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HAESE HARRIS PUBLICATIONS&

9

Core SkillsMathematics

Robert Haese

Sandra Haese

Derk Kappelle

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Y:\...\ALT_09\ALT09_00\001AL900.CDRMon Nov 03 11:42:26 2003

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CORE SKILLS MATHEMATICS 9

Robert Haese B.Sc.Sandra Haese B.Sc.Derk Kappelle Adv.Dip.T., M.Ed.

Haese & Harris Publications3 Frank Collopy Court, Adelaide Airport SA 5950Telephone: (08) 8355 9444, Fax: (08) 8355 9471email:web:

National Library of Australia Card Number & ISBN 1 876543 44 2

© Haese & Harris Publications 2003

Published by Raksar Nominees Pty Ltd, 3 Frank Collopy Court, Adelaide Airport SA 5950

First Edition 2003

Cartoon artwork by John Martin.Artwork by Piotr Poturaj and David PurtonCover design by Piotr Poturaj.Cover photograph: Copyright © Digital Vision® Ltd.Computer software by David Purton and Eli SieradzkiTypeset in Australia by Susan Haese (Raksar Nominees). Typeset in Times Roman 10 /11

. Except as permitted by the Copyright Act (any fair dealing for thepurposes of private study, research, criticism or review), no part of this publication may bereproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic,mechanical, photocopying, recording or otherwise, without the prior permission of the publisher.Enquiries to be made to Haese & Harris Publications.

: Where copies of part or the whole of the book are madeunder Part VB of the Copyright Act, the law requires that the educational institution or the bodythat administers it has given a remuneration notice to Copyright Agency Limited (CAL). Forinformation, contact the Copyright Agency Limited.

\Qw_ \Qw_

This book is copyright

Copying for educational purposes

[email protected]


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mailto:[email protected]://www.haeseandharris.com.au/

FOREWORD

[email protected]

We have written this book to provide an alternative course for Year 9 students. Our particular aimwas to show the everyday applications of mathematics in the context of the real world.We have deliberately presented the content in short chapters and because mastery of the basicskills is so important, sixteen chapters cover the core skills that we feel are appropriate for Year 9students who may go on to complete a Mathematical Applications or a Foundation Mathematicscourse in the future.In addition, there are four ‘option’ chapters which are entirely focused on particular themes:

17 Overseas travel18 Park improvement19 Maths in fitness and health20 Mathematics in design and art

In these chapters, the emphasis is on relating the theory and application of mathematics to reallife.Throughout the book, there are worked examples, exercises and activities, and with the support ofthe interactive Student CD, students are given the structure and content for independent practiceand revision.We hope that by presenting a course of mathematics in this way, students may understand therelevance and importance of mathematics in everyday life, but we also caution that no single bookshould be the sole resource for any classroom teacher.

We welcome your feedback. Email:Web:

RCHSHH

DWK


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mailto:[email protected]://www.haeseandharris.com.au/

1 INTEGERS AND FRACTIONS 9

2 GEOMETRY (POLYGONS AND CONSTRUCTIONS) 25

3 ALGEBRA 45

4 DECIMALS AND PERCENTAGES 67

5 ROUNDING, ESTIMATION AND ERRORS 85

A Integers 10B Order of operations 13C Using your calculator with integers 15D Operations with fractions 16

Review of Chapter 1 23

A Review of geometrical facts 27B Triangles 32C Quadrilaterals and other polygons 34D Constructing a triangle given its sides 36E Bisecting angles 38F Constructing perpendiculars 39G Angle construction 42


A The language of mathematics 47B Changing words to symbols 48C Evaluation 51D Collecting like terms 52E Algebraic simplification and index form 53F Index laws 56G Zero and negative indices 57H Number patterns and rules 58I Equations 59


A Decimals 68B Fraction and decimal conversions 71C Percentage 72D Working with percentages 74E Unitary method in percentage 76F Percentage increase and decrease 78G Standard form (scientific notation) 79


A Rounding numbers 86B Rounding money 88C One figure approximations 91

TABLE OF CONTENTS

4 TABLE OF CONTENTS


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5TABLE OF CONTENTS

D Estimation of numbers of objects 93E Rounding decimal numbers 94F Rounding with a calculator 96G Significant figure rounding 98H Error in measurement 100


A Length 105B Perimeter 107C Circumference 111D Areas of polygons 114E Areas of circles 120F Volume 123G Capacity 127H Mass 130I Time 133J 24-hour time 135


A Ratio 138B Simplifying ratios 139C Equal ratios 141D The unitary method for ratios 143E Using ratios to divide quantities 144F Rates 145G Rate graphs 148H Travel Graphs 150


A Interpreting graphs 154B Interpreting tables 159C Life expectancy and insurance 162D General information 164


A Wages and salaries 170B Overtime, holiday leave loading and bonuses 174C Deductions from earnings 177D Income tax 178


6 MEASUREMENT 103

7 RATIO AND RATES 137

8 INTERPRETING GRAPHS AND CHARTS 153

9 EARNING AN INCOME 169


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6 TABLE OF CONTENTS

10 CATEGORICAL DATA AND TIME SERIES 183

11 BUSINESS MATHEMATICS 199

12 COORDINATES, LINE GRAPHS AND MAPS 213

13 NUMERICAL DATA 235

14 INTEREST AND DEPRECIATION CALCULATIONS 255

15 PROBABILITY 267

A Statistics 185B Graphing categorical data 187C Data collection and misleading graphs 191D Time series data 194


A Profit and loss 200B Percentage profit, percentage loss 202C Discount 204D Goods and services tax (GST) 207


A The number plane 214B Linear relationships 217C Plotting linear graphs 221D Non-linear graphs 223E Bearings and directions 224


A Numerical data 236B Grouped discrete data 240C Continuous data 243D Measuring the centre 244E Measuring the spread of discrete data 247F Using technology 250G Comparing and reporting discrete data 251


A Simple interest 257B Compound interest 261C Depreciation 263


A Probability by experiment 269B Probability by symmetry 271C Probabilities from a grid 274D Simulating random events 276E Expectation 279


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7TABLE OF CONTENTS

F Odds 280Review of Chapter 15 282

A Reflection 285B Rotation 287C Translation 289D Enlargement 290E Properties of transformations 292F Similarity 293Review of Chapter 16 297

A Foreign exchange 300B Air fares 303C Accomodation and food costsD Travel insurance and other costs 305E A holiday in Fiji 306

A Order of tasks 313B Draw the plan 314C The fountain 314D The watering system 315E Brick paving 316F Rotary hoeing the lawn and garden areas 317G The total cost 317

A Heart rate 320B Body-mass index 322C Respiration 323D Stopping distance 325

A Making stellated solids 332B Making boxes and containers 333C Tiling patterns 337

304

16 TRANSFORMATIONS AND SIMILARITY 283

17 OVERSEAS TRAVEL 299

18 PARK IMPROVEMENT 311

19 MATHS IN FITNESS AND HEALTH 319

20 MATHEMATICS IN DESIGN AND ART 331

ANSWERS 340

INDEX 367


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Integers andfractions

Integers andfractions

11Chapter

�

�

�

�

�

�

add and subtract integersmultiply and divide integersperform operations using BEDMASuse a calculator to simplifyexpressions

operate with fractions with andwithout a calculatorsolve problems involving fractions

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

By the end of this chapter you should be able to

Outcomes · · · · · · · · · · · · · · · · · · · · · · · · · ·


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Y:\...\ALT_09\ALT09_01\009AL901.CDRThu Oct 30 16:39:32 2003


OPENING PROBLEM BECKY’S PROBLEMS

�

10 INTEGERS AND FRACTIONS (CHAPTER 1)

We all use numbers every day. We make machines to crunch numbers. We live in numberedstreets and have telephone numbers and tax-file numbers. We have registration numbers forour dogs and our cars. We use numbers to measure distance, time and mass.

An understanding of numbers and how to operate with them is an essential part of daily living.

Consider the following questions:1 a How much does Becky have in the bank after 6 weeks?

b What fraction of the money was her starting deposit?

2 a If Becky wanted to withdraw $65, how much would be left in the bank?b If Becky wanted to withdraw $120, by how much would the account be over-

drawn?c If Becky wanted to withdraw 2

5of her money, how much money would be left

in the bank?

The integers are all the natural numbers and zero,i.e., ::::::¡ 5, ¡4, ¡3, ¡2, ¡1, 0, 1, 2, 3, 4, 5 ::::::

We can show these numbers on a number line. Zero is neither positive nor negative.

