ICEEM
Mathematics Peter Brown Brian Dorofaeff Andy Edwards Michael Evans Garth Gaudry David Hunt Janine McIntosh Bill Pender
1B
Secondary
ICMEM Mathematics Secondary 1B Includes index. For junior secondary school students. ISBN 9780977525430. ISBN 0 9775254 3 0. 1. Mathematics  Textbooks. I. Evans, Michael (Michael Wyndham). II. Australian Mathematical Sciences Institute. III. International Centre of Excellence for Education in Mathematics. 510 Cover designed by Designgrant Layout designed by Rose Keevins Typeset by Claire Ho
This project is funded by the Australian Government through the Department of Education, Science and Training. The views expressed herein do not necessarily represent the views of the Australian Government Department of Education, Science and Training or the Australian Government.
© The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICEEM) 2006 All rights reserved Printed in Australia by McPherson’s Printing Group Other than as permitted under the Copyright Act, no part of this book may be used or reproduced in any manner or by any process whatsoever without the prior permission of The University of Melbourne. Requests for permission should be addressed to
[email protected], or Copyright Enquiries, AMSI, 111 Barry Street, University of Melbourne, Victoria 3010.
Contents
Chapter 11
Books in this series
vi
Student CDROM
vi
Preface
vii
Acknowledgements
ix
Integers
1
11A
Negative integers
2
11B
Addition and subtraction of a positive integer
7
11C
Addition and subtraction of a negative integer
10
11D
Multiplication and division involving negative integers
17
11E
Indices and order of operations
22
Review exercise
27
Challenge exercise
29
Chapter 12 Algebra and the number plane
31
12A
Substitution with integers
32
12B
The number plane
36
12C
Completing tables and plotting points
43
12D
Finding rules
48
Review exercise
56
Challenge exercise
58
Chapter 13 Triangles and constructions
61
13A
Review of geometry
61
13B
Angles in triangles
65
13C
Circles and compasses
75
13D
Isosceles and equilateral triangles
80
13E
Constructions with compasses and straight edge
88
13F
Quadrilaterals
95
13G Further constructions
99
Review exercise
101
Challenge exercise
105
iii
Contents Chapter 14 Negative fractions 14A
Addition and subtraction of negative fractions
108
14B
Multiplication and division of negative fractions
112
14C
Negative decimals
116
14D
Substitution involving negative fractions and decimals
119
Challenge exercise
Chapter 15 Percentages
122
123
15A
Percentages, fractions and decimals
123
15B
Expressing one quantity as a percentage of another
130
15C
Percentage of a quantity
132
Review exercise
135
Challenge exercise
136
Chapter 16 Solving equations
137
16A
An introduction to equations
137
16B
Equivalent equations
140
16C
Solving equations involving more than one operation
146
16D
Equations with integers
151
16E
Expanding brackets and equations
153
16F
Collecting like terms and solving equations
157
16G Equations with pronumerals on both sides
162
Solving problems using equations
164
16H
Review exercise
169
Challenge exercise
172
Chapter 17 Probability
iv
107
173
17A
An introduction to probability
173
17B
Experiments and counting
176
17C
Empirical probability
184
Review exercise
190
Challenge exercise
192
Chapter 18 Transformations and symmetry
193
18A
Translation
194
18B
Rotation
199
18C
Reflection
205
18D
The three transformations
210
18E
Symmetry
213
18F
Regular polygons
217
18G Combined transformations
222
Review exercise
226
Challenge exercise
230
Chapter 19 The five Platonic solids
233
19A
Building the regular tetrahedron
235
19B
Building the regular hexahedron or cube
240
19C
Building the regular octahedron
244
19D
Building the regular dodecahedron
247
19E
Building the regular icosahedron
249
Challenge exercise
251
Chapter 20 Review and problem solving
257
20A
Review
257
20B
Tessellations
267
20C
Sets and Venn diagrams
272
Answers to exercises
301
Books in this series Upper primary
Secondary
Transition 1A
Transition 1B
Transition 2A
Transition 2B
Secondary 1A
Secondary 1B
Secondary 2A
Secondary 2B
Secondary 3A
Secondary 3B
Secondary 4A
Secondary 4B
Student CDROM An electronic (PDF) version of this book is provided on the CDROM attached to the inside back cover.
vi
Preface ICEEM Mathematics is a new program for students in Years 5 to 10 throughout Australia. The program is being developed by the International Centre of Excellence for Education in Mathematics (ICEEM). ICEEM is managed by the Australian Mathematical Sciences Institute and funded by the Australian Government through the Department of Education, Science and Training. The program comprises a series of textbooks, teacher professional development, multimedia materials and continuing teacher support. ICEEM has developed the program in recognition of the increasing importance of mathematics in modern workplaces and the need to enhance the mathematical capability of Australian students. Students who complete the program will have a strong foundation for work or further study. ICEEM Mathematics is an excellent preparation for Years 11 and 12 mathematics. ICEEM Mathematics is unique because it covers the core requirements of all Australian states and territories. Beginning in upper primary school, it provides a progressive development and smooth transition from primary to secondary school. The writers are some of Australia’s most outstanding mathematics teachers and mathematics subject experts. Teachers throughout Australia who have taken part in the Pilot Program in 2006 have contributed greatly, through their suggestions, to the final version of the textbooks. The textbooks are clearly and carefully written. They contain background information, examples and worked problems, so that parents can assist their children with the program if they wish. There is a strong emphasis on understanding basic ideas, along with mastering essential technical skills. Students are given accessible, practical ways to understand what makes the subject tick and to become good at doing mathematics themselves. Mental arithmetic and other mental processes are given considerable prominence. So too is the development of spatial intuition and logical reasoning. Working and thinking mathematically pervade the entire ICEEM Mathematics program.
vii
Preface The textbooks contain a large collection of exercises, as do the classroom exercise sheets, classroom tests and other materials. Problem solving lies at the heart of mathematics. Since ancient times, mathematics has developed in response to a need to solve problems, whether in building, navigation, astronomy, commerce or a myriad other human activities. ICEEM Mathematics gives students a good variety of different kinds of problems to work on and helps them develop the thinking and skills necessary to solve them. The challenge exercises are a notable feature of ICEEM Mathematics. They contain problems and investigations of varying difficulty, some quite easy, that should catch the imagination and interest of students who wish to explore the subject further. The ICEEM Mathematics materials from Transition 1 and 2 to Secondary 1 are written so that they do not require the use of a calculator. Calculator use, in appropriate contexts, is introduced in Secondary 2B. This is a deliberate choice on the part of the authors. During primary school and early secondary years, students need to become confident at carrying out accurate mental and written calculations, using a good variety of techniques. This takes time and effort. These skills are essential to students’ further mathematical development, and lead to a feeling of confidence and mathematical selfreliance. Classroom practice is, of course, the prerogative of the teacher. Some teachers may feel that it is appropriate for their students to undertake activities that involve calculator use. While the ICEEM Mathematics program is comprehensive, teachers should use it flexibly and supplement it, where necessary, to ensure that the needs of their students, or local requirements, are met. This is one of the key messages of the ICEEM professional development program. The ICEEM Mathematics website at www.icemaths.org.au provides further information about the program, as well as links to supplementary and enrichment materials. New and revised content is being added progressively. ICEEM Mathematics textbooks can be purchased through the site as well as through normal commercial outlets.
viii
Acknowledgements We are grateful to Professor Peter Taylor, Director of the Australian Mathematics Trust, for his support and guidance as chairman of the Australian Mathematical Sciences Institute Education Advisory Committee. We gratefully acknowledge the major contribution made by those schools that participated in the Pilot Program during the development of the ICEEM Mathematics program. We also gratefully acknowledge the assistance of: Sue Avery Robyn Bailey Richard Barker Raoul Callaghan Gary Carter Claire Ho Jacqui Ramagge Nikolas Sakellaropoulos Michael Shaw James Wan Andy Whyte HungHsi Wu
ix
Chapter 11 Integers You have probably come across examples of negative numbers already. They are the numbers that are less than zero. For example, they are used in the measurement of temperature. The temperature 0°C is the temperature at which water freezes, known as freezing point. The temperature that is 5 degrees colder than freezing point is written as –5°C. In some Australian cities, the temperature drops to low temperatures. Canberra has a lowest recorded temperature of –10°C. The lowest recorded temperature in Australia is –23°C, recorded at Charlotte’s Pass in NSW. Here are some other lowest recorded temperatures: Alice Springs Paris London
–7°C –24°C –16°C
Negative numbers are also used to record heights below sea level. For example, the surface of the Dead Sea in Israel is 417 metres below sea level. This is written as –417 metres. This is the lowest point on land anywhere on Earth. The lowest point on land in Australia is at Lake Eyre, which is 15 metres below sea level. This is written as –15 metres.
Chapter 11 Integers
Brahmagupta, an Indian mathematician, wrote important works on mathematics and astronomy, including a work called Brahmasphutasiddhanta (The Opening of the Universe), which he wrote in the year AD 628. This book is believed to mark the first appearance of negative numbers in the way we know them today. Brahmagupta gives the following rules for positive and negative numbers in terms of fortunes (positive numbers) and debts (negative numbers). By the end of this chapter, you will be able to understand his words. A debt subtracted from zero is a fortune. A fortune subtracted from zero is a debt. The product of zero multiplied by a debt or fortune is zero. The product of zero multiplied by zero is zero. The product or quotient of two fortunes is a fortune. The product or quotient of two debts is a fortune. The product or quotient of a debt and a fortune is a debt. The product or quotient of a fortune and a debt is a debt.
11A Negative integers The whole numbers, together with the negative whole numbers, are called the integers. These are: …, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, … The numbers 1, 2, 3, 4, 5, … are called the positive integers. The numbers …, –5, –4, –3, –2, –1 are called the negative integers. The number 0 is neither positive nor negative.
ICEEM Mathematics Secondary 1B
The number line The integers can be represented by points on a horizontal line called a number line. The line is infinite in both directions, with the positive integers to the right of zero and the negative integers to the left of zero. The integers are equally spaced.
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
An integer a is less than another integer b if a lies to the left of b on the number line. The symbol < is used for less than. For example, –3 < –1, since –3 is to the left of –1. An integer b is greater than another integer a if b lies to the right of a on the number line. The symbol > is used for greater than. For example, 1 > –5, since 1 is to the right of –5.
a
b
a < b and b > a. A practical illustration of this is that a temperature of –8°C is colder than a temperature of –3°C, and –8 < –3.
Example 1 a List all the integers less than 10 and greater than 2. b List all the integers less than 5 and greater than –3. c List all the integers less than 2 and greater than –5. d List all the integers less than –2 and greater than –9.
Solution a 3, 4, 5, 6, 7, 8, 9
b –2, –1, 0, 1, 2, 3, 4
c –4, –3, –2, –1, 0, 1
d –8, –7, –6, –5, –4, –3
Chapter 11 Integers
Example 2 a Arrange the following integers from smallest to largest. –6, 6, 0, 100, –1000, –5, –100, 8 b Arrange the following integers from largest to smallest. –25, 1000, –500, –26, 53, 100, 56
Solution a –1000, –100, –6, –5, 0, 6, 8, 100 b 1000, 100, 56, 53, –25, –26, –500
Example 3 a Draw a number line and mark on it with dots all the integers less than 6 and greater than –5. b Draw a number line and mark on it with dots all the integers less than 2 and greater than –4. c Draw a number line and mark on it with dots all the integers greater than –6 and less than 3.
Solution a –6
–5 –4
–3
–2
–1
0
1
2
3
4
5
–6
–5 –4
–3
–2
–1
0
1
2
3
4
5
–6
–5 – 4
–3
–2
–1
0
1
2
3
4
5
b c
ICEEM Mathematics Secondary 1B
Example 4 a The sequence 10, 5, 0, –5, –10, … is ‘going down by fives’. Write down the next four numbers, and mark them on the number line. b The sequence –16, –14, –12, … is ‘going up by twos’. Write down the next four numbers, and mark them on the number line.
Solution a The next four numbers are –15, –20, –25, –30.
–45 –40 –35 –30 –25 –20 –15 –10
–5
0
5
10
2
4
6
b The next four numbers are –10, –8, –6, –4.
–16 –14 –12 –10 –8
–6
–4
–2
0
The opposite of an integer The number –2 is the same distance from 0 as 2, but lies on the opposite side of zero. We call –2 the opposite of 2. Similarly, the opposite of –2 is 2. The operation of forming opposites can be visualised by putting a pin in the number line at 0 and rotating the number line by 180˚. The opposite of 2 is –2. –6
–5 –4
–3
–2
–1
0
1
2
3
4
5
The opposite of –2 is 2. Notice that the opposite of the opposite is the number we started with. For example, –(–2) = 2. Note: The opposite of 0 is 0.
Chapter 11 Integers
Exercise 11A Example 1
Example 2
Example 3
1
a List the integers less than 3 and greater than –5.
b List the integers greater than –8 and less than –1.
c List the integers less than –4 and greater than –10.
d List the integers greater than –132 and less than –123.
2
a Arrange the following integers from smallest to largest. –10, 10, 0, 100, –100, –6, –1000, 5
b Arrange the following integers from largest to smallest. –30, 45, –45, –550, –31, 26, –26, 55
3
a Draw a number line and mark the numbers –2, –4, –6 and –8 on it.
b Draw a number line and mark the numbers –1, –3, –5 and –7 on it. c Draw a number line and mark the whole numbers less than 0 and greater than –8 on it. d Draw a number line and mark the whole numbers less than 3 and greater than –3 on it.
Example 4
4
The sequence –15, –13, –11, … is ‘going up by twos’. Give the next three terms. (Draw a number line to help you.)
5
The sequence 3, 1, –1, … is ‘going down by twos’. Give the next three terms. (Draw a number line to help you.)
6
The sequence –50, –45, –40, … is ‘going up by fives’. Give the next three terms. (Draw a number line to help you.)
7
Give the opposite of each integer.
a 5
b –4
c –10
d –12
e 7
f –8
g –4
h –3
8
Simplify:
a –(–2)
b –(–7)
c –(–20)
d –(–(–10))
e –(–(–30))
f –(–(–(–40)))
ICEEM Mathematics Secondary 1B
9
Insert the symbol > or < in each box to make a true statement.
a 3
5
b 3
c –7
–5
–4
d 2
–(–3)
10 Give the readings for each of the thermometers shown below. a b c d
°C 50 40 30 20 10 0 –10 –20 –30 –40 –50 –60
°C 50 40 30 20 10 0 –10 –20 –30 –40 –50 –60
°C 50 40 30 20 10 0 –10 –20 –30 –40 –50 –60
°C 50 40 30 20 10 0 –10 –20 –30 –40 –50 –60
11B Addition and subtraction of a positive integer If a submarine drops to a depth of –250 m and then rises by 20 m, its final position is –230 m. This can be written –250 + 20 = –230 m. Joseph has $3000 and he spends $5000. He now has a debt of $2000, so it is natural to interpret this as 3000 – 5000 = –2000. These are examples of adding and subtracting a positive integer.
The number line and addition The number line provides a useful picture for the addition and subtraction of integers.
Chapter 11 Integers
Addition of a positive integer When you add a positive integer, move to the right along the number line.
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
For example, to calculate –3 + 4, start at –3 and move to the right 4 steps. We see that –3 + 4 = 1. A practical situation such as money: I start with a debt of $3 but I then earned $4. I now have $1.
Subtraction of a positive integer We will start by thinking of subtraction as taking away. When you subtract a positive integer, move to the left along the number line. For example, to calculate 2 – 5, start at 2 and move to the left 5 steps. We see that 2 – 5 = –3.
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
The same question can be posed in a practical way: I had $2 and I spent $5. I now have a debt of $3.
Example 5
Ex
Write the answers to these additions. a –5 + 6
b –7 + 12
c –11 + 20
Solution
Ex
a –5
– 4 –3
–5 + 6 = 1
–2
–1
0
1
(Start at –5 on the number line and move 6 steps to the right.)
ICEEM Mathematics Secondary 1B
b –7 + 12 = 5
(Start at –7 on the number line and move 12 steps to the right.)
c –11 + 20 = 9
(Start at –11 on the number line and move 20 steps to the right.)
Example 6 Find the value of: a –2 – 3
b 6 – 9
c –4 – 11
Solution a –6
–5 – 4
–3
–2
–1
0
1
2
3
4
5
Start at –2 and move three to the left. We see that –2 – 3 = – 5. b 6 – 9 = –3
(Start at 6 and move 9 to the left.)
c –4 – 11= –15
(Start at –4 and move 11 to the left.)
Exercise 11B Example 5
1
Calculate these additions.
a –5 + 7 e –12 + 16 i –11 + 4 m –64 + 14
Example 6
2
b –2 + 3 f –5 + 2 j –12 + 4 n –8 + 42
c –5 + 10 g –6 + 12 k –32 + 50 o –71 + 6
d –1 + 4 h –5 + 10 l –50 + 11 p –37 + 42
Calculate these subtractions.
a 5 – 6 e –7 – 16 i –11 – 4 m 85 – 100
b 6 – 12 f –5 – 2 j –12 – 5 n –32 – 68
c –5 – 10 d –11 – 100 g –6 – 2 h 5 – 10 k –10 – 90 l –990 – 1000 o –100 – 1100 p –24 – 9
Chapter 11 Integers
3
Work from left to right to calculate:
a 15 – 6 – 8
e –7 – 16 – 20 f 5 – 2 – 10
b 6 – 12 – 5
c –8 – 10 – 11 d –11 + 100 – 200 g –6 – 2 – 20
h 5 – 10 + 20
i –11 – 4 – 30 j –12 – 5 + 20 k –20 – 30 – 10 l –5 + 6 – 7 4
Work from left to right to calculate:
a 11 – 10 – 20 –15
c 20 – 9 – 7 – 4
b –2 – 3 – 4 –5 d –11 + 1 + 2 + 8 + 1
e –20 – 2 – 4 + 6
5
a Johanne has a total amount of $3400 and spends $5000. What is Johanne’s debt?
b Francis has a debt of $4670 but earns $3456 and pays off a portion of the debt. How much does Francis owe now?
c A submarine is at a depth of –320 m and then rises by 40 m. What is the new depth of the submarine? d The temperture in a freezer is –17°C. The freezer is turned off and in 10 minutes the temperture has risen by 8°C. What is the temperature of the freezer now? e David has a debt of $3760 but earns $4000 and pays off the debt. How much does David have now?
11C Addition and subtraction of a negative integer In the previous section, we considered addition and subtraction of a positive integer. In this section, we will add and subtract negative integers.
Addition of a negative integer Adding a negative integer to another integer means that you take a certain number of steps to the left on a number line.
10
ICEEM Mathematics Secondary 1B
The result of the addition 4 + (–6) is the number you get by moving 6 steps to the left, starting at 4.
–6
–5
–4
–3
–2
–1
0
1
1
2
2
3
4
5
4 + (–6) = –2
Example 7 Work out the answer to –2 + (–3).
Solution –6
–5 – 4
–3
–2
–1
0
3
4
5
–2 + (–3) is the number you get by moving 3 steps to the left, starting at –2. That is, –5. Notice that –2 – 3 is also equal to –5.
All additions of this form can be completed in a similar way. For example: 4 + (–7) = –3
and note that
4 – 7 = –3
–11 + (–3) = –14
and note that
–11 – 3 = –14
This suggests the following rule.
To add a negative integer, subtract its opposite. For example: 4 + (–10) = 4 – 10 – 7 + (–12) = – 7 – 12 = – 6 = –19
Chapter 11 Integers
11
Subtracting a negative integer We have already seen that adding –2 means taking 2 steps to the left. For example:
0
5
7
7 + (–2) = 5 We want subtracting –2 to be the reverse of the process of adding –2. So to subtract –2, we take 2 steps to the right. For example:
0
7
7 – (–2) = 9 There is a very simple way to state this rule:
To subtract a negative number, add its opposite. For example: 7 – (–2) = 7 + 2 = 9
Example 8 Evaluate: a 12 + (–3)
b –3 + (–7)
c 6 – (–18)
d –12 – (–6)
Solution a 12 + (–3) = 12 – 3 b –3 + (–7) = –3 – 7 = 9 = – 10 c 6 – (–18) = 6 + 18 d –12 – (–6) = –12 + 6 = 24 =–6
12
ICEEM Mathematics Secondary 1B
9
Example 9 Calculate: a 4 – (–15)
b –25 – (–3)
Solution a 4 – (–15) = 4 + 15
= 19
b –25 – (–3) = –25 + 3
= –22
Example 10 Calculate: a 6 – (–3) + (–8)
b –14 + (–7) – (–15).
Solution a 6 – (–3) + (–8) = 6 + 3 – 8 =9–8 =1 b –14 + (–7) – (–15) = –14 – 7 + 15 = –21 + 15 = –6
Example 11 The minimum temperature on Saturday was –13ºC and the maximum temperature was –2ºC. Calculate the difference (minimum temperature – maximum temperature).
Solution minimum temperature – maximum temperature = –13ºC – (–2ºC) = –13ºC + 2ºC = –11ºC
Chapter 11 Integers
13
Example 12
Ex
Evaluate: a 347 – 625
b 456 – (–356)
c –234 + 568
d –120 – (–105)
Solution a 347 – 625 = –278
b 456 – (–356) = 456 + 356 = 812
c –234 + 568 = 568 + (– 234) d –120 – (–105) = –120 + 105 = 568 – 234 = – 15 = 334
Exercise 11C Example 7,8a,b
1
Write the answers to these additions.
a 5 + (–2)
b –6 + 2
c –5 + 10
d –11 + (–4)
e –12 + 16
f –5 + (–2)
g –6 + (–2)
h –5 + (–10)
j –12 + 4
k –20 + (–30)
l –110 + 100
i 11 + (–4) Example 8c,d.9
2
Write the answers to these subtractions.
a 5 – (–6)
b 6 – (–12)
c –5 – (–10)
d 11 – (–4)
e –12 – (–16)
f –5 – (–2)
g –6 – (–2)
h 5 – (–10)
j –12 – (–4)
k –15 – (–20)
l –30 – (–100)
i –11 – 4 Example 10
Ex
3
Evaluate:
a 15 – 26 + (–25)
b –10 – 12 + 8
c –39 + 54 – 1
d 31– 41 – (–9)
e 6 + 12 – 16
f –28 – (–35) – (–2)
g –36 – 17 + 26
h 5 – (–21) + 45
i 16 + (–4) – (–4)
j –92 + 54 – (–82) k –900 + 1000 – (–100) l –500 + 2000 – (–50)
14
ICEEM Mathematics Secondary 1B
Example 12
4
Write the answers to these subtractions.
a 234 – (–200) b –789 – (–560) c –654 – (–789) d –9856 – (–3455)
5
Evaluate:
a 45 – 50
b 30 – (–5)
c 60 – (–5)
d 4 – 11 – 21 + 40
e 12 – 20 + 30
f 7 – 10 – 20
g 7 – (–15) + 20
h –11 – 10 – (–4)
i –30 + 50 – 45 – (–6)
j –34 + 60 – (–5) + 10
k 43 + 50 – (–23)
l –10 – 45 + 30
6
a What is the distance on the number line between the points in each of the following pairs? (Draw a number line and mark the points on it as part of your answer.)
i 3, –5
ii –4, –12
b Verify that –5 < 3, and that the difference 3 – (–5) is equal to the distance between the points –5 and 3 on the number line. c Verify that –12 < –4, and that the difference –4 – (12) is equal to the distance between –12 and –4 on the number line. d By choosing other pairs of numbers and marking them on the number line, verify that the following statements are always true. Make sure you include some negative numbers in your choices. • If we subtract a smaller number from a larger one, then the answer is the distance between them on the number line. • If we subtract a larger number from a smaller number, then the answer is –(the distance between them on the number line). Example 11
7
The temperature in Moscow on a winter’s day went from a minimum of –19°C to a maximum of –2°C. By how much did the temperature rise?
8
The temperature in Ballarat on a very cold winter’s day went from –3°C to 7°C. What was the change in temperature?
Chapter 11 Integers
15
9
The table below shows minimum and maximum temperatures for a number of cities. Complete the table. Minimum (°C)
Maximum (°C)
–10
5
–15
5
–25
–3
–20
–15
–7
–5
–11
2
–13
5
Increase (°C)
10 A meat pie in the microwave rises in temperature by about 9°C for each minute of heating. If you take a frozen meat pie out of the freezer, where it has been stored at –14°C, how long does it have to be in the microwave before it reaches 40°C? 11 The temperature in Canberra on a very cold day went from 11°C to –3°C. What was the change in temperature? 12 The table below shows the temperatures inside and outside a building on different days. Day
Temperature inside (°C)
Temperature outside (°C)
M
20
25
T
13
18
W
24
20
T
10
−5
F
−5
−10
S
3
−6
For each day, calculate (Temperature inside – Temperature outside). What does it mean if the result of this calculation is negative?
