the triangles are similar.

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Exercise 8F

All angles in a equilateral triangle are 60°.

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Example 9

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6 Two rectangular picture frames are similar in shape. A corresponding pair of sides are of length 50 cm and 75 cm. The other side length on the smaller frame is 70 cm. Find the perimeter of the larger frame. 7 7 Two similar triangles have a scale factor of . 3 If the larger triangle has a side length of 35 cm, find the length of the corresponding side on the smaller triangle.

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8 One circle has a perimeter of 10π cm and another has an area of 16π cm2. Find the scale factor of the diameters of the two circles. 9 A square photo of area 100 cm2 is enlarged to an area of 900 cm2. Find the scale factor of the side lengths of the two photos. WO

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11 a I f a regular polygon such as this regular octagon is enlarged, do the interior angles change? b Are all polygons with the same number of sides similar? Give reasons. 12 If a square is similar to another by a scale factor of 4, what is the ratio of the area of small square to the area of large square?

Enrichment: Square designs 13 A square design consists of a series of squares as shown. An inner square is formed by joining the midpoints of the sides of the next largest square. a Use Pythagoras’ theorem to find the side length of: i the second largest square ii the third largest square b Find the scale factor of a pair of consecutive squares, e.g. first and 4 cm second or second and third. c If the side length of the outside square was x cm, show that the scale factor of consecutive squares is equal to the result found in part b. d What is the scale factor of a pair of alternate squares, e.g. first and third or second and fourth?

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Measurement and Geometry

8G Similar triangles

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E X T E N S I ON

Finding the approximate height of a tree or the width of a gorge using only simple equipment is possible without actually measuring the distance directly! Similar triangles can be used to calculate distances without the need to climb the tree or cross the gorge. It is important, however, to ensure that if the mathematics of similar triangles is going to be used, then the two triangles are in fact similar. We learnt earlier that there were four tests that help to determine if two triangles are congruent. Similarly there are four tests that help establish whether or not triangles are similar. Not all side lengths or angles are required to prove that two triangles are similar.

Similar triangles should be used to ﬁnd the width of a gorge before crossing it.

Let’s start: How much information is enough?

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Key ideas

Given a certain amount of information, it may be possible to draw two triangles that are guaranteed to be similar in shape. Decide if the information given is enough to guarantee that the two triangles will always be similar. If you can draw two triangles (ABC and DEF) that are not similar, then there is not enough information provided. • ∠A = 30° and ∠D = 30° • ∠A = 30°, ∠B = 80° and ∠D = 30°, ∠E = 80° • AB = 3 cm, BC = 4 cm and DE = 6 cm, EF = 8 cm • AB = 3 cm, BC = 4 cm, AC = 5 cm and DE = 6 cm, EF = 8 cm, DF = 10 cm • AB = 3 cm, ∠A = 30° and DE = 3 cm, ∠D = 30° • AB = 3 cm, AC = 5 cm, ∠A = 30° and DE = 6 cm, DF = 10 cm, ∠D = 30°

Two triangles are similar if: – Two angles of a triangle are equal to two angles of another triangle. 85° 60°

85° 60°

Cambridge University Press

Chapter 8 Transformations and congruence

Key ideas

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– Three sides of a triangle are proportional to three sides of another triangle. C

F3m

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DE EF DF = = =2 AB BC AC

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E DF EF = =3 AC BC

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If two triangles ABC and DEF are similar we write ABC ||| DEF. Abbreviations such as AAA are not used for similarity in NSW.

Example 11 Deciding if two triangles are similar Explain why these pairs of triangles are similar. a C F

b

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D 75° 70°

5 cm 10 cm A

50° B 2 cm

F 70° E

50° 4 cm

75°

A

B

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SOLUTION a

EXPLANATION

ED 4 = =2 AB 2 ∠B = ∠E = 50° EF 10 = =2 BC 5 ∴ ABC is similar to DEF, because two sides of one triangle are proportional to two sides of the other and the included angles are equal.

Work out the ratio of the two pairs of corresponding sides to see if they are equal. Note that the angle given is the included angle between the two given pairs of corresponding sides.

Two pairs of equal angles are given. This b ∠A = ∠D implies that the third pair of angles are also ∠B = ∠E ∴ ABC is similar to DEF, because two angles equal and that the triangles are similar. of one triangle are equal to two angles of the other.

Example 12 Finding a missing length Given that these pairs of triangles are similar, find the value of the pronumerals.

5

y

15

9

4 x SOLUTION 15 =3 5

First find the scale factor.

x = 4 × 3 = 12 y=9÷3=3

Multiply corresponding side by the scale factor to find a side on the larger triangle. Otherwise divide.

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1 Write a similarity statement for these pairs of similar triangles, e.g. ABC ||| DEF. Be careful with the order of letters and make sure each letter matches the corresponding vertex. a C b E F B D

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2 For the pairs of triangles in Question 1, which of the four similar triangle tests would be used to show their similarity? See the Key ideas for a description of each test. There is one pair for each test.

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5 Two triangular tracks are to be used for the junior and open divisions for a school event.

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a Work out the scale factor for each of the three corresponding pairs of sides. b Are the two tracks similar in shape? Give a reason. c How many times would a student need to run around the junior track to cover the same distance as running one lap of the senior track?

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7 In a game of minigolf, a ball bounces off a wall as shown. The ball proceeds and hits another wall at point A. How far down the right side wall does the ball hit?

3m

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A 8 Trees on a river bank are to be used as markers to form two triangular shapes. Each tree is marked by a red dot as shown in the diagram.

River 8m 6m

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a Are the two triangles similar? b Calculate the scale factor. c Calculate how far it is across the river.

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10 Give reasons why the pairs of triangles in these diagrams are similar. If an angle or side is common to both triangles, use the word ‘common’ in your reasoning. a C b A B

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∆ ABC and ∆ ∆FED FE FED 11 Show how Pythagoras’ theorem can be used to prove that these two triangles are similar. B

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Enrichment: Tricky unknowns 12 The pairs of triangles here show a length x cm, which is not the length of a full side of a triangle. Find the value of x in each case. a b 3.6 cm x cm

3 cm 3 cm

1 cm 2 cm c

x cm

3 cm 5 cm

5 cm

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8H Using congruent triangles to establish properties

of quadrilaterals The properties of special quadrilaterals, (parallelogram, rhombus, rectangle, square, trapezium and kite) can be examined more closely using congruence. By drawing the diagonals and using the tests for the congruence of triangles we can prove many of the properties of these special quadrilaterals.

Let’s start: Do the diagonals bisect each other?

Key ideas

Here is a parallelogram with two pairs of parallel sides. Let’s assume also that opposite sides are equal (a basic property of a parallelogram which we prove later in this exercise). D C • First locate ABE and CDE. E • What can be said about the angles ∠BAE and ∠DCE? • What can be said about the angles ∠ABE and ∠CDE? A B • What can be said about the sides AB and DC? • Now what can be said about ABE and CDE? Discuss the reasons. • What does this tell us about where the diagonals intersect? Do they bisect each other and why? (To bisect means to cut in half.)

This is a summary of the properties of the special quadrilaterals. ■ Kite: A quadrilateral with two pairs of adjacent sides equal – Two pairs of adjacent sides of a kite are equal – One diagonal of a kite bisects the other diagonal – One diagonal of a kite bisects the opposite angles – The diagonals of a kite are perpendicular ■ Trapezium: A quadrilateral with at least one pair of parallel sides – At least one pair of sides of a trapezium are parallel ■ Parallelogram: A quadrilateral with both pairs of opposite sides parallel – The opposite sides of a parallelogram are parallel – The opposite sides of a parallelogram are equal – The opposite angles of a parallelogram are equal – The diagonals of a parallelogram bisect each other ■ Rhombus: A parallelogram with two adjacent sides equal in length – The opposite sides of a rhombus are parallel – All sides of a rhombus are equal – The opposite angles of a rhombus are equal – The diagonals of a rhombus bisect the vertex angles – The diagonals of a rhombus bisect each other – The diagonals of a rhombus are perpendicular

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Key ideas

Measurement and Geometry

Rectangle: A parallelogram with a right angle – The opposite sides of a rectangle are parallel – The opposite sides of a rectangle are equal – All angles at the vertices of a rectangle are 90° – The diagonals of a rectangle are equal – The diagonals of a rectangle bisect each other Square: A rectangle with two adjacent sides equal – Opposite sides of a square are parallel – All sides of a square are equal – All angles at the vertices of a square are 90° – The diagonals of a square bisect the vertex angles – The diagonals of a square bisect each other – The diagonals of a square are perpendicular

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Example 13 Proving that diagonals of a parallelogram bisect each other Prove that the diagonals of a parallelogram bisect each other and give reasons.