The rules for addition and subtraction of integers are:

+ (positive number) = (positive number)

¡ (positive number) = (negative number)

+ (negative number) = (negative number)

¡ (negative number) = (positive number)

Becky opens bank account. Her starting deposit isEvery week for weeks Becky deposits in

the account.

a$ . $50 106

INTEGERSA

Simplify:a 4 + ¡9 b 4 ¡ ¡9 c ¡3 + ¡5 d ¡3 ¡ ¡5

a 4 + ¡9 b 4 ¡ ¡9 c ¡3 + ¡5 d ¡3 ¡ ¡5

= 4 ¡ 9 = 4 + 9 = ¡3 ¡ 5 = ¡3 + 5

= ¡5 = 13 = ¡8 = 2

Example 1

�5 �4 �3 �2 �1 0 1 2 3 4 5negative integers

zeropositive integers


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Y:\...\ALT_09\ALT09_01\010AL901.CDRTue Oct 21 12:32:17 2003


INTEGERS AND FRACTIONS (CHAPTER 1) 11

EXERCISE 1A.1

1 Find the value of:a ¡(+2) b +(¡2) c ¡(¡2) d +(¡3)

e ¡(¡4) f ¡(+7) g ¡(¡8) h +(¡1)

2 Find the value of:a 13 ¡ 8 b 13 + ¡8 c 13 ¡ ¡8 d ¡13 + 8

e ¡13 ¡ 8 f ¡13 ¡ ¡8 g 8 ¡ 13 h 13 + 8

i 16 + 25 j 16 ¡ 25 k 16 + ¡25 l 16 ¡ ¡25

m ¡16 + 25 n ¡16 + ¡25 o ¡16 ¡ ¡25 p 25 ¡ 16

3 The temperature dropped 9oC to the present temperature of 4oC. What was the originaltemperature?

4 A missile is fired from a submarine, 200 m below sea level. It hits a low-flying enemyplane at 243 m above sea level. How many metres does the missile climb in total?

5 The temperature rose from ¡6oC to 10oC. How many degrees did the temperature in-crease?

6 Nicky had $213 in her cheque account and wrote a cheque for $350. By how much wasthe cheque account overdrawn?

7 John’s cheque account was overdrawn by $73, so he paid $120 into it. How much moneywas in the account then?

8 The following temperatures were recorded yesterday in the cities named:Rome ¡4oC, London 1oC, Paris ¡3oC,Helsinki ¡18oC, Moscow ¡15oC, Hobart 13oC.

a Arrange the city names in order of coldest to warmest for yesterday.b Find the difference in temperature between:

i Rome and London ii Paris and Helsinkic i How much warmer is Hobart than Moscow?

ii How much colder than Moscow is Helsinki?

The rules for multiplication of integers are:

(positive) £ (positive) = (positive)

(positive) £ (negative) = (negative)

(negative) £ (positive) = (negative)

(negative) £ (negative) = (positive)

Find the value of:a 3 £ 4 b 3 £ ¡4 c ¡3 £ 4 d ¡3 £ ¡4

a 3 £ 4

= 12

b 3 £ ¡4

= ¡12

c ¡3 £ 4

= ¡12

d ¡3 £ ¡4

= 12

Example 2

DEMO

DEMO


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EXERCISE 1A.2

1 Find the value of:

a 6 £ 7 b 6 £ ¡7 c ¡6 £ 7 d ¡6 £ ¡7

e 5 £ 8 f 5 £ ¡8 g ¡5 £ 8 h ¡5 £ ¡8

i 7 £ 9 j ¡7 £ 9 k 7 £ ¡9 l ¡7 £ ¡9

m 12 £ 11 n ¡12 £ ¡11 o 12 £ ¡11 p ¡12 £ 11

2 Find the value of:a ¡52 b (¡5)2 c (¡1)3

d ¡13 e 3 £ ¡2 £ 5 f ¡3 £ 2 £ ¡5

g ¡3 £ ¡2 £ ¡5 h 2 £ (¡3)2 i ¡2 £ (¡3)2

j ¡24 k (¡2)4 l (¡3)2 £ (¡2)2

The rules division with integers are:

(positive) ¥ (positive) = (positive)

(positive) ¥ (negative) = (negative)

(negative) ¥ (positive) = (negative)

(negative) ¥ (negative) = (positive)

3 Find the value of:a 15 ¥ 3 b 15 ¥ ¡3 c ¡15 ¥ 3 d ¡15 ¥ ¡3

e 24 ¥ 8 f 24 ¥ ¡8 g ¡24 ¥ 8 h ¡24 ¥ ¡8

i4

8j

¡4

8k

4

¡8l

¡4

¡8

Simplify:a ¡42 b (¡4)2 c ¡23 d (¡2)3

a ¡42 b (¡4)2 c ¡23 d (¡2)3

= ¡(4 £ 4) = ¡4 £ ¡4 = ¡(2 £ 2 £ 2) = ¡2 £ ¡2 £ ¡2

= ¡16 = 16 = ¡8 = ¡8

Example 3

Find the value of:a 14 ¥ 2 b 14 ¥ ¡2 c ¡14 ¥ 2 d ¡14 ¥ ¡2

a 14 ¥ 2 b 14 ¥ ¡2 c ¡14 ¥ 2 d ¡14 ¥ ¡2

= 7 = ¡7 = ¡7 = 7

Example 4

for


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ORDER OF OPERATIONS RULES:

² Work out the part in brackets first.² Work out any exponents.² Start from the left. Perform all divisions

and multiplications as you come to them.² Start from the left. Perform all additions

and subtractions as you come to them.

BracketsExponentsDivision andMultiplicationAddition andSubtraction

EXERCISE 1B

1 Simplify:a 5 + 8 ¡ 3 b 5 ¡ 8 + 3 c 5 ¡ 8 ¡ 3

d 2 £ 10 ¥ 5 e 10 ¥ 5 £ 2 f 5 £ 10 ¥ 2

2 Simplify:a 4 + 7 £ 3 b 7 £ 2 + 8 c 13 ¡ 3 £ 2

d 5 £ 4 ¡ 20 e 33 ¡ 3 £ 4 f 15 ¡ 6 £ 0

g 2 £ 5 ¡ 5 h 60 ¡ 3 £ 4 £ 2 i 15 ¥ 3 + 2

j 8 £ 7 ¡ 6 £ 3 k 5 + 2 + 3 £ 2 l 7 ¡ 5 £ 3 + 2

Simplify: a 23 ¡ 10 ¥ 2 b 3 £ 8 ¡ 6 £ 5

a 23 ¡ 10 ¥ 2 b 3 £ 8 ¡ 6 £ 5

= 23 ¡ 5 f¥ before ¡g = 24 ¡ 30 f£ before ¡g= 18 = ¡6

Example 6

Simplify: a 3 + 7 ¡ 5 b 6 £ 3 ¥ 2

a 3 + 7 ¡ 5 fWork left to right as only + and ¡ are involved.g= 10 ¡ 5

= 5

b 6 £ 3 ¥ 2 fWork left to right as only £ and ¥ are involved.g= 18 ¥ 2

= 9

Example 5

ORDER OF OPERATIONSB

The wordmay help you

remember this order.

BEDMAS

DEMO


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3 Simplify:a 12 + (5 ¡ 2) b (12 + 5) ¡ 23 c (8 ¥ 4) ¡ 2

d 82 ¥ (4 ¡ 2) e 84 ¡ (12 ¥ 6) f (84 ¡ 12) ¥ 62

g 32 + (8 ¥ 2) h 32 ¡ (8 + 14) ¡ 7 i (32 ¡ 8) + (14 ¡ 7)

j (16 ¥ 8) ¥ 2 k 16 ¥ (8 ¥ 2) l 18 ¡ (6 £ 3) ¡ 4

4 Simplify:a 8 ¡ [(2 ¡ 3) + 4 £ 3] b [16 ¡ (9 + 3)] £ 2

c 13 ¡ [(8 ¡ 3) + 6] d [16 ¡ (12 ¥ 4)] £ 3

e 200 ¥ [4 £ (6 ¥ 3)] f [(12 £ 3) ¥ (12 ¥ 3)] £ 2

5 Simplify:

a75

5 £ 5b

21

16 ¡ 9c

18 ¥ 3

14 ¡ 11d

7 + 9

6 ¡ 2

e53 ¡ 21

9 ¡ 5f

3 £ 8 + 6

6g

57

7 ¡ (2 £ 3)h

(3 + 8) ¡ 5

3 + (8 ¡ 5)