13 Jane has just received her first credit card, and has already used it to buy some clothes. The balance is –$140. She spends another $70 at the grocery store the next day. At the end of the week, she will be paid $280. If she uses this to pay off her credit card, how much will Jane have left?
16
ICEEM Mathematics Secondary 1B
11D Multiplication and division involving negative integers Multiplication with negative integers 5 × (–3) can be defined as 5 lots of –3 added together. This means that 5 × (–3) = (–3) + (–3) + (–3) + (–3) + (–3) = –15. Just as 8 × 6 = 6 × 8, we will take –3 × 5 to be the same as 5 × (–3). All products like 5 × (–3) and –3 × 5 are treated in the same way. For example: –6 × 3 = 3 × (–6) = –18 –15 × 4 = 4 × (–15) = –60 –5 × 10 = 10 × (–5) = –50 The question remains as to what we might mean by multiplying two negative integers together. We first investigate this by looking at a multiplication table. In the lefthand column below, we are taking multiples of 5. The products go down by 5 each time. In the righthand column, we are taking multiples of –5. The products go up by 5 each time.
3 × 5 = 15
3 × (–5) = –15
2 × 5 = 10
2 × (–5) = –10
1 × 5 =
5
1 × (–5) = – 5
0 × 5 =
0
0 × (–5) =
0
–1 × 5 = –5
–1 × (–5) =
?
–2 × 5 = –10
–2 × (–5) =
?
Chapter 11 Integers
17
The pattern suggests that it would be natural to take –1 × (–5) to equal 5 and –2 × (–5) to equal 10 so that the pattern continues in a natural way. All products like –5 × (–2) and –5 × (–1) are treated in the same way. For example: –6 × (–2) = 12 –3 × (–8) = 24 –20 × (–5) = 100 So, we have the following rules.
The sign of the product of two integers • The product of a negative number and a positive number is a negative number. For example, – 4 × 7 = –(4 × 7) = –28. • The product of two negative numbers is a positive number. For example, – 4 × (–7) = 4 × 7 = 28.
Example 13 Evaluate each of these products. a 3 × (–20)
b –6 × 10
c –25 × (–30)
d 15 × (–40)
e –12 × 8
f –40 × (–8)
Solution a 3 × (–20) = –60 b –6 × 10 = –(6 × 10) = –60 c –25 × (–30) = 25 × 30 d 15 × (–40) = –(15 × 40) = 750 = –600 f –40 × (–8) = 40 × 8 e –12 × 8 = –(12 × 8) = –96 = 320
18
ICEEM Mathematics Secondary 1B
Division involving negative integers Every multiplication statement, for nonzero numbers, has an equivalent division statement. For example, 7 × 3 = 21 is equivalent to 21 ÷ 3 = 7. We will use this fact to establish the rules for division involving integers. Here are some more examples: 7 × 6 = 42 is equivalent to 42 ÷ 6 = 7.
7 × (–6) = –42 is equivalent to –42 ÷ (–6) = 7.
×6 7
× (–6) 42
7
÷6
÷ (–6)
–7 × 6 = –42 is equivalent to –42 ÷ 6 = –7.
–7 × (–6) = 42 is equivalent to 42 ÷ (–6) = –7.
×6 –7
–42
× (–6) –42
–7
÷6
42
÷ (–6)
The sign of the quotient of two integers • The quotient of a positive number and a negative number is a negative number. For example, 28 ÷ (–7) = –4. • The quotient of a negative number and a positive number is a negative number. For example, –28 ÷ 7 = –4. • The quotient of two negative numbers is a positive number. For example, –28 ÷ (–7) = 4. Chapter 11 Integers
19
Notice that the rules for the sign of a quotient are the same as the rules for the sign of a product.
Ex
Example 14 Evaluate each of these divisions. a –45 ÷ 9
b –20 ÷ (–4)
c 63 ÷ (–9)
Solution
Ex
a –45 ÷ 9 = –5
b –20 ÷ (–4) = 5
c 63 ÷ (–9) = –7
As before, we use another way of writing division. For example, –16 ÷ 2 can be written as –16 . 2
Example 15 Evaluate: a –45 9
b –36
c 60
b –36 = 9
c 60 = –5
–4
–12
Solution a –45 = –5 9
–4
–12
Exercise 11D Example 13
1
Calculate each multiplication.
a 5 × (–2)
b 6 × (–2)
c 5 × (–1)
d 11 × (–4)
e 12 × (–16)
f –5 × 2
g –6 × 2
h –5 × 10
i –11 × 4
j –12 × 4
k –20 × (–6)
l 16 × (–3)
m –7 × (–18)
q –17 × (–9)
20
n –13 × (–13) o –19 × 8
p 15 × (–4)
r –6 × (–17)
t –12 × (–15)
ICEEM Mathematics Secondary 1B
s –14 × 20
Example 14
2
Calculate each division.
a –15 ÷ 3
b –26 ÷ 2
c –35 ÷ 7
d –21 ÷ 3
e –120 ÷ 3
f 15 ÷ (–3)
g 36 ÷ (–2)
h 45 ÷ (–5)
i 21 ÷ (–7)
j 456 ÷ (–1)
k –51 ÷ (–3)
l –72 ÷ (–12)
m –100 ÷ (–50) n –121 ÷ 11
o –64 ÷ (–4)
p –144 ÷ (–6)
q –39 ÷ (–13) r –500 ÷ (–10) s –162 ÷ 6 Example 15
3
Evaluate:
a 5
4
Calculate each division.
a –48
e 132
c 6
d 8
b –52
c –60
d –112
f –600
g –225
h 292
i –80
j 696
k –196
l –1000
m –144 6 5 Evaluate:
n 256
o –98
p –288
–1 e –1 –1 i –10 2
–12
b –5
t –396 ÷ 11
–1 f 1 –1 j 12 –3
–2 g –50 –1 k –9 –3
13
–4
5
10
–24 –4
12
15
14
–7
–4 h –2 1 l –6 6
–8
–4
–100 –16
a 3 × (–2) × (–6)
b –4 × (–7) × (–6)
c 60 × (–4) × (–10)
d –45 × (–7) × 20
e 45 × (–10) ÷ 3
f 6 × (–10) ÷ 5
g 45 × (–3) × (–20)
h –34 × (–3) × (–2)
i –10 × 20 ÷ (–5)
k 16 ÷ (–8) × (–25)
l –36 ÷ (–9) × (–12)
j –5 × 12 ÷ (–6) 6
Copy and complete these multiplications and divisions.
a 2 × … = –30
b 5 × … = –65
c –7 × … = 42
d … × (–8) = 56
e 50 ÷ … = –10
f –45 ÷ … = 9
g –312 ÷ … = 3
h 5664 ÷ … = –708
i 2685 ÷ … = –895
k … × 9 = –126
l … ÷ (–13) = –13
j … × (–15) = 255
Chapter 11 Integers
21
11E Indices and order of operations You need to be particularly careful with the order of operations when working with negative integers. For example, –4 2 = –16 and (–4)2 = 16. In the first case, 4 is first squared and then the opposite is taken. In the second case, –4 is squared. Notice how different the two answers are. Remember that multiplication is done before addition. For example: and –3 × (5 + 2) = –3 × 7 –3 × 5 + 2 = –15 + 2 = –21 = –13 The same general rules that we have previously found for whole numbers also apply when dealing with negative integers.
Order of operations • Evaluate expressions inside brackets first. • In the absence of brackets, carry out operations in the following order: – powers – multiplication and division from left to right – addition and subtraction from left to right.
Example 16 Evaluate:
22
a (–6)2
b –62
d 6 × (–2) + 8
e –2 × (–6) + 10 – 15
f –20 ÷ 2 × 10
g 2 × (–4) ÷ 8
ICEEM Mathematics Secondary 1B
c –6 – 5 + 4
Solution a (–6)2 = –6 × (–6) b –6 2 = –(6 × 6) = 36 = –36 c –6 – 5 + 4 = –11 + 4 d 6 × (–2) + 8 = –12 + 8 = –7 = –4 e –2 × (–6) + 10 – 15 = 12 + 10 – 15 = 22 – 15 =7 f –20 ÷ 2 × 10 = –10 × 10 g 2 × (–4) ÷ 8 = –8 ÷ 8 = –100 = –1
Example 17 Evaluate: a 3 × (–6 + 8)
b –3 + 6 × (7 – 12)
c 6 – (5 + 4)
d 6 × ( –2 + 8)
e –2 × (–6 + 10) – 15
f 3 × (–6) + 3 × 8
g 3 × (–62) + 2 × 21
Solution a 3 × (–6 + 8) = 3 × 2 b –3 + 6 × (7 – 12) = –3 + 6 × (–5) = 6 = –3 + (–30) = –33 c 6 – (5 + 4) = 6 – 9 d 6 × (–2 + 8) = 6 × 6 = –3 = 36 e –2 × (–6 + 10) – 15 = –2 × 4 – 15 = –8 – 15 = –23 f 3 × (–6) + 3 × 8 = –18 + 24 =6 g 3 × (–62) + 2 × 21 = 3 × (–36) + 42 = –108 + 42 = –66 Chapter 11 Integers
23
Example 18 Evaluate: a 4 × (–6) ÷ 2 + 3
b –7 + 36 ÷ (–2)2 + 4
Ex
Solution a 4 × (–6) ÷ 2 + 3 = –24 ÷ 2 + 3 = –12 + 3 =–9 b –7 + 36 ÷ (–2)2 + 4 = –7 + (36 ÷ 4) + 4 =–7+9+4 =6
Exercise 11E Example 16
1
Evaluate:
a (–8)2
e (–10)2 × –(3)2 f (–12)2
i (–2)5
b –(11)2
c 2 × (–4)2
d –9 × (–3)2
g (–5)3
h (–2)4
j (–2)6
k (–1)3
l (–1)4
2
Evaluate:
a 2 × (–2)6
b 3 × (–2)5
c 4 × (–4)3
d 5 × (–2)2
e 3 × (–4)2
f 2 × (–1)5
g 4 × (–3)3
h 7 × (–1)23
3
Evaluate:
a –6 + 20 – 15
b –4 – (–10) + 20
c –6 + 12 – 15
d –4 + 11 – (–15)
e –15 + 7 – 8
f 65 – (–34) + 50
g –12 + 20 – 50
h –50 – 23 – 47
i –20 – (–25) + 60
4
Evaluate:
a –3 × (–16) + 8
c –6 – 18 × 4
24
ICEEM Mathematics Secondary 1B
b –4 + 6 × 11 – 14 d –6 × (–3) + 12
Ex
Example 17
e –2 × (–6 + 16) – 25
f –15 + 5 × (–3) + 12
g –11 + 5 × 12 + (–15)
h –18 – 4 × 26 – (–12)
5
Evaluate:
a –(3 – 17)
b –(27 – 54)
d – 43 + (6 – 11)
e 15 – 21 + 4 × (–3) f –3 × (56 – 87)
g –14 × (2 – 11)
h 5 × (13 – 41)
j (34 + 34) – (–5) × (–120) 6
Evaluate:
a –3 × (–16 + 8)
c –6 – (15 + 4)
i –7 × (11 – 18)
k (50 + 70) × (–3) – 5 × (–2)
b –4 + 6 × (11 – 12) d –6 × (–2 + 12)
e –2 × (–6 + 16) – 20
f –89 + 5 × (–32 + 12)
g –71 + 5 × (51 + (–35))
h –18 – 4 × (26 – (–12))
7
Evaluate:
a 40 ÷ (–5) ÷ 8
b 80 × (–3) ÷ 10
c 50 ÷ 10 × 2
d 60 × (–5) ÷ 25
8
Evaluate:
a (–10)2 + 2 × (–10)
c 2 × (–10)3 + 102 Example 18
c 12 + (4 – 16)
b (–10)2 × (–10)3 d –2 × (–10)2 × (–10)
9
Evaluate:
a 3 × (–12) ÷ 4 + 1
b –5 + 49 ÷ (–7)2 + 2
c –4 × 6 ÷ 8 – 5
d 3 – 50 ÷ (3 – 8)2 – 2
e 14 – 3 × 6 ÷ (–2)
f 7 – 32 × (1 – 3)2
g 5 × (–14) ÷ (–7) – 3
h 16 + 12 ÷ (–2)2 – 4
10 A shop manager buys 200 shirts at $16 each and sells them for a total of $3000. Calculate the total purchase price, and subtract this from the total amount gained from sales. What does this number represent?
Chapter 11 Integers
25
11 A man puts $1000 into a bank account every month for 12 months. Initially, he had $3000 in the bank. a How much does he have in the account at the end of 12 months, given that he has not withdrawn any money? b At the end of the 12 months, he writes a cheque for $20 000. How much does he now have left? 12 A pizza delivery van costs $200 a day to deliver pizzas from the pizza shop to its customers. Each pizza costs $3 to make and sells for $9. a If the pizza shop delivers 90 pizzas in a day, how much money does the pizza shop make? b The price of a pizza is increased to $10 and the cost of making a pizza is unchanged. How much money does the pizza shop make if 90 pizzas are delivered? c If the price of a pizza is decreased to $8 and the cost of making it increases to $4, how much does the pizza shop make or lose if it delivers 45 pizzas in a day? 13 The local council is planning to run a fair, and are trying to decide how much to charge for entry. The hall where they are planning to hold it will cost them $500 to rent for the day. They plan to charge $5 per person for entry, and to give each person a show bag that costs $2 to produce. a If 120 people come to the fair, how much money will the council make? b If they decide instead to charge $8 per person, and 120 people attend, how much will they make or lose? c If they charge $5 per child and $8 per adult, and 60 children and 60 adults attend, how much will they make or lose?
26
ICEEM Mathematics Secondary 1B
Review exercise 1
Complete each addition.
a 25 + (–2)
b –36 + 22
c –35 + 50
d –51 + (–44)
e –32 + 16
f –45 + (–23)
g –160 + (–20)
h –50 + (–10)
i 110 + (–40)
j –120 + 40
k 35 + (–3)
l –72 + 22
m –75 + 50
n –91 + (–44)
o –65 + 59
p –60 + (–25)
q –165 + (–25)
r –55 + (–10)
s 115 + (–45)
t –125 + 43
u –332 + (–215)
2
In an indoor cricket match, a team has made 25 runs and lost 7 wickets. What is the team’s score? (A run adds 1 and a wicket subtracts 5.)
3
The temperature in June at a base in Antarctica varied from a minimum of –60°C to a maximum of –35°C. What was the value of:
a maximum temperature – minimum temperature?
b minimum temperature – maximum temperature?
4
The temperature in Canberra had gone down to –3°C. The temperature in a heated house was a cosy 22°C. What was the value of:
a inside temperature – outside temperature?
b outside temperature – inside temperature?
5
Complete each multiplication.
a 125 × (–2)
b –36 × 11
c –35 × 50
d –51 × (–40)
e –3 × 16
f –50 × (–23)
g –160 × (–20)
h –50 × (–10)
i 11 × (–40)
k –20 × (–5)
l –25 × (–4)
j –120 × 20
Chapter 11 Integers
27
6
Complete each division.
a 125 ÷ (–5)
b –36 ÷ 9
c –35 ÷ 5
d –51 ÷ (–3)
e –16 ÷ (–4)
f –50 ÷ (–10)
g –160 ÷ (–20)
h –1500 ÷ (–10)
i 110 ÷ (–40)
k –196 ÷ (–14)
l 625 ÷ (–25)
j –120 ÷ 20 7
Evaluate each expression.
a –4 × (6 – 7)
b 7 × (11 – 20)
c –3 × (5 + 15)
d –6 × (–4 – 6)
e –12 × (–6 + 20)
f –(–4)2
g (3 – 7) × (11 – 15) h (10 – 3) × (–3 + 10) i (–5 – 10) × (10 – 4)
8
Start with the number –5, add 11 and then subtract 20. Multiply the result by 4. What is the final result?
9
Start with –100, subtract 200 and then add –300. Divide the result by 100. What is the final result?
10 Evaluate:
a (–8)2
c –11 – 15 + 14
b –82 d 16 × (–2) + 10
e –3 × (–8) + 100 – 150
f –200 ÷ 2 × 10
g 2 × (–6) ÷ 8
h 4 × (–6) ÷ (–3)
11 Evaluate:
a 5 × (–7 + 18)
c 16 – (15 + 14)
28
b –13 + 16 × (7 – 12) d 16 × (–12 + 8)
e –3 × (–16 + 20) – 25
f 3 × (–8) + 3 × 18
g 5 × (–72) + 3 × 42
h 7 × (–3)2 + 3 × (–4)
ICEEM Mathematics Secondary 1B
Challenge exercise 1
What is the smallest product you could obtain by multiplying any two of the following numbers:
–8, –6, –1, 1 and 4?
2
Evaluate:
a (–1)1000
3
The integers on the edges of each triangle below are given by the sum of integers which are to be placed in the circles. Find the numbers in the circles.
a
b (–1)1001
–1
b
–6
–1
–9
–6
c
0
–18
d
–18
–4
–34
4
–7
–10
Put the three numbers 4, –2 and –7 into the boxes below +
–
=
so that the answer is:
a –1
b 9
c –13
Chapter 11 Integers
29
5
Put the three numbers 5, –5 and –4 into the boxes below
+
–
=
so that the answer is:
a 6
6
Find the number that must be placed in the box to make the following statement true.
3 –
7
Place brackets in each statement below to make the statement true.
a 5 + (–3) × 3 + 4 = 14
b 5 + (–3) × 3 + 4 × 2 = 4
c 5 – 5 × 6 + 7 × 6 – 5 = 37
8
This is a magic square. All rows, columns and diagonals have the same sum. Complete the magic square. 2
b –14
c 4
+ (–5) = 0
–5
0
–4
9
a Find the value of 2 – 4 + 6 – 8 + 10 – 12 by:
i working from left to right ii pairing the numbers ((2 – 4) + (6 – 8) + (10 – 12)). b Evaluate 2 – 4 + 6 – 8 + 10 – 12 + 14 – 16 + … + 98 – 100. 10 Evaluate 100 + 99 – 98 – 97 + … + 4 + 3 – 2 – 1. 11 The average of five numbers was 2. If the smallest number is deleted, the average is 4. What is the smallest number? 12 Find the value of:
30
a (1 – 3) + (5 – 7) + (9 – 11) + (13 – 15) + (17 – 19)
b 1 – 3 + 5 – 7 + 9 – 11 + … + 101 – 103 ICEEM Mathematics Secondary 1B
Chapter 12 Algebra and the number plane This chapter deals with the substitution of negative numbers into algebraic expressions. The following example illustrates why this is important. In the United States, temperature is measured on the Fahrenheit scale, while in Australia we use the Celsius scale. It is useful to be able to convert from one scale to the other. For example, if the temperature in a town in the US is –5°C, what is the temperature in degrees Fahrenheit (°F)? The rule for converting to °F is to multiply the Celsius temperature 9
value by 5 and add 32 to the result. 9 We can write this in algebraic notation as F = 5 C + 32, where F and C are the temperature values in the Fahrenheit and Celsius scales, respectively. Substituting C = –5 gives 9
F = 5 × (–5) + 32 = –9 + 32 = 23. We have used substitution to find that a Celsius temperature of –5°C corresponds to a Fahrenheit temperature of 23°F. In this chapter, we will also see how such situations can be illustrated on the number plane.
Chapter 12 Algebra and the number plane
31
12A Substitution with integers In Chapter 3, we substituted positive whole number values for pronumerals. In Chapter 7, we substituted positive fractions. We will now look at how to substitute negative integer values, as illustrated in the following examples.
Example 1 Evaluate each expression for x = –5. a 4x + 3
b –4x + 3
c 4(x + 3)
d –4(x + 3)
e –4x2
f (–4x)2
Solution a 4x + 3 = –20 + 3 b –4x + 3 = 20 + 3 = –17 = 23 c 4(x + 3) = 4 × (–2) d –4(x + 3) = –4 × (–2) = –8 =8 e –4x2 = –4 × 25 f (–4x)2 = (20)2 = –100 = 400
Example 2 Evaluate each expression for m = –5, n = 6 and p = –10. a m + n
b m + p
c m – p
d mp
e np
p
f m
Solution a m + n = –5 + 6 b m + p = –5 + (–10) c m – p = –5 – (–10) = –15 = –5 + 10 = 1 =5 p
f m = –10 d mp = –5 × (–10) e np = 6 × (–10) –5 = 50 = –60 = 2
32
ICEEM Mathematics Secondary 1B
Example 3 Angelo has $100 in a bank account. He takes $x from the bank account every day. How much does he have in the account after: a 1 day?
b 4 days?
Solution a Amount left = $(100 – x) b Amount left = $(100 – 4x)
Example 4 The temperature is now 12°C. a What is the new temperature if the temperature drops by 15°C? b What is the new temperature if the temperature drops by x°C? c Find the new temperature if: i x = 10
ii x = 20
Solution a New temperature = 12°C – 15°C = –3°C b New temperature = (12 – x)°C c i If x = 10, new temperature = 12°C – 10°C = 2°C ii If x = 20, new temperature = 12°C – 20°C = –8°C
Chapter 12 Algebra and the number plane
33
Example 5
Ex
Christina has $100 in a bank account. She takes $x from the bank account every day. a How much money does she have in the account after 4 days? b How much does she have left in the account after 4 days if: i x = 10?
ii x = 20?
iii x = 25?
c Interpret the outcome in words if x = 30.
Solution a Amount left = $(100 – 4x) b i Amount left = 100 – 40 = $60 ii Amount left = 100 – 80 = $20 iii Amount left = 100 – 100 = $0 c Amount = 100 – 120 = –$20 Christina has overdrawn her account by $20.
Exercise 12A Example 1
34
1
Evaluate each expression for x = –2.
a 2x e 2x + 3 i 3 – x
2
Evaluate each expression for x = –30.
a 2x + 3
b –x + 6
c 2x – 4
d 5 – x
e 6 – 2x
f 5 – 4x
g – x
h x – 4
b –x f x3 j 3 – 2x
ICEEM Mathematics Secondary 1B
c x + 2 g –x3 k 5 + 2x
2
d x – 3 h (–x)2 l 2 – 5x
2
Example 2
3
Substitute m = –4, n = 3 and p = –24 to evaluate:
a m + n
b m + p
e np
f m
4
Given that m = –15, n = 6 and p = –5, evaluate:
a m + n d mp
5
Evaluate each expression for x = –2.
a 5x + 4 d –5(x + 4)
6
Evaluate each expression for a = –3.
a 5 + 2a d 4 – a g (–2a)2
7
Evaluate each expression for x = –10.
a 5x + 6 d –5(x + 6)
8
Evaluate each expression for x = –1.
a 6 – x
b 6 + x
c x3
d x5
e –x2
f (–2x)2
g –2x2
h 5 – 2x
9
Substitute a = –5, b = –2 and c = 5 to evaluate:
a a + b d bc
p
c m – p
d mp
g mnp
h n
b m + p e np
p
c m – p p f m
b –5x + 4 e –5x2
c 5(x + 4) f (–5x)2
b 6 – 3a e (–a)3 h (–2a)3
c 2a + 3 f –a 3 i a3 + 2
b –5x + 6 e –5x2
c 5(x + 6) f –4(x – 5)
b c + a e ac
c b – c f ac
10 Evaluate each expression for z = –3.
a –3z d (2z)3
b z 4 e (–z)3
c 5 – 2z f –2z 2
11 Substitute m = –20, n = –10 and p = 50 to evaluate:
a m + n
b p + m
c n – p
d mn
e mp
f m p
g m
p
h mnp
Chapter 12 Algebra and the number plane
35
12 Evaluate each expression for w = 10.
a –40w d (–2w)3
b –w 4 e (–w)3 + w2
c 10 – 2w f w 3 – 10w 2
13 Evaluate each expression for w = –2.
Example 3,5
a –20w d (–2w)3
b –w 4 e (–w)3 + w2
c 10 – 2w f w 3 – 10w 2
14 Buffy has $1000 in a bank account. She takes $x from the bank account every day. a How much money does she have in the account after: i 1 day?
ii 5 days?
b Find the value of her bank account after 5 days if: i x = 100 Example 4
ii x = 200
iii x = 250
15 The temperature in a room drops by x°C every hour. The temperature in the room at 12:00 pm is 25°C. a What will the temperature be at: i 1:00 pm?
ii 6:00 pm?
b If x = 6, what will the temperature be at 5:00 pm?
12B The number plane We have previously represented numbers as points on the number line. This idea can be extended by using a plane called the number plane. We start with two perpendicular straight lines. They intersect at a point O called the origin. Each of the lines is called an axis. The plural of axis is axes.
