solving 7:08; D&D Ch7 percentages MT 1, 2, 3;
operating on PA 7:08
integers MT 1; 2:01D, D:01–D:08
A:01–A:04
intercepts and ratio 3:02
x- MT 8;
Inv 8:02 perimeter MT 2, 12; D&D Ch12
y- MT 8;
Inv 8:02 12:01; Inv 12:01
intersection of two lines 8:04
inverse operation MT 7 and area Inv 12:06
isosceles triangle MT 10
of sectors 12:02
knots Inv 15:03
of composite figures 12:02;
language 1:01
of mathematics 14:01 Challenge 12:04
of chance MT 4
plot MT 8
like terms 8:05
lines Chapter 8; D&D polygons MT 10
8:03
gradient–intercept 8:04 regular MT 10; 10:09
form of 11:03
11:03 angle sum of Inv 10:09A;
graphing straight 8:03
horizontal Chapter 11 D&D Ch10
intersection of
parallel exterior angle sum of Inv 10:09B
perpendicular
vertical frequency and
locus
cumulative 13:02; 13:05
regular and tessellations Inv 10:09C
types of 10:09
power MT 6
prisms
closed 12:05
open 12:05
probability Chapter 14; D&D;
MT 14; Tech;
CW 14:03; FS 14:04;
FS 13:04
problem solving using
equations 9:06; Inv 2:02
mean 13:02 locus 11:05
measurement 2:01E; S 2:01
median 13:03 problem, translating into
median class MT 13;
13:05 equations 9:05; RM 9:05
mode 13:03
multiplication of algebraic pronumeral MT 4; 9
4:04B
fractions MT 14; 14:04 proportional change 3:03
mutually exclusive events
Pythagoras’ theorem Chapter 5
and perimeter 12:01
and speed Inv 5:04
negative indices MT 6; 6:03; D&D quadrant MT 8;
Ch6 quadrilateral 8:01; Tech 8:01
networks Chapter 15 MT 10; 10:08; D&D
number plane MT 8; D&D Ch8 Ch10; Tech 10:08
8:01; FS 8:01
numerator MT 4 random MT 14
range 13:03
obtuse-angled triangle MT 10 rates 1:07; D&D Ch1;
ogive MT 13; MT 2; 2:01A
13:05 ratio MT 2; 2:01B;
origin MT 8 Chapter 3
outcomes MT 14, 13 and proportion Chapter 3
and rational numbers Inv 3:01
parallel lines MT 10 dividing a quantity in
parentheses MT 4 2:01C
a given ratio
increasing and 3:02
1:06; D&D Ch1
decreasing
simplifying
INDEX 481
rational numbers MT 1; 1:04 subtraction of algebraic 4:04A
recurring decimals MT 1; 1:05; fractions MT 10
Challenge 1:05 MT 12
regular polygon MT 10; 10:09 supplementary angles 12:05; D&D Ch12
repeating decimals see recurring decimals surface area 12:07
right-angled triangle Ch5; MT 10 12:06
rounding off see approximation of prisms MT 14
practical applications
sample MT 14 of composite solids MT 1
scalene triangle MT 10 survey
sector MT 12 10:08
significant figures MT 1; 1:08; terminating decimal 10:08
D&D Ch1 tests 10:08
simplifying MT 14; 14:03
ratio 1:06 for a parallelogram Inv 14:01
algebraic expressions 4:04 for a rhombus 15:03; MT 15
MT 14 for a rectangle RM 14:02
simulation 8:05 theoretical probability MT 10
slope MT 7 throwing dice 15:01B
solution MT 7 topology 15:01B; MT 15
solve tossing a coin MT 10; 10:06;
solving Chapter 7; D&D transversal D&D Ch10
2:02 trees 10:08
equations 2:01 minimum spanning
non-routine problems Inv 9:03; triangles 2:03; Challenge 2:03
routine problems Challenge 14:03 15:01A; MT 15
spreadsheet formula Challenge 12:04 types of quadrilaterals 8:03; FS 8:03;
Chapter 13; D&D; D&D Ch8
staggered starts Tech 9:02 Venn diagrams
statistics MT 9 vertex MT 10; 10:02
14:01 vertical lines MT 2
subject of a formula MT 7
subjective probability vertically opposite
substitute MT 9; D&D Ch9 angles
substitution into algebraic Tech 9:02; 9:02
volume
expressions
482 INTERNATIONAL MATHEMATICS 3
Acknowledgments
We would like to make the following acknowledgements for permission to reproduce copyright
material in this book.