6 Using £, ¥, + or ¡ only, insert symbols to make correct equations:a 9 2 3 2 2 = 8 b 9 2 3 2 2 = 25 c 9 2 3 2 2 = 5

Simplify:3 + (11 ¡ 7) £ 2

3 + (11 ¡ 7) £ 2

= 3 + 4 £ 2 fwork the brackets firstg= 3 + 8 f£ before +g= 11

Example 7

Simplify:[12 + (9 ¥ 3)] ¡ 11

[12 + (9 ¥ 3)] ¡ 11

= [12 + 3] ¡ 11 fwork the inner brackets firstg= 15 ¡ 11 fouter brackets nextg= 4

Example 8

Simplify:

3 £ (12 ¡ 8)

11 ¡ 5

3 £ (12 ¡ 8)

11 ¡ 5

=3 £ 4

11 ¡ 5

fbrackets firstg

=12

6

= 2

Example 9

simplify numeratorsimplify denominator


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7 Insert brackets, if necessary, to make the following true:a 8 ¡ 6 £ 3 = 6 b 120 ¥ 4 £ 2 = 60

c 5 £ 7 ¡ 3 ¡ 1 = 15 d 5 £ 7 ¡ 3 ¡ 1 = 33

You should learn how to operate your calculator. Different calculatorsdo not have exactly the same keys. It is important that you have aninstruction booklet for your calculator, and use it.

Most modern calculators have the Order of Operationsrules built into them.

For example, consider 5 £ 3 + 2 £ 5:

If you key in 5 3 2 5 the calculatorgives an answer of 25, which is correct as 5 £ 3 + 2 £ 5

= 15 + 10= 25

Now consider12

4 + 2. This means 12 is divided by the sum 4 + 2,

i.e., 12 is divided by (4 + 2).

To get the correct answer you must use the two grouping symbols keys:

Left hand bracket Right hand bracket

You must key in 12 4 2 :

You then obtain the answer 2, which is correct.

Calculate: a 12 + 32 ¥ (8 ¡ 6) b75

7 + 8

a 12 32 8 6 Answer: 28

b 75 7 8 Answer: 5

EXERCISE 1C

1 Use your calculator to find:a 17 + 23 £ 15 b (17 + 23) £ 15 c 128 ¥ 8 + 8

d 128 ¥ (8 + 8) e84

3 + 9f (21 + 7) ¥ 7

USING YOUR CALCULATOR WITH INTEGERSC

+

+

+

+

× × =

=

=

=

÷

÷

÷

(

(

(

(

)

)

)

)

Try to rememberBEDMAS!

Example 10

�


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Y:\...\ALT_09\ALT09_01\015AL901.CDRMon Oct 13 15:34:32 2003



THE SIGN CHANGE KEY

The or key is used to enter negative numbers into the calculator.

To enter ¡5 into the calculator, key in 5 or 5 and ¡5 will appear onthe display.

2 Use your calculator to find:a ¡23 £ ¡16 b 34 £ ¡8 c ¡64 ¥ ¡16

d89 + ¡5

¡7 £ 3e ¡25 + 32 ¥ ¡4 f

¡15 ¡ 5

8 ¥ 4

A fraction consists of two whole numbers, a numerator and a denominator, separated by abar symbol.

For example,

The numerator and the denominator are integers.

TYPES OF FRACTIONS

4

5is a proper fraction fas the numerator is less than the denominatorg

7

6is an improper fraction fas the numerator is greater than the denominatorg

234

is a mixed number fas it is really 2 + 34g

1

2, 36

are equivalent fractions fas both fractions represent the same amountg

The fraction key on your calculator is .

To enter the fraction 45

, press 4 5 and the display shows 4 5 .

To enter the fraction 234

, press 2 3 4 and the display shows 2 3 4 .

Calculate: a 41 £ ¡7 b ¡18 £ 23

a 41 7 Answer: ¡287

b 18 23 Answer: ¡414

Example 11

+/–

+/–

×

×

=

=

+/–

+/–

OPERATIONS WITH FRACTIONSD

4

5

numerator

denominatorbar (which also means )divide

DEMO


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EXERCISE 1D

1 Draw the calculator display for these fractions:

a 38

b 47

c 711

d 245

e 414

f 51119

2 Give the fraction for these calculator displays:a b c d1 4 7 12 2 1 5 3 5 6

3 Simplify:

a 68

b 610

c 1420

d 721

e 2430

f 4060

g 3248

h 75125

4 Simplify:

a 74

b 107

c 165

d 198

5 Write as a mixed number:

a 54

b 73

c 94

d 125

ADDITION AND SUBTRACTION

To add (or subtract) two fractions we write them as equivalent fractions with thesame denominator and add (or subtract) the new numerators.

Simplify 912

. 912

= 9¥312¥3

= 34

Calculator: 9 12

Example 12

=

Simplify 125

. 125

= 10+25

= 105

+ 25

= 225

Calculator: 12 5

Example 13

=

DEMO


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6 Find:

a 12

+ 12

b 34

+ 14

c 16

+ 16

d 314

+ 614

e 716

+ 216

f 211

+ 511

g 1115

+ 215

h 123

+ 13

7 Find:

a 12

+ 14

b 15

+ 12

c 23

+ 15

d 16

+ 23

e 25

+ 16

f 23

+ 12

g 34

+ 13

h 58

+ 34

8 Find:

a 112

+ 12

b 112

+ 13

c 234

+ 114

d 358

+ 213

e 334

+ 156

f 518

+ 223

Find: a 34

+ 56

b 123

+ 358

a 34

+ 56

fLowest Common Denominator, LCD = 12g

= 3£34£3

+ 5£26£2

fto achieve a common denominator of 12g

= 912

+ 1012

= 1912

= 1 712

Calculator: 3 4| {z } 5 6| {z } 1 7 123

4+ 5

6= 1 7

12

b 123

+ 358

fLowest Common Denominator = 24g

= 4 + 23

+ 58

fadding the whole numbers firstg

= 4 + 2£83£8

+ 5£38£3

fto get a common denominator of 24g

= 4 + 1624

+ 1524

fsimplifyingg

= 4 + 3124

fadding the fractionsg

= 4 + 1 724

= 5 724

Calculator: 1 2 3| {z } 3 5 8| {z } 5 7 24123

+ 358

= 5 724

+

+

=

=

Example 14


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9 Find:

a 79

¡ 29

b 45

¡ 35

c 23

¡ 13

d 34

¡ 34

e 58

¡ 35

f 78

¡ 14

g 611

¡ 13

h 79

¡ 23

10 Find:

a 1 ¡ 18

b 2 ¡ 38

c 5 ¡ 318

d 3 ¡ 212

e 112

¡ 12

f 112

¡ 13

g 223

¡ 14

h 334

¡ 16

11 Find:a 21

3¡ 11

4b 13

4¡ 11

3c 12

3¡ 3

4

d 314

¡ 112

e 234

¡ 56

f 235

¡ 134

Find: a 34

¡ 15

b 323

¡ 145

a 34

¡ 15


= 3£54£5

¡ 1£45£4

fto get a denominator of 20g

= 1520

¡ 420

= 1120

Calculator: 3 4| {z } 1 5| {z } 11 203

4¡ 1

5= 11

20

b 323

¡ 145


= 113

¡ 95

fwrite as improper fractionsg

= 11£53£5

¡ 9£35£3

fto get a denominator of 15g

= 5515

¡ 2715

= 2815

= 11315

Calculator: 3 2 3| {z } 1 4 5| {z } 1 13 15323

¡ 145

= 11315

�

�

=

=

Example 15

DEMO


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MULTIPLICATION

To multiply two fractions, we multiply the numbers on the top together and multiplythe numbers on the bottom together.

For example, 23

£ 57

= 2£53£7

multiplying numeratorsmultiplying denominators

= 1021

12 Find:

a 12

£ 12

b 12

£ 13

c 12

£ 14

d 13

£ 14

e 25

£ 13

f 35

£ 34

g 23

£ 15

h 45

£ 25

13 Find:a 11

2£ 1

3b 2

3£ 11

4c 11

2£ 11

2d 21

4£ 13

5

DIVISION

To divide one fraction by another, we interchange the top and bottom numbers ofthe second fraction and then multiply this fraction by the first fraction.