D E A

SOLUTION

C

B

EXPLANATION

∠BAE = ∠DCE (alternate angles in parallel lines) ∠ABE = ∠CDE (alternate angles in parallel lines) AB = CD (opposite sides of a parallelogram) ∴ ABE ≡ CDE (AAS) ∴ BE = DE and AE = CE ∴ Diagonals bisect each other.

D

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The two triangles are congruent using the AAS test. Since ABE and CDE are congruent the corresponding sides are equal.

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128°

70° a°

3 SSS is one test for congruence of triangles. Write down the other three. 4 Name the side (e.g. AB) that is common to both triangles in each diagram. a D

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B 5 Which of the four tests for congruence of triangles would be used to prove that each pair of triangles is congruent? c

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Step 1. List the pairs of equal angles in ABE and CDE giving reasons why they are equal.

Step 2. List the pairs of equal sides in ABE and CDE giving reasons why they are equal.

Step 3. Write ABE ≡ CDE and give the reason SSS, SAS, AAS or RHS.

Step 4. State that BE = DE and AE = CE and give a reason.

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6 Prove, giving reasons, that the diagonals in a parallelogram bisect each other. You may assume here that opposite sides are equal so use AB = CD. Complete the proof by following these steps.

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Example 13

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10 For this square assume that MQ = QO and NQ = PQ. a Give reasons why MNQ ≡ ONQ. b Give reasons why ∠MQN = ∠OQN = 90°. c Give reasons why ∠QMN = 45°.

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9 For this rhombus assume that VW = WT. a Give reasons why VWU ≡ TWU. b Give reasons why ∠VWU = ∠TWU = 90°.

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M 11 Use the information in this kite to prove these results. a ABD ≡ CBD b ∠DAB = ∠DCB c ∠ADB = ∠CDB

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8 A parallelogram ABCD has two pairs of parallel sides. a What can be said about ∠ABD and ∠CDB and give a reason? b What can be said about ∠BDA and ∠DBC and give a reason? c Which side is common to both ABD and CDB? d Which congruence test would be used to show that ABD ≡ CDB? e If ABD ≡ CDB, what can be said about the opposite sides of a parallelogram?

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7 Prove (as in Example 13) that the diagonals in a square EFGH bisect each other.

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13 Use the steps outlined in Question 6 to show that opposite sides of a rectangle ABCD are equal. Give all reasons.

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15 Use the information given for this kite to give reasons for the following properties. You may assume that ∠ADE = ∠CDE as marked. a ∠DAE = ∠DCE b AED ≡ CED c ∠AED = ∠CED = 90°

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Enrichment: Prove the converse 16 Prove the following results, which are the converse (reverse) of some of the proofs completed earlier in this exercise. a If the opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram; i.e. show that the quadrilateral has two pairs of parallel sides. D

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b If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. D

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c If the diagonals of a parallelogram are equal, then the parallelogram is a rectangle. D C

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AC = BD B

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Investigating congruence with dynamic geometry software Computer dynamic geometry is ideal for these activities but geometrical instruments could also be used.

Conditions for congruent triangles 1 SSS a Construct a triangle with side lengths 4 cm, 7 cm and 9 cm. Measure the given side lengths using 1 decimal place and if using dynamic geometry, adjust the triangle to suit. Use the measure tool for the side lengths. b Construct a second triangle with the same side lengths c Do your two triangles look congruent? Measure the angles to help check. d Can you draw another triangle given the same information so that it does not look congruent to the others? 7.0 cm

9.0 cm

4.0 cm 4.0 cm 9.0 cm

7.0 cm

2 SAS a Construct a triangle with side lengths 4 cm and 7 cm and an included angle of 30°. Measure the given side lengths and the angle using 1 decimal place and if using dynamic geometry, adjust the triangle to suit. b Construct a second triangle with the same side lengths and included angle. c Do your two triangles look congruent? Measure the missing angle and side lengths to help check. d Can you draw another triangle given the same information so that it does not look congruent to the others? 3 AAS a Construct a triangle with two angles 30° and 50° and one side length 6 cm. Measure the side given length and the angles using 1 decimal place and if using dynamic geometry, adjust the triangle to suit. b Construct a second triangle with the same side length and angles. c Do your two triangles look congruent? Measure the missing side length and angle to help check. d Can you draw another triangle given the same information so that it does not look congruent to the others?

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Investigation

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Chapter 8 Transformations and congruence

4 RHS a Construct a triangle with a 90° and 50°, hypotenuse of length 6 cm and other side of length 4 cm. Measure the two given side lengths and the angle using 1 decimal place and if using dynamic geometry, adjust the triangle to suit. b Construct a second triangle with the same two side lengths and angle. c Do your two triangles look congruent? Measure the missing side length and angles to help check. d Can you draw another triangle given the same information so that it does not look congruent to the others?

Exploring SSA With the SAS test, the included angle must be between the two given sides. This investigation will explore a triangle using SSA. 1 Construct a triangle with sides 5 cm and 3 cm and one angle 30° (not an included angle). 2 Can you construct a second triangle using the same information that is not congruent to the first? 3 Consider this diagram. Use it to help explain why SSA is not a test for congruence of two triangles.

3 cm

3 cm

30° 5 cm

Properties of quadrilaterals 1 Construct a parallelogram including its diagonals as shown. Use the parallel line tool to create the parallel sides. D

C E

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2 Measure the sides and angles of ΔABE and ΔCDE. What do you notice? 3 Now drag one of the vertices of the parallelogram. Does this change you conclusion from part b? 4 Explore other quadrilaterals by constructing their diagonals and looking for congruent triangles. Write up a brief summary of your findings.

Cambridge University Press

1 How many similar triangles are there in this figure? Give reasons.

2 Can you fit the shapes in the two smaller squares into the largest square? What does this diagram illustrate? Try drawing or constructing the design and then use scissors to cut out each shape.

3 A strip of paper is folded five times in one direction only. How many creases will there be in the original strip when it is folded out?

4 Use congruent triangles to find the radius r in these diagrams. a b 3

r

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4 r r r

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4 5 What is the side length of the largest square that can be cut out of this triangle? Use similar triangles.

12 5 13

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Chapter summary

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Chapter 8 Transformations and congruence

Translation 2 Vector –1 2

Rotation

( )

Clockwise by 90° 1 C

Congruent figures A

Reflection

Transformation and congruence

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∆ABC ≡ ∆DEF ∆ABC ∆ AB = DE, BC = EF, AC = DF ∠ = ∠D, ∠B = ∠E, ∠C = ∠F ∠A Tests: SSS, SAS, AAS, RHS

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Measurement and Geometry

Multiple-choice questions 1 The number of lines of symmetry in a regular pentagon is: A 10 B 5 C 2 D 1

E 0

2 Which vector describes a translation of 5 units to the left and 3 units up? A

( ) 3

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−5

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6 3 The point A(−3, 4) is translated to the point A′ by the vector . The coordinates of point A′ −4 are: A (3, 8) B (−9, 8) C (3, 0) D (0, 3) E (−9, 0) 4 An anticlockwise rotation of 125° about a point is the same as a clockwise rotation about the same point of: A 235° B 65° C 55° D 135° E 245° 5 Point A(1, 3) is rotated clockwise about C by 90° to A′. The coordinates of A′ are: A (3, 0) B (3, −1) C (−3, 1) D (−1, 3) E (3, 1)

already

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Questions 6, 7, 8 and 9 relate to this pair of congruent triangles. C