36
ICEEM Mathematics Secondary 1B
O
Next we mark off segments of unit length along each axis, and mark each axis as a number line with 0 at the point O. The arrows are drawn to show that the axes extend infinitely, in both directions 4 3 2 1
–4 –3 –2 –1 0 –1
O 1
2
3
4
5
–2 –3 –4
The axes are called the coordinate axes or sometimes the Cartesian coordinate axes. They are named after the French mathematician and philosopher René Descartes (1596–1650). He introduced coordinate axes to show how algebra could be used to solve geometric problems. Although the idea is simple, it revolutionised mathematics. Now we add vertical and horizontal lines to the diagram through the integer points on the axes. We can describe each point where the lines meet by a pair of integers. This pair of integers is called the coordinates of the point. The first number is the horizontal coordinate and the second number is the vertical coordinate. For example, the coordinates of the point labelled A below are (1, 4). This is where the line through the point 1 on the horizontal axis and the line through the point 4 on the vertical axis meet. We move 1 unit to the right of the origin and 4 units up to reach A. The point D has coordinates (2, –4). We move 2 units to the right of the origin and 4 units down to get D. The point B has coordinates (–2, 1). We move 2 units to the left of the origin and 1 unit up to get B. The point C has coordinates (–4, –3). We move 4 units to the left of the origin and 3 units down.
A
4 3 2
B –4 –3 –2 –1
1 0
1
2
3
4
–1 –2
C
–3 –4
D
Chapter 12 Algebra and the number plane
37
Example 6 On a number plane, plot the points with the given coordinates. a A(–2, 2)
b B(–3, 0)
c C(1, –1)
d D(–3, 3)
e E(0, 3)
f F(3, –2)
Solution y 4
D
3
A
E
2 1
B
–4 –3 –2 –1 0 –1 –2
1
2
3
4
x
C F
–3 –4
Remember, the first coordinate tells us where to go from the origin in the horizontal direction. If it is negative, we go to the left of the origin; if it is positive, we go to the right of the origin. The second coordinate tells us where to go from the origin in the vertical direction. If it is negative, we go below the origin; if it is positive, we go above the origin. 4
First coordinate (xcoordinate)
3
(4, 2)
2
Second coordinate (ycoordinate)
1
–4 –3 –2 –1 0 O 1 –1
2
3 4
5
–2 –3 –4
The first coordinate is usually called the xcoordinate and the second coordinate is usually called the ycoordinate.
38
ICEEM Mathematics Secondary 1B
Example 7 Plot the following points on the grid and join them in the order they are given to complete the picture. y 4 3 2 1 –4 –3 –2 –1 0
1
2
3
4
x
–1 –2 –3 –4
Shape 1: Join (2, 4) to (4, –1) to (2, –1) to (2, 4). Shape 2: Join (1, 4) to (1, –1) to (–4, –1) to (1, 4). Shape 3: Join (5, –2) to (4, –3) to (–2, –3) to (–5, –2) to (5, –2).
Solution y 4 3 2 1 –5 –4 –3 –2 –1 0
1
2
3
4
5
x
–1 –2 –3 –4
Chapter 12 Algebra and the number plane
39
Example 8
Ex
Write down the coordinates of the points labelled A to G in the following diagram of a house. y 3
A G
2
F
1
–6 –5 –4 –3 –2 –1
0
1
2
–1
E
B C 3
4
x
5
D
Ex
–2 –3 –4 –5
Solution A(2, 2), B(3, 2), C(3, 1), D(5, –1), E(–6, –1), F(–4, 1), G(2, 1)
Exercise 12B 1
Give the coordinates of points A to G.
y 6
F
5 4
C
A
3 2 1
D
–5 –4 –3 –2 –1
0 –1
E
1
2
3
4
6
B
–2 –3 –4
40
5
ICEEM Mathematics Secondary 1B
G
7
x
Ex
Example 6
Example 7
2
On a number plane, plot the points with the given coordinates.
a A(5, 1)
b B(–2, 4)
c C(–3, –3)
d D(3, –1)
e E(2, 4)
f F(0, –4)
g G(–5, 1)
h H(–5, –2)
3
On a number plane, plot the points with the given coordinates.
a A(–4, –1)
b B(–2, –3)
c C(2, –2)
d D(4, –1)
e E(–1, 4)
f F(–3, –2)
g G(–3, 2)
h H(0, –4)
4
For each of the following, plot the points on a grid and join them in the order they are given to draw a picture. a (2, 3), (2, 2), (−1, 2), (−1, 0), (1, 0), (1, −1), (−1, −1), (−1, −3), (2, −3), (2, −4), (−2, −4), (−2, 3), (2, 3). b (5, 0), (2, 1), (3, 3), (1, 2), (0, 5), (−1, 2), (−3, 3), (−2, 1), (−5, 0), (−2, −1), (−3, −3), (−1, −2), (0, −5), (1, −2), (3, −3), (2, −1), (5, 0). c Shape 1: Join (2, 4) to (2, 2) to (0, 2) to (0, 4) to (2, 4). Shape 2: Join (3, 0) to (2, 1) to (2, −1) to (0, −1) to (0, 1) to (−1, 0) to (−2, 1) to (1, 2) to (4, 1) to (3, 0). Shape 3: Join (2, −1) to (3, −4) to (2, −4) to (1, −2) to (0, −4) to (−1, −4) to (0, −1) to (2, −1).
Example 8
5
Write down the coordinates of the points labelled A to G in the following diagrams.
a
b
y
E F
y
4 3
G G
A
1
D
1
2
3
4
B x
F
1
–4 –3 –2 –1 0
–1
–1
–2
–2
–3 –4
B
3
A
2
2
–4 –3 –2 –1 0
4
C
E
1
2
D
3
4
x
C
–3 –4
Chapter 12 Algebra and the number plane
41
c
y 4
y 4
B
3 2
d
2
A
1
–1 –2
F
–3
1
2
3
4
C D
G E
–4
42
1
G
–4 –3 –2 –1 0
A
3
x
–4 –3 –2 –1 0
F
E
–1
1
B
2
3
4
x
C
–2 –3
D
–4
6
Draw coordinate axes and mark the integer points on it 1 cm apart.
a Plot the points A(0, 1), B(3, 1), C(3, 4) and D(0, 4), and join them to form AB, BC, CD and DA. Describe the shape formed and evaluate its area.
b Plot the points A(–2, 0), B(4, 0) and C(1, 4), and join them to form AB, BC and CA. Describe the shape formed and evaluate its area.
c Plot the points A(–4, –4), B(7, –4), C(7, 1) and D(–4, 1), and join the points to form AB, BC, CD and DA. Describe the shape formed and evaluate its area.
d Plot the points A(–6, 4), B(–1, 4) and C(–6, 1), and join them to form AC, CB and BA. Describe the shape formed and evaluate its area.
e Plot the points A(0, 2) and B(1, 4), and draw the line passing through them. Now plot the points C(–2, 5) and D(0, 4), and draw the line passing through these points. Describe the relationship between the lines.
f
Plot the points A(0, 1), B(4, 3), C(10, 3) and D(6, 1), and join them to form AB, BC, CD and DA. Describe the shape formed and calculate its area.
ICEEM Mathematics Secondary 1B
7
a On a grid, join (0, 0) to (3, 1) to (4, 2) to (4, 4) to (2, 4) to (1, 3) to (0, 0) to draw one petal of a flower. b Complete, and then plot, the following list of points to form a second petal the same shape as the first. Join (0, 0) to (3, −1) to (__, −2) to (4, __) to (__, __) to (__, __) to (0, 0). c Draw the remaining two petals of the same shape to complete the flower. d Write down the ordered list of points required to draw each of the petals in part c.
12C Completing tables and plotting points The following example shows how an understanding of the number plane can help us with algebra, and vice versa. Two students play a simple game to improve their multiplication of integers. Liam gives a number and Andrea multiplies it by 2. Liam starts at –2 and gives Andrea each integer up to 2. Their results are recorded in a table. We can write the rule as Andrea’s number = 2 × Liam’s number, and the table is: Liam’s number
–2
–1
0
1
2
Andrea’s number
–4
–2
0
2
4
If we denote Liam’s number by x and Andrea’s number by y, then we can write the rule as y = 2x, and the table can now be written as: x
–2
–1
0
1
2
y
–4
–2
0
2
4
Chapter 12 Algebra and the number plane
43
We can also plot the points in the table on the number plane, as shown below. y (2, 4)
4 3
(1, 2)
2 1
(0, 0)
–4 –3 –2 –1 0
1
2
3
4
x
–1
(–1, –2)
–2 –3
(–2, –4)
–4
The points (–2, –4), (–1, –2), (0, 0), (1, 2) and (2, 4) are plotted. What do you notice about these points? A line can be drawn through all of the points. Try it! We can follow the same kind of procedure for any similar rule – a table can be formed and the corresponding points plotted.
Example 9 For each given rule, complete the table, list the coordinates of the points, and plot the points on a number plane. a y = –x x
–3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
y
b y = x + 1 x
–3
y
44
ICEEM Mathematics Secondary 1B
Solution a y = –x x
–3
–2
–1
0
1
2
3
y
3
2
1
0
–1
–2
–3
The points are (–3, 3), (–2, 2), (–1, 1), (0, 0), (1, –1), (2, –2) and (3, –3). y 4 3
(–3, 3) (–2, 2) (–1, 1)
2 1
(0, 0)
–4 –3 –2 –1 0
1
–1 –2 –3
2
3
4
x (1, –1) (2, –2) (3, –3)
–4
b y = x + 1 x
–3
–2
–1
0
1
2
3
y
–2
–1
0
1
2
3
4
The points are (–3, –2), (–2, –1), (–1, 0), (0, 1), (1, 2), (2, 3) and (3, 4).
y 4 3 2
(–1, 0)
1
–4 –3 –2 –1 0
(3, 4) (2, 3) (1, 2) (0, 1) 1
2
3
4
x
–1 (–2, –1) (–3, –2)–2 –3 –4
Chapter 12 Algebra and the number plane
45
Exercise 12C Example 9
1
For each given rule, complete the table, list the coordinates, and plot the corresponding set of points on a number plane. Check that each set of points lies on a line. a y = 3x x
–3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
y
b y = –2x x
–3
y
c y = x – 2 x
–3
y
d y = x + 2 x
–3
y
e y = 2x + 1 x
–3
y
f y = 1 – x x
–3
y
g y = 3 – 2x x
–3
y
46
ICEEM Mathematics Secondary 1B
2
For each given rule, complete the table, list the coordinates, and plot the corresponding set of points on a number plane. a y = x + 1 2
–3
x
–2
–1
0
1
2
3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
y
b y = x – 1 2
–3
x y
c y = 2x + 1 2
–3
x y
d y = –x + 1 2
–3
x y
3
Complete the table for each given rule. a y = 5x – 7 0
x
1
2
3
4
5
6
–3
–2
–1
0
1
2
y
b y = 9 – 4x –4
x
3
4
y
4
Complete the table for the rule y = x2, list the coordinates, and plot the corresponding set of points on a number plane. Note that they do not lie on a line. x
–3
–2
–1
0
1
2
3
y
Chapter 12 Algebra and the number plane
47
12D Finding rules In the previous section we looked at completing tables and plotting the corresponding points. In this section we will find a rule corresponding to a table or a plot of points.
Example 10 Fill in the boxes to find a rule for each of the following tables. a
b
x
1
2
3
4
5
t
–2
–1
0
1
2
y
5
10
15
20
25
d
–4
–1
2
5
8
y=
×x+
d =
×t+
Solution a Pick two pairs to find a rule. When x = 1, y = 5, so
5 = 5 × 1 + 0
When x = 2, y = 10, so 10 = 5 × 2 + 0 In both cases,
y = 5 × x + 0
(Note that we have found a rule where we put a ‘5’ in the first box and a ‘0’ in the second box. Use the other pairs from the table to check that the rule is y = 5x.) When x = 3, y = 5 × 3 = 15. When x = 4, y = 5 × 4 = 20. When x = 5, y = 5 × 5 = 25. A rule for the table is y = 5x. b Pick two pairs to find a rule.
48
When t = 1, d = 5, so
5 = 3 × 1 + 2
When t = 2, d = 8, so
8 = 3 × 2 + 2
In both cases,
d = 3 × t + 2
ICEEM Mathematics Secondary 1B
(Note that we have found a rule where we put a ‘3’ in the first box and a ‘2’ in the second box. Use the other pairs from the table to check that the rule is d = 3t + 2.) When t = −2, d = 3 × (−2) + 2 = −4. When t = −1, d = 3 × (−1) + 2 = −1. When t = 0, d = 3 × 0 + 2 = 2. A rule for the table is d = 3t + 2.
Example 11 Tiles are formed into the letter ‘X’ as shown below.
Diagram 1
Diagram 2
Diagram 3
a Copy and complete the table below, where n is the number of the diagram. Diagram number (n)
1
2
3
Number of tiles (t)
5
9
13
4
5
6
b How does the number of tiles increase as we move from one diagram to the next? c Plot the points (n, t) for values of n from 1 to 6, using your table of values. d Write a rule that tells us the number of tiles we need for the nth diagram. (continued on next page)
Chapter 12 Algebra and the number plane
49
Solution a
Diagram number (n)
1
2
3
4
5
6
Number of tiles (t)
5
9
13
17
21
25
b We need an extra four tiles each time we make a bigger ‘X’. c
t 25 20 15 10 5
0
(6, 25) (5, 21) (4, 17) (3, 13) (2, 9) (1, 5) n 1 2 3 4 5 6
d Pick two pairs to find a rule. When n = 1, t = 5, so
5 = 4 × 1 + 1
When n = 2, t = 9, so
9 = 4 × 2 + 1
In both cases,
t = 4 × n + 1
(Use the other pairs to check that the rule is t = 4n + 1.) When n = 3, t = 4 × 3 + 1 = 13. When n = 4, t = 4 × 4 + 1 = 17. When n = 5, t = 4 × 5 + 1 = 21. When n = 6, t = 4 × 6 + 1 = 25. A rule for the number of tiles is t = 4n + 1.
Example 12 Plot the points (–2, 4), (–1, 2), (0, 0), (1, –2) and (2, –4) on a number plane and give a rule connecting the ycoordinate to the xcoordinate.
50
ICEEM Mathematics Secondary 1B
Ex
Solution y (–2, 4)
4 3
(–1, 2)
2 1
–4 –3 –2 –1 0
(0, 0) 1
2
3
4
x
–1
(1, –2)
–2 –3
(2, –4)
–4
The rule is y = –2x.
Exercise 12D Example 10
1
Fill in the boxes to give a rule for each of the following tables.
a
b
x
1
2
3
4
5
t
–2
–1
0
1
2
y
2
3
4
5
6
d
0
3
6
9
12
y =
×x+
d =
c
×t+
d
x
1
2
3
4
5
t
–2
–1
0
1
2
y
4
6
8
10
12
d
12
9
6
3
0
y =
×x+
d =
e
×t+
f
m
1
2
3
4
5
x
–2
–1
0
1
2
n
0
3
6
9
12
y
3
1
–1
–3
–5
n =
×m–
y=
×x–
Chapter 12 Algebra and the number plane
51
Example 11
2
A pile of matchsticks is used to make the following pattern of shapes. The first diagram uses three matches to form one triangle. The second diagram uses five matches to form two triangles.
Diagram 1 Diagram 2
Diagram 3
Diagram 4
a Count the number of matches used to make each diagram, and complete the table below. Number of triangles (t)
1
2
3
Number of matches (m)
3
5
7
4
5
6
b How many matches do we add each time to create an extra triangle? c Plot the points (t, m) for values of t from 1 to 6, using your table of values. d Write a rule that tells us the number of matches we need to make any number of triangles. 3
The first diagram shows four chairs placed around one square table. The second diagram shows six chairs placed around two square tables. The third diagram shows eight chairs placed around three square tables. Consider the number of chairs needed each time an extra table is added to the row.
Diagram 1
Diagram 2
Diagram 3
a Count the number of chairs used to make each diagram, and complete the table below.
52
Number of tables (t)
1
2
3
Number of chairs (c)
4
6
8
ICEEM Mathematics Secondary 1B
4
5
6
b What is the difference in the number of chairs each time a table is added? c Plot the points (t, c) for values of t from 1 to 6, using your table of values. d Write a rule that tells us the number of chairs we need to place around any number of tables. 4
Tommy the terrible two year old emptied the kitchen cupboards and used all the cans of food to make a tower as in Diagram 1. When his mother discovered what he had done, she noticed the tower was in the shape of an ‘L’. Not wanting to miss an opportunity to teach Tommy the alphabet, she proceeded to pack the cans away four at a time as shown in the following diagrams.
Diagram 1
Diagram 2
Diagram 3
a How many cans of food did Tommy use to build his first tower? b Count the number of cans used to make each tower, and complete the table below. Diagram number (n)
1
2
3
Number of cans (c)
40
36
32
4
5
6
c Plot the points (n, c) for values of n from 1 to 6, using your table of values. d Write a rule that tells us the number of cans needed to create each ‘L’. e How many different ‘L’s can Tommy’s mother make before the tower loses its ‘L’ shape? Chapter 12 Algebra and the number plane
53
Example 12
5
For each of the following, plot the points on a number plane and give a rule connecting the ycoordinate to the xcoordinate.
a (–2, –2), (–1, –1), (0, 0), (1, 1), (2, 2)
b (–2, –1), (–1, 0), (0, 1), (1, 2), (2, 3)
c (–2, –4), (–1, –2), (0, 0), (1, 2), (2, 4)
d (–2, –4), (–1, –3), (0, –2), (1, –1), (2, 0)
6
Sarah is given $1000 for her 18th birthday. She decides to use it to sponsor a child in Africa at a cost of $20 each month. a Complete the table below to show how much money Sarah has left at the end of each month. Month, m
0
1
Dollars, d
1000
980
2
3
4
5
6
b Write a rule to show how many dollars, d, Sarah has left after m months. c How much money will Sarah have after 10 months? d For how many months can Sarah sponsor the child? 7
Frank recently turned 16 and got his learner’s permit. His mother supervises him driving the family car to and from school each day, a trip which takes him 30 minutes each way. Frank is keeping a log of the total hours he has driven. The table shows the total number, h, of hours Frank has driven after w weeks. Week number (w)
1
2
3
Number of driving hours (h)
5
10
15
4
5
6
a Complete the table. b Write a rule that tells us the number of driving hours, h, after w weeks. c How many hours of driving will Frank have done after 12 weeks? d How many weeks driving will Frank need to do to complete 45 hours of driving?
54
ICEEM Mathematics Secondary 1B
8
Match each diagram with the correct rule from the list below.
y = 3x y = –2x y = x + 3 y = –2x – 1 y = –2x + 1 y = x – 4
a
y
b
y
8
8
6
6
4
4
2
2
–6 –4 –2 0 2 4 6
x
–6 –4 –2 0 2 4 6
–2
–2
–4
–4
–6
–6
–8
–8
c
y
e
d
8
6
6
4
4
2
2
x
–6 –4 –2 0 2 4 6
–2
–2
–4
–4
–6
–6
–8
–8
f
y
8
6
6
4
4
2
2
x
x
y
8
–6 –4 –2 0 2 4 6
y
8
–6 –4 –2 0 2 4 6
x
–6 –4 –2 0 2 4 6
–2
–2
–4
–4
–6
–6
–8
–8
x
Chapter 12 Algebra and the number plane
55
Review exercise 1
Given that m = –1, n = 2 and p = –6, evaluate:
a m + n b m + p
2
On a number plane, plot each of the points whose coordinates are given below.
a A(1, 1) d D(–4, 0)
3
Give the coordinates of each of the points A to G.
c m – p
d mp
p
e np
b B(2, –3) e E(–4, –2)
f m
c C(0, 6) f F(–4, 5)
y
A
4
B
3 2
G –4 –3 –2 –1
E
1 0
1
2
3
4
x
–1
C
–2
D
–3 –4
F
4
Make up a table with x values from –3 to 3 for each of the rules given below. List the corresponding coordinates and plot the points.
a y = 3 – x
5
Evaluate each expression for x = –3.
a 10 – x e –x2
6
David has $600 in a bank account. He takes $x from the account every week.
b 10 + x f (–5x)2
b y = 2x – 3
c x2 g –25x2
d x3 h 5 – 5x
a How much money does he have in the account after: i 1 week?
56
ii 5 weeks?
ICEEM Mathematics Secondary 1B
b Find the value of his bank account after 5 weeks if: i x = 100 7
ii x = 200
The temperature in a freezer drops by 2x˚C every hour after 6:00 pm until it reaches –5˚C. The temperature at 6:00 pm is 20˚C. a What will the temperature be at:
i 7:00 pm?
ii 11:00 pm?
b If x = 1 , what will the temperature be in 8 hours? 2
c If x = 2, what will the temperature be in 5 hours? d If x = 2 1 , when does the temperature reach –5°C? 2
8
ABCD is a square. The coordinates of A, B and C are (0, 0), (0, 6) and (6, 6), respectively. What are the coodinates of D?
9
ABCD is a rectangle. The coordinates of A, B and C are (1, –6), (1, 2) and (7, 2). What are the coordinates of D?
10 Fill in the boxes to give a rule for each of the following tables.
a
b
x
1
2
3
4
5
t
–2
–1
0
1
2
y
3
4
5
6
7
d
4
6
8
10
12
y =
×x+
d =
c
×t+
d
m
1
2
3
4
5
p
–2
–1
0
1
2
n
0
4
8
12
16
q
–5
–3
–1
1
3
n =
×m–
q =
e
×p–
f
t
1
2
3
4
5
x
–2
–1
0
1
2
d
12
9
6
3
0
y
–1
–4
–7
–10
–13
d =
×t+
y =
×x–
Chapter 12 Algebra and the number plane
57
Challenge exercise 1
David and Angela have 10 CDs to divide between them. a Copy and complete the following table showing how the CDs can be divided. CDs for David
0
1
10
CDs for Angela
10
9
0
Let x be the number of CDs that David has and y be the number of CDs that Angela has. b Write coordinates corresponding to each column of the table. c Plot these points on a number plane with the x and yaxes labelled from 0 to 10. d Write a rule for y in terms of x. 2
David and his twin brother Andrew are to share 10 CDs with Angela in such a way that the twins receive CDs in pairs, and have at least one pair of CDs. a Copy and complete the table below, showing how the CDs can be divided. Pairs of CDs for David and Andrew
5
4
Single CDs for Angela
0
2
Let x be the number of pairs of CDs that the twins receive, and y be the number of CDs that Angela receives. b Write coordinates corresponding to each column of the table.
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ICEEM Mathematics Secondary 1B
c Plot these points on a number plane, with the x and yaxes labelled from 0 to 10. d Write a rule for y in terms of x. 3
a ABCD is a square. A has coordinates (4, 5), D has coordinates (8, 5) and C has coordinates (4, 9). Find the coordinates of B. b OXYZ is a square. O is the origin and X is the point with coordinates (0, 5). Give the possible coordinates for the points Y and Z. c ABCD is a square. A has coordinates (0, 0) and B has coordinates (4, 4). Find the possible coordinates of C and D.
Note: In questions 4, 5 and 6, the number plane axes have markers at 1 cm intervals, that is, the point (1, 0) is 1 cm from the origin etc. 4
AB is an interval on the number plane. A has coordinates (5, 0) and B has coordinates (10, 0). Describe the points C such that triangle ABC has area 20 cm2.
5
AB is an interval on the number plane. A has coordinates (0, 4) and B has coordinates (0, 10). Points C and D are such that ABCD is a square of area 36 cm2. Find the possible coordinates of C and D.
6
AB is an interval on the number plane. A has coordinates (0, 4) and B has coordinates (0, 10). Points C and D are such that ABCD is a rectangle of area 42 cm2. Find the possible coordinates of C and D.
7
A very large garden grows pineapples and mangoes. The manager of the garden insists that the fruit is stacked as follows. • Mangoes are placed in stacks of 10. • Pineapples are placed in stacks of 5. a List all the different ways you can choose 30 pieces of fruit. Stacks of mangoes
Stacks of pineapples
3
0
…
…
0
6
Chapter 12 Algebra and the number plane
59
Let x be the number of mango stacks and y be the number of pineapple stacks. b List the coordinates (stacks of mangoes, stacks of pineapples). c Plot these points on a number plane.
d Write a rule for y in terms of x.
8
The admission prices to an agriculture show are
Adults: $9 each Children: $2 each.
A group of people arrives at the ticket counter and pays a total of $90.