For example, 23

¥ 57

= 23

£ 75

= 1415

finterchanging the top and bottomnumbers of the second fractiong

Find: a 23

£ 45

b 134

£ 213

a 23

£ 45

= 2£43£5

= 815

Calculator:

2 3| {z } 4 5| {z }2

3£ 4

5=

b 134

£ 213

= 74

£ 73

fwrite as improper fractionsg

= 7£74£3

= 4912

= 4 112

Calculator:

1 3 4| {z } 2 1 3| {z }134

£ 213

=

×

×

=

=

Example 16

DEMO


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14 Find:

a 34

¥ 56

b 34

¥ 23

c 712

¥ 34

d 23

¥ 4

e 2 ¥ 34

f 112

¥ 58

g 34

¥ 212

h 213

¥ 334

Anna scored 35

of her team’s goals in a netball match.How many goals did she shoot if the team shot 70 goals?

Anna shot 35

of 70

= 351

14£ 70

= 42 goals.

Calculator: 3 5 70

Remember that‘of ’ means ‘ ’£

× =

Find: a 3 ¥ 23

b 12

¥ 3 c 213

¥ 23

a 3 ¥ 23

b 12

¥ 3

= 31

¥ 23

= 12

¥ 31

= 31

£ 32

= 12

£ 13

= 92

= 16

= 412

Calculator: Calculator:

3 2 3| {z }3 ¥ 2

3=

1 2| {z } 31

2¥ 3 =

c 213

¥ 23

= 73

¥ 23

= 73

1

1£ 3

2

= 72

= 312

Calculator:

2 1 3| {z } 2 3| {z }213

¥ 23

=

=

=

=÷

÷

÷

Example 17

Example 18

DEMO


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15 Find:a 1

2of 30 b 1

4of 16 c 1

3of 24

d 78

of 40 e 25

of 25 f 34

of 12

16 Solve the following problems:

a If 35

of a class of 30 students went swimming, how many went swimming?

b Twelve apples were cut into thirds. How many pieces of apple are there now?

c Bob eats 14

of a pie and later eats 25

of the pie.i What fraction does he eat?ii What fraction remains?

d Andrea has 5 m of coloured ribbon for wrapping presents. On the first present sheuses 11

3m and on the second 11

4m.

i How much ribbon does she use?ii How much ribbon does she have left?

e Three watermelons were shared equally between 20 people. What fraction of awatermelon did each person receive?

f 27 bags of peanuts each containing 14

kg were emptied into a container. What wasthe total weight of the peanuts?

g How many boxes of chocolates weighing 12

kg could be packed from 42 kg ofchocolates?

h Jim the carpenter needs two pieces of timber for wall framing.One piece is 21

5m long and the other is 11

4m long.

Can he cut these from one plank that is 312

m long?

i The price of a beach coat is 57

of the price of a matching swimsuit.What does the beach coat cost if the swimsuit sells for $52:50?

j Keri keeps a rain gauge.On Monday 21

2mm of rain falls, on Thursday 41

4mm and on Friday 32

3mm falls.

Find the total rainfall over this period.

k The cost of a Holden is 213

of the cost of a Rolls Royce.If the current price of a Rolls Royce is $211 250, what is the cost of a Holden?

l The house rates came to $125 for the month.Janet paid 2

5of this, Janice paid 1

3and Beatrice

paid the rest.i What fraction did Janet and Janice pay?ii How much did Janet and Janice pay?iii How much did Beatrice pay?


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REVIEW OF CHAPTER 1


1 Find the value of:a ¡8 ¡ ¡3 b ¡7 £ 10 c 32 ¥ ¡8

2 Becky had $110 in her bank. If she withdrew $120, by how much would the accountbe overdrawn?

3 The temperature in Moscow was ¡15oC and inParis it was ¡3oC.

a Which city was warmer?b By how many degrees was it warmer?

4 Simplify:a 5 + 9 £ 2 b 40 ¥ 10 ¡ 2 c (8 + 6) £ 5

d42

16 ¡ 9e

7 ¡ 3

1 + 3f [21 ¡ (6 + 5)] £ 3

5 Find:a 21

5+ 3

5b 23

4+ 34

5c 14

5¡ 3

4

6 Find:a 3

4£ 1

2b 2

3¥ 2 c 2 ¥ 2

3

7 A painter paints 13

of a house on one day and1

4of it on the next day.

What fraction of the house is yet to be painted?

8 Share 34

of a pizza equally between two children.What fraction does each child receive?

9 James earned $28 mowing lawns.He spent 3

4of that amounnt.

How much did he spend?

10 There was 13

of a cake remaining and Jenny ate onehalf of it. What fraction of the cake did Jenny eat?

11 Find the total mass of 12 bags of apples each bag weighing 212

kilograms.


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Y:\...\ALT_09\ALT09_01\023AL901.CDRFri Oct 24 14:05:10 2003



12 At the beginning of a summer a water tank was7

8full. There was no rain, and at the end of the

summer it was only 110

full.

What fraction of a tank of water had been used?


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Geometry(polygons andconstructions)

Geometry(polygons andconstructions)

22Chapter

�

�

�

�

�

�

�

�

�

�

use correct terminology for anglepairs, triangles and other polygonsfind unknown angles for parallel linestest whether or not two lines areparalleluse the properties of triangles to findunknown anglesuse the exterior angles property ofpolygons

construct a triangle given its sidelengthsbisect an angle by constructionmethodsconstruct perpendiculars at a point,on a line and from a point to a lineconstruct a bisector of a line segmentconstruct angles of 60°, 30° and 45°

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·


Outcomes · · · · · · · · · · · · · · · · · · · · · · · · · ·


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Y:\...\ALT_09\ALT09_02\025AL902.CDRFri Oct 31 09:54:00 2003


OPENING PROBLEM SQUARING THE SLAB

0

26 GEOMETRY (POLYGONS AND CONSTRUCTIONS) (CHAPTER 2)

THEOREMS

We can make discoveries in geometry by drawing accurate figures and then making precisemeasurements of sizes of angles and lengths of sides.Look at these triangles and find the sum of their angles.

The sum is 180o for each triangle.

In fact: “The sum of the interior angles of any triangle is 180o.”

This result is called a theorem.

They use two boards 4 m long and two othersthat are 3 m long.They have no large set square but have a longtape measure.For you to consider:

² If the framing is ‘out of square’, whatis its shape?

² How do they make sure that the framinghas right angled corners?

² What are the problems if the concrete block is not square?

ANGLE PAIRS

² ]PAQ and ]QAR are adjacent angles.They have the same vertex and share a commonarm.

² For intersecting lines, angles which are directlyopposite each other are called verticallyopposite angles.

53° 60°

67° 29° 26°

125°

16°

82° 82°

41° 54°

85°

A

PQ

R

Sarah and Ray are boarding up for shed floor Concrete to depth ofmm is to be used.

a . a10

DEMO


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GEOMETRY (POLYGONS AND CONSTRUCTIONS) (CHAPTER 2) 27

LINE TERMINOLOGY

² Line AB is the endless straight line passing throughthe points A and B.

² Line segment AB is the part of the straight line con-necting A to B.

² Concurrent lines are three or more lines all passingthrough a common point.

² Perpendicular lines intersect at right angles.

² Parallel lines are lines which never intersect. Arrowheads show that lines are parallel.

² A transversal is a line which crosses over two otherlines.

ANGLE TYPES

REVIEW OF GEOMETRICAL FACTSA

Term Meaning

Right angle angle = 90o

Straight angle angle = 180o

A revolution angle = 360o

Acute angle 0o < angle < 90o

Obtuse angle 90o < angle < 180o

Reflex angle 180o < angle < 360o

F eigur

A

B

A

B


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TRIANGLE CLASSIFICATION

By sides:

By angles:

GEOMETRIC THEOREMS

We discovered and used these theorems (special results) in Year 8.

Title Theorem Figure

Angles at a point The sum of the sizes of theangles at a point is 360o:

a+ b+ c = 360

Adjacent angles on a The sum of the sizes of thestraight line angles on a line is 180o.

a+ b = 180

a°b°

c°

a° b°

Figure Term Meaning

Scalene triangle triangle with noequal sides

Isosceles triangle triangle with at leasttwo equal sides

Equilateral triangle triangle with threeequal sides

Figure Term Meaning

Acute angled triangle All angles of thetriangle are acute.

Obtuse angled triangle One angle of thetriangle is obtuse.

Right angled triangleOne of the anglesof the triangle is a

right angle.