Karly Abery © Pearson Education Australia: p.91
Courtesy the Library, American Museum of Natural History: p.26
Australian Picture Library: pp.2, 33 top, 86, 125, 155, 164, 185, 315, 350
Australian Taxation Office, 2003 © Commonwealth of Australia reproduced with permission: p.252
Larry Bagnell: pp.40 top, 119, 168
Getty Images: p.224
Sven Klinge: pp.36, 40 bottom, 41 top, 41 bottom, 45, 89, 102, 109, 117, 122, 135, 161, 199, 201
bottom, 202, 206, 277, 291, 292, 294, 297, 317, 323, 331, 369, 372
© Department of Lands, Panorama Avenue Bathurst 2795, www.lands.nsw.gov.au: p.105
Courtesy of NASA: p.158
Kim Nolan © Pearson Education Australia: pp.301, 406
Photolibrary.com: pp.52, 200
Lisa Piemonte © Pearson Education Australia: pp.42 bottom, 322
Photograph courtesy of Redpath Ideal Greenhouses Pty Ltd: p.201 top
From Police Life courtesy of Victorian Police Centre: p.38
Reproduced with permission of UBD © Universal Press Pty Ltd: p.34
The authors would also like to sincerely thank Sven Klinge for the many photographs he has taken
for this book.
Every effort has been made to trace and acknowledge copyright. However, should any infringement
have occurred, the publishers tender their apologies and invite the copyright owners to contact
them.
INDEX 483
Foundation Worksheet
5:01 Pythagoras’ Theorem A
Name: Class:
Example c
Measure each side of the triangle. a
Work out a2, b2 and c2.
Check that c2 = a2 + b2 b
a b c a2 b2 c2
14 48 50 196 2304 2500
2500 = 196 + 2304 True.
Exercise
Measure each side of the triangle, in millimetres. Work out a2, b2 and c2. Does c2 = a2 + b2?
1 2 3c
a
ca b
bc
b a 6
4 5a
cb c
a
cb
b
a 8b 9 a
7a a c
c
bc b
b
10 c
a
Answers can be found in the Interactive Student CD. 12 © Pearson Education Australia 2008.
This page may be photocopied for classroom use.
INTERNATIONAL MATHEMATICS FOR THE MIDDLE YEARS 3
Foundation Worksheet
5:02 Pythagoras’ Theorem B
Name: Class:
Examples The hypotenuse is the
longest side.
Find the length of the hypotenuse in each case.
c2 = 32 + 52
1 c2 = 62 + 82 2 5 = 34
c
c 6 = 100 3 c = 34
= 5·83 (2 dec. pl.)
c = 100
8 = 10 (exactly)
Exercise
Find the length of the hypotenuse (exactly, or correct to two decimal places).
1 c2 = 22 + 32 2 c2 = 42 + 92 3 5 c2 = 52 + 122
= 13 = 97 = 169
c
2 ∴c=…
=…
3 ∴ c = 13 c 9 ∴ c = 97 12 c
=… =…
4
4 10 c2 = 82 + 102 5 c2 = 92 + 122 6 6 c2 = 62 + 72
=… =…
8c c =… 7
∴c=… 12 ∴ c = … c ∴c=…
=… 9 9 =…
=…
83 10
7 c2 = …
12 = … 11 c 5
16 c
c ∴c=…
=…
10 6 11 8 12 c
c 11
6 c 15 10
Fun Spot 3:02 | What do cats put in soft drinks? U 232
Complete these calculations. Match the letters with the answers.