Let x be the number of adults and y be the number of children in the group. a List the coordinates (x, y) which satisfy the rule 9x + 2y = 90. b Plot these points on a number plane. c Find the number of children and the number of adults if the total number of people in the group is: ii 31.
i 38 9
A square has vertices with coordinates O(0, 0), A(a, 0), B(a, a), C(0, a). a State the area S of the square in terms of a.
b Complete the table of values. a
1 2
1
1
1 2
2
2
1 2
S
c Plot these points on a number plane. Note that they do not lie on a line.
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ICEEM Mathematics Secondary 1B
Chapter 13 Triangles and constructions Geometrical constructions are an enjoyable and practical part of geometry. They have been used for centuries by builders and others. The constructions are based on results about triangles, so we begin with the geometry of triangles. The study of triangles was undertaken by the Babylonians as early as 3000 BC. They knew of methods of working out the areas of some triangles. The ancient Egyptians also worked on the measurement of side lengths and areas of triangles. In this chapter, we see that measurements of the lengths of the sides and the area of a triangle are not the only things to consider when studying triangles. It was the ancient Greeks who introduced a remarkably effective way of thinking about geometry, which is still important today.
13A Review of geometry Chapter 6 introduced angles and parallel lines. You will need to know the definitions of acute, obtuse and reflex angles and you should revise these before starting the chapter. Here is a quick review of the methods used in the problems of that chapter. Remember that using correct arguments in geometry is just as important as getting the right answers. Be as specific as possible, naming the relevant points or parallel lines. Chapter 13 Triangles and constructions
61
Angles at a point • Adjacent angles can be added. • Angles in a revolution add to 360˚. • Angles in a straight angle are supplementary (meaning that they add to 180˚). • Vertically opposite angles are equal.
Angles across transversals to parallel lines When a transversal crosses two parallel lines, pairs of corresponding, alternate and cointerior angles are formed.
Example 1 Find the value of the pronumeral. a
A 40˚
P
b
i
E B
A
130˚
P B
Q
c
H
F
d
B A
b
H
62
65˚ A
80˚ Q
P
Q
a
ICEEM Mathematics Secondary 1B
P
B
c b a
Q
Solution a i = 40˚ (corresponding angles, AB  PQ) b a = 130˚ (alternate angles, AB  PQ) c b + 80˚ = 180˚ (cointerior angles, AB  PQ), so b = 100˚ d c = 65˚ (vertically opposite), a = 65˚ (corresponding angles, AB  PQ), b = 115˚ (supplementary), Note: a and c are alternate angles. b and c are cointerior angles.
Transversals and angles Suppose that a transversal crosses two other lines. • If the lines are parallel, then the corresponding angles are equal. • If the lines are parallel, then the alternate angles are equal. • If the lines are parallel, then the cointerior angles are supplementary.
Proving that two lines are parallel The converses of these three results can be used to prove that two lines are parallel.
Proving that two lines are parallel Suppose that a transversal crosses two other lines. • If the corresponding angles are equal, then the lines are parallel. • If the alternate angles are equal, then the lines are parallel. • If the cointerior angles are supplementary, then the lines are parallel.
Chapter 13 Triangles and constructions
63
Exercise 13A Example 1
1
Find the value of the pronumeral. Give reasons for all your statements.
a
A
P
b
B 85˚
110˚ Q
a
P Q
B
F A
B
A
i
70˚
c
E
Q
b
P
H
H
2
Find a, b, c and/or i in each diagram below. Give reasons for all your statements.
a
b
C 50˚
B
P
M
O
L
d
S
25˚
e
S
72˚
Q
E G
U
g
A
h
35˚ N
b
B
i
A c
30˚
Z
64
i
J
R
40˚ O
E
ICEEM Mathematics Secondary 1B
L
S
82˚
T
105˚
a
62˚ a
b
c G
P
M
I
a i A
L
N
S
L
T
F
f 118˚
b
45˚
F
c R
X
b
100˚
a 20˚
G
58˚
i
A
c
T
M
H K
L
a b N c 30˚ O
3
In each diagram, identify two parallel lines, giving reasons. Hence find the size of the marked angle, ∠CAT, again giving reasons.
a
C
b
B
A
D
c
G
d
102˚ 107˚
C 50˚
A 50˚
A
C
B
102˚ D
38˚ N
T
A
125˚ T
O
K
C
E 125˚
80˚ T
70˚
D
T
13B Angles in triangles In this section, we will prove two useful results about the angles of any triangle. You may have seen these two results already, but proving them will probably be new to you.
Triangles A triangle is formed by taking any three noncollinear points A, B and C and joining the three intervals AB, BC and CA. These intervals are called the sides of the triangle, and the three points are called its vertices (the singular is vertex.) A
The triangle to the right is called ‘triangle ABC ’. This is written in symbols as ABC. B
C
Investigating the interior angles of a triangle The first important result about triangles is that the sum of the three interior angles of a triangle is always 180˚, whatever the triangle may look like. Here are three ways of checking this result. Chapter 13 Triangles and constructions
65
• Draw a number of differentlooking triangles, measure their three angles and check that their sum is 180˚. If you have set squares, they provide excellent examples of triangles.
60˚
45˚
30˚
45˚
90˚ + 45˚ + 45˚ = 180˚
90˚ + 60˚ + 30˚ = 180˚
• Cut out a triangle. Tear the corners off and place them together so that they form a straight angle.
b
a
c
b
a
c
• Cut out a triangle. Fold it without any tearing to demonstrate that the three interior angles form a straight angle.
b Fold up
a
c
a
b
c
Proving that the sum of the interior angles is 180˚ Doing measurements and experiments on any number of different triangles does not prove a general result – however many triangles you check, there are always more. Here is an argument that establishes the result for any triangle. The result and its proof are set out rather formally in the manner traditional for geometry. The statement of the result is called a theorem. This is a Greek word meaning ‘a thing to be gazed upon’ or ‘a thing contemplated by the mind’ – our word ‘theatre’ comes from the same root.
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ICEEM Mathematics Secondary 1B
Theorem: The sum of the interior angles of a triangle is 180˚. Given:
Let ABC be a triangle. Let ∠BAC = a, ∠B = b and ∠C = c.
Aim:
To prove that a + b + c = 180˚
Draw the line XAY parallel to BC through the vertex A. Proof: X A Y a
B
b
c C
Then ∠XAB = b (alternate angles, XY  BC ), and ∠YAC = c (alternate angles, XY  BC). Hence a + b + c = 180˚ (straight angle at A).
A shorter notation for angles In the Given section of the above proof, we referred to ∠B and ∠C, rather than to ∠ABC and ∠CBA. We can use this shorter notation because there is only one nonreflex angle at each of the vertices B and C. There are several angles at the vertex A, however, so we have to use the longer forms, like ∠BAC and ∠XAB, to show precisely which one we mean.
The exterior angles of a triangle Let ABC be a triangle, with the side BC produced to D. (The word ‘produced’ means ‘extended’.) Then the marked angle ∠ACD formed by the side AC and the extension CD is called an exterior angle of the triangle. The angles ∠A and ∠B are called the opposite interior angles, because they are opposite the exterior angle at C.
A
B
C
D
An exterior angle and the interior angle adjacent to it are adjacent angles on a straight line, so they are supplementary: ∠ACD + ∠ACB = 180˚ (straight angle at C ). Chapter 13 Triangles and constructions
67
There are two exterior angles at each vertex, as shown in the diagram below. Because the two angles are vertically opposite, they are equal in size: A ∠ACD = ∠BCE (vertically opposite angles at C ). B
D
C E
The exterior angle theorem The vital fact about exterior angles is that an exterior angle of a triangle is equal to the sum of the two interior opposite angles. This theorem can be proven using the angle sum of the triangle. The argument on the left below gives a particular case, while the argument on the right gives the general case. A
A
a
50˚ 70˚ B
b C
D
In the diagram above,
B
C
D
In the diagram above,
∠ACB + a + b = 180˚ ∠ACB + 50˚ + 70˚ = 180˚ (angle sum of ABC ) (angle sum of ABC ) ∠ACB = 180˚ − (50˚ + 70˚)
∠ACB = 180˚ − (a + b)
so ∠ACD = a + b so ∠ACD = 50˚ + 70˚ (straight angle at C ) (straight angle at C ). = 120˚. The theorem can also be proven without using the angle sum of a triangle result by drawing a parallel line. This is done below. Theorem: An exterior angle of a triangle equals the sum of the interior opposite angles.
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ICEEM Mathematics Secondary 1B
Given:
Let ABC be a triangle, with the side BC produced to D. Let ∠A = a and ∠B = b.
Aim:
To prove that ∠ACD = a + b
Proof:
Draw the ray CZ through C parallel to BA. A
Z
a
b B
C
D
Then ∠ZCD = b (corresponding angles, BA  CZ), and ∠ACZ = a (alternate angles, BA  CZ). Hence ∠ACD = a + b (adjacent angles at C).
Using these two theorems in problems These two theorems can now be used in geometrical problems.
Two theorems about the angles of a triangle • The sum of the interior angles of a triangle is 180˚. • An exterior angle of a triangle is equal to the sum of the interior opposite angles. Always be specific and name the triangles and angles involved.
Example 2 Find ∠A in the triangle opposite.
A 70˚
C
20˚ B
(continued on next page)
Chapter 13 Triangles and constructions
69
Solution
Ex
∠A + 20˚ + 70˚ = 180˚ (angle sum of ABC ), so
∠A = 90˚.
Example 3 Find i and a in the diagrams. a
A
b
P a
60˚ B
80˚ Q
i
65˚
120˚ R
S
C D
Solution a i = 60˚ + 80˚ (exterior angle of so i = 140˚. b a + 65˚ = 120˚ (exterior angle of so a = 55˚.
ABC ),
PQR),
Exercise 13B
70
1
a Draw a large triangle ABC. Then produce (extend) the side AB to D.
b Measure the three interior angles of the triangle and confirm that their sum is 180˚.
c Measure the exterior angle ∠CBD and confirm that it is the sum of ∠A and ∠C.
ICEEM Mathematics Secondary 1B
Ex
Example 2
2
Use the interior angle sum of a triangle to find a, b, c or i in each diagram below.
a
b
A
c
I
20˚
a 70˚
b B
30˚
B
e
J
M
a
i
T
81˚
E 60˚ Q
c T
60˚
P 37˚ I
L F
R
i A
B
R 41˚ G
i
C A
S
100˚
R
I L
T
O
Q
h
L
U
B
35˚ B
g
f
N
U
b
A
3
35˚
T
72˚ 85˚
Example 3
c
C
d
C
49˚
P C
E A 60˚
Q
b
a R
c
70˚
S
B
M
Use the exterior angle theorem to find a, b, c or i in each diagram below. a
b
A
E
60˚ 60˚
d A
70˚ 43˚
c
R
G
E
f
M W a
T
B
A
b
71˚ R I
O
S
E
N
37˚
i D
L
D
e
R
E
b
24˚
a C
N 121˚
T B
c
X
20˚
T S
G
81˚
L
Chapter 13 Triangles and constructions
71
g
h
T c 157˚
140˚
E G
Y P
J K L
C
77˚
20˚ i
M
4
Explain your answers to these questions.
a Can a triangle have two obtuse angles?
b Can a triangle have two right angles?
c What is the minimum number of acute angles a triangle can have?
d Can a triangle have an acute exterior angle?
e Can a triangle have two acute exterior angles?
5
Use the exterior angle theorem to find a, b, c and/or i in each diagram below. Give reasons.
a
b
P
O 112˚
c Y
b
d W
S
M a 135˚ L
40˚
f
Y
30˚ i
T c H
A Z
C 119˚ B X
151˚
L a
A b G
I a
L N 115˚
30˚
T b
V
h
F b G
20˚ G
H I
72
N
130˚
A b
i 65˚ H
e
120˚ A C
g
G
T
L
a 108˚ N
I
c
X
ICEEM Mathematics Secondary 1B
E a 20˚
125˚
T 32˚ D
F
63˚ 24˚
S
S
6
Find a, b, c or i in each diagram below.
a
b
A
c
P
X
b + 10˚
c 110˚ R
2a
a B
C
b
Q c
c
Y
d
e
F
2a A
W 100˚
i I
f T 40˚ Y
C
i
S
45˚
35˚ H
D
R I b
g
h
L
9c
3i
i K
J
120˚ S R
c T
7
Find a, b, c and/or i in each case, giving reasons.
a
H
b
O 35˚ a i
b
Y
a
L 70˚
50˚ c
115˚
A
C
25˚ a i
55˚
R T
e
T
P
f
R 32˚ c
b
c
D
U
b a A
I
75˚ A b
70˚ R
P
c
V L
E S
d
b S
U
2i
Z
F
R
70˚
65˚
G
a 48˚
A
I
115˚
c
b H
Chapter 13 Triangles and constructions
73
g F
a
h
G
40˚
S
95˚ c b I
T
65˚ M
8
Find the size of the marked angle, ∠AVB, in each case, giving reasons.
a
b
V
B
A
78˚
P
V 50˚
f
B
P
28˚
65˚
g
V
A
h
P
Q
60˚ 135˚ V 70˚
Q
P
A
Q
V
B 135˚
22˚
O
70˚
B
P
A
Q
Q
70˚
27˚
V B
R
Q
V
65˚
e
A
P
25˚
34˚ Q
25˚
R
A
A
d
B
60˚ 75˚
c
V P
80˚ Q
P
i V
P A
70˚ R
A
ICEEM Mathematics Secondary 1B
25˚
Q
40˚
B B
74
a A
30˚
W
R b c T
B
C S
13C Circles and compasses From this section on, you will also need compasses for your geometrical constructions. Make sure that your pencil is very sharp. ‘Compasses’ is a plural word. We use ‘a pair of compasses’, just as we wear ‘a pair of trousers’ and use ‘a pair of scissors’. The singular word ‘compass’ is the instrument that navigators use to find magnetic north.
Using compasses to draw a circle You are probably used to drawing circles with a pair of compasses, but here is an exercise just to get the language sorted out. Copy or trace the interval AB and point O shown below into your exercise book. Open your compasses to the length of the interval AB. Then place the point of your compasses firmly into the point O, called the centre. Holding the compasses only by the very top, draw a circle.
A
B
O
This is called drawing a circle with centre O and radius AB. Notice that every point on the circle is the same distance from the centre O, because the distance between the point and the pencil lead never changes.
Parts of circles Here we will identify some important parts of a circle. We start by drawing a circle with centre O. Radius Draw an interval from any point A on the circle to the centre O. This interval AO is called a radius of the circle. Every radius of the circle has the same length, because the setting of the compasses remained the same while the circle was being drawn.
A O
Chapter 13 Triangles and constructions
75
The word ‘radius’ is used both for the interval AO and for the length of the interval AO. ‘Radius’ is a Latin word meaning a spoke of a wheel. Its plural is ‘radii’. Diameter Draw a line through the centre O, cutting the circle at A and B. The interval AB is called a diameter of the circle. Every diameter has length twice that of any radius, because a diameter consists of two radii put together.
A B
O
The word ‘diameter’ is used both for the interval AB and for the length of the interval AB. It comes from Greek and means ‘to measure through’. Chord P
Choose any two distinct points P and Q on the circle, and join the interval PQ. This interval is called a chord (from a Greek word meaning ‘a cord or string’). O
A diameter is thus a chord passing through the centre. Arc
Q
Choose two distinct points P and Q on the circle. These two points cut the circle into two curved parts called arcs. O
Q
Constructing a triangle with given measurements In the exercises, you will construct triangles with different side lengths and angle sizes. Here is a simple way to construct a triangle whose three side lengths are given.
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ICEEM Mathematics Secondary 1B
M ajor arc
Minor arc
There are two arcs PQ. The larger arc is called the major arc PQ and the smaller arc is called the minor arc PQ. (‘Major’ and ‘minor’ are the Latin words for ‘larger’ and ‘smaller’; ‘arc’ is from a Latin word for a bow or arch.)
P
Example 4 Construct a triangle ABC in which AB = 8 cm, AC = 5 cm and BC = 6 cm.
Solution C 5 cm
6 cm
A
B 5 cm
6 cm C´
Step 1: Draw an interval AB of length 8 cm. Step 2: Draw a circle with centre A and radius 5 cm. Step 3: Draw a circle with centre B and radius 6 cm. Let C and C´ be the two points where the circles intersect. Then the triangles ABC and ABC´ satisfy the conditions, because C and C´ were constructed to be 5 cm from A and 6 cm from B.
Circles and constructions • A circle is drawn by opening the arms of the compasses to some given radius and placing the point on some given centre. • Some words associated with circles: centre, radius, diameter, chord, major and minor arcs • All radii of a circle have equal length. • Every diameter of a circle is twice the length of the radius of that circle. • Compasses can be used to construct triangles with given measurements. Chapter 13 Triangles and constructions
77
Exercise 13C 1
In the diagram below, give the correct names of: A P
O B Q
2
a the point O
b the interval OP
c the interval PQ
d the interval AB
e the curve PAB
f the curve PQB.
In the diagram below, name: U
O
R
78
T
S
a the centre of the circle
c two chords that are not diameters d four radii
e a major arc (there are four)
3
a Set your compasses to a radius of 8 cm. Then choose a point O in the middle of your page and draw a circle with centre O and radius 8 cm.
b Change the setting of the compasses to 6 cm and draw a circle with the same centre O and radius 6 cm.
c Repeat this process, drawing circles with centre O and radii 4 cm and 2 cm.
These four circles are called concentric circles because they all have the same centre. ICEEM Mathematics Secondary 1B
b two diameters of the circle
f a minor arc (there are four).
Ex
d Draw a horizontal line through the centre O. From left to right, label the eight points where the line intersects the circles with the letters A, B, C, D, E, F, G and H. e Use your compasses to demonstrate that the eight intervals AB, BC, CD, DO, OE, EF, FG and GH all have the same lengths.
Example 4
4
a Set your compasses to a radius of 3 cm, and do not change the radius again until the final part of this question.
b Choose a point O in the middle of your page and draw a circle with centre O.
c Choose any point A on the circle, and draw a second circle with centre A.
d Let the two circles intersect at B and F.
e With centre B, draw a third circle, cutting the first circle at A and C.
f With centre C, draw a fourth circle, cutting the first circle at B and D.
g With centre D, draw a fifth circle, cutting the first circle at C and E.
h With centre E, draw a sixth circle – this circle should cut the first circle at D and F.
i With centre F, draw a seventh circle – this circle should cut the first at E and A.
j Change the setting of the compasses to 6 cm. Now draw a circle with centre O that just touches the outsides of the six outer circles.
5
Construct a triangle ABC in which the three side lengths are AB = 9 cm, AC = 6 cm and BC = 5 cm.
a Draw an interval AB of length 9 cm.
b With radius 6 cm and centre A, draw an arc above the interval AB.
c With radius 5 cm and centre B, draw another arc above the interval AB.
d Let the two arcs meet at C, and join the intervals AC and BC.
e Measure the sizes of the three angles with your protractor.
Chapter 13 Triangles and constructions
79
6
a Construct a triangle ABC in which:
• two of the side lengths are AB = 7 cm and AC = 4 cm • the angle between these two sides is ∠A = 110˚.
b Join the interval BC and measure its length.
7
a Construct a triangle ABC in which:
• two of the angles are ∠A = 40˚ and ∠B = 30˚ • the side joining these vertices has length AB = 8 cm.
b Explain why the third angle ∠C is 110˚.
c Measure the lengths of the sides AC and BC.
13D Isosceles and equilateral triangles Triangles with two or three sides equal have some interesting properties. At this stage, however, we can only give informal proofs of the results of this section.
Isosceles triangles
A
An isosceles triangle is a triangle with two (or more) sides equal. • The equal sides AB and AC of the isosceles triangle ABC to the right are called the legs. They have been marked with double dashes to indicate that they are equal in length. • The vertex A where the legs meet is called the apex.
legs
base B
C base angles
• The third side BC is called the base. • The angles ∠B and ∠C at the base are called base angles. The word isosceles is a Greek word meaning ‘equal legs’ – iso means ‘equal’, and sceles means ‘leg’.
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ICEEM Mathematics Secondary 1B
apex
Constructing an isosceles triangle using a circle Any two radii AB and AC of the circle opposite are equal. Thus they form the equal legs of the isosceles triangle ABC. The chord BC is the base and the centre A is the apex. Using two radii of a circle is an easy way to construct an isosceles triangle. It is used in the investigation below.
A
B
C
The base angles of an isosceles triangle are equal The base angles of an isosceles triangle are equal, regardless of the size and shape of the triangle. This is illustrated in the investigation described below. Step 1: Draw a large circle with centre A. Step 2: Draw any two radii AB and AC, and join the chord BC. Step 3: Measure the base angles ∠B and ∠C (∠B and ∠C should be equal)
∠B = . . . . . . . . . . . . .
∠C = . . . . . . . . . . . . .
A
B
C
Instead of measuring the base angles, you could also cut the triangle out and fold it so that the side AB falls on the side AC. The angles ∠B and ∠C would then fall exactly on top of each other. Any isosceles triangle can be folded like this – the two equal sides fall exactly on top of each other, and the base angles coincide exactly. This provides an informal proof of the result.
Chapter 13 Triangles and constructions
81
A test for an isosceles triangle If a triangle has two equal angles, then the two sides opposite those angles are equal and the triangle is isosceles. Here is an investigation to illustrate the result. You will need a protractor. A
35˚
35˚
B
C
Step 1: Draw a long interval BC. Step 2: Use a protractor to construct angles of 35˚ at B and at C on the same side of the interval. (Any acute angle will do.) Step 3: Let the arms of these angles meet at A. Step 4: Use compasses to confirm that the sides AB and AC opposite the equal angles have equal length. Instead of measuring the sides, you could cut the triangle out and fold it so that ∠B falls on ∠C. The sides AB and AC would then fall exactly on top of each other. A
B
M
C
Any triangle with two equal angles can be folded like this – the two equal angles fall exactly on top of each other, and the opposite sides coincide exactly. Again, this provides an informal proof of the result. It is possible to give a proper proof of this result now, provided that we use the previous result that the base angles of an isosceles triangle are equal. See the Challenge exercise at the end of this chapter for the details.
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ICEEM Mathematics Secondary 1B
Converses of the two results The two results discussed so far are converses of each other – each result is the other read backwards. The first (slightly reworded) says:
If two sides of a triangle are equal, then the opposite angles are equal.
The second says:
If two angles of a triangle are equal, then the opposite sides are equal.
Isosceles triangles • An isosceles triangle is a triangle with two (or more) sides equal. – The equal sides are called the legs – the legs meet at the apex – the third side is the base – the angles opposite the legs are called base angles. • Two radii of a circle and the chord joining them form an isosceles triangle. • The base angles of an isosceles triangle are equal. • Conversely, if two angles of a triangle are equal, then the sides opposite those angles are equal.
Constructing an equilateral triangle using two circles An equilateral triangle is a triangle in which all three sides have equal length. The diagram to the right shows an equilateral triangle ABC. Notice that it is an isosceles triangle in three different ways, because the base could be taken as AB, BC or CA.
A
B
C
The word equilateral comes from Latin – equi means ‘equal’ and latus means ‘side’.
Chapter 13 Triangles and constructions
83
Here is a construction of an equilateral triangle with sides of length 3 cm, using only compasses and ruler. It is exactly the same construction used in Section 13C to construct a triangle with three given side lengths – because all the sides are equal, we don’t need to change the setting of the compasses. C
3
1 3 cm
A
2 B
Step 1: Open the compasses to a radius of 3 cm and draw a circle with any centre A. Step 2: Choose any point B on the circle, and draw a second circle with centre B and the same radius. Step 3: Choose C to be one of the two points where the circles intersect. Step 4: Join up the triangle
ABC. This triangle is equilateral.
Can you explain why all the sides of this triangle ABC are 3 cm long?
The angles of an equilateral triangle are all 60˚ Use a protractor to confirm that all three angles of the equilateral triangle that you drew above are 60˚. We can prove this result, provided that we use the earlier result that the base angles of an isosceles triangle are equal. Any two angles in the triangle must be equal because they are opposite equal sides. Therefore all the angles must be equal. Since they add up to 180˚, each angle must be 180˚ ÷ 3 = 60˚.
84
ICEEM Mathematics Secondary 1B
Equilateral triangles • An equilateral triangle is a triangle with all three sides equal. • An equilateral triangle can be constructed using two circles, as described above. • The interior angles of an equilateral triangle are all 60˚.
Example 5 Find ∠B in the diagram opposite. O 110˚ A
B
Solution First, AO = BO (radii). Hence ∠B = 35˚ (base angles of isosceles
ABO).
Example 6 P
Find the length of PQ in the diagram opposite.
6 cm
a Q
a R
Solution PQ = 6 cm (Opposite angles ∠Q and ∠R are equal.)