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D 15 cm

And 1cm

8 A congruent statement with ordered vertices for the triangles is: A ABC ≡ FED B ABC ≡ EDF C ABC ≡ DFE D ABC ≡ DEF E ABC ≡ EFD

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9 Which of the four tests would be chosen to show that these two triangles are congruent? A AAS B RHS C SAS D AAA E SSS 10 ABCD is a parallelogram. Based on the information supplied in the diagram, which test could be most easily used to prove that ABC ≡ CDA. A AAS B RHS C SAS D AAA E SSS

A

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Short-answer questions 1 Copy these shapes and draw the reﬂected image over the mirror line. a b c

d

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2 The triangle A(1, 2), B(3, 4), C(0, 2) is reﬂected in the given axis. State the coordinates of the image points A′, B′ and C′. a x-axis b y-axis 3 How many lines of symmetry do these shapes have? a Square b Isosceles triangle c Rectangle d Kite e Regular hexagon f Parallelogram 4 Write the vectors that translate each point A to its image A′. a A(2, 5) to A′(3, 9) b A(−1, 4) to A′(2, −2) c A(0, 7) to A′(−3, 0) d A(−4, −6) to A′(0, 0)

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Measurement and Geometry

5 Copy these shapes and draw the translated image using the given translation vector. a

( ) 3

−1

b

( ) −2 2

c

y

O –1 –2

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2 1 –2

( ) y

2 1 1 2

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6 What is the order of rotational symmetry of these shapes? a Equilateral triangle b Parallelogram

–2 –1–1O –2

1 2

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c Isosceles triangle

7 Rotate these shapes about the point C by the given angle. a Clockwise 90°

C b Clockwise 180°

C c Anticlockwise 270° C

8 For these congruent quadrilaterals, name the object in quadrilateral EFGH that corresponds to the given object in quadrilateral ABCD. A H B C

D a i B b i AD c i ∠C

F E

G ii C ii BC ii ∠A

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9 Which of the tests SSS, SAS, AAS or RHS would you choose to explain the congruence of these pairs of triangles? a b

c

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10 Find the values of the pronumerals for these congruent triangles. a 3 cm a° 25° x cm b 18° xm

5m a°

11 This quadrilateral is a parallelogram with two pairs of parallel sides. You can assume that AB = DC as shown. a Prove that ABE ≡ CDE. b Explain why BD and AC bisect each other. 12 This quadrilateral is a square with AE = EC. a Prove that ABE ≡ CBE. b Explain why the diagonals intersect at right angles.

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Measurement and Geometry

Extended-response question 1 The shape on this set of axes is to be transformed by a succession of transformations. The image of the first transformation is used to start the next transformation. For each set of transformations write down the coordinates of the vertices A′, B′ and C′ of the final image.

Parts a, b and c are to be treated as separate questions. a Set 1 i Reﬂection in the x-axis −2 ii Translation by the vector 1 iii Rotation about (0, 0) by 180°. b Set 2 i Rotation about (0, 0) clockwise by 90°. ii Reﬂection in the y-axis 5 iii Translation by the vector 3 c Set 3 i Reﬂection in the line y = −x 4 ii Translation by the vector 2 iii Rotation about (1, 0) by 90° clockwise

( )

y C

5 4 3 2 1

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–5 –4 –3 –2 –1–1O –2 –3 –4 –5

B

1 2 3 4 5

()

()

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Chapter 8 Transformations and congruence

Chapter

8

Transformations and congruence

What you will learn

8A 8B 8C 8D 8E 8F 8G 8H

Reﬂection Translation Rotation Congruent ﬁgures Congruent triangles Similar ﬁgures EXTENSION Similar triangles EXTENSION Using congruent triangles to establish properties of quadrilaterals

Cambridge University Press

Measurement and Geometry

NSW Syllabus

for the Australian Curriculum Strands: N umber and Algebra

Measurement and Geometry Substrands: L INEAR RELATIONSHIPS PROPERTIES OF GEOMETRICAL FIGURES

Outcomes A student creates and displays number patterns; graphs and analyses linear relationships; and performs transformations on the Cartesian plane. (MA4–11NA) A student classiﬁes, describes and uses the properties of triangles and quadrilaterals, and determines congruent triangles to ﬁnd unknown side lengths and angles. (MA4–17MG)

Symmetrical architecture Geometry is at the foundation of design and architecture. In modern times, many public buildings as well as private residences have been designed and built with a strong sense of geometric expression. In the ancient Roman world, the famous architect Marcus Vitruvius said ‘For without symmetry and proportion no temple can have a regular plan’. The Pantheon is about 2000 years old and is one of the oldest and best kept buildings in Rome. Its rectangular portico is supported by eight monolithic cylindrical columns, and these combined with the triangular roof provide perfect line symmetry at the entrance of the building. Inside the main dome the symmetry changes. At the centre of the dome roof is a

large oculus (hole) that lets in the sunlight. This natural light allows visitors to see the perfect rotational symmetry within. The height of the dome is the same as its width, meaning that a sphere of the same diameter would ﬁt perfectly under the dome. It is no wonder that people travel to the Pantheon to pray.

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Pre-test

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1 Complete the simple transformations of the given shapes as instructed. a Reﬂect this shape over the mirror line.

b Shift this shape 2 units to the right and 1 unit down.

c Rotate this shape 180° around point C. C

D 2 Consider this pair of quadrilaterals. a Name the vertex, E, F, G or H, that matches (corresponds to) vertex: i B ii D b Name the angle, ∠A, ∠B, ∠C or ∠D, that matches (corresponds to) angle: A i ∠E ii ∠G E c Name the side, EF, FG, GH or HE, that matches (corresponds to) side: i CD ii BC

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3 Find the unknown angle in these triangles. a b 40° a° 110° a° 80°

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4 Which of the special quadrilaterals definitely fit the description? A Square B Rectangle C Rhombus D Parallelogram E Kite F Trapezium a Opposite sides are of equal length. b It has at least one pair of equal opposite angles. c Diagonals are of equal length. d Diagonals intersect at right angles.

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Measurement and Geometry

8A Reﬂ ection When an object is shifted from one position to another, rotated about a point, reﬂected over a line or enlarged by a scale factor, we say the object has been transformed. The names of these types of transformations are reﬂection, translation, rotation and dilation. The first three of these transformations are called isometric transformations because the object’s geometric properties are unchanged and the transformed object will be congruent to the original object. The word ‘isometric’ comes from the Greek words isos meaning ‘equal’ and metron meaning ‘measure’. Dilation (or enlargement) results in a change in the size of an object to produce a ‘similar’ figure and this will be studied later in this chapter. The first listed transformation, reﬂection, can be thought of as the creation Reﬂection creates an image reversed as in a of an image over a mirror line. mirror or in water.

Let’s start: Visualising the image This activity could be done individually by hand on a page, in a group using a whiteboard or using dynamic geometry projected onto a white board. • Draw any shape with straight sides. • Draw a vertical or horizontal mirror line outside the shape. • Try to draw the reﬂected image of the shape in the mirror line. • If dynamic geometry is used, reveal the precise image (the answer) using the Reﬂection tool to check your result. • For a further challenge, redraw or drag the mirror line so it is not horizontal or vertical. Then try to draw the image.

Dynamic geometry software provides a reﬂection tool.

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Reﬂection is an isometric transformation in which the size of the object is unchanged. The image of a point A is denoted A′. Each point is reﬂected at right angles to the mirror line. The distance from a point A to the mirror line is equal to the distance from the image point A′ to the mirror line. When a line of symmetry is used as a reﬂection line, the image looks the same as the original figure. When a point is reﬂected across the x-axis, – the x-coordinate remains unchanged – the y-coordinate undergoes a sign change, e.g. (5, 3) becomes (5, −3); that is, the y values are equal in magnitude but opposite in sign When the point is reﬂected across the y-axis, – the x-coordinate undergoes a sign change – the y-coordinate remains unchanged, e.g. (5, 3) becomes (−5, 3)

C

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Example 1 Drawing reflected images Copy the diagram and draw the reﬂected image over the given mirror line. a A b C B C

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EXPLANATION

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Reﬂect each vertex point at right angles to the mirror line. Join the image points to form the final image.