DEMO


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Title Theorem Figure

Adjacent angles in a The sum of the sizes of theright angle angles in a right angle is 90o:

a+ b = 90

Vertically opposite Vertically opposite angles areangles equal in size.

a = b

Corresponding angles When two parallel lines are cutby a third line, then angles incorresponding positions areequal in size.

a = b

Alternate angles When two parallel lines are cutby a third line, then angles inalternate positions are equal insize.

a = b

Co-interior angles When two parallel lines are cut(also called allied) by a third line, co-interior angles

add to 180o.

a+ b = 180

Angles of a triangle The sum of the interior anglesof a triangle is 180o:

a+ b+ c = 180

Exterior angle of a The size of the exterior angletriangle of a triangle is equal to the sum

of the interior opposite angles.c = a+ b

The sum of the interior anglesAngles of aquadrilateral of a quadrilateral is 360o:

a+ b+ c+ d = 360

Isosceles triangle In an isosceles triangle theangles opposite the equalsides are equal in measure.

i.e., x = y

a°

b°

a°

b°

a°b°

a°

b°

c°

a° b°

c°

a°

b°

a°

b°

b° c°

d°a°

x°

y°

DEMO

DEMO

DEMO

DEMO

DEMO


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EXERCISE 2A

1 State whether the following are true (T) or false (F):a An angle measuring 89o is an acute angle.b An angle measuring 91o is an obtuse angle.c The size of an angle depends on the lengths of its arms.d When a vertical line crosses a horizontal line the angle formed is a straight angle.e 3

4of a straight angle is an acute angle.

f A right angle is neither an acute angle nor an obtuse angle.

2 a Use your protractor to measure angle ABC to the nearest degree.

b i Measure angles ABC and DBE.What do you notice?

ii Measure angles PQR and RQS.What do you notice?

3 Find the value(s) of the variable(s) and give brief reasons:a b c

d e f

i ii iii

A B C

iv

A

B

C

A

B

C

A

B

C

A

BC

D

E

PQ

R

S

x° 37°x°

108°x°

311°

56°x°

19°x°37°110°

95° x°

PRINTABLE

WORKSHEET


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b°a°125°

s°

r°75°

c°b° a°120°d�°

60°160°b°

c°c°

a°

b°72°

61°y°

103°x°

127°x°

y°39°

120°

b°

a°

29°d°

i°

92°42°

b°

a°

39°f °

57°

c°

70°

115°b°

a° 130°

g h i

j k l

m n o

p q r

s t u

v w x

y z

80°50°

70°76°

x°x°

x°


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4 State whether KL is parallel to MN, giving a brief reason for your answer. Thesediagrams have not been drawn accurately.

a b c

95°

95°

K

L

M

N

95°

95°K L

M N

95° 95°

K

L

M

N

TRIANGLESB

Find the unknown in the following, giving brief reasons:a b

a x = 180 ¡ 38 ¡ 19 fangle sum of a triangleg) x = 123

b y = 39 + 90 fexterior angle of a triangleg) y = 129

Example 1

38° 19°

x° 39°y°

A triangle is a polygon which has three sides.² The sum of the interior angles of a triangle is 180o.² Any exterior angle is equal to the sum of the interior opposite angles.² The longest side is opposite the largest angle.² The triangle is the only rigid polygon.

DISCUSSION

Bridges and other specialised structures often have triangular supports rather than rectan-gular ones. The reason for this is that “ ”.What is meant by “rigid polygon”? Is the statement true?

the triangle is the only rigid polygon

DEMO

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EXERCISE 2B

1 Find the unknown in the following, giving brief reasons:a b c

3 Find the unknown in the following giving a brief reason:a b c

d e f

102°

17°a°

32°

18°

c°60°68°

b°

d e f

2

a b c

d e f

25°

30°f °

64°

e°

41°

79°

d�°

80°

90°10°A

B

C 58°

102°

A

B

C

19°81°

C

B

A

27°

16°

A

B

C

28°

28°

C

A

B

21°

100°

B

C

A

The longest sideis opposite thelargest angle.

The following triangles are drawn to scale to try to trick you. State the longest sideof each triangle.

not

x°

68°

x°

x°

x°

40°

x°

60°x° x°

56°


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Y:\...\ALT_09\ALT09_02\033AL902.CDRWed Oct 29 16:34:58 2003



A quadrilateral is a four-sided polygon.² The sum of the interior angles of a quadrilateral is 360o.

Polygons are closed figures which have straight lines for sides.² The sum of the interior angles of an n-sided polygon is S = (n¡ 2) £ 180o.

For example, a hexagon has 6 sides, so n = 6.We split the hexagon into 6 ¡ 2 = 4triangles each having an angle sum of 180o.

So the sum of the angles of a hexagon is 4 £ 180o = 720o.

A regular polygon has all sides equal in length and all angles equal in size.

This hexagon is not regular although all of its sidesare equal in length. Why?

EXERCISE 2C.1

1 Copy and complete:a A quadrilateral can be split into .... triangles. So the sum of its interior angles is

.... £ 180o = ......o.b A pentagon can be split into .... triangles. So the sum of its interior angles is

.... £ 180o = ......o.c A 9-gon can be split into .... triangles. So the sum of its interior angles is

.... £ 180o = ......o.

2 These figures have not been drawn accurately. Find the value of the unknown:a b c

QUADRILATERALS AND

OTHER POLYGONSC

DEMO

DEMO

Find the value of the unknown in:

The angles of a quadrilateral add to 360o

) x = 360 ¡ 89 ¡ 90 ¡ 119

) x = 62

Example 2

x°

119°

89°

113°

70°x°

131° 118°

68°x°

x°

x°

x°

x°


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INVESTIGATION EXTERIOR ANGLES OF POLYGONS


d e f

3 Copy and complete the following table.

Regular polygon Number of sides Number of angles Size of each angle

equilateral trianglesquare

pentagonhexagonoctagondecagon

4 Copy and complete:² the sum of the angles of an n-sided polygon is ......² the size of each angle µ, of a regular n-sided polygon is µ = ::::::

What to do:

Step 1: Draw any triangle. Measure oneexterior angle from each vertex (a,b and c in the figure alongside).

Find the sum of these angles, i.e.,ao+bo+co. Repeat this procedurewith two other triangles of yourchoice.

What do you notice?

Step 2: Draw any quadrilateral. Measure one exterior an-gle from each vertex. Find the sum of the exteriorangles. What do you notice?

Step 3: From your results for triangles and quadrilater-als, suggest a value for the sum of the exteriorangles of any polygon. Draw any pentagon andhexagon. Check your value by measuring eachof the exterior angles and calculating their sum.

In fact: “the sum of the exterior angles of any polygon is 360o”.

119°

x°

x°

x°

x° x°

x°

a°

a° a°

a°

a°a°

a°

b°

c°

d°

a°

b°

c°

GEOMETRY

PACKAGE


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Y:\...\ALT_09\ALT09_02\035AL902.CDRSun Nov 02 09:08:05 2003



EXERCISE 2C.2

1 Find x if:a b c

2 Copy and complete:

Regular Sum of exterior Size of each Size of eachpolygon angles exterior angle interior angle

e.g. equilateral triangle 360o 360o ¥ 3 = 120o 180o ¡ 120o = 60o

squarepentagonhexagonoctagon

Check that your answers agree with the interior angle sizes found in question 3 ofExercise 2C.1

CONSTRUCTIONS

A construction must be drawn with a straight edge (usually a ruler) and a drawing ‘compass’only.

No other drawing aids such as a T-square, set square or protractor can be used.

118° 100°

x°

96°

x°

x°

x°

87°

71°

x°

010

100

20

110

90

180

6070

160

80

17

0150

30

40

130

50

140

120

protractor

set square

geoliner

3030 1515

CONSTRUCTING A TRIANGLE

GIVEN ITS SIDESD


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A pencil is essential.

Make sure that your pencil point is very sharp.

Construct a triangle ABC which has sides AB = 4 cm, BC = 3 cmand AC = 2 cm.

Step 1: Draw a line segment the length of AB.(Choose the largest side.) Use this asthe base of the triangle.

Step 2: Open your compass to a radius equalto the length of one of the other sides,say AC. Using this radius draw an arcfrom A.

Step 3: Now open the compass to a radiusequal to the length of the other side,BC. Draw another arc from B to inter-sect the first arc.

Step 4: Join A and B to the point of intersec-tion of the two arcs, C, to form triangleABC.

EXERCISE 2D

1 Construct triangle ABC in which AB = 5 cm, BC = 6 cm and AC = 4 cm.Classify the triangle.

2 Construct triangle PQR in which PQ = 4 cm, QR = 5 cm and PR = 8 cm.Classify the triangle.

3 Construct:a an equilateral triangle with sides of 4 cmb an isosceles triangle with sides of 5 cm, 5 cm and 2 cm.

4 Try to construct a triangle DEF where DE = 8 cm, EF = 4 cm and DF = 3 cm. Whatdo you notice?

It is necessary toshow all your

construction lines.Do not erase them.