Chapter 13 Triangles and constructions
85
Example 7 A
Find i in the diagram opposite.
i B
C
D
Solution First, ∠ACB = 60˚ (equilateral
ABC ).
Hence i + 60˚ = 180˚ (straight angle at C ), so
i = 120˚.
Ex 5,
Exercise 13D 1
In the isosceles triangle to the right, PQ = PR.
P
a Name the apex of the isosceles triangle. b Name the legs and measure their lengths. c Name the base of the isosceles triangle. d Name the base angles and measure their sizes. Q
R
e What property of isosceles triangles do these angle sizes illustrate? 2
Determine whether each triangle Give reasons.
a
b
A
ABC below is an isosceles triangle.
A
c
A
55˚
63˚ B
86
63˚
62.5˚ C B
ICEEM Mathematics Secondary 1B
C
40˚ B
C
d
e
A
f
A
A
159˚ B
74˚
27˚
54˚ C
42.5˚
B
105˚ C
B
C
3
In each part, draw freehand diagrams, and mark equal intervals and the given sizes of angles. a In ABC, AB = AC. Draw D, a point on BC on the opposite side of C from B. ∠ACD = 110˚. Calculate the size of ∠ABC. b P and Q are points on a circle with centre O such that ∠POQ = 56˚. Calculate the size of ∠OPQ.
Example 5,6,7
4
Find the values of the pronumerals in these diagrams. Measurements are in degrees and millimetres.
a
b
A 94˚ B
45˚ 5
e
A
C
B
5 a
30˚
x a
a 6
b
x a 120˚ N
b a M
6
c
C
30˚
h
g
L 6
6
B
6
f
120˚
i
8
x 5
4 45˚
M
C
R
A x
b
b a
15˚ Q
45˚
A
C M
15˚
x
a
y
B
a
a
d
c
P
45˚
5
15˚
a
15˚ Chapter 13 Triangles and constructions
87
5
Find the values of the pronumerals in these diagrams. Measurements are in degrees and millimetres.
a
b
A
D
a C
M
D
E
36˚
20˚ A B
E
B
c
b
165˚ C F
a
A
B
C
e
d A 110˚ 39˚ B
f C
23˚ O a
C E
A
D
120˚ C
b
b D
a
A
B B
13E Constructions with compasses and straight edge Careful constructions with compasses and straight edge have always been an essential part of geometry. These constructions are based on a fundamental fact about circles: All radii of a circle are equal. This is obvious whenever you draw a circle using compasses, because as the pencil lead moves, it always remains exactly the same distance from the fixed point.
88
ICEEM Mathematics Secondary 1B
Construction – Bisecting an angle The diagram below shows the steps to follow to bisect a given angle ∠AOB. A
A
A
O
B
Step 1
O
B
Step 2
O
B
Step 3
The two arcs in step 2 can have a different radius from the arc in step 1. Folding the paper along the constructed line provides an informal proof that the construction works. The arms OA and OB then fall exactly on top of each other, so ∠AOB has been cut into two equal pieces. The line you have constructed also bisects the reflex angle ∠AOB. (Can you prove this?)
Construction – A right angle at the endpoint of an interval A right angle is half a straight angle. Thus bisecting a straight angle using the previous construction will give a right angle. We begin by producing (extending) the interval BA.
A
Step 1
B
A
Step 2
B
A
B
Step 3
The two arcs in step 2 will need to have a larger radius than the arc in step 1.
Chapter 13 Triangles and constructions
89
Construction – An angle of 60˚ at the endpoint of an interval The angles of an equilateral triangle are all 60˚. Thus constructing an equilateral triangle will give an angle of 60˚.
A
B
A
B
Step 1
A
Step 2
B
Step 3
This time the arcs in steps 1 and 2 must have the same radius.
Construction – Further angles by bisection Many other angles can now be constructed by applying the angle bisection construction to angles already constructed. For example: • Bisecting 90˚ will give 45˚. – Bisecting again will give 22 1 ˚ and 67 1 ˚. 2
2
• Bisecting 60˚ will give 30˚. – Bisecting again will give 15˚ and 75˚.
Construction – The perpendicular bisector of an interval Use these steps to bisect a given interval AB.
A
B
A
B
A
B
Step 1
Step 2
The arcs in steps 1 and 2 must have the same radius.
90
ICEEM Mathematics Secondary 1B
Step 3
Informally speaking, the diagram is symmetric about the line you have constructed, as you can see by folding the paper along it. This means that the line bisects AB and is perpendicular to AB. Note: You can use the construction above to produce the midpoint of an interval, even if you don’t actually need the perpendicular bisector.
Construction – Dropping a perpendicular from a point to a line Use these steps to construct a line passing through a given point F and perpendicular to a given line AB. F
F
A
B
Step 1
A
F
B
Step 2
A
B
Step 3
The two arcs in step 2 can have a different radius from the arc in step 1. Informally speaking, the diagram is symmetric about the line you have constructed, so the line is perpendicular to AB.
Chapter 13 Triangles and constructions
91
Exercise 13E Note: Make an accurate copy of these diagrams in your workbook. Use only compasses and straight edge and a sharpened pencil. 1
Use the angle bisection construction to bisect each angle.
a
b
c
2
Use the equilateral triangle construction to construct an angle of 60˚ at the endpoint A of each interval AB. b
a
B
A
92
B A
ICEEM Mathematics Secondary 1B
3
Use the construction for bisecting a straight angle to construct a right angle at the endpoint A of each interval AB. (First produce the interval BA to a point X.) b
a
B
B
A A
4
Each part below requires a combination of constructions. a Construct angles of 30˚ and 15˚ at A.
A
B
b Construct angles of 45˚ and 22 1 ˚ at A. 2
A
B
c Construct an angle of 150˚ at A.
A
B
Chapter 13 Triangles and constructions
93
d Construct an angle of 135˚ at A.
A
B
5
Construct the perpendicular bisector of each interval.
a
b
c
A
B
A B
A B
6
Use the construction for dropping a perpendicular to construct the line through P perpendicular to AB. a
b P
A
c
A
B
A
B
P
P B
7
a
Construct the bisectors of A ∠AOB and ∠BOC. Check that the two bisectors are perpendicular. D
B
O
C
94
ICEEM Mathematics Secondary 1B
b Use the angle bisection construction seven times to divide ∠AOB into eight equal parts.
A
B O
13F Quadrilaterals A quadrilateral
A
is a plane figure bounded by four straight sides.
B
C
A B
D D
C
The quadrilateral on the left above is called a convex quadrilateral because none of its four interior angles is a reflex angle.
The quadrilateral on the right is called a nonconvex quadrilateral because one of its interior angles is a reflex angle.
Note: We do not allow any interior angle of a quadrilateral to be 180˚ – such a figure is best described as a triangle. The word quadrilateral comes from Latin – quadri means ‘four’ and latus means ‘side’. A quadrilateral has two diagonals that join opposite vertices. A
B
C
A B
D C
The lefthand quadrilateral above is convex – notice that both diagonals are inside the q uadrilateral.
D
The righthand quadrilateral above is nonconvex – notice that one diagonal is inside the quadrilateral and the other is outside. Chapter 13 Triangles and constructions
95
The interior angles of a quadrilateral have sum 360˚ The interior angles of a quadrilateral always add to 360˚. Here are two ways to check this result for a particular quadrilateral. • Construct quadrilaterals of different shapes and measure their angles. • Cut out a quadrilateral. Tear off the four corners and show that they fit together to form a revolution. It is much better, however, and a lot less trouble, to give a proper proof. Theorem: The sum of the interior angles of a quadrilateral is 360˚. Proof:
There are two kinds of quadrilaterals, as shown in the diagrams below. Join a pair of opposite vertices to form two triangles as shown. The angle sum of each triangle is 180˚. Hence the angle sum of the quadrilateral is 360˚.
Ex
Example 8 Find the size of ∠C in the diagram opposite.
D
55˚
110˚
A
C
60˚ B
Solution ∠C + 110˚ + 55˚ + 60˚ = 360˚ so
96
∠C + 225˚ = 360˚, ∠C = 135˚.
ICEEM Mathematics Secondary 1B
(angle sum of quadrilateral ABCD)
Quadrilaterals • A quadrilateral is a plane figure that is bounded by four straight sides and has four vertices. • A quadrilateral is convex if all of its interior angles are less than 180˚. Both diagonals of a convex quadrilateral lie inside the figure. • A quadrilateral is nonconvex if one of its interior angles is greater than 180˚. One diagonal of a nonconvex quadrilateral lies outside it; the other inside. • The sum of the interior angles of a quadrilateral is 360˚.
Exercise 13F Example 8
1
Find the values of the pronumerals in these diagrams.
a
b
C
R
a
B 120˚
b
B 130˚
c
75˚
Q
D
68˚
120˚
C a 58˚ D
A
70˚ A
d
R Q
110˚ 120˚
e
P
f
C B
a
S
U 51˚
110˚
S
a P
120˚ a
a
D
A
X
V
W
2
In each part, three angles of a quadrilateral are given. Calculate the size of the fourth angle of the quadrilateral. Draw a quadrilateral with these angle sizes.
a 90˚, 90˚, 120˚
c 70˚, 150˚, 55˚
b 60˚, 90˚, 120˚ d 75˚, 75˚, 75˚ Chapter 13 Triangles and constructions
97
3
Find the values of the pronumerals in these diagrams. b a B
C
b
a
B a
60˚
65˚
A
Q
R
a
d
B 110˚
b
62˚
c
P
D
75˚ A
D
c
C
70˚
S
a
A
D
4
Find the values of the pronumerals in these diagrams.
a
D
b
c
C
C B a 92˚
B 115˚ F E
A
d C 125˚
i
a b
c
43˚
D
W
110˚
b P
140˚
R
b
S
A
S
C
B
54˚
73˚
b
D
O
T
Find the values of the pronumerals in these diagrams.
a
b
B
R
32˚
c
Q
Q
D
a
a
D
C
90˚ A
74˚ 162˚
b C
98
c
D
f
Q a
338˚
R
E
P a E
Q a
50˚
60˚ A
e
225˚
128˚ a A B
5
C a
P
ICEEM Mathematics Secondary 1B
S
P
QP = QR
R
d
e
C
120˚
108˚ B
a
a
6
M
N
P
A b
R 3a
D
108˚ A
f
Q
D
2a
37˚ E
B 24˚ a 24˚
a C
S
E
The diagram below is a map of the first floor of Tony’s house.
Kitchen
Balcony Dining room Shelves
Recreation room
Living room TV table
a Tony would like to build some wooden shelves at the end of the recreation room. Each shelf would be an isosceles triangle. Evaluate the internal angles of these shelves. b Tony would also like to construct a TV table in the shape of an isosceles triangle. Evaluate the internal angles of the TV table.
13G Further constructions Various interesting constructions can be done using the simple constructions of the previous section. Two things are required for the questions in this exercise. • You must be able to perform the constructions of the previous section. • You must be able to read carefully and carry out exactly the steps described.
Chapter 13 Triangles and constructions
99
Exercise 13G Note: In this exercise, use a straight edge and compasses. 1
a On a new page of your exercise book, construct a circle and label the centre A. b Construct any two radii AB and AC and join the isosceles triangle ABC. c Use compasses and straight edge to bisect the apex angle ∠BAC. d Check that the bisector meets the base BC at right angles.
2
a On a new page of your exercise book, draw a large triangle ABC. b Construct the angle bisectors of the three interior angles at A, B and C. c These bisectors should be concurrent, meeting at a single point I. d Open your compasses so that you can draw the circle with centre I just touching each of the three sides. (This circle is called the incircle of the triangle, and the point I is called the incentre.)
3
a On a new page of your exercise book, draw a large triangle ABC.
b Construct the perpendicular bisectors of the three sides.
c These perpendicular bisectors should also be concurrent, meeting at a single point O. d Open your compasses so that you can draw the circle with centre M passing through all three of the vertices. (This circle is called the circumcircle of the triangle, and the point O is called the circumcentre.)
4
a On a new page of your exercise book, draw a large rectangle ABCD. Use compasses and straight edge to construct the right angles at its vertices. b Join the two diagonals and let them meet at M. c Construct the circumcircle with centre M passing through all four of the vertices.
100
ICEEM Mathematics Secondary 1B
Review exercise 1
Find a, b, c and/or i in each diagram, giving reasons.
a
B b c
c
Q
c
b
C I
F 110˚
a
115˚
30˚
C
a
D
b
P
A
d
b
50˚
K
J
a
B
L
M G
A
i
N
20˚
O
H
2
Find a, b, c and/or i in each diagram, giving reasons.
a
R
b U
c
V 40˚
30˚
B
100˚
b T
W
3b
a
R
Y 80˚
S
d X
T
X
i P
e A i
F
B
125˚
P c
70˚
40˚
f
K 140˚
C Q
J
L
2i 3i M
E
D Chapter 13 Triangles and constructions
101
3
Use the diagram opposite to prove that the sum of the interior X angles of a triangle is 180˚.
Y A a C
c b
B
4
Find a, b, c and/or i in each diagram, giving reasons.
a
A
B b
b
c
F
A
24˚
a
i 110˚
25˚
C
B
O
H
c
G
d
e
Q 20˚
S
f
E D 100˚
i
O P
H
98˚ A X D P 105˚
a 70˚
b
F
F
G
Y
92˚ B C
Q i
G
I
R
5
Find the length x in each diagram, giving reasons.
a
b
A
c 110˚
x
10 cm
8 cm
70˚
70˚
C
S
B
102
ICEEM Mathematics Secondary 1B
70˚
40˚ 5 cm
x 55˚
F
P
T
R M
Q G
x
6
Identify any pairs of parallel lines in each diagram, giving reasons.
a A
b
B
M
60˚ N C
120˚
110˚
D
c
P
20˚
X P
E 70˚ Q
Q
a A
7
B
C
a Use ruler and compasses to construct a triangle with side lengths 6 cm, 7 cm and 8 cm. b Use ruler, compasses and protractor to construct an isosceles triangle with legs of length 6 cm and an apex angle of 110˚.
8
a Use ruler and compasses to construct an equilateral triangle of side length 5 cm.
b Explain why all the angles of this triangle are 60˚.
9
Use ruler and compasses to construct angles of:
a 60˚
b 30˚
c 15˚
d 90˚
e 45˚
f 22 1 ˚ 2
10 a Draw a large obtuseangled triangle. b Use ruler and compasses to construct the bisector of each vertex angle. c These bisectors should be concurrent. Construct the circle with centre at this point that just touches each of the three sides. 11 a Draw a large triangle. b Use ruler and compasses to construct the perpendicular bisector of each side. c These bisectors should be concurrent. Construct the circle with centre at this point that passes through each vertex. Chapter 13 Triangles and constructions
103
12 Find the values of the pronumerals. a
M N
L
K
b
B
A
38˚
a
b a 36˚
32˚
T
R S
c
J
K
86˚
O
70˚
G
F
d
D N 120˚
O
N
140˚ T
70˚ E
L
a
M L
e
Z
X c 70˚ N
f
I
B
G a b E a
34˚
C
H
D
Y
g
E
h
X
60˚ a
K
120˚
T
110˚ U
Q
W 86˚
D
a 100˚ A
I Z
104
J 45˚ F
A
65˚ O
C
A a
ICEEM Mathematics Secondary 1B
Y
Challenge exercise 1
This question provides a proof of the following theorem:
If two angles of a triangle are equal, then the opposite sides are equal.
It uses the method of proof by contradiction, and relies on the theorem that the base angles of an isosceles are equal.
If the sides AB and AC are not equal, then one of them is longer. Suppose for example that AB is longer than AC. Mark the point D on AB so that AD = AC. Mark the angles in the two triangles as shown in the diagram opposite. A
a Use angle ∠ADC as an exterior angle of BCD to show that β > α. b Show that ∠ACD is also equal to β. Now look at the diagram to see that β < α. c Explain how the theorem follows from this contradiction. 2
D
b
a
a
B
C
a Construct a large circle with centre O. b Using ruler and compasses only, construct an equilateral triangle ABC in which all three vertices A, B and C lie on the circle.
3
a Construct a large circle with centre O and draw a diameter AOB. b Choose any point P, different from A and B, on the circle and join PA and PB. c Prove that ∠APB is a right angle. Do this by drawing the radius OP and working with the angles in the two isosceles triangles AOP and BOP. Chapter 13 Triangles and constructions
105
4
a On a new page of your exercise book, draw a large quadrilateral ABCD. Don’t make it any special sort of quadrilateral. b Use straight edge and compasses to construct the midpoints of the four sides. c Join these four midpoints up to form a new, smaller quadrilateral inside ABCD. d Use compasses to check that the opposite sides of the smaller quadrilateral are equal.
5
a Construct a large circle with centre O. b Draw two diameters AOB and POQ, not at right angles. c Join up the quadrilateral APBQ. d Prove that each angle of APBQ is a right angle.
6
a Construct a large circle with centre O. b Using ruler and compasses only, construct a square ABCD with all four vertices A, B, C and D on the circle.
7
a Draw an interval AB. b Using ruler and compasses only, construct an isosceles triangle ABC in which the angle ∠ACB is a right angle.
8
a Construct a large circle with centre O. b Mark any four points A, B, C and D going clockwise around the circle and join up the quadrilateral ABCD. Make sure the centre O is inside ABCD. c Prove that these opposite angles are supplementary. Do this by drawing the radii OA, OB, OC and OD, and working with the angles in the four isosceles triangles they form. d Repeat part b but this time choose points A, B, C and D so that point O is outside the quadrilateral ABCD. Now repeat part c. The proof this time will be a little different, because you will need to take differences as well as sums of angles.
106
ICEEM Mathematics Secondary 1B
Chapter 14 Negative fractions In Chapter 4 and 5, we reviewed fractions. In the same way that we extended the positive integers to negative integers in Chapter 11, we can also extend the positive fractions to the negative fractions.
We can now look at negative fractions as well. They lie to the left of 0 on the number line. Each positive fraction has an opposite fraction. For example, 5 has as its 4 opposite – 5 = –1 1 = –1.25 4
4
A situation where we see negative fractions is in temperatures. A temperature of –4.5°C or – 9 °C is 4.5°C below zero. 2
Several negative fractions are marked on the number line below.
–378
–4
–6
–5
–3
–2
– 43
–1
– 12
4 3
0
1 2
1
2
3
378 4
5
The arithmetic for the integers, which we have considered in the previous sections, also extends to the fractions.
Chapter 14 Negative fractions
107
Example 1 a Arrange the numbers –2 1 , 1 , 4 2 and –3 3 from smallest to 2 3 3 4 largest. b Draw a number line from –5 to 5 and mark on it the numbers –2 1 , 1 , 4 2 and –3 3 . 2 3
3
4
Solution a –3 3 , –2 1 , 1 , 4 2 4
2 3
b
–334
–5
3 –2 12
–4
–3
13
–2
–1
4 23
0
1
2
3
4
5
14A Addition and subtraction of negative fractions We use techniques for addition and subtraction of fractions similar to the techniques used for the integers.
Example 2 Write the answer to these additions. a – 1 + 2
b –1 1 + 2
2
2
c – 1 + 2 3
d – 4 + 2
3
5
5
Solution a – 1 + 2 = 1 1
b –1 1 + 2 = 1
c – 1 + 2 = 1
d – 4 + 2 = – 2
2 3
2
3
3
2
5
2
5
5
When adding or subtracting fractions, use the LCM of the denominator.
108
ICEEM Mathematics Secondary 1B
Example 3 Write the answer to these additions. a – 2 + 1 3
b – 3 + – 1
5
5
3
c 1 + – 2 4 5
d – 9 + 1 5
3
Solution a – 2 + 1 = – 10 + 3 3
5
=
b – 3 15 15 5 7 – 15
+ – 1 = – 9 + – 5 3
15 =–9 15 = – 14 15
–
15 5 15
c 1 + – 2 = 5 + – 8 4 5 20 20
d – 9 + 1 = – 27 + 5 5 3 15 15 = 5 – 8 = – 22 20 20 15 3 =– = –1 7 20 15
Example 4 Write the answer to these subtractions. a – 2 – 1 3
b – 2 – – 1
3
3
2
c – 3 – – 2 5 3
d – 3 – 7 4
8
Solution a – 2 – 1 = –1 3
3
b – 2 – – 1 = – 4 – – 3 3
2
c – 3 – – 2 = – 9 – – 10 5
3
d – 3 15 15 4 = – 9 + 10 15 15 1 = 15
6 = –4 6 = –1 6
+
6 3 6
– 7 = – 6 – 7 8
8 8 = – 13 8 = –1 5 8
Chapter 14 Negative fractions
109
Example 5
Ex
Write the answer to each of these additions and subtractions. a –2 1 – 3 3 2
b –2 1 – –4 1
4
3
c –2 3 + –4 3
2
7
4
Solution a –2 1 – 3 3 = –2 2 – 3 3 2
4
b –2 1 – –4 1 = –2 2 – –4 3
4 4 5 = –5 4 = –6 1 4
3
2
= =
6 –2 2 6 21 6
Ex
6
+
43 6
c –2 3 + –4 3 = –2 12 – 4 21 7
4
28 –6 33 28 –7 5 28
=
=
28
Exercise 14A Example 1a
1
Arrange each set of numbers from smallest to largest.
a –2, 1 , –1, – 7 , 1 , 1 4
b – 7 , – 11 , 1 , 7
5 2
4
c – 2 , – 4 , – 11 , 12 3
Example 1b
2
110
4 2
d –1 11 , – 15 , –2, –1
12 13
13
Ex
13
Draw a number line from –5 to 5 and mark on it the numbers –3 1 , –4 3 , –3 1 , –1 1 and 1 1 . 2
Example 2
5
5
4
4
2
2
3
Write the answer to these additions.
a – 1 + 1
b –2 1 + 5
1 2
11 2
2
e –1 +
2
f –4 +
ICEEM Mathematics Secondary 1B
c – 2 + 1
5 g – 2 3
+
5 1 3
d – 2 + 5 7 h – 2 5
+
7 4 5
Example 3
4
Write the answer to these additions.
a – 1 + 1
b – 1 + 3
2 1 e + – 1 5 2 i – 3 + 1 5 2 Example 4
4
2 f – 2 3 j – 5 6
6
–4 7 1 4
+ +
g – 2 5 k – 1 3
+ +
d – 1 + 1
4 1 3 2 5
7 3 4 h – + – 2 5 9 l 2 + – 1 5 3
5 Write the answer to these subtractions.
a – 1 – 1
b – 1 – 4
e – 1 – – 1
4
2
5
2
5
Evaluate:
a – 1 + 3
2 e – 3 4
–
d – 1 – 1
f – 2 – – 3
g – 2 – – 1
h – 4 – – 1
j – 5 – – 2
k – 1 – – 3
l – 2 – 1
b –1 1 + 4
c – 2 – 1
d – 2 + 1
5
3
2
6
c – 3 – 1
2
i – 3 – 1
Example 5
c 1 + – 3
5
4
5
6
5
5
f – 2 3
+
3
5
5
3 g – 3 4
4 5
7
4
3
2
3 4
6
8
5
3 + 1 4
3 h – 1 5
3
–
3 2 5
i – 3 + 3
j – 3 – 3
k –1 1 + – 7
l 3 – 2 1
m –3 1 + 3 2 3 q – – 3 5 4
n –2 1 + 4 2 2 r – + 1 3 4
o – 2 5 s – 3 4
p – 3 4 t – 1 5
u – 3 + 3 2
v – 3 – 3 2
w – 5 + –3 5 x 3 – – 2
4
4
5
9
5
2
7
9
7
– 1 3 + 1 5
13
7
2 + 1 3 2 – 7
8
7
7 Write the answer to each of these additions and subtractions. b 2 1 – –2 3
c –2 1 – 2 3
d –3 2 – 2 3
e –2 1 + 1 1
f –1 1 – 3 4
g –1 3 + 2 1
h –2 1 – 4 1
i –3 1 5
j –1 2 3
k –4 2 5
l –2 4 5
a – 3 – –3 2 4
11
4
2
+
–2 1 2
m –2 3 – 3 1 5
5
2
4
2
5
5
–
4
4
–5 3 5
5
6
+
–1 1 4
7
8 3
– –3 1 8
n –1 5 – –2 2 o –2 1 – –1 3 p –1 2 + 3 1 6
5
3
5
5
3
Chapter 14 Negative fractions
111
14B Multiplication and division of negative fractions The methods we have used with the positive fractions also work with the negative fractions. In addition we use the rules for multiplying positive and negative numbers.