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E′ E Reﬂect points A, B and C at right angles to the mirror line to form A′, B′ and C′. Note that A′ is in the same position as A as it is on the mirror line. Join the image points to form the image triangle.

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Example 2 Using coordinates State the coordinates of the vertices A′, B′ and C′ after this triangle is reﬂected in the given axes. a x-axis b y-axis

y 4 3 2 1

B C

–4 –3 –2 –1–1O A1 2 3 4 –2 –3 –4 SOLUTION

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a A′ = (1, 0) B′ = (2, −3) C′ = (4, −2)

y 4 B 3 C C′′ C 2 b 1 A′ A x –4 –3 –2 –1–1O 1 2 3 4 a –2 C′ C′ –3 B′ –4 B′

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b A′ = (−1, 0) B′ = (−2, 3) C′ = (−4, 2)

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8 A point (2, 4) is reﬂected in the given horizontal or vertical line. State the coordinates of the image point. As an example, the graph of the mirror lines x = 1 and y = −2 are shown here. The mirror lines are: a x = 1 b x = 3 c x = 0 d x = −1 e x = −4 f x = −20 g y = 3 h y = −2 i y = 0 j y = 1 k y = −5 l y = −37

Regular octagon y (2, 4)

4 3 2 1 –4 –3 –2 –1–1O –2 –3 –4

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6 State the coordinates of the vertices A′, B′, C′ and D′ after this rectangle is reﬂected in the given axes. a x-axis b y-axis y

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5 State the coordinates of the vertices A′, B′ and C′ after the triangle (right) is reﬂected in the given axes. a x-axis b y-axis

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9 A shape with area 10 m2 is reﬂected in a line. What is the area of the image shape? Give a reason for your answer.

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10 How many lines of symmetry does a regular polygon with n sides have? Write an expression. 11 A point is reﬂected in the x-axis then in the y-axis and finally in the x-axis again. What single reﬂection could replace all three reﬂections? 12 Two important lines of reﬂection on the coordinate plane are the line y = x and the line y = −x as shown. a Draw the coordinate plane shown here. Draw a y=x y triangle with vertices A(−1, 1), B(−1, 3) and C(0, 3). Then complete these reﬂections. i Reﬂect triangle ABC in the y-axis. ii Reﬂect triangle ABC in the x-axis. iii Reﬂect triangle ABC in the line y = x. x iv Reﬂect triangle ABC in the line y = −x. O b Draw a coordinate plane and a rectangle with vertices A(−2, 0), B(−1, 0), C(−1, −3) and D(−2, −3). Then complete these reﬂections. i Reﬂect rectangle ABCD in the y-axis. ii Reﬂect rectangle ABCD in the x-axis. y = –x – iii Reﬂect rectangle ABCD in the line y = x. iv Reﬂect rectangle ABCD in the line y = −x. 13 Points are reﬂected in a mirror line but do not change position. Describe the position of these points in relation to the mirror line. 14 Research to find pictures, diagrams, cultural designs and artwork which contain line symmetry. Make a poster or display electronically.

Enrichment: Reﬂ ection through a point 15 Rather than completing a reﬂection in a line, it is possible to reﬂect an object through a point. An example of a reﬂection through point P is shown here. A goes to A′, B goes to B′ and C goes to C′ through P. a Draw and reﬂect these shapes through the point P. i ii iii P

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b Like line symmetry, shapes can have point symmetry if they can be reﬂected onto themselves through a point. Decide if these shapes have any point symmetry. i ii iii

c How many special quadrilaterals can you name that have point symmetry? © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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8B Translation Translation is a shift of every point on an object in a given direction and by the same distance. The direction and distance is best described by the use of a translation vector. This vector describes the overall direction using a horizontal component (for shifts left and right) and a vertical component (for shifts up and down). Negative numbers are used for translations to the left and down. Designers of animated movies translate images in many of their scenes. Computer software is used and translation vectors help to define the specific movement of the objects and characters on the screen.

Animated characters move through a series of translations.

Let’s start: Which is further? Consider this shape on the grid. The shape is translated by the vector 3 (3 to the right and 2 down). −2

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Now consider the shape being translated by these different vectors. a

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Translation is an isometric transformation that involves a shift by a given distance in a given direction. a A vector can be used to describe the distance and direction of a translation. b Right 2 2 Vector Left 1 −3

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Vector

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Up 4

– If a is positive you shift to the right. – If a is negative you shift to the left. – If b is positive you shift up. – If b is negative you shift down. The image of a point A is denoted A′.

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−3 −4 0 4 • By drawing and looking at the image from each translation, which vector do you think takes the shape furthest from its original position? • Is there a way that you can calculate the exact distance? How?

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Example 3 Finding the translation vector State the translation vector that moves the point A(−1, 3) to A′(2, 0). y A

3 2 1

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A′ 1 2 3

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SOLUTION

EXPLANATION

( )

To shift A to A′ move 3 units to the right and 3 units down.

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Example 4 Drawing images using translation

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Draw the image of the triangle ABC after a translation by the vector − 3 . 2 y 3 2 1

A O –3 –2 –1–1 –2 B –3

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SOLUTION

EXPLANATION

First translate each vertex, A, B and C, left 3 spaces, then up 2 spaces.

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1 Use the words left, right, up or down, to complete these sentences. 2 a The vector means to move 2 units to the ______ and 4 units _____. 4 −5 b The vector means to move 5 units to the _____ and 6 units _____. 6 3 c The vector means to move 3 units to the _____ and 1 unit _____. −1

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−10 means to move 10 units to the _____ and 12 units _____. −12

2 Write the vector that describes these transformations. b 2 units to the left and 6 units down d 9 units to the right and 17 units up

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4 Write the vector that takes each point to its image. Use a grid to help you. a A(2, 3) to A′(3, 2) b B(1, 4) to B′(4, 3) c C(−2, 4) to C′(0, 2) d D(−3, 1) to D′(−1, −3) e E(−2, −4) to E′(1, 3) f F(1, −3) to F′(−2, 2) g G(0, 3) to G′(2, 0) h H(−3, 5) to H′(0, 0) i I(5, 2) to I′(−15, 10) j J(−3, −4) to J′(−12, −29)

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Example 3

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6 Write the coordinates of the image of the point A(13, −1) after a translation by the given vectors. 2 8 0 −4 a b c d 3 0 7 3 −2 −10 −2 6 e f g h −1 −5 −8 −9 12 −26 −4 −21 i j k l −3 14 18 −38

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2 8 A car makes its way around a city street grid. A vector represents travelling 200 m east and 3 300 m north. a Find how far the car travels if it follows these vectors: 2 −5 3 −2 , , and 3 1 −3 −4

()( )( ) ( )

b What vector takes the car back to the origin (0, 0), assuming it started at the origin? 9 A point undergoes the following multiple translations with these given vectors. State the value of x and y of the vector that would take the image back to its original position. 3 −1 x 2 −7 −1 x a , , b , , , 4 −2 y 5 2 −3 y 0 7 −4 x −4 12 −36 x c , , , d , , , 4 0 −6 y 20 0 40 y

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( ) ( )( )

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Enrichment: How many options for the rabbit?

Hole

12 Hunters spot a rabbit on open ground and it has 1 second to find a hole before getting into big trouble with the hunter’s gun. It can run a xm 4m maximum of 5 metres in one second. a Use Pythagoras’ theorem to check that the distance x m in this diagram is less than 5 m. Rabbit b The rabbit runs a distance and direction described by the 2m −4 translation vector . Is the rabbit in trouble? 3 c The rabbit’s initial position is (0, 0) and there are rabbit holes at every point that has integers as its coordinates, e.g. (2, 3) and (−4, 1). How many rabbit holes can it choose from to avoid the hunters before its 1 second is up? Draw a diagram to help illustrate your working.

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10 A reverse vector takes a point in the reverse direction by the same distance. Write the reverse vectors of these vectors. 3 −5 x −x a b c d −2 0 y −y 11 These translation vectors are performed on a shape in succession (one after the other). What is a single vector that would complete all transformations for each part in one go? 2 −3 0 6 6 −11 a c −a a , , b , , c , , 1 −4 3 4 −2 0 b −a a – c

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8C Rotation When the arm of a crane moves left, right, up or down, it undergoes a rotation about a fixed point. This type of movement is a transformation called a rotation. The pivot point on a crane would be called the centre of rotation and all other points on the crane’s arm move around this point by the same angle in a circular arc.