B 400 C 182 E 152 I 289 M 162 S 361
256 17 324 225 324 529 20 225 19
Answers can be found in the Interactive Student CD. 13 © Pearson Education Australia 2008.
This page may be photocopied for classroom use
INTERNATIONAL MATHEMATICS FOR THE MIDDLE YEARS 3
Foundation Worksheet
5:03 Pythagoras’ Theorem
Name: Class:
Examples When finding a shorter side,
subtract the squares.
Find the value of the pronumeral.
9 b2 + 92 = 152
1a a2 + 102 =122 2 b2 + 81 = 225
∴ b2 = 144 (225 − 81)
a2 + 100 =144
12 10 ∴ a2 = 44 (144 − 100) b 15
a = 44 b = 144
= 12 (exactly)
= 6·63 (2 dec. pl.)
Exercise
Find the value of the pronumeral.
1 5 b2 + 52 = 72 2 a a2 + 52 = 132 3b b2 + 172 = 202
13 a2 + … = …
b2 + 25 = 49 5 b2 + … = …
10 ∴ a2 = …
b7 ∴ b2 = 24 8 a= … 20 17 ∴ b2 = …
12 =…
b = 24 b= …
b2 + 82 = 102
=… =…
4 a2 + 402 = 412 5 6 a2 + 202 = 302
a2 + … = …
41 40 ∴ a2 = … b 30 20
a a= … 8 a
=…
7 22 b2 + 222 = 252 a 96
b
25 x 72
10 9 2 12 15
x 11 x x 25
12 17 15
Fun Spot 3:03 | Which side of a chicken has the most feathers?
Complete these calculations. Match the letters with the answers.
D 102 − 52 E 192 − 132 H 152 + 172 I 72 + 62
O 32 + 52 S 122 − 52 T 232 − 132 U 612 − 602
360 514 192 34 121 360 119 85 75 192
Answers can be found in the Interactive Student CD. 14 © Pearson Education Australia 2008.
This page may be photocopied for classroom use.
INTERNATIONAL MATHEMATICS FOR THE MIDDLE YEARS 3
Challenge Worksheet
14 : 03 Probability: An Unusual Case
Name: Class:
Bradley Efron, a mathematician at Stanford
University, invented a dice game that involves
unexpected probabilities.
The faces of four dice are numbered as shown
below.
Die A (used by Andrew) 1, 2, 3, 9, 10, 11
Die B (used by Ben) 0, 1, 7, 8, 8, 9
Die C (used by Cassie) 5, 5, 6, 6, 7, 7
Die D (used by Diane) 3, 4, 4, 5, 11, 12
• Each player throws their die once. Die A
1 2 3 9 10 11
• Andrew beats Ben if the number on his die is 0 AAAAAA
1 TAAAAA
higher than the number on Ben’s die. 7 BBBAAA
8 BBBAAA
• The table of outcomes on the right can be 8 BBBAAA
9 BBBTAA
used to determine who has the greater
• A means die A wins.
probability of success. 11 Die B • B means die B wins.
P(A wins) = 2----2- • T means a tie occurs.
9 10
36
P(B wins) = 1----2- 8
36 87
P(tie) = --2---
36
Clearly, Andrew has the greater probability of
winning.
Exercise
1 Complete the tables of outcomes below.
a Die B b Die C c Die D
017889 556677 3 4 4 5 11 12
1
53 2
3
54 9
10
64 11
Die C
Die D
Die A
65
7 11
7 12
2 a Should Andrew beat Ben? b Should Ben beat Cassie?
c Should Cassie beat Diane? d Should Diane beat Andrew?
3 Explain why this situation is so unusual. b Should Ben beat Diane?
4 a Should Andrew beat Cassie?
Answers can be found in the Interactive Student CD. 5 © Pearson Education Australia 2008.
This page may be photocopied for classroom use.
INTERNATIONAL MATHEMATICS FOR THE MIDDLE YEARS 3