Example 6 Carry out the following multiplications. a – 2 × 1
b – 3 × – 1
d – 3 × 16
e – 5 × – 33
3 4
5
5
21
c – 2 × – 1
2
11
3
2
50
Solution a – 2 × 1 = – 2 b – 3 × – 1 = 3 × 1 3 5 15 5 2 5 2 3 = 10
4
3 16 d – 3 × 16 = – 4 × 21 4 21 7
c – 2 × – 1 = 2 × 1 3 2 3 2
= 1
3
5
= –4 7
3
33
e – 5 × – 33 = 11 × 50 11 50 10 = 3
10
Example 7 Carry out the following divisions. a –2 ÷ 1 4
112
c 1 ÷ – 3
e 7 ÷ – 14
f – 5 ÷ – 25
5
d – 3 ÷ – 1 8
b 3 ÷ – 2 4
11
ICEEM Mathematics Secondary 1B
33
4
5
8
24
Solution a –2 ÷ 1 = – 2 × 4 4
b 3 ÷ – 2 = – 3 ÷ 2
1 1 8 – 1
=
= –8
c 1 ÷ – 3 = – 1 ÷ 3 4
5
d – 3 4 5 8 1 5 = – × 4 3 = – 5 12
5
1 5 3 =– × 5 1 2 15 =– 2 = –7 1 2
÷ – 1 = 3 ÷ 1 4
=
11
33
=
=
=
f – 5 11 33 8 3 7 33 – 11 × 14 2 – 3 2 –1 1 2
4 ×4 1
=3 2
= 11
e 7 ÷ – 14 = – 7 ÷ 14
8 3 82
2
÷ – 25 = 5 ÷ 25 24
= =
8 5 8 3 5
×
24 3 24 255
In earlier work we saw that 16 ÷ 2 can be written as 16 and, similarly, 3 ÷ 4 2
can be written as 3 . 4
This notation can also be used with negative numbers. So, –3 ÷ 4 can be written as –3 4
–16 ÷ (–2) can be written as –16 –2
3 ÷ (–4) can be written as 3
–4
We note that:
–3 4
= 3 = – 3 and –12 = 12 = –4 –4
and –5 = 5 –8
8
4
3
–3
and –16 = 16 = 8 –2
2
Chapter 14 Negative fractions
113
Example 8 Carry out the following divisions. a 36
b –8
c –72
e –11
f –27
g –20
–4
4
–2
d –14
–12
54
16
–25
Solution a 36 = –9
b –8 = –2
–4
c –72 = 72
4
–12
d
–14 = 16
=
12
=6
7
– 14 e –11 = 11 f –27 = – 27 168 –2 2 54 542 7 1 – = 5 = – 1 8 2 2
g –20 = 20 –25
=
25 4 5
Example 9
Ex
Carry out each of these multiplications and divisions. a – 1 × 2
b –1 1 ÷ 2
2
c –1 2 × 3
2
3
d –1 1 ÷ –2 1
5
4
5
e –2 1 ÷ –1 1 f –5 2 × –2 3 g –4 2 ÷ 2 7 2
4
7
4
3
8
Solution a – 1 × 2 = – 2 2
114
2
b – 1 1 ÷ 2 = – 3 × 1
= – 1
ICEEM Mathematics Secondary 1B
2
=
2 3 – 4
2
Ex
c –1 2 × 3 = – 5 × 3
d –1 1 ÷ –2 1 = 5 ÷ 11 4 5 4 5 5 = –1 = × 5
3
5
3
5
=
4 25 44
11
e –2 1 ÷ –1 1 = 5 ÷ 5 2
4
f –5 2 × –2 3 = 37 × 11 7 4 7 4 2 407 5 4 = × = 28 2 5 = 2 = 14 15 2
4
28
g –4 2 ÷ 2 7 = – 14 ÷ 23 3 8 3 8 14 =– × 8
=
=
3 23 112 – 69 –1 43 69
Exercise 14B Example 6
1
Complete the following multiplications.
a – 1 × 2
b – 2 × 1
c – 3 × 3
d – 3 × 5
e – 3 × 10
f – 3 × 4
g 2 × – 3
h 5 × – 7
i 2 × – 1
j 1 × – 4
k – 3 × – 1
l – 1 × – 3
m – 2 × – 1
n – 5 × – 33
4 5
5
21
3
7
3
Example 7
2
7
5
8
4
9
5
11
9
11
5
Complete the following divisions.
a –3 ÷ 1
d 2 ÷ – 14 9
27
7
5
11
6
2
13
3
5
50
2
4
5
b 5 ÷ – 3
c 1 ÷ – 5
e – 5 ÷ – 1
f – 2 ÷ – 38
7
6
12
2
7
13
39
Chapter 14 Negative fractions
115
Example 8
3
Complete the following divisions.
a 28
b –15
–7 –18 e –4 i –34 5 Example 9
c –81
5 –14 f 42 j – 456 –7
d –24
–9 g –9 –24 k – 345 20
32 –52 h 7 l 87 –7
4
Write the answers to these multiplications.
a 2 × – 3
b – 3 × – 5
c – 11 × – 5
e 5 × – 3
f 11 × – 4
g –2 7 × –1 7 h –1 1 × 2 1 8 11 3 5
i 12 × (–4) × 3
j –5 × 7 × –2
k 1 2 × –2 1 × 1 1
l 1 1 × –3 5 × 2
5
Write the answers to these divisions.
a 2 ÷ – 2
8
8
10
9
12
13
3
8
3
2
12
9
6
4
2
d 4 × – 3 × 5 8
9
8
5
b –2 1 ÷ –1 2
c –3 1 ÷ 2 2
d –1 4 ÷ –1 2
e –6 1 ÷ –1 2
f –1 1 ÷ –2 4
g –4 2 ÷ 3 2
h 5 1 ÷ –2 3
i –1 6 ÷ –3 5
9
5 5
3
3
3
9
3
4
3
4
9
8
11
6 7
9
6
14C Negative decimals The methods we have used with positive decimals also work with negative decimals.
Example 10 Complete the following calculations. Give your answers as decimals. a –0.2 + 2
b –0.2 – 0.4
c 4.25 – (–1.3)
d –0.6 × 0.7
e –0.2 × (–0.4)
f –1.2 ÷ 0.3
g –12.6 ÷ (–0.3)
116
ICEEM Mathematics Secondary 1B
Solution a –0.2 + 2 = 1.8
b –0.2 – 0.4 = –0.6
c 4.25 – (–1.3) = 4.25 + 1.3
d –0.6 × 0.7 = – 6 × 7
= 5.55
=
e –0.2 × (–0.4) = – 2 × – 4
10 10 – 42 100
= –0.42
f –1.2 ÷ 0.3 = – 12 ÷ 3
10 10 8 100
=
= 0.08
=–
10 4 12 10
10 × 10 3
= –4
g –12.6 ÷ (–0.3) = – 126 ÷ – 3
10 10 126 10 × 10 3 126 3
=
=
= 42
Example 11 Complete the following calculations. Give your answers as decimals. a (–0.2)3
b (–0.1)3
c (–0.2)3 + (–0.1)3
Solution a (–0.2)3 = –0.2 × (–0.2) × (–0.2) b (–0.1)3 = –0.1 × (–0.1) × (–0.1)
= –(0.2 × 0.2 × 0.2)
= –(0.1 × 0.1 × 0.1)
= – 2 × 2 × 2
=– 1 × 1 × 1
=
=
= –0.008
10 10 10 8 – 1000
10 10 1 – 1000
10
= –0.001
c (–0.2)3 + (–0.1)3 = –0.008 + (–0.001) = –0.008 – 0.001 = –0.009
Chapter 14 Negative fractions
117
Exercise 14C Example 10a
Example 10b,c
1
Complete the following calculations. Give your answers in decimal form.
a –0.4 + 3
b –6 + 11.25
c –4.8 + 6
d –7.2 + 6
e –3.5 + 7
f –2.5 + 3.2
g –4.6 + 2.1
h –3.75 + 2.2
2
Complete the following calculations. Give your answers in decimal form.
a –0.5 – 0.2
b –3.2 – 5
c 3.62 – (–2.5) d 4.6 – 5
f –4.2 – 6.1
g 3.72 – (–1.21) h 7.16 – (–2.31)
e –2.7 – 3.1 Example 10d,e
3
Complete the following calculations. Give your answers in decimal form.
a –0.3 × 0.8
Example 10f,g
f –0.01 × (–0.2)
d –0.2 × (–0.91)
e 2.5 × (–0.3)
g –0.12 × (–0.2)
h –2.1 × (–0.3)
4
Complete the following calculations. Give your answers in decimal form.
d 12.5 ÷ (–0.5)
g –5.5 ÷ (–5)
118
c –0.3 × (–0.8)
a –2.5 ÷ 0.5
Example 11
b –0.4 × (–0.7)
5
b –16.4 ÷ (–0.4)
c –12.3 ÷ (–0.3)
e –6 ÷ 10
f –6.3 ÷ (–3)
h –21.7 ÷ (–0.07)
Complete the following calculations. Give your answers in decimal form.
a (–0.3)3
b (–0.4)3
c (–0.3)3 + (–0.4)3
d (–0.1)3 + (–0.1)2
e (–0.2)3 + (–0.1)5
f (–0.2)3 + (–0.2)4
6
Complete the following calculations. Give your answers in decimal form.
a –56 ÷ 11
b –26.3 + (–4.1)
c –15.72 + (–0.63)
d –2.7 + (–5.06)
e –2.07 + (–0.96)
f –17.01 + 2.34
g –23.56 – 2.7
h 67 ÷ (–11)
i –4.5 ÷ 3 + 6
7
Complete the following calculations. Give your answers in decimal form.
a –0.07 × (–0.3)
b –0.025 × (–0.3)
c –0.525 × (–0.4)
d –0.525 × 0.05
e –6.25 × (–0.05)
f –5.75 × 0.001
ICEEM Mathematics Secondary 1B
14D Substitution involving negative fractions and decimals When we considered plotting points on the number line, we found that we could plot points which were not integers, and, of course, fractions will be involved throughout our work in algebra. This section provides practice in the important skill of substitution of both positive and negative fractions.
Example 12 Evaluate each expression for x = – 3 . 4
a 4x + 3 d –6 x + 4 7
b –2x + 4
c 5(x + 2)
e – 1 x2
f – 1 x
2
2
2
Solution a 4x + 3 = 4 × – 3 + 3 4
b –2x + 4 = – 2 × – 3 + 4 3 2
4
= –3 + 3
=
= 0
= 51
c 5(x + 2) = 5 × – 3 + 2 4
+4 2
d –6 x + 4 = –6 × – 3 + 4 7
= 5 × 11
= –6
= 5 × 5
= –6
= 25
= 15
=
=
4
4
4 6 1 4 2
e – 1 x2 = – 1 × – 3 2 2 4
= –1 × 9
=
7 + 16 28
14 11 14
2
4 × – 21 28 × – 5 28
2
f – 1 x = – 1 × – 3 2 2 4
2 16 9 – 32
2 = 3 8
= 9
64
Chapter 14 Negative fractions
119
Example 13 Evaluate each expression for m = – 1 , n = 1 3 and p = – 2. 2
4
a m + n
b m + p
c m – p
d mp
e np
f m
p
Solution a m + n = – 1 + 1 3
b m + p = – 1 + (–2) 2 4 2 1 = 1 = –2 1 4 2 c m – p = – 1 – (–2) d mp = – 1 × (–2) 2 2 1 = 1 = 1 2 p e np = 1 3 × (–2) f m = –2 ÷ – 1 4 2 1 = –3 = 4 2
Exercise 14D Example 12
Example 13
120
1
Evaluate each expression for x = – 1 .
a 2x
b –x
c x + 2
d x – 3
g –x3
h (–x)2
i 3 – x
j 3 – 2x k 5 + 2x
2
Substitute m = –4, n = 1 and p = – 1 to evaluate:
a m + n
b m + p
e np
f m
3
Given that m = –1 1 , n = 5 and p = –5, evaluate:
a m + n b m + p
4
Evaluate each expression for x = –0.1
a 5x + 4 d –5(x + 4)
2
3
l 5 + 6x
2
p
5
e 2x + 3 f x3
c m – p
d mp
g mnp
h n
p
6
c m – p
b –5x + 4 e –5x2
ICEEM Mathematics Secondary 1B
d mp
e np
c 5(x + 4) f (–5x)2
p
f m
5
Evaluate each expression for x = – 2 .
a 5x + 6 d –5(x + 6)
6
Evaluate each expression for x = – 3 .
a 6 – x e –x2
7
Substitute a = –0.1, b = –0.9 and c = –5 to evaluate:
a a + b
8
Substitute m = –4 1 , n =5 1 and p = – 1 1 to evaluate:
a m + n
b m + p
e np
f m
9
Substitute a = –0.2, b = –0.11 and c = –4 to evaluate:
a a + b
5
b –5x + 6 e –5x2
c 5(x + 6) f x5 4
b 6 + x f (–2x)2
b c + a
c x3 g –2x2
c b – c
6
3
p
b c + a
d bc
c b – c
d x5 h 5 – 2x a
e ac
f c
2
c m – p
d mp
g mnp
h n
d bc
p
a
e ac
f c
10 Evaluate each expression for x = –1.1.
a 10 – x e –x2
b 10 + x f (–4x)2
c x3 g –5x2
d x5 h 10 – 2x
11 Given that a = –2 1 , b = 1 1 and c = –20, evaluate: 5
a a + b e ac
4
b a – b f c ÷ a
c b – a g a ÷ b
d bc h a ÷ c
12 Substitute a = –0.4, b = –0.15 and c = –4 to evaluate:
a a + b + c
b a – b – c
d abc
e c
ab
c c – b – a bc
f a
13 Given that p = –0.001, q = 0.02 and r = –0.14, evaluate:
a p + q – r
d p
qr
b p – q + r r
e pq
pr
c q q
f p + r
Chapter 14 Negative fractions
121
Challenge exercise 1
Complete the following magic square, in which each row, column and diagonal must add to the same sum. –3 –1
2 3 –
–4
2
1 3
Find two fractions between – 1 and – 1 . 3
2
3
Arrange the following numbers in ascending order. . . . .. . . a –3 7 , –3.6 3, –3.63, –3. 6 b –1.435, –1.435, –1 4 , –1. 435 11
4
9
The average of four numbers is –5 1 . The average of the same four numbers and a fifth number is
5
62. 3
3
Find the fifth number.
Place the numbers – 7 , – 3 , – 8 , – 17 , – 9 , – 19 in the circle so that the 5
2
sum along any of the lines
5
10 57 is – . 10
5
10
– 11 10
– 13 10
122
ICEEM Mathematics Secondary 1B
–6 5
Chapter 15 Percentages We encounter percentages often and they can be very useful in many aspects of our lives. For example, we see discounts offered in shops everywhere – as smart shoppers, we definitely need to know how to do percentage calculations if we want the best deal. The word ‘percentage’ comes from the Latin per centum, meaning ‘per hundred’. A percentage is another way of writing a fraction with a denominator of 100. The symbol for percentage is %. For example: 8% =
8 , 100
25% =
25 , 100
99% =
99 , 100
150% =
150 100
In this chapter, you will learn how to do several different types of practical calculations involving percentages.
15A Percentages, fractions and decimals To understand percentages, we need go no further than to look at the meaning: ‘out of a hundred’. A square that has been divided into 100 smaller squares can be used to model percentages. If we colour in three of them, we say that ‘three out of a hundred’ or ‘3 per cent’ are coloured in. Chapter 15 Percentages
123
If we colour in 50 of them, we say that ‘50 out of a hundred’ or ‘50 per cent’ are coloured in. In this case, the fraction coloured in is 50 = 1 , 100 2 so half of the smaller squares are coloured in.
Converting percentages to fractions A percentage is a fraction that has a denominator of 100. To convert a percentage to its fraction equivalent, write it as a fraction with a denominator of 100 and then simplify. For example: 65% = 65
=
150% = 150
100 13 20
=
12 1 % = 12.5
100 1 1 2
2
=
100 1 8
Converting percentages to decimals A percentage can easily be converted to a decimal by first writing it as a fraction with a denominator of 100, and then converting this to a decimal. 65% = 65
150% = 150
100
= 0.65
37.5% = 37.5
100
100
= 1.5
= 0.375
Converting fractions and decimals to percentages Fractions with denominator 100 convert easily to percentages. For example: 2 100
= 2%
37 100
= 37%
175 100
= 175%
100 100
= 1 = 100%
Equivalent fractions can be used for some fractions whose denominators are not 100: 2 = 20 10 100
3 = 15 20 100
3 = 60 5 100
5 250 = 2 100
= 20% = 15% = 60% = 250% To convert a fraction or a decimal to a percentage, multiply by 100%, which is the same as multiplying by 1: 2 = 2 × 100% 0.6 = 0.6 × 100% 3.2 = 3.2 × 100% 5 5 = 2 × 100% = 60% = 320% 5 1 200 = % 5
= 40%
124
ICEEM Mathematics Secondary 1B
Here are some commonly used percentages and their fraction equivalents. It is very useful to know these.
Fraction
Percentage
1 2
50%
1 4
25%
3 4
75%
1 5
20%
2 5
40%
3 5
60%
4 5
80%
1 8
12.5%
3 8
37.5%
5 8
62.5%
7 8
87.5%
Percentages • A percentage is another way of writing a fraction that has a denominator of 100. For example, 5% = 5 . 100
• To convert a percentage to a fraction, write the percentage as a fraction with a denominator of 100 and then simplify. • To convert a percentage to a decimal, write the percentage as a fraction with a denominator of 100 and then convert to a decimal. • To convert a fraction or a decimal to a percentage, multiply by 100%.
Chapter 15 Percentages
125
Example 1 Convert each percentage to a fraction. a 4%
b 85%
c 235%
Solution a 4% = 4
=
b 85% = 85
100 1 25
=
c 235% = 235
100 17 20
100 = 2 35 100 =27 20
Example 2 Convert each percentage to a decimal. a 55%
b 5%
c 235%
Solution a 55% = 55
b 5% = 5
100
c 235% = 235
100
= 0.55
= 0.05
=
100 2 35 100
= 2.35
Example 3 Express each fraction as a percentage. a 33 100
b 3 10
c 7
d 23
20
20
Solution a 33 = 33% 100
b 3 = 3 × 10 10
=
10 × 10 30 100
= 30%
126
ICEEM Mathematics Secondary 1B
c 7 = 7 × 5 20
or
7 7 = 20 20
20 × 5 35 100
=
5
× 100 % 1
= 35%
= 35% d 23 = 23 × 5 20
or
20 × 5 115 100
=
23 20
5
23 100 = 20 × 1 %
=115%
= 115%
Example 4 Write each decimal as a percentage. a 0.35
b 0.07
c 1.4
d 1.03
Solution a 0.35 = 35
100
= 35%
c 1.4 = 14
10 14 × 10 10 × 10 140 100
=
=
= 140%
d 1.03 = 103 100
b 0.07 = 7
100
or
or
= 7%
1.4 = (1.4 × 100)%
= 140%
1.03 = (1.03 × 100)%
= 103%
=103%
Chapter 15 Percentages
127
Exercise 15A
Ex 3c
1
What percentage of each square is shaded?
a
b
c
Ex
e
d
Example 1
2
Convert each percentage to a fraction or mixed number.
a 8%
b 15%
c 75%
d 60%
e 30%
f 80%
g 120%
h 165%
j 450%
k 125%
l 448%
i 210% Example 2
3
Convert each percentage to a decimal.
a 7%
b 25%
c 75%
d 6%
e 35%
f 90%
g 127%
h 168%
j 460%
k 170%
l 496%
i 410% Example 3a,b
4
Express each fraction as a percentage.
a 27
b 97
c 13
d 127
e 229
f 7
g 9
h 7
j 527
k 987
l 328
100 100
i 53 1000
128
100 10
1000
ICEEM Mathematics Secondary 1B
100 10
1000
100
1000 1000
Example 3c,d
5
Express each fraction as a percentage.
a 3
5 23 e 50 i 53 50 Example 4
b 17
c 11
f
g
j
20 19 25 52 25
k
25 19 20 87 20
d 7
20 h 17 5 l 38 25
6
Write each decimal as a percentage.
a 0.35
b 0.27
c 0.7
d 0.3
e 0.73
f 1.3
g 5.6
h 7.89
j 0.125
k 0.375
l 0.875
i 1.29 7
Write each percentage as a fraction.
a 2 1 %
8
Complete the table, using decimal, fraction and percentage equivalents for each value.
2
Decimal
b 5 1 % 2
Fraction
0.5
c 6 1 % 4
d 87 1 % 2
Percentage 50%
1 4 75% 0.4 100% 0.457 23 100 0.403 5 8
Chapter 15 Percentages
129
15B Expressing one quantity as a percentage of another Sometimes we want to describe one quantity as a percentage of another.
Example 5 There are 50 people in a swimming club, and those who go to squad training number 35. Calculate the number of people who go to squad training as a percentage of the number of swimming club members.
Solution 35 of the 50 or 35 go to squad training. 50
Percentage going to squad = 35 × 100% 50 35 50
× 100 %
=
= 70%
1
So 70% of the swimming club members go to squad training.
• To express one quantity as a percentage of another, write the first as a fraction of the second, and then convert to a percentage by multiplying by 100%.
Example 6 Express the first number as a percentage of the second. a 30, 50
b 35, 40
Ex
Solution a 30 = 60
30 50
30
2
100
= 50 × 1 % 50 100 1 = 60% = 60%
130
or
ICEEM Mathematics Secondary 1B
b 35 = 7 40
or
8
35 40
= 87 1 % 2
35
5
100
= 40 × 1 % 2 = 175 % 2
= 87 1 % 2
Example 7 Express each quantity as a percentage of the second. a 70 cents, $1
b 350 m, 1 km
c 340 grams, 2 kg
Solution a (100 cents = $1) 70 = 70% 100
b (1 km = 1000 m) 350 = 35 1000
100
or
350 = 350 1000 1000
× 100 %
340 = 340 2000 2000
× 100 %
= 35%
1
= 35%
c (1000 grams = 1 kg) 340 2000
= 17 100
or
= 17%
1
= 17%
Exercise 15B Example 6
1
Express the first number as a percentage of the second.
a 40, 50
b 25, 40
c 21, 75
d 17, 20
e 3, 10
f 15, 40
g 100, 80
h 75, 80
i 80, 40
j 56, 64
k 30, 150
l 25, 8
Chapter 15 Percentages
131
Example 7
2
Express each quantity as a percentage of the second.
a 60 cents, $1
b 750 m, 1 km
c 340 grams, 4 kg
d $1.20, $5
e 23 mm, 5 cm
f $1.50, $10
g $4.80, $20
h $3.60, $8.00
i 1250 m, 2 km
k $23 500, $100 000
l $34.20, $60
j 200 cm2, 5 m2 Example 5
3
Express each amount as a percentage of the total. a 18 boys in a class with a total of 25 students b 160 women in a train carrying a total of 200 people c 15 people out of a total of 125 people in a restaurant ordering lasagne d 54 marks out of a total of 80 marks on a test e 100 marks out of a total of 160 marks on a test
15C Percentage of a quantity In Chapter 5, we found fractions of a quantity. In this section, we use percentages for the same purpose.
Example 8 25% of people interviewed watched the tennis final. Calculate how many people watched the tennis final if: a 300 people were interviewed
b 252 people were interviewed.
Solution a If 300 people were interviewed, then 25 people for every 100 people interviewed watched the final. That is, 25 × 3 = 75 people watched it. This can be set out as follows: 25% of 300 people interviewed were watching the final.
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ICEEM Mathematics Secondary 1B
Number of people watching = 25% of 300
= 25 × 300
= 75
100
1
b 25% of 252 people interviewed were watching the final. Number of people watching = 25% of 252
= 25 × 252
=
= 63
100 1 1 252 × 4 1
Example 9 Calculate: a 20% of 415
b 63% of 197
b 150% of 362
d 200% of 5.2
Solution a 20% of 415 = 20 × 415
b 63% of 197 = 63 × 197
100 1 1 415 × 5 1
=
= 83
c 150% of 362 = 150 × 362
100 1 12 411 100
=
= 124.11
d 200% of 5.2 = 200 × 52
100 1 3 362 × 2 1
=
= 543
=
100 10 2 26 × 1 5
= 10.4
Chapter 15 Percentages
133
Exercise 15C Example 8
1
Johan spends 25% of his wages on rent. He earns $888 a week. How much rent does he pay a week?
Example 9
2
Calculate:
a 20% of 100
b 20% of 200
c 20% of 1000
d 50% of 300
e 50% of 30
f 50% of 3
g 150% of 100
h 150% of 500
i 150% of 30
j 200% of 100
k 200% of 118
l 200% of 3.5
3
Use your knowledge of equivalent fractions to given percentages (see the table on page 125) to calculate these amounts.
a 50% of 368
b 25% of 144
c 12.5% of 328
d 33 1 % of 621
e 20% of 750
f 40% of 255
g 37.5% of 128
h 87.5% of 640
i 67.5% of 328
4
Celine receives $200 as a gift from her grandparents. She decides that she will spend 13% of the money purchasing CDs. How much does she spend on CDs.