A B4 cm

2 cm

A B

4 cm

DEMO

A B4 cm

3 cm

A B4 cm

C

2 cm 3 cm

Example 3

This example shows how to construct trianglewhen you are given the lengths of the threesides.

a


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When we bisect an angle with a straight line we divide it into two angles of equal size.This construction shows angle bisection with a compass and straight edge only.

EXERCISE 2E

1 Use your protractor to draw an angle ABC of size 50o:a Bisect angle ABC, using a compass and straight edge only.b Check with your protractor the size of each of the two angles you constructed.

2 Draw an acute angle XYZ of your own choice.a Bisect the angle, without using a protractor.b Check your construction using your protractor.

3 a Draw any obtuse angle and bisect it using construction techniques.b Repeat a for any reflex angle of your choice.

4 a With a ruler draw any triangle with sides greater than 5 cm.b Bisect each angle using a compass and straight edge only.c What do you notice about the three angle bisectors?

BISECTING ANGLESE

Bisect angle ABC.

Step 1:

Step 2: With Q as centre, draw an arc withinthe angle ABC.

Step 3: Keep the same radius.With centre P draw another arc to in-tersect the previous one at M.

Step 4: Join B to M. BM bisects angle ABC,i.e., ]ABM = ]CBM.

Example 4

DEMO

A

B C

A

B C

P

QA

B C

P

Q

A

B C

P

Q

M

A

B C

P

Q

M

WORKSHEET

W BP

ith centre draw an arc which cutsBA at and BC at Q.


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d

e Click on the icon for a demonstration of the angle bisectingproperties of triangles.

PERPENDICULAR AT A POINT ON A LINE

Look at this example.

Construct an angle of 90o at Pon the line segment XY.

Step 1: Draw a line segment XY. Draw a semi-circle with centre P which cuts XY atM and N.

Step 2: With centre M and radius larger thanMP, draw an arc above P.

Step 3: Keep the same radius. With centre Ndraw an arc to cut the first one at W.

Step 4: Draw the line from P through W.Angles WPY and WPX are 90o.

GEOMETRY

PACKAGE

Perpendicular linesare at right angles to

a given line.

given line

perpendicularline

P

CONSTRUCTING PERPENDICULARSF

DEMO

X YP

X YPM N

X YPM N

X YPM N

W

X YPM N

W

Example 5

Is the same result true for an obtuse angled triangle? Drawone of your choice to check it out.

Suppose we have line orline segment which containspoint How do we construct

perpendicular to the givenline, which passes through P?

aa

P.a


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EXERCISE 2F.1

1 a

b Construct a perpendicular to line AB passing through C.

2 a Draw a line segment PQ which is 5 cm long.b Construct a perpendicular to PQ passing through P.c Locate R on the perpendicular so that PR is 4 cm long.d Join QR and measure its length to the nearest mm.

(If you have done this accurately QR should be + 64 mm.)

3 Use the construction method given to construct:a a square with 4 cm sidesb a rectangle which is 6 cm by 4 cm.

PERPENDICULAR FROM A POINT TO A LINE

Construct a line perpendicular to the lineAB from a point P, not on the line.

Step 1: With centre P, draw an arc tocut line AB in two points X andY.

Step 2: With X and Y as centres andthe same radius, draw arcs tomeet at Q on the other side ofthe line from P.

Step 3: Join P to Q. Then line PQ isperpendicular to line AB.

WORKSHEETDraw line segment AB which is cm long. Use your rulerto locate point on AB which is cm from A.

aC

52

Remember thatall construction

lines shouldremain visible.

P

DEMO

P

A B

P

X YA B

P

X YA B

QP

X YA B

Q

Example 6

This time we are given line or line segmentand point which does not lie on the line.The example shows how we constructperpendicular from to the given line.

aP

aP


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EXERCISE 2F.2

1 a

b Construct a perpendicular from P to line segment AB, meeting AB at R.c Measure the length of PR.d Measure the length from P to several other points on the line AB.

Can you find a distance shorter than PR?e Copy and complete: The shortest distance from a point to a line is the length

of the ...................... to the line from the point.

2 a Draw any acute angled triangle with sides less than 6 cm.Let the triangle have vertices A, B and C.

b From A construct the perpendicular to BC. From B construct the perpendicular toAC. From C construct the perpendicular to AB.

c Copy and complete: The perpendiculars from the vertices of an acute angledtriangle all ......... ..... ..... .......... .

d Draw an obtuse angled triangle and repeat steps a and b.Does the result of c also apply to obtuse angled triangles?

PERPENDICULAR BISECTORS OF LINE SEGMENTS

Suppose we have a line segment. It has tohave a line drawn perpendicular to it andpassing through its midpoint. How do weconstruct such a line?

WORKSHEET

Draw line segment AB of length cm. Mark point above AB,near the middle of AB.

7 a P

perpendicularbisector

midpoint

Perpendicularly the line segment AB.bisect

Step 1: With A as centre and radius morethan half the length of AB, draw anarc.

Step 2: Keep the same radius. With centre Bdraw an arc to intersect the previousarc at C and D.

Step 3: Draw a line through C and D tointersect AB at M.M is the midpoint of AB.CD is the perpendicularbisector of AB.

A B

A B

C

D

A B

DEMO

A B

C

D

M

Example 7


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EXERCISE 2F.3

1

a measuring AM and BMb using a protractor.

2 a Draw a line segment PQ which is 6 cm long. Perpendicularly bisect PQ.b Choose any point X on the perpendicular bisector of PQ and join PX and QX.c Measure PX and QX with a ruler. What do you notice?

3 a Draw any triangle ABC with sideslarger than 6 cm.

b Bisect side AB at P.c Bisect side BC at Q.d Bisect side AC at R.e Draw line segments

AQ, BR and CP.

f Copy and complete: The medians of a triangle all passthrough ...... ...... .......... .

g Click on the icon for a demonstration.

Some special angles can be constructed. For example, angles measuring 60o.

We could also construct 30o and 15o angles by bisecting an angle of 60o then bisecting anangle of 30o.

To construct 45o and 2212

o angles we could first construct a 90o angle and also use bisectiontechniques.

Construct an angle of 60o at X on a line segment XY.

Step 1: Draw a line segment XY. With centreX draw an arc which cuts XY at Z.

Step 2: Keep the same radius. With centre Zdraw an arc to cut the first one at W.

A B

C

P

QR

Note:

median

Any line segment from vertex to themidpoint of the opposite side is called

aa .

ANGLE CONSTRUCTIONG

X Y

X YZ

X YZ

W

Example 8

DEMO

PRINTABLE

WORKSHEET

DEMO

Draw line segment AB of length cm. Using compass and straightedge only, perpendicularly AB at M. Check your construction by:

a a7bisect


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REVIEW OF CHAPTER 2


Step 3: Draw the line from X through W.Angle WXY is 60o.

EXERCISE 2G

1 a Draw a line segment AB which is 4 cm long. Construct a 60o angle at A and letthis angle be ]BAC.

b Construct a 30o angle, ]BAX, by bisecting angle BAC.

2 Construct a 45o angle by first constructing a 90o angle and then bisecting it.

3 a Explain how to construct a 15o angle.b Explain how to construct a 221

2

o angle.

1 State whether the following are true or false:a A square is a parallelogram.b The sum of the interior angles of an n-sided polygon is n£ 180o.c A rhombus is a regular polygon.

2 Find x in:a b c

d e f

3 Draw XY 6 cm long and construct its perpendicular bisector.

4 Draw any acute angle ABC and bisect it.

DISCUSSION

Explain why angle WXZ in Example 8 measures 60o.Explain how to construct an angle which measures 30o.

X YZ

W

67° x°x°

x° x°

83°

47°x°

123°

x°

110°142°

x°

126° x°


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5 a Copy and complete:“The sum of the exterior angles of any polygon is ........”

b Find x in:i ii

6 Find the value(s) of the unknown(s) in each figure, giving a brief reason for youranswer:

a b c

7 Find the value of the unknown in each triangle, giving a brief reason for your answer:a b c

8 Classify each figure giving brief reasons for your answer:a b c

9 a Draw a line segment AB which is 6 cm long. Construct a perpendicular to theline segment at A.

b Draw a diagram like the oneshown. Show how to construct aperpendicular from B to the linesegment.