Let’s start: Parallelogram centre of rotation This activity will need a pencil, paper, ruler and scissors.

Every point on this crane’s arm rotates around the pivot point when it moves.

Key ideas

• Accurately draw a large parallelogram on a separate piece of paper and cut it out. • Place the tip of a pencil at any point on the parallelogram and spin the shape around the pencil. • At what position do you put the pencil tip to produce the largest circular arc? • At what position do you put the pencil tip to produce the smallest circular arc? • Can you rotate the shape by an angle of less than 360° so that it exactly covers the area of the shape in its original position? Where would you put the pencil to illustrate this?

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Rotation is an isometric transformation about a centre point and by a given angle. An object can be rotated clockwise or anticlockwise . Each point is rotated on a circular arc about the centre of rotation C. This diagram shows a 90° anticlockwise rotation about the point C.

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A shape has rotational symmetry if it can be rotated about a centre point to produce an exact copy covering the entire area of the original shape. – The number of times the shape can make an exact copy in a 360° rotation is called the order of rotational symmetry. If the order of rotation is 1, then it is said that the shape has no rotational symmetry. – This equilateral triangle has rotational symmetry of order 3.

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Example 5 Finding the order of rotational symmetry Find the order of rotational symmetry for these shapes. a b

SOLUTION

EXPLANATION

a Order of rotational symmetry = 2

1

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Example 6 Drawing a rotated image Rotate these shapes about the point C by the given angle and direction. a Clockwise by 90° b 180°

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SOLUTION

EXPLANATION

a

Take each vertex point and rotate about C by 90°, but it may be easier to visualise a rotation of some of the sides first. Horizontal sides will rotate to vertical sides in the image and vertical sides will rotate to horizontal sides in the image.

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5 Rotate these shapes about the point C by the given angle and direction. a Clockwise by 90° b Anticlockwise by 90° c Anticlockwise by 180° C

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9 Which capital letters of the alphabet (A, B, C, . . . , Z) have rotational symmetry of order 2 or more? 10 Draw an example of a shape that has these properties. a Rotational symmetry of order 2 with no line symmetry b Rotational symmetry of order 6 with 6 lines of symmetry c Rotational symmetry of order 4 with no line symmetry

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11 What value of x makes these sentences true? a Rotating x degrees clockwise has the same effect as rotating x degrees anticlockwise. b Rotating x degrees clockwise has the same effect as rotating 3x degrees anticlockwise.

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13 Make a copy of this diagram and rotate the shape anticlockwise by 135° around point C. You will need to use compasses and a protractor as shown in Question 12. C

Enrichment: Finding the centre of rotation 14 Finding the centre of rotation if the angle is known involves the calculation of an angle inside an isosceles triangle. For the rotation shown, the angle of rotation is 50°. The steps are given as: i Calculate ∠CAA′ and ∠CA′A. (2x + 50 = 180, so x = 65) ii Draw the angles ∠AA′C and ∠A′AC at 65° using a protractor. iii Locate the centre of rotation C at the point of intersection of AC and A′C. a On a sheet of paper draw two points A and A′ about 4 cm apart. Follow the steps above to locate the centre of rotation if the angle of rotation is 40°. b Repeat part a using an angle of rotation of 100°. c When a shape is rotated and the angle is unknown, there is a special method for accurately pinpointing the centre of rotation. Research this method and illustrate the procedure using an example.

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8D Congruent ﬁ gures In mathematics, if two objects are identical we say they are congruent. If you ordered 10 copies of a poster from a printer, you would expect that the image on one poster would be congruent to the image on the next. Even if one poster was ﬂipped over, shifted or rotated you would still say the images on the posters were congruent.

Let’s start: Are they congruent?

Each brochure being printed is identical, so the images on them are congruent.

Here are two shapes. To be congruent they need to be exactly the same shape and size.

Key ideas

• Do you think they look congruent? Give reasons. • What measurements could be taken to help establish whether or not they are congruent? • Can you just measure lengths or do you need to measure angles as well? Discuss.

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A ﬁgure is a shape, diagram or illustration. Congruent ﬁgures have the same size and shape. They also have the same area. They are identical in every way.

The image of a figure that is reﬂected, translated or rotated is congruent to the original figure. Corresponding (matching) parts of a figure have the same geometric properties. – Vertex C corresponds to vertex E. D E C – Side AB corresponds to side FD. – ∠B corresponds to ∠D. F A congruence statement can be written using the symbol ≡, e.g. ABC ≡ FDE. A – The symbol ≅ can also be used for congruence. B – In a congruence statement vertices are named in matching order, e.g. ABC ≡ FDE not ABC ≡ DEF because B matches D. Two circles are congruent if they have equal radii.

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Example 7 Naming corresponding pairs These two quadrilaterals are congruent. Name the objects in quadrilateral EFGH that correspond to these objects in quadrilateral ABCD. a C b AB c ∠C A G F

H B E

SOLUTION

EXPLANATION

a G

C sits opposite A and ∠A is the smallest angle. G sits opposite E and ∠E is also the smallest angle.

b EH

Sides AB and EH are both the longest sides of their respective shapes. A corresponds to E and B corresponds to H.

c ∠G

∠C and ∠G are both the largest angle in their corresponding quadrilateral.

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1 In this diagram ABC has been reﬂected to give the image triangle DEF. a Is DEF congruent to ABC; i.e. is DEF ≡ ABC? C b Name the vertex on DEF which corresponds to: i A ii B iii C A B c Name the side on DEF which corresponds to: i AB ii BC iii AC d Name the angle in DEF which corresponds to: i ∠B ii ∠C iii ∠A

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3 In this diagram ABC has been rotated to give the image triangle DEF. a Is DEF congruent to ABC; i.e. is DEF ≡ ABC? A b Name the vertex on DEF which corresponds to: i A ii B iii C c Name the side on DEF which corresponds to: i AB ii BC iii AC d Name the angle in DEF which corresponds to: i ∠B ii ∠C iii ∠A

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7 Using the lines in the diagram as the sides of triangles, how many triangles are there that have: 1 a area ? b area 1? c area 2? 2 d area 4? e area 8?

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9 Write the pairs of corresponding vertices for these congruent shapes, e.g. (A, E), (B, D). a C E A B Z W D b

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10 An isosceles triangle is cut as shown, using the midpoint of AB. a Name the two triangles formed. b Will the two triangles be congruent? Explain.

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11 A kite ABCD has diagonals AC and BD. a The kite is cut using the diagonal BD. i Name the two triangles formed. ii Will the two triangles be congruent? Explain. b The kite is cut using the diagonal AC. i Name the two triangles formed. ii Will the two triangles be congruent? Explain.

C

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12 If a parallelogram is cut by either diagonal will the two triangles be congruent?

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13 What is the condition for two circles to be congruent?

Enrichment: Combined congruent transformations 14 Describe the combination of transformations (reﬂections, translations and/or rotations) that map each shape to its image under the given conditions. The reﬂections that are allowed include only those in the x- and y-axes and rotations will use (0, 0) as its centre. a A to A′ with a reﬂection and then a translation b A to A′ with a rotation and then a translation c B to B′ with a rotation and then a translation d B to B′ with 2 reﬂections and then a translation e C to C′ with 2 reﬂections and then a translation f C to C′ with a rotation and then a translation y A

5 4 3 2 1

–5 –4 –3 –2 –1–1O –2 –3 B C′ –4 –5

C

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A′ 1 2 3 4 5

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8E Congruent triangles Imagine the sorts of design and engineering problems we would face if we could not guarantee that two objects such as window panes or roof truss frames were not the same size or shape. Further, it might not be possible or practical to measure every length and angle to test for congruence. In the case of triangles, it is possible to consider only a number of pairs of sides or angles to test whether or not they are congruent.

This wall at Federation Square in Melbourne includes many congruent triangles.