5
Daniel earns $400 a week. He pays 15% of this in tax. How much tax does he pay every week?
6
Nick works out that there are 168 hours in each week. He spends 37.5% of this time sleeping, 20% at school, 2% at his parttime job and 5% doing homework; and the rest is leisure time.
3
a How many hours each week does Nick work at his parttime job? b How many hours homework does he do each week? c What percentage of his week does Nick have as leisure time?
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ICEEM Mathematics Secondary 1B
Review exercise 1
Convert each percentage to a fraction or mixed number.
a 25%
b 56%
c 3%
d 140%
e 108%
f 999%
g 2 1 %
h 12 1 %
2
Convert each percentage to a decimal.
a 34%
b 99%
c 23%
d 2%
e 9%
f 123%
g 250%
h 383%
3
Convert each of these numbers to a percentage.
a 2
b 0.45
c 98
d 0.2
e 23
f
g 1.8
h
i 3.7
j
k 0.03
l 2
m
n 9
o 0.0004
4
Calculate:
a 33% of 100
b 89% of 100
c 24% of 50
d 12% of 150
e 15% of 120
f 50% of 18
g 15% of 488
h 39% of 285
i 87% of 34
k 2 1 % 2
l 37.5 % of 4000
10 100 10
40
j 12.5% of 1888 5
2
100 4 5 1 4
4
4
of 1200
10 8 20
a Express $2.40 as a percentage of $5.00 b Express 250 metres as a percentage of 4.5 kilometres. c Express $23 678 as a percentage of $200 000. d Express 345 grams as a percentage of 4 kg.
6
Twenty three of the students in a maths class of 25 students completed their test. What percentage is this?
7
At the final of a tennis tournament, 15% of the reserved seats are vacant. If there is a total of 860 reserve seats, how many are vacant?
8
There are 750 students in the school and 200 are in year 7. What percentage of the school population is in year 7? Chapter 15 Percentages
135
Challenge exercise 1 Raylee gives 60% of her weekly wage to her mother, and 25% of the remainder to her brother. She still has $240 left. How much does she earn in one week? 2 Felicity spent 20% of her savings on a bicycle and 15% of the remainder on a book. What percentage of her savings did she have left? 3 Calculate:
a 20% of 30% of 100
b 10% of 50% of 100
c 5% of 30% of 200
4
Ana went on a shopping spree. She spent $24 of her pocket money on a dress and 20% of the remainder on a shirt. She still had 2 of her 3 money left. How much did she have before she began spending?
5
Judith’s salary is $800 per week. She receives a pay increase of 12% in January but then has a pay decrease in July of 8%. What is her salary after this decrease?
6
In a lottery, only 0.0008% of tickets won prizes. If there were 5 prizes how many tickets were sold?
7 Gary makes up a drink so that 10% of the drink is pure orange juice. The remaining 90% of the fluid has no orange juice in it. Gary has 400 mL of the drink in a jug. He wants to have 12% of the drink to be orange juice. How much extra orange juice does he need to add to the jug?
136
ICEEM Mathematics Secondary 1B
Chapter 16 Solving equations In the introduction to Chapter 12 we discussed the use of algebra to convert a temperature from the Celsius to the Fahrenheit scale. 9 5
We substituted C = –5 in the expression C + 32. The result was 9 × (–5) + 32 = 23. In Australia where we use the Celsius scale, it 5
is more likely we would want to convert a temperature from the Fahrenheit to the Celsius scale. For example, what does a temperature of 86°F in the United States mean to us in Australia? We ask what 9 5
value of C makes C + 32 = 86 a true statement? This statement is called an equation, and 30 is the value of C which makes it a true statement. This value is called the solution of the equation. So 86°F corresponds to 30°C, a fairly warm day! The solution of equations is central to the study of mathematics. The Arab mathematician Musa alKhwarizmi, who we mentioned in Chapter 3, developed the idea of equations as we know them today.
16A An introduction to equations Joe has a pencil case that contains a number of pencils. He has three other pencils, and in total he has 11 pencils.
Chapter 16 Solving equations
137
Let x be the number of pencils in the pencil case. We know that x + 3 = 11. This statement is called an equation. The solution is x = 8, because 8 + 3 = 11 and no other number works.
Finding a solution by trial and error Here is another equation: 2x + 4 = 10 We can find a solution to this equation by trying a few numbers as values of x. For example: 2×1+4=6 2×2+4=8
Ex
The solution of this equation is x = 3, as 2 × 3 + 4 = 10. We have found this solution by trial and error. It turns out, however, that only very simple equations can be solved in this way. (You would quickly find this out if you tried to use trial and error to write down solutions to the equations in Exercise 16A.) In the rest of this chapter, we begin to develop systematic methods for solving equations.
Example 1 Use trial and error to solve each of these equations mentally. a x + 4 = 6 c 6x + 7 = 19
b 10 – a = 6 d x = 5 3
Solution a For x + 4 = 6, the solution is x = 2, as 2 + 4 = 6. b For 10 – a = 6, the solution is a = 4, as 10 – 4 = 6. c For 6x + 7 = 19, the solution is x = 2, as 6 × 2 + 7 = 19. d For x = 5, the solution is x = 15, as 15 = 5. 3
138
ICEEM Mathematics Secondary 1B
3
Exercise 16A 1
Three pencil cases each have x pencils in them, and there are four loose pencils. We know that there are 31 pencils in total.
x pencils
x pencils
x pencils
a Write down an equation for x. b Write down the solution. Example 1
2
Solve each of these equations for x mentally.
a x + 3 = 10
b x – 4 = 11
c x – 5 = 10
e x – 14 = 7
f x = 7
g 2x + 3 = 11 h 17 – x = 10
j 18 – 2x = 0
k x = 6
i 2x + 6 = 15
3
d 2x = 10 l x = 20
4
5
3
Solve each of these equations for a mentally.
a a + 3 = 11
b a – 7 = 23
e a = 7
f 5a + 7 = 22 g 15 – a = 8
4
Twentythree boxes of chocolates each have a chocolates in them. There are also 17 loose chocolates. In total there are 707 chocolates.
a Write an equation for a.
5
Tom has 10 bags of marbles, each containing x marbles. He tidies his room and finds 23 more marbles under his bed. He knows that he now has a total of 523 marbles.
a Write an equation for x.
6
A company runs minibuses, each of which carries n passengers. If there are six minibuses and they hold a total of 72 passengers:
a write an equation for n.
5
c 2a = 30
d 5a = 30 h 2a + 10 = 30
b Solve the equation for a.
b Solve the equation for x.
b solve the equation for n.
Chapter 16 Solving equations
139
7
Craig has six packets of chocolates, each of which holds c chocolates. He also has seven chocolates that are not in a packet. He has a total of 127 chocolates.
a Write an equation for c.
8
Sally has 19 stickers and gives x stickers to her sister. She counts the remaining stickers and discovers she now has 12.
a Write an equation for x.
9
Peter owes his father $10 for a movie ticket. He pays his father and has $24.65 left. How much money did he start with?
b Solve the equation for c.
b Solve the equation for x.
10 Tania has $6.50 in her purse. She collects her wages of $200 and buys groceries at the supermarket. She counts her money and finds she has $120.70 left. How much did Tania spend at the supermarket? 11 If y = 3x, what is the value of x when y = 33? 12 If y = 2x + 1, what is the value of x when y = 11? 13 If y = x – 5, what is the value of x when y = 19?
16B Equivalent equations Consider these equations: 2 x + 3 = 9 2x + 5 = 11
(1) (2)
Equation (2) is obtained from equation (1) by adding 2 to each side of the equation. Equation (1) is obtained from equation (2) by subtracting 2 from each side of the equation. Equations (1) and (2) are said to be equivalent equations.
2x = 6 x = 3
(3) (4)
Equation (3) is obtained from equation (2) by subtracting 5 from each side of the equation.
140
ICEEM Mathematics Secondary 1B
Equation (4) is obtained from equation (3) by dividing each side of the equation by 2. You can obtain equation (3) from equation (4) by multiplying each side by 2. Once again, we say that equations (3) and (4) are equivalent. So equations (1), (2), (3) and (4) are all equivalent.
Intepretation with scales Imagine a pair of balanced scales, as shown in the diagrams below. Two different weights are used: 1 and x. You could think of the numbers as representing weights in kilograms, so 1 means 1 kg. When we have simple scales like the ones shown, in which the arms are of equal length, then they are balanced when the weights are equal. The equation x + 3 = 7 can be used to represent the fact that the scales in the first diagram are balanced. 1
x
1
1
1
1
1
1
1
1
1
Subtract 3 from both sides to show that x = 4 is the solution to the equation x + 3 = 7. 1 1 1 x
1
Consider another example. In the diagram below, the equation 2x = 4 represents the fact that the scales are balanced.
x
x
1
1
1
1
Chapter 16 Solving equations
141
Divide both sides by 2 to show that x = 2 is the solution of the equation 2x = 4.
1 x
1
From the examples above, we can see the following general rules.
Equivalent equations • If we add the same number to, or subtract the same number from, both sides of an equation, the new equation is equivalent to the original equation. • If we multiply or divide both sides of an equation by the same nonzero number, the new equation is equivalent to the original equation. • Equivalent equations have the same solution.
Example 2 Solve each equation for x. a x + 3 = 5 d
x 4
= 7
b x – 4 = 7
c 3x = 23
e x + 7 = 13
f x – 6 = 11
Solution a
x + 3 = 5
– 3 x + 3 – 3 = 5 – 3 x = 2 b
x – 4 = 7
+ 4 x – 4 + 4 = 7 + 4 x = 11
142
(Subtract 3 from both sides of the equation.)
ICEEM Mathematics Secondary 1B
(Add 4 to both sides of the equation.)
c
3x = 23
3x ÷ 3 = 23 3 3
x = 23
= d x 4
x = 4
3 72 3
7
× 4 × 4 = 7 × 4 x = 28 e
(Divide both sides of the equation by 3.)
(Multiply both sides of the equation by 4.)
x + 7 = 13
– 7 x + 7 – 7 = 13 – 7 x = 6 f
(Subtract 7 from both sides of the equation.)
x – 6 = 11
+ 6 x – 6 + 6 = 11 + 6 x = 17
(Add 6 to both sides of the equation.)
The notation × 4 , – 7 and so on is recommended as a way of checking your work. You do not need both the box and the comment in parentheses.
Example 3 In each case below, write an equation and solve it. a A number x has 7 added to it and the result is 35. b A number x is multiplied by 15 and the result is 165. c A number x has 13 subtracted from it and the result is 25. d A number x has 12 added to it and the result is 13. e A number x has 10 subtracted from it and the result is 3. f A number x is divided by 9 and the result is 12. (continued on next page)
Chapter 16 Solving equations
143
Solution a
x + 7 = 35
–7 x = 35 – 7 = 28
(Subtract 7 from both sides of the equation.)
Ex 2a
The number is 28. b
15x = 165
÷ 15 x = 165 15 = 11
Ex
(Divide both sides of the equation by 15.)
The number is 11. c x – 13 = 25 + 13 x = 25 + 13 = 38
Ex
(Add 13 to both sides of the equation.)
The number is 38. d x + 12 = 13 – 12 x = 13 – 12 = 1
(Subtract 12 from both sides of the equation.)
The number is 1. e x – 10 = 3 + 10 x = 3 + 10 = 13
(Add 10 to both sides of the equation.)
The number is 13. x f = 12 9 ×9 x = 12 × 9 = 108
The number is 108.
144
ICEEM Mathematics Secondary 1B
(Multiply both sides of the equation by 9.)
Ex
Exercise 16B Example 2a,b,e,f
Example 2c
Example 2d
Example 3
1
Solve each equation for m.
a m + 3 = 6
b m + 5 = 16
c m + 8 = 10
d m – 5 = 11
e m – 6 = 11
f m – 10 = 6
g m – 6 = 3
h m + 5 = 6
i m – 11 = 5
2
Solve each equation for n.
a 2n = 6
b 3n = 9
c 5n = 25
d 3n = 16
e 12n = 100
f 18n = 46
g 5n = 17
h 6n = 51
i 3n = 17
3
Solve each equation for x.
a x = 12
3 d x 10
= 2
4
Solve each equation for x.
a x + 3 = 7
b 5x = 27
c x = 5
d x – 4 = 5
e 3x = 36
f x + 2 = 6
g 2x = 18
h x + 5 = 11
i x = 2
j x – 3 = 11
k x – 5 = 2
l 3x = 7
m 4x = 9
n 7x = 15
o 6x = 17
5
b x = 15
c x = 16
e
f
2 x 3
= 6
5 x 4
=8
4
5
In each case below, write an equation and solve it. a A number x has 5 added to it and the result is 21. b A number x is multiplied by 7 and the result is 35. c A number x is multiplied by 5 and the result is 37. d A number x is multiplied by 5 and the result is 50. e A number x is divided by 6 and the result is 10. f A number x is divided by 3 and the result is 23. g A number x has 15 subtracted from it and the result is 37. Chapter 16 Solving equations
145
6
In each case below, write an equation and solve it to find the number. a A number z has 7 added to it and the result is 12. b Twelve is added to a number z to give 19. c Six is subtracted from a number z and the result is 14. d A number z is taken away from 9 and the result is 6. e A number z is multiplied by 3 and the result is 5. f A number z is multiplied by 7 and the result is 37. g Five is multiplied by a number z and the result is 45. h A number z is multiplied by 7 and the result is 20. i A number z is divided by 6 and the result is 7.
16C Solving equations involving more than one operation To solve the equation 2x + 3 = 7 we need an extra step. First subtract 3 from both sides of the equation.
2x + 3 – 3 = 7 – 3 2x = 4 Now, divide both sides of the equation by 2.
2x = 4 2 2
x = 2
A similar method is used to solve the equation 3x – 5 = 13. First add 5 to both sides of the equation.
3x – 5 + 5 = 13 + 5 3x = 18 Now, divide both sides of the equation by 3.
146
3x = 18 3 3
x = 6
ICEEM Mathematics Secondary 1B
Example 4 Solve each equation for x. a 2x – 3 = 11
b 4x + 3 = 10
Solution a 2x – 3 = 11 + 3 2x – 3 + 3 = 11 + 3 2x = 14
(Add 3 to both sides of the equation.)
x = 14 ÷ 2 2 = 7
(Divide both sides of the equation by 2.)
Check: LHS = 2 × 7 – 3 = 11 = RHS b 4x + 3 = 10 – 3 4x + 3 – 3 = 10 – 3 4x = 7 7
x= ÷ 4 4 = 1 3 Check:
(Subtract 3 from both sides of the equation.) (Divide both sides of the equation by 4.)
4 3 LHS = 4 × 1 + 3 4 =4× 7 +3 4
= 10 = RHS
Chapter 16 Solving equations
147
Example 5 Solve each equation for x. a 2x – 11 = 7
b 6 + 4x = 22
c x + 7 = 22 5
Solution a
2x – 11 = 7
+ 11 2x = 7 + 11 2x = 18
(Add 11 to both sides of the equation.)
÷2
(Divide both sides of the equation by 2.)
b
x = 9 6 + 4x = 22
–6 4x = 16
(Subtract 6 from both sides of the equation.)
÷4
(Divide both sides of the equation by 4.)
c
x = 4 x 5
+ 7 = 22
x = 15 –7 5
(Subtract 7 from both sides of the equation.)
×5 x = 15 × 5 = 75
(Multiply both sides of the equation by 5.)
Ex 4,
Example 6 In each case below, write an equation and solve it. a A number x is multiplied by 15, and 5 is added to it. The result is 170. b A number x is divided by 11, and 6 is added to it. The result is 13.
148
ICEEM Mathematics Secondary 1B
Ex
Solution a
15x + 5 = 170
–5 15x = 170 – 5 15x = 165
(Subtract 5 from both sides of the equation.)
÷ 15 x = 165 15 = 11
(Divide both sides of the equation by 15.)
The number is 11. x 11
b
+ 6 = 13
–6
x = 11 x = 11
× 11
13 – 6
(Subtract 6 from both sides.)
7
x = 11 × 7
(Multiply both sides by 11.)
= 77 The number is 77.
Exercise 16C Example 4,5a,b
Example 5c
1
Solve each equation for x. Check your solutions to c, f and i.
a 2x + 1 = 7
b 5x – 1 = 11
c 7x + 3 = 17
d 4x + 2 = 18
e 1 + 5x = 21
f 5 + 20x = 100
g 2 + 10x = 44
h 5x – 11 = 30
i 10x + 23 = 100
2
Solve each equation for z. Check your solutions to c and f.
a z + 5 = 11
4 d z – 7 = 10 11 z g + 13 = 16 9 z + 7 = 18 j 15
b z – 5 = 10
3 e z + 11 = 20 8 z h – 8 = 17 7 z k + 55 = 68 6
c z – 7 = 8
4 f z – 12 = 8 8 z i + 73 = 84 12 z l – 42 = 58 17
Chapter 16 Solving equations
149
3
Solve each equation for x.
a x + 3 = 7
b 5x = 27
c 3x – 4 = 11
d 5x + 6 = 14
e 2x + 11 = 20
f x – 4 = 11
4
Solve:
a 2x – 4 = 5
b 3a – 6 = 36
c 3z – 7 = 17
d 11b + 4 = 121
e 4x + 18 = 30
f x – 2 = 8
g 2b – 6 = 12
h 3a – 16 = 19
i 2a + 8 = 23
j 12z – 18 = 12
k 11k + 22 = 43
l 12n – 15 = 15
m 10a + 16 = 42
n x – 2 = 15
o x + 20 = 32
q
r
p Example 6
5
x 16
– 3 = 12
12 z + 7
6 = 18
3
7
7 p 18
– 2 = 12
In each case below, write an equation and solve it to find the number. a A number x is multiplied by 7, and 6 is then added. The result is 20. b A number x is divided by 6, and 3 is then added. The result is 7. c A number x is divided by 4, and 11 is then added. The result is 20. d A number x is divided by 3, and 4 is then subtracted. The result is 23. e A number x is multiplied by 4, and 6 is then added. The result is 30. f A number x is divided by 7, and 4 is then subtracted. The result is 50. g A number x is multiplied by 7, and 4 is then added. The result is 68. h A number a is divided by 11 and 6 is added to it. The result is 30. i A number m is divided by 7 and 11 is subtracted from it. The result is 20. j A number z is divided by 4 and 13 is added to it. The result is 14. k A number m is multiplied by 6 and 14 is subtracted from it. The result is 10.
150
ICEEM Mathematics Secondary 1B
16D Equations with integers The methods we have developed in the previous section can be used to solve equations involving negative numbers.
Example 7 Solve each equation for x. a x + 3 = –2
b x – 4 = –6
d x = –5
e 2x + 5 = –6
3
c –2x = 10
Solution a
x + 3 = –2
–3 x = –2 – 3 = –5 b
(Subtract 3 from both sides of the equation.)
x – 4 = –6 +4
x = –6 + 4
(Add 4 to both sides of the equation.)
= –2 c
–2x = 10
10 ÷ (–2) x = –2 = –5
x = 3
d
×3
(Divide both sides of the equation by –2.)
–5
x = –5 × 3
(Multiply both sides of the equation by 3.)
= –15 (continued on next page)
Chapter 16 Solving equations
151
e
2x + 5 = –6
–5 2x = –6 – 5 2x = –11
÷2
x = –11
=
(Subtract 5 from both sides of the equation.) (Divide both sides of the equation by 2.)
2
–5 1 2
Exercise 16D Example 7a,b
1
Solve:
a x + 3 = –6
b x – 5 = –11
e x – 5 = –5
f x – 15 = –10 g x – 6 = –10
h x + 10 = 10
j x + 2 = –5
k x – 6 = –15
l m – 15 = –15
n x + 20 = 5
o m + 15 = –30 p z + 6 = –8
a 2x = –6
b –3x = 15
c –3x = –9
d 5x = –5
e x = –4
f x = –10
g x = –4
h x = –10
i –5x = –15
j –3m = –27
k 7m = –98
l 11x = –55
m z = –25
n
i x – 3 = –9
m x + 6 = 3 Example 7c,d
2
Solve:
3
4
Example 7e
152
c x + 6 = –11 d x – 10 = –5
5
a = –9 8
7
o
p = –11 9
11
p
b = –15 6
3
Solve:
a 2x + 8 = –6
b 3x – 6 = –15
c –3x + 12 = –9
d 5x – 10 = –5
e 4x + 27 = 21
f –5x – 4 = –12
g 20 – 4x = 30
h 5x – 11 = –60
i 2x + 20 = –34
j 15 – 2x = –30
k 7 – 12x = –77
l –12m – 14 = 34
m –15x + 30 = –45 p –5m + 7 = 13
n 23x + 13 = –33
o 100 – 10x = 50
q 3p + 19 = 12
r –4z + 18 = –84
ICEEM Mathematics Secondary 1B
4
Solve:
a x = –4
3 x d + 12 = –16 8 g m – 15 = –30 12 j m + 12 = 6 3
b x = –6
c x – 4 = –5
e
f
2 x 7
+ 10 = 5
3 x 6
– 5 = –11
h 10 – m = 20
i – m – 15 = 17
k 15 –
l
3 n = 4
5
6 p + 16
20 = 5
16E Expanding brackets and equations In Chapter 1, we saw that: 2(3 + 5) = 2 × 3 + 2 × 5 This can be illustrated with a diagram of a rectangle, broken into two rectangles as shown below. 3
5
2
The total area can be found by multiplying 2 by (3 + 5) or by finding the area of each of the smaller rectangles and adding them, 2 × 3 + 2 × 5. This idea can also be used in algebra. For example: 2(x + 5) = 2 × x + 2 × 5
= 2x + 10
Again, we can use a diagram with rectangles to illustrate this. x
5
Area = 2(x + 5) = 2x + 10
2x 10 2
We call the process of obtaining the righthand side of the equation from the lefthand side ‘expanding the brackets’. Chapter 16 Solving equations
153
Example 8 Expand the brackets. a 4(x + 5)
b 3(x – 4)
c 6(2x – 4)
d x(x + 2)
Solution a 4(x + 5) = 4 × x + 4 × 5
b 3(x – 4) = 3 × x – 3 × 4
= 4x + 20
c 6(2x – 4) = 6 × 2x – 6 × 4
= 3x – 12
d x(x + 2) = x × x + x × 2
= 12x – 24
= x2 + 2x
Brackets can appear in equations. In this section, we expand the brackets first, then solve the equation.
Example 9 Expand the brackets and solve the equations for z. a 2(z + 4) = 11
b 2(z – 5) = 13
Solution a
2(z + 4) = 11
2z + 8 = 11
–8 2z = 11 – 8 2z = 3 ÷2
154
z = 3 2
= 1 1 2
ICEEM Mathematics Secondary 1B
(Expand the brackets.) (Subtract 8 from both sides of the equation.) (Divide both sides of the equation by 2.)
b
2(z – 5) = 13
2z – 10 = 13
(Expand the brackets.)
+ 10 2z = 13 + 10 2z = 23 z = 23
÷2
2
(Add 10 to both sides of the equation.) (Divide both sides of the equation by 2.)
= 11 1 2
Example 10 In each case below, write an equation and solve it. a A number x has 5 is added to it and the result is multiplied by 5. The result of this is 32. b A number x has 3 subtracted from it and the result is multiplied by 7. The result of this is 47.
Solution a
5(x + 5) = 32
5x + 25 = 32
5x = 32 – 25 – 25 5x = 7
7
x = ÷5 5 = 1 2
(Expand the brackets.) (Subtract 25 from both sides of the equation.) (Divide both sides of the equation by 5.)
5
(continued on next page)
Chapter 16 Solving equations
155
b
7(x – 3) = 47
7x – 21 = 47
(Expand the brackets.)
7x = 47 + 21 + 21 7x = 68
(Add 21 to both sides of the equation.)
x = 68 (Divide both sides of the ÷7 7 equation by 7.) = 9 5
7
Exercise 16E Example 8
Example 9
Example 10
1
Expand the brackets.
a 4(x + 5)
b 3(x – 4)
c 6(2x – 4)
d 5(3x – 5)
e 7(2a – 4)
f 6(4 + 3x)
g 9(6x – 11)
h x(x – 3)
2
Expand the brackets and solve each equation for z.
a 3(z + 2) = 11
b 5(z – 6) = 21
c 3(2z – 11) = 10
d 4(z – 1) = 11
e 7(z + 6) = 13
f 2(3z + 1) = 15
3
Expand the brackets and solve each equation.
a 5(2x – 4) = 37
b 6(3m + 5) = 61
c 5(7m – 11) = 21
d 6(x – 7) = 12
e 12(12m – 4) = 52
f 7(4n – 10) = 11
g 7(2m – 11) = 30
h 14(7n + 1) = 100
i 11(5m + 2) = 30
4
In each case below, write an equation and solve it. a A number x has 4 added to it and the result is multiplied by 3. The result of this is 32. b A number x has 2 subtracted from it and the result is multiplied by 5. The result of this is 42. c A number x has 7 added to it and the result is multiplied by 3. The result of this is 50.