10 Construct angles of a 60o b 45o.

11 Construct a square with sides 5 cm.

12 Construct an equilateral triangle with sides 5 cm.

13 Redraw the illustrated triangle. Constructthe altitude from A to BC.

122°

117°

x°

x°

95°

80°

50°

x°40°

x° 65° x°

70° 70°

40°

130°e°

80°40°

60°b°

b°

a°

74°

B

A

B C


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AlgebraAlgebra

33Chapter

�

�

�

�

�

�

�

�

�

�

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use the correct language of algebraconvert written statements intoalgebraic formevaluate an algebraic expressionusing substitutionunderstand what a formula is andhow it is usedcollect like termsuse product and index notationuse index laws correctly

understand the meaning of zero andnegative indicesdiscover formulae from patternssolve simple equations by inspectionunderstand how expressions andequations are ‘built up’solve equations by doing the sameoperation on both sides of theequation

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·


Outcomes · · · · · · · · · · · · · · · · · · · · · · · · · ·


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OPENING PROBLEM THE POOL AND ITS SURROUNDINGS

�

46 ALGEBRA (CHAPTER 3)

Algebra is used in problem solving.

Mathematical expressions and formulae are used to describe complicated situations in asimple form.

For example, consider a rectangle with length represented by the unknown l, and widthrepresented by w.

If A represents its area, we write

A = l£w or simply A = lw.

This formula is used instead of writing down:

“the area of a rectangle is found by multiplying its length by its width”.

So, instead of writing “the perimeter of a rectangle is found by adding twice its length totwice its width” we write P = 2l+ 2w.

So, Algebra is a form of arithmetic where we use unknowns.

Alf decides to put a rectangular pool, 7 m by 3 m in his backyard. Aconcrete path is to be laid around the pool with a border of lawn. Thelawn will be double the width of the path. The whole area needs to befenced and Alf has 50 m of fencing available.

Think about the following:² Alf has some problems. What do you

think they could be?² Suggest ways in which Alf could solve

his problems.² To solve the problem using algebra,

what length would you choose to berepresented by the unknown x?

² What is the length of the fence in termsof x?

² Can you find x by trial and error sub-stitution, by using a spreadsheet, or bysolving an equation?

² Can you draw a plan of the enclosedpool area showing dimensions?

² If Alf has 68 m of fencing availablehow would this affect the answer?

pool

grass

fence

concrete path

7 m3 m

w

l


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THE LANGUAGE OF MATHEMATICSA

ALGEBRA (CHAPTER 3) 47

Here are some key words used in algebra.

Word Meaning Example(s)

unknown is a letter or symbol used to C = 2¼r has unknownsrepresent a number C and r

expression consists of numerals(numbers), unknowns and 2x+ y ¡ 7,

2a+ b

coperation signs

equation is an algebraic form which 3x+ 8 = ¡1,x¡ 1

2= ¡4

contains an “=” sign

terms are separated by + or ¡ signs, 3x¡ 2y + xy ¡ 7 has four terms.the signs being included These are 3x, ¡2y, xy and ¡7.

like terms are terms with exactly the In 4x+ 3y ¡ 3x:same unknown form ² 4x and ¡3x are like terms

² 4x and 3y are unlike terms

constant term is a term which does not In 3x¡ y2 + 7 + x3,contain an unknown 7 is a constant term.

coefficient is the numeral factor of an In 4x¡ 2xy:algebraic term ² 4 is the coefficient of x

² ¡2 is the coefficient of xy

Consider 4y2 ¡ 6x+ 2xy ¡ 5 + x2:a Is this an equation or an expression?b How many terms does it contain?c State the coefficient of i x ii x2.d What is the constant term?

a We have an expression, as there is no “=” sign present.

b The expression contains five terms (4y2, ¡6x, 2xy, ¡5 and x2).

c i The coefficient of x is ¡6. ii The coefficient of x2 is 1.

d The constant term is ¡5.

Example 1


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CHANGING WORDS TO SYMBOLSB


EXERCISE 3A

1 State the coefficient of x in the following:a 4x b ¡7x c x d ¡x

e 5 + 6x f xy ¡ 3x g x2 ¡ 2x h 2x3 ¡1

x+ x

2 State the coefficient of y in the following:a 2y b ¡2y c 11y d ¡3y

e 4x¡ y f 3x¡ 7y + 1 g y3 + 3y h 6y2 ¡ 13y + 8

3 How many terms are there in:

a 3x+ y ¡ 7 b x2 ¡ ab+ c¡ 11 c x4 + x2 + x¡ 6 +4

x?

4 Which of the following are equations and which are expressions?a a+ b¡ 2c b 3x¡ y ¡ 7 c 3x¡ y = 7

d x2 = 16 e x2 = 22 + 32 fx

2¡ 5 = x2

In algebra we can convert sentences into algebraic expressions or equations.

For example,

Twice a number| {z }, increased by 7| {z } is 5 less than the number| {z } becomes2x + 7 = x¡ 5:

Many algebraic statements use words such as sum, difference, product and quotient.

Word Meaning Example

sum The sum of two or more numbers is 3 + 7, a+ 4, b+ c+ dobtained by adding them. are sums.

difference The difference between two numbers 9 ¡ 5, d¡ 6 (if d > 6)is the larger one minus the smaller one. are differences.

product The product of two or more numbers is 3 £ 6, 3a, xyz are products.obtained by multiplying them.

quotient The quotient of two numbers is the first The quotient of a and b isa

b:

one mentioned divided by the second.

average The average of a set of numbers isobtained by dividing their sum by thenumber of numbers.

The average of a, b and c isa+ b+ c

3:


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EXERCISE 3B.1

1 Write expressions for the sum of:a 7 and 3 b 4 and y c t and 2p d a, b and c

2 Write expressions for the product of:a 7 and 3 b 4 and y c t and 2p d a, b and c

3 Write expressions for the quotient of:a 7 and 3 b 4 and y c t and 2p d a+ b and c

4 Write expressions for the average of:a 7 and 3 b 4 and y c t and 2p d a, b and c

5 Write expressions for the difference between:a 7 and 3 b 4 and y if 4 < y c 4 and y if 4 > y

6 Write down algebraic expressions for:a three times x is subtracted from b b the product of a and the square of bc the sum of a and five times b d 3 less than te 4 more than a f the product of the square of x and 9g the square of the product of c and d h the sum of the squares of x and y

GENERALISING ARITHMETIC

To find algebraic expressions for many real world situationswe first think in terms of numbers.

For example, suppose we are asked to find the total costof x pens where each pen costs $y. We could first find thetotal cost of 5 pens where each pen costs $4:So, the total cost is 5 £ $4 = $20.

We multiplied the two quantities, and so the total cost ofx pens at $y each is x£ $y = $xy:

a State the sum of 6 and a.b State the difference between c and d (d > c).c State the average of p, q and r.

a The sum of 6 and a is 6 + a.

b The difference is d¡ c. fas d is largerg

c The average isp+ q + r

3:

Example 2With products, we

leave out themultiplication signbetween unknownsand write them in

alphabetical order!

First work out howto do the problemusing numbers.


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EXERCISE 3B.2

1 Find the total cost (in cents) of buying:a 5 apples at 30 cents each b a apples at 30 cents eachc a apples at c cents each

2 Find the total cost (in dollars) of buying:a 8 apples at 50 cents each b a apples at 50 cents eachc a apples at y cents each

3 Find the change from $100 when buying:a 3 books at $11 each b x books at $11 eachc x books at $y each

4 Demi has $25 in his pocket. How much would he have if:a he spends $6 b he spends $x c he is given $y?

5 Brian went on a journey to see his friends. He travelled 5 km to see Jim, then anothery km to see Susan. Finally he travelled another p km to see James. He then drove the6 km directly home. How far did he travel on his journey?

6 John is now 14 years old. How old will he be in x years time?

7 Peta can run at 10 km/h. How far can she run in t hours?

8 You have an 8 m length of string and cut 3 lengths of x m from it. What length remains?

9 Graham buys p pencils and b books. Find the total cost in cents, if each pencil costs45 cents and each book costs 85 cents.

10 a A cyclist travels at an average speed of 20 km/hfor 5 hours. How far does the cyclist travel?

b How far does the cyclist travel at an averagespeed of s km/h for t hours?