Let’s start: How much information is enough? Given one corresponding angle (say 30°) and one corresponding equal side length (say 3 cm), it is clearly not enough information to say two triangles are congruent. This is because more than one triangle can be drawn with the given information. 30° 3 cm

30° 3 cm 30° 3 cm

Decide if the following information is enough to determine if two triangles are congruent. If you can draw two non-identical triangles, then there is not enough information. • ABC with AC = 4 cm and ∠C = 40° • ABC with AB = 5 cm and AC = 4 cm • ABC with AB = 5 cm, AC = 4 cm and ∠A = 45° • ABC with AB = 5 cm, AC = 4 cm and BC = 3 cm • ABC with AB = 4 cm, ∠A = 40° and ∠B = 60° © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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Key ideas

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Chapter 8 Transformations and congruence

■

Two triangles are congruent if one of these four sets of conditions is satisfied. – If the three sides of a triangle are respectively equal to the three sides of another triangle, then the two triangles are congruent (SSS test) – If two sides and the included angle of a triangle are respectively equal to two sides and the included angle of another triangle, then the two triangles are congruent (SAS test).

– If two angles and one side of a triangle are respectively equal to two angles and the matching side of another triangle, then the two triangles are congruent (AAS test) – If the hypotenuse and a second side of a right-angled triangle are respectively equal to the hypotenuse and a second side of another right-angled triangle, then the two triangles are congruent (RHS test)

Example 8 Choosing a test for congruence Which of the tests (SSS, SAS, AAS or RHS) would you choose to test the congruence of these pairs of triangles? a b 12 m 12 m 70° 60° 5 cm

70° 60°

5 cm

15 m

15 m

SOLUTION

EXPLANATION

a AAS

Two angles on one triangle are equal to two angles on the other triangle. The side that is 5 cm is opposite to the 70° angle on both triangles.

b RHS

Both triangles are right-angled. The hypotenuse and a second side are the same on both triangles.

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1 Pick the congruence test (SSS, SAS, AAS or RHS) that matches each pair of congruent triangles.

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2 Write a congruence statement (e.g. ABC ≡ DEF) for these pairs of congruent triangles. Try to match vertices.

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50° 80°

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4 cm © David Greenwood et al. 2014 ISBN: 9781107671812 Photocopying is restricted under law and this material must not be transferred to another party

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3 Which of the tests (SSS, SAS, AAS or RHS) would you choose to test the congruence of these triangles?

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Example 8

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4 These pairs of triangles are congruent. Find the values of the pronumerals.

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1 m

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ym

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5 Decide if these pairs of triangles are congruent. If they are, write the test (SSS, SAS, AAS or RHS). a

11 m

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6m 6m C 3m B c

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7 Decide if each pair of triangles is congruent. You may first need to use the angle sum of a triangle to help calculate some of the angles.

a

b 35°

100°

75°

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5 cm 30° 100°

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40 40°

35°° 10 cm

10 cm 65° 65

5 cm 8 Two adjoining triangular areas of land have the dimensions shown. Use Pythagoras’ theorem to help decide if the two areas of land are congruent. 5 km 12 km

13 km

5 km 9 a E xplain why SSA is not sufficient to prove that two triangles are congruent. Draw diagrams to show your reasoning. b Explain why AAA is not sufficient to prove that two triangles are congruent. Draw diagrams to show your reasoning.

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∠B = 30°, ∠C = 20°, EF = 3 m, FD = 3 m, AB = 6 cm, AC = 6 cm, ∠C = 20°, ∠A = 20° a A

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6 Decide which piece of information from the given list needs to be added to the diagram if the two triangles are to be congruent.

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Enrichment: Proof challenge 11 Write a logical set of reasons (proof) as to why the following are true. Refer to Question 10 for an example of how to set out a simple proof. A (Apex) a The segment joining the apex of an isosceles triangle to the midpoint M of the base BC is at right angles to the base, i.e. prove ∠AMB = ∠AMC = 90°. D b The segment AC = 2AB in this diagram. c A kite has one pair of equal opposite angles. D A

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10 Here is a suggested proof showing that the two triangles in this diagram are congruent. In ABC and EDC ∠CAB = ∠CED (alternate angles in parallel lines) A B ∠ACB = ∠ECD (vertically opposite angles) AC = EC (given equal and corresponding sides) C ABC ≡ EDC (AAS) D E Write a proof (similar to the above) for these pairs of congruent triangles. Any of the four tests (SSS, SAS, AAS or RHS) may be used.

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8F Similar figures

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E X T E N S I ON

When you look at an object through a telescope or a pair of binoculars you expect the image to be much larger than the one you would see with the naked eye. Both images would be the same in every way except for their size. Such images in mathematics are called similar figures. Images that have resulted in a reduction in size (rather than an enlargement in size) are also considered to be similar figures.

Let’s start: Are they similar? Here are two quadrilaterals.

An item seen through a telescope is similar to how it appears to the naked eye.

■

■ ■ ■

Similar ﬁgures have the same shape but can be of different size. All corresponding angles are equal. All corresponding sides are in the same ratio. The ratio of sides is often written as a fraction, shown here. It is often called the scale factor.

E

F

A

B C

G

D

H Scale factor =

EF FG GH HE = = = AB BC CD DA

Cambridge University Press

Key ideas

• Do they look similar in shape? Why? • How could you use a protractor to help decide if they are similar in shape? What would you measure and check? Try it. • How could you use a ruler to help decide if they are similar in shape? What would you measure and check? Try it. • Describe the geometrical properties that similar shapes have. Are these types of properties present in all pairs of shapes that are ‘similar’?

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Example 9 Identifying corresponding features For the following pairs of similar figures complete these tasks. y cm B a° C 2 cm 1 cm

F

A

G x cm

D a b c d

6 cm

E

50° 6 cm

H List the pairs of corresponding sides. List the pairs of corresponding angles. Find the scale factor. Find the values of the pronumerals.

SOLUTION

EXPLANATION

a (AB, EF), (BC, FG), (CD, GH), (DA, HE)

Pair up each vertex, noticing that G corresponds with C, H corresponds with D and so on.

b (∠A, ∠E), (∠B, ∠F), (∠C, ∠G), (∠D, ∠H)

∠G and ∠C are clearly the largest angles in their respective shapes. Match the other angles in reference to these angles.

c

HE 6 = =3 DA 2

HE and DA are corresponding sides both with given measurements.

d a = 50

x = 3 × 1 = 3 cm

y = 6 ÷ 3 = 2 cm

∠B and ∠F are equal corresponding angles. HE GH The scale factor should also equal 3. = 3 so DA CD Alternatively, say that GH is 3 times the length CD. Similarly, EF should be 3 times the length AB (y cm).

Example 10 Deciding if shapes are similar Decide if these shapes are similar by considering corresponding angles and the ratio of sides. a D b F G C C F E 2 cm 7 cm

7 cm

D A

A

3 cm

B

E

14 cm

5 cm

B

H

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SOLUTION

EXPLANATION

a All corresponding angles are equal. EH 14 = =2 BC 7 GH 7 1 = =2 AB 3 3

All angles are 90° and so corresponding angles are equal.

The scale factor needs to be equal if the shapes are to be similar.

Match pairs of sides to find scale factors.

Scale factors are not equal. Shapes are not similar.

b All angles are 60°. DE 2 = = 0.4 AB 5 EF 2 = = 0.4 BC 5 FD 2 = = 0.4 CA 5 Triangles are similar, because the corresponding sides are in the same ratio.

All scale factors are equal so the triangles are similar.

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1 List any pairs of shapes that look similar (i.e. same shape but different size).

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Exercise 8F

All angles in a equilateral triangle are 60°.

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C F 4 cm

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E a Name the angle on the larger triangle which corresponds to: i ∠A ii ∠B iii ∠C b Name the side on the smaller triangle which corresponds to: i DE ii EF iii FD c Find these side ratios. DE EF FD i ii iii AB BC CA d Would you say that the two triangles are similar? Why or why not?

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4m

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Example 10

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4 For the following pairs of similar figures, complete these tasks. i List the pairs of corresponding sides. ii List the pairs of corresponding angles. iii Find the scale factor. iv Find the values of the pronumerals.

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Example 9

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6 Two rectangular picture frames are similar in shape. A corresponding pair of sides are of length 50 cm and 75 cm. The other side length on the smaller frame is 70 cm. Find the perimeter of the larger frame. 7 7 Two similar triangles have a scale factor of . 3 If the larger triangle has a side length of 35 cm, find the length of the corresponding side on the smaller triangle.

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8 One circle has a perimeter of 10π cm and another has an area of 16π cm2. Find the scale factor of the diameters of the two circles. 9 A square photo of area 100 cm2 is enlarged to an area of 900 cm2. Find the scale factor of the side lengths of the two photos. WO

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11 a I f a regular polygon such as this regular octagon is enlarged, do the interior angles change? b Are all polygons with the same number of sides similar? Give reasons. 12 If a square is similar to another by a scale factor of 4, what is the ratio of the area of small square to the area of large square?

Enrichment: Square designs 13 A square design consists of a series of squares as shown. An inner square is formed by joining the midpoints of the sides of the next largest square. a Use Pythagoras’ theorem to find the side length of: i the second largest square ii the third largest square b Find the scale factor of a pair of consecutive squares, e.g. first and 4 cm second or second and third. c If the side length of the outside square was x cm, show that the scale factor of consecutive squares is equal to the result found in part b. d What is the scale factor of a pair of alternate squares, e.g. first and third or second and fourth?

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10 Are the following statements true or false? Give reasons for your answers. a All squares are similar. b All rectangles are similar. c All equilateral triangles are similar. d All isosceles triangles are similar. e All rhombuses are similar. f All parallelograms are similar. g All kites are similar. h All trapeziums are similar. i All circles are similar.

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Measurement and Geometry

8G Similar triangles

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E X T E N S I ON

Finding the approximate height of a tree or the width of a gorge using only simple equipment is possible without actually measuring the distance directly! Similar triangles can be used to calculate distances without the need to climb the tree or cross the gorge. It is important, however, to ensure that if the mathematics of similar triangles is going to be used, then the two triangles are in fact similar. We learnt earlier that there were four tests that help to determine if two triangles are congruent. Similarly there are four tests that help establish whether or not triangles are similar. Not all side lengths or angles are required to prove that two triangles are similar.

Similar triangles should be used to ﬁnd the width of a gorge before crossing it.

Let’s start: How much information is enough?

■

Key ideas

Given a certain amount of information, it may be possible to draw two triangles that are guaranteed to be similar in shape. Decide if the information given is enough to guarantee that the two triangles will always be similar. If you can draw two triangles (ABC and DEF) that are not similar, then there is not enough information provided. • ∠A = 30° and ∠D = 30° • ∠A = 30°, ∠B = 80° and ∠D = 30°, ∠E = 80° • AB = 3 cm, BC = 4 cm and DE = 6 cm, EF = 8 cm • AB = 3 cm, BC = 4 cm, AC = 5 cm and DE = 6 cm, EF = 8 cm, DF = 10 cm • AB = 3 cm, ∠A = 30° and DE = 3 cm, ∠D = 30° • AB = 3 cm, AC = 5 cm, ∠A = 30° and DE = 6 cm, DF = 10 cm, ∠D = 30°

Two triangles are similar if: – Two angles of a triangle are equal to two angles of another triangle. 85° 60°

85° 60°

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Chapter 8 Transformations and congruence

Key ideas

522

– Three sides of a triangle are proportional to three sides of another triangle. C

F3m

2m

A 1 m B

DE EF DF = = =2 AB BC AC

6m

4m d

2m

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– Two sides of a triangle are proportional to two sides of another triangle, and the included angles are equal. E F B C DE DF = = 1.5 AB AC 2 cm 3 cm 40° 3 cm 4.5 cm 40° A D – The hypotenuse and a second side of a right-angled triangle are proportional to the hypotenuse and second side of another right-angled triangle. B 4m C

12 m

F

E DF EF = =3 AC BC

5m A

15 m D

■ ■

If two triangles ABC and DEF are similar we write ABC ||| DEF. Abbreviations such as AAA are not used for similarity in NSW.

Example 11 Deciding if two triangles are similar Explain why these pairs of triangles are similar. a C F

b

C

E

D 75° 70°

5 cm 10 cm A

50° B 2 cm

F 70° E

50° 4 cm

75°

A

B

D

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SOLUTION a

EXPLANATION

ED 4 = =2 AB 2 ∠B = ∠E = 50° EF 10 = =2 BC 5 ∴ ABC is similar to DEF, because two sides of one triangle are proportional to two sides of the other and the included angles are equal.

Work out the ratio of the two pairs of corresponding sides to see if they are equal. Note that the angle given is the included angle between the two given pairs of corresponding sides.

Two pairs of equal angles are given. This b ∠A = ∠D implies that the third pair of angles are also ∠B = ∠E ∴ ABC is similar to DEF, because two angles equal and that the triangles are similar. of one triangle are equal to two angles of the other.

Example 12 Finding a missing length Given that these pairs of triangles are similar, find the value of the pronumerals.

5

y

15

9

4 x SOLUTION 15 =3 5

First find the scale factor.

x = 4 × 3 = 12 y=9÷3=3

Multiply corresponding side by the scale factor to find a side on the larger triangle. Otherwise divide.

EXTENSION

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1 Write a similarity statement for these pairs of similar triangles, e.g. ABC ||| DEF. Be careful with the order of letters and make sure each letter matches the corresponding vertex. a C b E F B D

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Scale factor =

EXPLANATION

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2 For the pairs of triangles in Question 1, which of the four similar triangle tests would be used to show their similarity? See the Key ideas for a description of each test. There is one pair for each test.

B

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80°

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26 m

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40° 18 m

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Example 11

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15° D 4 Given that these pairs of triangles are similar, find the value of the pronumerals. a b 20 6 12 y 10 6 x

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5 Two triangular tracks are to be used for the junior and open divisions for a school event.

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250 m Open

100 m 200 m

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150 m

a Work out the scale factor for each of the three corresponding pairs of sides. b Are the two tracks similar in shape? Give a reason. c How many times would a student need to run around the junior track to cover the same distance as running one lap of the senior track?

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Example 12

M AT I C A

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6 By using the angle sum of a triangle decide if the given pairs of triangles are similar. a b 35° 62° 100° 38°°

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28°° 37.5°

81°

7 In a game of minigolf, a ball bounces off a wall as shown. The ball proceeds and hits another wall at point A. How far down the right side wall does the ball hit?

3m

9m

2m ?

A 8 Trees on a river bank are to be used as markers to form two triangular shapes. Each tree is marked by a red dot as shown in the diagram.

River 8m 6m

20 m

a Are the two triangles similar? b Calculate the scale factor. c Calculate how far it is across the river.

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10 Give reasons why the pairs of triangles in these diagrams are similar. If an angle or side is common to both triangles, use the word ‘common’ in your reasoning. a C b A B

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∆ ABC and ∆ ∆FED FE FED 11 Show how Pythagoras’ theorem can be used to prove that these two triangles are similar. B

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Enrichment: Tricky unknowns 12 The pairs of triangles here show a length x cm, which is not the length of a full side of a triangle. Find the value of x in each case. a b 3.6 cm x cm

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9 Explain why only two pairs of angles are required to prove that two triangles are similar.

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Chapter 8 Transformations and congruence

8H Using congruent triangles to establish properties

of quadrilaterals The properties of special quadrilaterals, (parallelogram, rhombus, rectangle, square, trapezium and kite) can be examined more closely using congruence. By drawing the diagonals and using the tests for the congruence of triangles we can prove many of the properties of these special quadrilaterals.

Let’s start: Do the diagonals bisect each other?

Key ideas

Here is a parallelogram with two pairs of parallel sides. Let’s assume also that opposite sides are equal (a basic property of a parallelogram which we prove later in this exercise). D C • First locate ABE and CDE. E • What can be said about the angles ∠BAE and ∠DCE? • What can be said about the angles ∠ABE and ∠CDE? A B • What can be said about the sides AB and DC? • Now what can be said about ABE and CDE? Discuss the reasons. • What does this tell us about where the diagonals intersect? Do they bisect each other and why? (To bisect means to cut in half.)