Find: a the cost of x bananas at 30 cents eachb the change from $50 when buying y books at $6 each.

a Suppose we were buying 7 bananas at 30 cents each.Then the cost of 7 bananas at 30 cents each would be 7 £ 30 cents.) the cost of x bananas at 30 cents each is x£ 30 = 30x cents.

b Suppose we were buying 5 books at $6 each.Then the change when buying 5 books at $6 each would be 50 ¡ (5 £ 6) dollars.

) the change when buying y books at $6 each is 50 ¡ (y £ 6) dollars= 50 ¡ 6y dollars.

Example 3


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EVALUATIONC


We can find algebraic expressions for many real world situations.

For example, the total cost of buying five apples at a cents each and three bananas at b centseach would be 5a+ 3b cents.

If an apple cost 10 cents and a banana cost 15 cents, the totalcost would be

5 £ 10 + 3 £ 15 = 50 + 45 = 95 cents.

We evaluated (found the value of) the expression for particularvalues of a and b (a = 10 and b = 15).

EXERCISE 3C

1 If a = 3, b = 2, c = 5, and d = 6, find the value of:a a+ b b 2a c a+ 2b d 3c+ d

e c¡ d f ab g 3bc h 2c2

i 2(a+ b) j 3(c¡ b) k 4a2 l (4a)2

2 If m = 4, n = 2, g = 7, and h = 5, evaluate:a 3m+ n b 11 + 2h c mn¡ g d 3n2

e (3n)2 f 2h3 g (2h)3 h 3m¡ 5n

i m2 ¡ 3m j m(m¡ 3) k 2(g + h) l 2g + 2h

For a = 3, b = 2 and c = 4, evaluate: a 2a+ 5b b 2c2

a 2a+ 5b

= 2 £ 3 + 5 £ 2

= 6 + 10

= 16

b 2c2

= 2 £ 42

= 2 £ 16

= 32

Example 4

If a = 12, b = 3 and c = 5,

evaluate:a+ b

c

a+ b

c

=12 + 3

5

=15

5

= 3

Example 5


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COLLECTING LIKE TERMSD


3 If a = 4, b = 3, c = 5, d = 2 find the value of:

aa

db

b+ c

ac

a+ c

bd

a+ b+ c

b

e2a+ d

cf

a+ c

b¡ dg

4b+ 2a

c¡ bh

b2 + 3c

bd

USING FORMULAE

If we have a formula such as s =d

tand we have actual number values for d and t, we

can substitute them into the formula to find the value of the unknown s.

For example, if d = 100 and t = 20, then s =100

20= 5:

You will recall that:

Like terms are algebraic terms which contain the same unknowns to the same powers.

For example, 3bc and 2bc are like terms and x2 and 5x are not like terms.

Algebraic expressions can often be simplified by adding or subtracting like terms. This issometimes called collecting like terms.

Consider 3x+ 2x i.e., x+ x+ x| {z } + x+ x| {z }.3 lots of x 2 lots of x

In total we have 5 lots of x, thus 3x+ 2x = 5x:

4 At my work I get paid $25 per hour plus $10 a week for travel. My boss uses theformula W = 25h+ 10 (wages = 25 £ number of hours + 10) to find my wage.

a If I work for 10 hours, how much will I get paid?b If I work for 40 hours, how much will I get paid?

5 The perimeter of a rectangle is given by the formula P = 2l + 2w where l is itslength and w is its width. Use this formula to find the perimeter of a rectangle which is

a 4 m long and 3 m wide b 6:2 m long and 4:3 m wide

6 The cost of painting my house is given by theformula C = ah+bn. a is the hourly rate forlabour, h is the number of hours worked, b is thecost per metre2 for the paint and n is the numberof square metres.Find the total cost of painting my house. I amcharged $20 per hour for labour, it takes 45 hoursto do the painting, the area to be painted is 75 m2and it costs $5 per m2 for paint.

DEMO


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ALGEBRAIC SIMPLIFICATION

AND INDEX FORME


EXERCISE 3D

1 Simplify, where possible, by collecting like terms:a 3 + x+ 5 b 8 + 7 + x c p+ 3 + 7

d 12 + a+ a e b+ 3 + b f 2x+ x

g a+ 3 + a+ 7 h x+ 3x i 3x¡ 2x

j 3x¡ x k a2 + a2 l 7x+ 3

2 Simplify, where possible:a 11n¡ 11n b 11n¡ n c 11n¡ 11

d 3ab+ ba e xy + 2xy f 2p2 ¡ p2

g 3a+ 2 + a+ 4 h 2a+ 3a+ 4a i b+ 3 + 2b+ 4

j 3xy + 4yx k 2a+ b+ 3a+ b l 3x+ 2x¡ x

m n+ 2n¡ 3n n 3x+ 7x¡ 10 o 3x+ 7x¡ x

PRODUCT NOTATION

In algebra we agree:

² to leave out the “£” signs between any multiplied quantities provided that at leastone of them is an unknown (letter)

² to write numerals (numbers) first in any product² where products contain two or more letters, we write them in alphabetical order.

For example, 3b is used rather than 3 £ b or b33bc is used rather than 3cb.

WRITING SUMS AS PRODUCTS

Sums of identical terms are easy using product notation.

For example, 5 + 5 + 5 + 5 = 4 £ 5 f4 lots of 5g) a+ a+ a+ a = 4 £ a = 4a f4 lots of ag

Likewise, b+ b+ b = 3b and x+ x+ x+ x+ x+ x = 6x.

Simplify, where possible, by collecting like terms:a 3a+ 4a b 11b¡ b c 5 + x+ 2

a 3a+ 4a b 11b¡ b c 5 + x+ 2

= 7a = 11b¡ 1b = 5 + 2 + x

= 10b = 7 + x f5 and 2 are likeg

Example 6


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INDEX NOTATION

When the same number is multiplied two or more times we use index notation as a quickway of writing the product.

For example, 2 £ 2 £ 2 = 23

and x£ x£ x£ x£ x = x5 where x5 is read as “x to the fifth”.

Note: ² In 23, the 2 is called the base number and the 3 is the index number(or power or exponent).

² x2 is read “x squared”, and x3 is read “x cubed”.

EXERCISE 3E.1

1 Simplify:a x£ x£ x b a£ a£ a£ a c b£ b£ b£ b£ b

d 2 £ p£ p e 3 £ n£ n£ n£ n f 4 £ y £ y £ 3 £ y

2 Simplify:a a+ a+ a+ b b a+ a£ a£ a c b£ b+ b£ b

d 5 + 2 £ a£ a e 4 £ a¡ a£ a f 3 £ a+ a+ a+ a

SIMPLIFYING ALGEBRAIC PRODUCTS

3 £ 2x and a2 £ 2ab are algebraic products.

Often algebraic products can be simplified using these steps.

² Expand out any brackets.² Calculate the coefficient of the final product by multiplying all the numbers.² Simplify the unknowns by using index notation if appropriate. Two or more

unknowns should be in alphabetical order.

Simplify:a a+ a+ b+ b+ b

b a£ a+ b+ c+ c

a a+ a+ b+ b+ b b a£ a+ b+ c+ c

= 2a+ 3b = a2 + b+ 2c

Example 7

Simplify: a x2 £ x3 b 2a£ 5a2

a x2 £ x3 b 2a£ 5a2

= x£ x£ x£ x£ x = 2 £ a£ 5 £ a£ a

= x5 = 10a3

Example 8


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INVESTIGATION DISCOVERING INDEX LAWS


EXERCISE 3E.2

1 Simplify:a a£ a b b2 £ b c c£ c3

d n2 £ n2 e 4a£ 3b f 5c£ 2a

g m£m2 h k3 £ k2 i p2 £ p4

2 Simplify:

a (a2)2 b (s3)2 c (g2)3

d (ac)2 e (m4)2 f (3ab)2

Look carefully at any patterns as you complete the following:


a 22 £ 23 = (2 £ 2) £ (2 £ 2 £ 2) = 25

b 33 £ 31 = =

c a3 £ a4 = =

From the above examples, am £ an =


a25

23=

2 £ 2 £ 2 £ 2 £ 2

2 £ 2 £ 2= 22 b

34

31= =

ca5

a2= = d

x7

x4= =

From the above examples,am

an= :


a (23)2 = 23 £ 23 = (2 £ 2 £ 2) £ (2 £ 2 £ 2) =

b (32)4 = = =

c (a2)3 = = =

From the above examples, (am)n = :

Simplify: (x2)2 (x2)2

= x2 £ x2

= x£ x£ x£ x

= x4

Example 9

WORKSHEET


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INDEX LAWSF


From the Investigation, we can state the laws for positive indices.

If m and n are positive integers, then

² am £ a

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