This is a summary of the properties of the special quadrilaterals. ■ Kite: A quadrilateral with two pairs of adjacent sides equal – Two pairs of adjacent sides of a kite are equal – One diagonal of a kite bisects the other diagonal – One diagonal of a kite bisects the opposite angles – The diagonals of a kite are perpendicular ■ Trapezium: A quadrilateral with at least one pair of parallel sides – At least one pair of sides of a trapezium are parallel ■ Parallelogram: A quadrilateral with both pairs of opposite sides parallel – The opposite sides of a parallelogram are parallel – The opposite sides of a parallelogram are equal – The opposite angles of a parallelogram are equal – The diagonals of a parallelogram bisect each other ■ Rhombus: A parallelogram with two adjacent sides equal in length – The opposite sides of a rhombus are parallel – All sides of a rhombus are equal – The opposite angles of a rhombus are equal – The diagonals of a rhombus bisect the vertex angles – The diagonals of a rhombus bisect each other – The diagonals of a rhombus are perpendicular

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Key ideas

Measurement and Geometry

Rectangle: A parallelogram with a right angle – The opposite sides of a rectangle are parallel – The opposite sides of a rectangle are equal – All angles at the vertices of a rectangle are 90° – The diagonals of a rectangle are equal – The diagonals of a rectangle bisect each other Square: A rectangle with two adjacent sides equal – Opposite sides of a square are parallel – All sides of a square are equal – All angles at the vertices of a square are 90° – The diagonals of a square bisect the vertex angles – The diagonals of a square bisect each other – The diagonals of a square are perpendicular

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Example 13 Proving that diagonals of a parallelogram bisect each other Prove that the diagonals of a parallelogram bisect each other and give reasons.

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SOLUTION

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EXPLANATION

∠BAE = ∠DCE (alternate angles in parallel lines) ∠ABE = ∠CDE (alternate angles in parallel lines) AB = CD (opposite sides of a parallelogram) ∴ ABE ≡ CDE (AAS) ∴ BE = DE and AE = CE ∴ Diagonals bisect each other.

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The two triangles are congruent using the AAS test. Since ABE and CDE are congruent the corresponding sides are equal.

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3 SSS is one test for congruence of triangles. Write down the other three. 4 Name the side (e.g. AB) that is common to both triangles in each diagram. a D

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Step 1. List the pairs of equal angles in ABE and CDE giving reasons why they are equal.

Step 2. List the pairs of equal sides in ABE and CDE giving reasons why they are equal.

Step 3. Write ABE ≡ CDE and give the reason SSS, SAS, AAS or RHS.

Step 4. State that BE = DE and AE = CE and give a reason.

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6 Prove, giving reasons, that the diagonals in a parallelogram bisect each other. You may assume here that opposite sides are equal so use AB = CD. Complete the proof by following these steps.

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Example 13

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10 For this square assume that MQ = QO and NQ = PQ. a Give reasons why MNQ ≡ ONQ. b Give reasons why ∠MQN = ∠OQN = 90°. c Give reasons why ∠QMN = 45°.

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9 For this rhombus assume that VW = WT. a Give reasons why VWU ≡ TWU. b Give reasons why ∠VWU = ∠TWU = 90°.

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M 11 Use the information in this kite to prove these results. a ABD ≡ CBD b ∠DAB = ∠DCB c ∠ADB = ∠CDB

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8 A parallelogram ABCD has two pairs of parallel sides. a What can be said about ∠ABD and ∠CDB and give a reason? b What can be said about ∠BDA and ∠DBC and give a reason? c Which side is common to both ABD and CDB? d Which congruence test would be used to show that ABD ≡ CDB? e If ABD ≡ CDB, what can be said about the opposite sides of a parallelogram?

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7 Prove (as in Example 13) that the diagonals in a square EFGH bisect each other.

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13 Use the steps outlined in Question 6 to show that opposite sides of a rectangle ABCD are equal. Give all reasons.

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12 Use Pythagoras’ theorem to prove that the diagonals in a rectangle are equal in length. You may assume that opposite sides are equal.

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14 A trapezium ABCD has one pair of parallel sides. a Which angle is equal to ∠BAE? b Which angle is equal to ∠ABE? c Explain why ABE is not congruent to CDE.

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15 Use the information given for this kite to give reasons for the following properties. You may assume that ∠ADE = ∠CDE as marked. a ∠DAE = ∠DCE b AED ≡ CED c ∠AED = ∠CED = 90°

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Enrichment: Prove the converse 16 Prove the following results, which are the converse (reverse) of some of the proofs completed earlier in this exercise. a If the opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram; i.e. show that the quadrilateral has two pairs of parallel sides. D

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c If the diagonals of a parallelogram are equal, then the parallelogram is a rectangle. D C

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Investigating congruence with dynamic geometry software Computer dynamic geometry is ideal for these activities but geometrical instruments could also be used.

Conditions for congruent triangles 1 SSS a Construct a triangle with side lengths 4 cm, 7 cm and 9 cm. Measure the given side lengths using 1 decimal place and if using dynamic geometry, adjust the triangle to suit. Use the measure tool for the side lengths. b Construct a second triangle with the same side lengths c Do your two triangles look congruent? Measure the angles to help check. d Can you draw another triangle given the same information so that it does not look congruent to the others? 7.0 cm

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2 SAS a Construct a triangle with side lengths 4 cm and 7 cm and an included angle of 30°. Measure the given side lengths and the angle using 1 decimal place and if using dynamic geometry, adjust the triangle to suit. b Construct a second triangle with the same side lengths and included angle. c Do your two triangles look congruent? Measure the missing angle and side lengths to help check. d Can you draw another triangle given the same information so that it does not look congruent to the others? 3 AAS a Construct a triangle with two angles 30° and 50° and one side length 6 cm. Measure the side given length and the angles using 1 decimal place and if using dynamic geometry, adjust the triangle to suit. b Construct a second triangle with the same side length and angles. c Do your two triangles look congruent? Measure the missing side length and angle to help check. d Can you draw another triangle given the same information so that it does not look congruent to the others?

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Investigation

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Chapter 8 Transformations and congruence

4 RHS a Construct a triangle with a 90° and 50°, hypotenuse of length 6 cm and other side of length 4 cm. Measure the two given side lengths and the angle using 1 decimal place and if using dynamic geometry, adjust the triangle to suit. b Construct a second triangle with the same two side lengths and angle. c Do your two triangles look congruent? Measure the missing side length and angles to help check. d Can you draw another triangle given the same information so that it does not look congruent to the others?

Exploring SSA With the SAS test, the included angle must be between the two given sides. This investigation will explore a triangle using SSA. 1 Construct a triangle with sides 5 cm and 3 cm and one angle 30° (not an included angle). 2 Can you construct a second triangle using the same information that is not congruent to the first? 3 Consider this diagram. Use it to help explain why SSA is not a test for congruence of two triangles.

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Properties of quadrilaterals 1 Construct a parallelogram including its diagonals as shown. Use the parallel line tool to create the parallel sides. D

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2 Measure the sides and angles of ΔABE and ΔCDE. What do you notice? 3 Now drag one of the vertices of the parallelogram. Does this change you conclusion from part b? 4 Explore other quadrilaterals by constructing their diagonals and looking for congruent triangles. Write up a brief summary of your findings.

Cambridge University Press

1 How many similar triangles are there in this figure? Give reasons.

2 Can you fit the shapes in the two smaller squares into the largest square? What does this diagram illustrate? Try drawing or constructing the design and then use scissors to cut out each shape.

3 A strip of paper is folded five times in one direction only. How many creases will there be in the original strip when it is folded out?

4 Use congruent triangles to find the radius r in these diagrams. a b 3

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4 5 What is the side length of the largest square that can be cut out of this triangle? Use similar triangles.

12 5 13

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Puzzles and challenges

Measurement and Geometry

Chapter summary

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Chapter 8 Transformations and congruence

Translation 2 Vector –1 2

Rotation

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Reflection

Transformation and congruence

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∆ABC ≡ ∆DEF ∆ABC ∆ AB = DE, BC = EF, AC = DF ∠ = ∠D, ∠B = ∠E, ∠C = ∠F