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ICEEM Mathematics Secondary 1B
d A number x has 3 subtracted from it and the result is multiplied by 2. The result of this is 15. e A number x has 11 added to it and the result is multiplied by 4. The result of this is 23. f A number x has 6 subtracted from it and the result is multiplied by 3. The result of this is 16. g A number m has 10 subtracted from it and the result is multiplied by 5. The result of this is 26. h A number z has 5 added to it and the result is multiplied by 6. The result of this is 42. i A number n has 11 added to it and the result is multiplied by 12. The result of this is 150. j A number z has 8 subtracted from it and the result is multiplied by 6. The result of this is 100.
16F Collecting like terms and solving equations If Tim has 3 pencil cases with the same number, x, of pencils in each, he has 3x pencils in total. x pencils
x pencils
x pencils
If Sarah then gives him 2 more pencil cases with x pencils in each, then he has 3x + 2x = 5x pencils in total. This is because the number of pencils in each case is the same.
Like terms The terms 2x and 3x in the above example are called like terms and they have been collected together. (In contrast, the terms 2x and 3y are not like terms, because the pronumerals are different.) Like terms have been discussed in Chapter 3.
Chapter 16 Solving equations
157
Adding and subtracting like terms Like terms can be added and subtracted, as shown in the examples below.
Example 11 Simplify these expressions by adding or subtracting like terms. a 2x + 3x + 5x
b 3x + 2y + 5x + 7y
c 2 + 3x + 4x + 7y – 3y
d 2x + 1 + 3y – 3 – x + 4y
Solution a 2x + 3x + 5x = 10x
b 3x + 2y + 5x + 7y = 8x + 9y
c 2 + 3x + 4x + 7y – 3y = 2 + 7x + 4y d 2x + 1 + 3y – 3 – x + 4y = x + 7y – 2
Example 12 Simplify these expressions by first expanding brackets and then adding or subtracting like terms. a 2(x + 4) + 3x
b 4(5x – 3) + 10
c 5(2x + 3) + 4(x + 6)
d 2(3x – 4) + 3(4x – 7)
Solution a 2(x + 4) + 3x = 2x + 8 + 3x
b 4(5x – 3) + 10 = 20x – 12 + 10
= 5x + 8
c 5(2x + 3) + 4(x + 6) = 10x + 15 + 4x + 24
= 14x + 39
d 2(3x – 4) + 3(4x – 7) = 6x – 8 + 12x – 21
158
= 18x – 29
ICEEM Mathematics Secondary 1B
= 20x – 2
Example 13 Collect like terms and solve each equation for x. a 5x + 3x + 4 = 36
b 10x – 4x + 6 = 24
c 5x + 2x – 3 = 34
Solution a
5x + 3x + 4 = 36
8x + 4 = 36
–4 8x = 32 ÷8 x = 4 b
10x – 4x + 6 = 24
6x + 6 = 24
(Collect like terms.) (Subtract 4 from both sides of the equation.) (Divide both sides of the equation by 8.)
(Collect like terms.)
–6 6x = 18
(Subtract 6 from both sides of the equation.)
÷6 x = 3
(Divide both sides of the equation by 6.)
c
5x + 2x – 3 = 34
7x – 3 = 34
(Collect like terms.)
7x = 37 +3
(Add 3 to both sides of the equation.)
÷7 x = 37 7
(Divide both sides of the equation by 7.)
= 5 2 7
Chapter 16 Solving equations
159
Example 14 Expand the brackets, collect like terms and solve each equation for z. a 2(3z + 4) + 5 = 20
Ex
b 4(z + 2) + 2z = 24
c 5z + 2(z – 4) = 20
Solution a
2(3z + 4) + 5 = 20
6z + 8 + 5 = 20
(Expand brackets.)
6z + 13 = 20
(Collect like terms.)
– 13 6z = 7
(Subtract 13 from both sides of the equation.)
÷ 6 z = 7 6 = 1 1
(Divide both sides of the equation by 6.)
Ex
6
b
4(z + 2) + 2z = 24
4z + 8 + 2z = 24
(Expand brackets.)
6z + 8 = 24
(Collect like terms.)
–8 6z = 16 (Subtract 8 from both sides of the equation.) ÷ 6 z = 16 6 = 2 2
Ex
(Divide both sides of the equation by 6.)
3
c
5z + 2(z – 4) = 20
5z + 2z – 8 = 20
(Expand brackets.)
7z – 8 = 20
(Collect like terms.)
+8
7z = 28
÷7 z = 4
(Add 8 to both sides of the equation.) (Divide both sides of the equation by 7.)
Checking your answer in the original equation is advised.
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ICEEM Mathematics Secondary 1B
Ex
Exercise 16F Example 11
1
Simplify each expression by adding or subtracting like terms.
a 3x + 4x + 7x
c 2x + 3x + 4x + 7x – 3x
Example 12
Example 13
d 2x + 3x – 2x + 4x
e 11x – 3x + 12x – 4x
f 2 + 3x + 5 + 6x
g 5x – 2x + 2 + 7x
h 11x + 2y + 3x + 5y
i 4x + 3x + 3y – y
j x + 2x + 5x + 7y – y
k 3x + 2 + 3y + 5 + 4x + 7y
l 4x + 6 – 2x – y + 5 + 3y
2
Simplify each expression by first expanding brackets and then adding or subtracting like terms.
a 3(x + 2) + 4x
b 6(4x – 3) + 12
c 4(5x + 2) + 7(x + 2)
d 5(x + 6) – 5x
e 7(6x + 2) + 4x
f 3(2x + 6) + 5(x – 6)
g 4(2x + 3) + 5(x – 4)
h 3(4x + 2) + 7(x + 4)
i 2(x – 1) + 4(2x – 3)
j 3(3x – 2) + 5(2x – 6)
3
Collect like terms and solve each equation for x.
a 3x + 2x + 6 = 36
b 11x – 5x + 7 = 25
c 6x + x – 2 = 35
d 4x + 11x + 3 = 33
e 2x – 7x + 12 = 36
f 7x – 2 + 8x = 28
g 6x + 9x – 4 = 41
h 8x – 11x + 12x – 18 = 0
i 8x – 2x + 24 = 60 Example 14
b 2x + 3x + 7x + 4x
j 14x + 3 – 6x – 4 = 11
4
Expand the brackets, collect like terms and solve each equation for z.
a 4(3z + 4) + 10 = 40
c 5z + 2(z – 4) = 27
e 5(2z – 3) + 6 = 23
b 4(z + 4) + 2z = 36 d 3(5z + 1) + 1 = 25 f 6 + 7(5z – 3) = 25
Chapter 16 Solving equations
161
5
Solve:
a 5m + 2(3m – 4) = 25
b 6(3m – 4) + 10 = 12
c 4x + 5(3x + 8) = 15
d 3m + 5(2 + 4m) = 30
f 5n + 15n + 3(n + 4) = 50
e 6x + 4x + 2(x – 4) = 100
16G Equations with pronumerals on both sides When there are pronumerals on both sides of the equation, the first step is to get all terms involving the pronumeral on the same side. We then solve as before.
Example 15 Solve the equation 5x + 5 = 2x + 14.
Solution
5x + 5 = 2x + 14 – 2x 5x + 5 – 2x = 2x + 14 – 2x (Subtract 2x from both sides of the equation.) 3x + 5 = 14 3x = 9 –5
(Collect like terms.)
x = 3 ÷ 3
(Divide both sides of the equation by 3.)
(Subtract 5 from both sides of the equation.)
Example 16 Solve these equations by collecting like terms.
162
a 2x + 3 = 5x + 1
b 15x – 10 = 6x + 3
c 20 – 8x + 12x = 22
d 6x – 4 – 4x – 4 = 2
ICEEM Mathematics Secondary 1B
Solution a 2x + 3 = 5x + 1 – 2x 2x + 3 – 2x = 5x + 1 – 2x 3 = 3x + 1 –1 2 = 3x ÷ 3
x = 2
3
(Subtract 2x from both sides of the equation.) (Collect like terms.) (Subtract 1 from both sides.) (Divide both sides by 3.)
15x – 10 = 6x + 3 b – 6x 15x – 10 – 6x = 6x + 3 – 6x (Subtract 6x from both sides of the equation.)
9x – 10 = 3
(Collect like terms.)
9x = 13 + 10
(Add 10 to both sides of the equation.)
÷9 x = 13 9 x = 1 4
(Divide both sides of the equation by 9.)
9
c
20 – 8x + 12x = 22
20 + 4x = 22
(Collect like terms.)
– 20 4x = 2
(Subtract 20 from both sides of the equation.)
x = 1 ÷4 2
(Divide both sides of the equation by 4.)
d
(Collect like terms.)
6x – 4 – 4x – 4 = 2 2x – 8 = 2
+8 2x = 10 ÷2 x = 5
(Add 8 to both sides of the equation.) (Divide both sides of the equation by 2.)
Chapter 16 Solving equations
163
Exercise 16G Example 15,16a,b
1
Solve each equation for x. Check your answer for c, f, i and l.
a 2x + 6 = x + 9
b 3x + 4 = x + 10
c 5x + 6 = 9x + 3
d 6x – 4 = 4x + 6
e 2x + 10 = x + 12
f 5x + 4 = x + 9
g 2x + 6 = 11x – 3
h 6x – 6 = 2x + 6
i 2x – 8 = x + 12
k 2x + 6 = 11x – 12
l 6x – 4 = 2x + 6
j 3x – 4 = x + 12
Example 16c,d
2
Solve:
a 2m + 5 = m
b 13z – 2 = 11z
c 20n + 5 = 10n
d 7m – 4 = 20m
e 20z + 6 = 8z
f 10m – 6 = 5m
3
Solve these equations by first collecting like terms.
a 50 – 4x + 10x = 150
b 6x – 5 – 2x – 6 = 12
c 60 – 2x + 8x = 4x + 100
d 5x – 2x + 6x = 3x + 12
e 20x + 30x – 10x = 5x + 100
f 17x + 10 = 13x + 20
g 60 = 7x + 45 – 4x
h 2x + 14 = 3x – 26 + 4x
i 5x – 12 = –6x + 23 + 4x
j 11x + 57 + 3x = 32x + 41 + 2x
k 9x – 125 + 16x = 8x – 100
l 12x – 36 + 10x = 15x + 36 – 3x
16H Solving problems using equations The algebra introduced so far can be used to solve problems that would otherwise be quite difficult. In each problem, we can use the following procedure: Step 1: Use a pronumeral to represent an unknown. Step 2: Construct an equation from the given information. Step 3: Use the methods we have learnt to solve the equation. Step 4: Check your answer with the given information.
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ICEEM Mathematics Secondary 1B
The three examples below show how to apply this procedure.
Example 17 A number is multiplied by 11 and 6 is added to the result. The final result is 24. What is the number?
Solution Let x be the number. The resulting equation is: –6
11x + 6 = 24 11x = 18 (Subtract 6 from both sides of the equation.) x = 18 (Divide both sides of the equation by 11.)
÷ 11
=
11 17 11
The number is 1 7 . 11
Check:
LHS = 11 × 1 7 + 6
= 11 ×
11 18 + 11
6
= 18 + 6 = 24 = RHS
Example 18 The length of a rectangle is twice its width. The perimeter of the rectangle is 45 cm. Find the length and width of the rectangle. (continued on next page)
Chapter 16 Solving equations
165
Solution Let the width of the rectangle be x cm.
2x cm
x cm
The length of the rectangle is 2 × x = 2x cm. The perimeter is 2x + 2x + x + x = 6x cm.
Ex
But it is given that the perimeter is 45 cm, so ÷6
6x = 45. x = 45 6 15 2
=
(Divide both sides of the equation by 6.)
The width of the rectangle is 15 = 7 1 cm. 2
2
The length of the rectangle is 2 × 15 = 15 cm. 2
Example 19 One number is 6 less than another number. The sum of the two numbers is 8 1 . Find each of the numbers. 4
Solution Let the larger number be x. Then the other number is x – 6. Therefore x + x – 6 = 8 1 . +6
2x
4 – 6 = 8 1 4 2x = 14 1 4 x = 7 1 8
÷2
The larger number is
166
71 8
(Collect like terms.) (Add 6 to both sides of the equation.) (Divide both sides of the equation by 2.)
and the smaller number is
ICEEM Mathematics Secondary 1B
71 8
–6=
11. 8
Ex
Exercise 16H For each question, introduce a pronumeral for the unknown and follow the steps given on page 164. Example 17
1
In each part, find the number. a A number has 7 added to it and the result is 15. b A number has 11 subtracted from it and the result is 23. c A number is divided by 7 and the result is 14. d A number is multiplied by 4 and the result is 48. e A number is multiplied by 7 and the result is 31. f A number is multiplied by 6, and 3 is added to the result. The result of this is 39. g A number is multiplied by 7, and 6 is subtracted from the result. The result of this is 23. h A number is multiplied by 3, and 11 is subtracted from the result. The result of this is 45. i A number is multiplied by 7, and 12 is added to the result. The result of this is 98. j A number has 3 added to it and the result is multiplied by 4. The result of this is 40. k A number has 7 subtracted from it and the result is multiplied by 8. The result of this is 32. l A number has 2 is added to it and the result is multiplied by 6. The result of this is 50.
Example 18
2
A square has perimeter 84 cm. Find the length of each side of the square.
3
The width of a rectangle is three times the length of the rectangle. The perimeter of the rectangle is 96 cm. Find the length and the width of the rectangle. Chapter 16 Solving equations
167
Example 19
4
One number is 3 more than another number. The sum of the two numbers is 33. Find the numbers.
5
The sum of two consecutive even numbers is 110. Find the numbers.
6
The difference of two numbers is 3 . The sum of the two numbers is 9 1 . 4 4 Find the numbers.
7
An equilateral triangle has perimeter 96 cm. Find the length of each side.
8
A crop of 2181 bananas is packed into a number of cases. There are 27 cases, each holding the same number of bananas. There are also 21 loose bananas. How many bananas are there in each case?
9
a Multiplying a number by 4 and subtracting 3 gives the same result as multiplying the number by 6 and subtracting 7. Find the number. b Multiplying a number by 7 and adding 12 gives the same result as subtracting the number from 14. Find the number. c Multiplying a number by 11 and subtracting 14 gives the same result as multiplying the number by 9 and adding 6. Find the number. d Multiplying a number by 3 and subtracting 8 gives the same result as subtracting twice the number from 12. Find the number. x + 25
10 Find x.
8x – 10
11 The perimeter of the isosceles triangle is 20. Find x.
x
12 Find x. x˚
168
x+5
(x – 100)˚
ICEEM Mathematics Secondary 1B
Review exercise 1
Twelve recycling boxes have x old newspapers in them. Three current newspapers are loose on the table. There are 363 newspapers in total.
a Write an equation for x.
2
Solve:
a x – 7 = 5
b 5a = 27
c 3z – 5 = 119
d 11b + 22 = 121
e 5x = 12
f x + 1 = 11
g n – 17 = –3
h 2p = 6
j
k x = –2
l x – 11 = 29
n 7x = 29
o 3x – 7 = 11
p 5x + 4 = 12
r 9 – m = –5
s 2(x + 6) = 23 t 5(3x – 5) = 17
4
i –4y = –18
m x + 7 = 10
7 x 6
= 18
b Solve the equation for x.
10
4
q 5x = 25
u 7(4 + 2x) = 50
v 6(3x – 4) = 13
w 4(2m + 3) = 50
x 7(2p – 11) = 77
3
3
In each case below, write an equation and solve it to find the number. a A number x has 11 added to it and the result is 23. b A number x is multiplied by 7 and the result is 25. c A number x is multiplied by 6 and then 2 is added. The result is 44. d A number x is divided by 7, and 3 is subtracted from it. The result is 6.
4
A rectangle is four times as long as it is wide. The perimeter of the rectangle is 120 cm. Find the length and the width of the rectangle.
5
The difference between two numbers is 12. The sum of the two numbers is 44. Find the two numbers.
6
Solve:
a x + 4 = –12
b x – 9 = –14
c x + 8 = 8
d 3x = –12
e –8x = –24
f –9x = 9
g x 4
= –7
j 5x – 3 = –14
h
–3x = 2
–8
i 4x + 1 = –8
k –2x + 10 = –12
l 4x – 6 = –8 Chapter 16 Solving equations
169
7
Solve each equation for x.
a 3x + 4 = 2x + 5
b 4x + 7 = x + 13
c 7x + 2 = 3x + 1
d 5x – 2 = 8x + 1
e x + 14 = 4x + 10
f 13x + 4 = 2x + 7
8
Solve:
a 5(2x – 4) + 3x = 10
b 6x – 2x + 7x = 42
c 32 + 4(x + 5) = 60
d 4(x – 5) + 26 = 52
f 10(2x – 10) + 15 = 84
e 50 + 2(4x – 5) = 100
In each of the following questions, first introduce a pronumeral and then solve the equation. 9
I think of a number and divide it by 5. The answer is 4. What is the number?
10 In 29 years a man will reach the retiring age of 65 years. How old is he now? 11 I have read 163 pages of a book. How many pages are left to be read if the book contains 390 pages? 12 A man is three times as old as his son. In eight years’ time the man will be 44. How old is his son now? 13 A boy throws away onethird of the cakes he has cooked. If he has 34 left, how many did he cook in the first place? 14 The perimeter of a rectangle is 40 cm. If the width is 9 cm, find the length. 15 The nth odd number is 81. Find n. 16 Multiplying a number by 6 and subtracting 6 gives the same result as multiplying the number by 3 and subtracting 4. Find the number. 17 $400 is to be shared among three friends. David gets $8 more than Jennifer who gets $16 more than Brent. How much does each of the friends receive? 18 Onequarter of the Smarties in a packet are blue. There are 69 Smarties in the packet that are not blue. How many Smarties are there in the packet?
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ICEEM Mathematics Secondary 1B
19 The two rods AB and CD have the same length and overlap by 5 cm. Find the length of each rod. 55 cm A
B C
D 5 cm
20 Three athletes are training for a race. In a particular week Jennifer ran 12 km less than Anne, and Carl ran half as far as Anne and the same distance as Jennifer. How far did each of the athletes run? 21 I think of a number. I subtract 7 from the number and multiply the difference by 6. The result is 48. Find the number. 22 a Multiplying a number by 8 and adding 6 gives the same result as multiplying the number by 3 and subtracting 2. What is the number? b Subtracting 10 from 6 times a number gives the same result as multiplying the number by 7 and adding 6. What is the number? c Multiplying a number by 11 and adding 7 gives the same result as subtracting the number from 10. What is the number? 23 The perimeter of the isosceles triangle is 50. Find x.
x + 10
x
24 Find the value of x.
(x + 20)˚ (x + 10)˚
x˚
25 Four hundred and fifty dollars is to be divided among three students, David, Isabel and Jason. David receives three times as much as Jason, and Isabel receives twice as much as Jason. How much does each student receive?
Chapter 16 Solving equations
171
Challenge exercise 1
a Find the perimeter of the shape shown opposite in terms of x. (All angles are right angles.)
b If the perimeter is 64 cm, find the value of x.
2
Think of a number. Let this number be x.
2x cm
3x cm
a Write the following using algebra to see what you get. • Add 6 to the number, then • multiply the result by 10, then • subtract the result from 20, then • multiply this result by 2. b If this final result is twice the original number, find the original number, x. c If the number obtained in a is 1000 less than the original number, find the original number x. 3 Humphrey leaves Tantown at 2:00 pm and travels at 60 km/h along the Trans Highway. He passes through Sintown which is 16 km from Tantown along the Trans Highway. Petro leaves Sintown at 3:00 pm and travels at 80 km/h along this highway in the same direction that Humphrey is travelling in. When and where does Petro catch up to Humphrey? 4
172
Let n be the age of a student (in years). The student’s sister is three years older, and the student’s father is 25 years older than the student. The sum of the ages of the student and his sister is 11 less than the age of their father. Find the ages of the student, his sister and their father.
ICEEM Mathematics Secondary 1B
Chapter 17 Probability In this chapter, we introduce probability, which deals with how likely it is that something will happen. This is an area of mathematics with many diverse applications. The study of probability began in 17thcentury France, when the two great French mathematicians Blaise Pascal (1623–1662) and Pierre de Fermat (1601–1665) corresponded about two problems from games of chance. Problems like those that Pascal and Fermat solved continued to influence the early development of the subject. Nowadays probability is used in areas ranging from weather forecasting and insurance (where it is used to calculate risk factors and premiums), to predicting the risks and benefits of new medical treatments and forecasting the effects of global warming.
17A An introduction to probability Most people would agree with these statements:
Hello!
???
• It is certain that the sun will rise tomorrow. • If I toss a coin, getting a head and getting a tail are equally likely. • There is no chance of finding a plant that speaks English. Chapter 17 Probability
173
Everyday language Using everyday language to discuss probabilities can cause problems, because people do not always agree on the interpretation of words such as ‘likely’, ‘probable’ and ‘certain’. Consider these two examples: • Two farmers are discussing the prospects of getting a good wheat crop this year. Farmer Bill says, ‘I don’t think it is likely to rain for the next two weeks. I’m not going to plant wheat yet.’ Farmer Tony says, ‘I reckon you’re wrong. I’m certain we’ll have rain. It can’t go on the way it has. I’m getting the tractor out tomorrow.’ • Alanna is captain of the Platypus Netball Team. They are going to play the Echidnas, whose captain is Maria. Each captain says to her team before the match, ‘I think we’ll probably win. Just follow the plans we’ve practised all week.’ Alanna and Maria cannot both be right! It would be futile to try to assign a probability that the Echidnas (or the Platypuses) are going to win on the basis of what the captains told their teams. There are many situations in which it would be useful to be able to measure how likely (or unlikely) it is that an event will occur. We can do this in mathematics by using the idea of probability, which we define as a number between 0 and 1 that we assign to any event we are interested in. Then: • a probability of 1 represents an event that is ‘certain’ or ‘guaranteed to happen’ • a probability of 0 represents an event that we would describe as ‘impossible’ or one that ‘cannot possibly occur’ • an event that has a probability 1 is as likely to occur as not to occur 2
• an event that has a probability closer to 0 than to 1 is unlikely to occur • an event that has a probability closer to 1 than to 0 is likely to occur. 0 Impossible
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ICEEM Mathematics Secondary 1B
1 2
1 Certain
Using these ideas, let us look at the three statements at the start of this section and express them in the new language of probability. Original statement
Statement in terms of probability
It is certain that the sun will rise tomorrow.
The probability that the sun will rise tomorrow is 1.
If I toss a coin, getting a head and getting a tail are equally likely.
If I toss a coin, the probability 1 and the of getting a head is 2 1 probability of getting a tail is . 2
There is no chance of finding a plant that speaks English.
The probability of finding a plant that speaks English is 0.
Example 1 A TV quiz contestant is shown three closed doors and told that there is a prize behind only one of the doors. If the contestant opens one of the doors, what is his probability of winning the prize? 1
2
3
Solution From the point of view of the quiz contestant the prize is equally likely to be behind each of the doors. So the probability of the contestant winning the prize is 1 . 3
Chapter 17 Probability
175
Exercise 17A Example 1
1
Complete each of the following probability statements.
a It is certain that an iceblock will melt in the sun in Sydney. The probability that an iceblock will melt in the sun in Sydney is …….
b There is no chance of finding a dog that speaks German. The probability of finding a dog that speaks German is …….
c If there are three red discs and three blue discs in a bag and you take one out without looking in the bag, you are equally likely to get a blue disc or a red disc. If there are three red discs and three blue discs in a bag and you take one out without looking in the bag, the probability of getting a blue disc is …… and the probability of getting a red disc is …….
d If it is Thursday today, tomorrow will be Friday. If it is Thursday today, the probability that it will be Friday tomorrow is …….
e If today is the 31st of January, there is no chance that tomorrow will be the 1st of May. If today is the 31st of January, the probability that tomorrow is the 1st of May is …….
2
Make up some statements for which there is a corresponding probability statement involving the probabilities 0, 1 or 1 . 2
17B Experiments and counting A standard dice is a cube with each face marked with a number from 1 to 6 or with marks as shown in the diagram. Suppose we toss a standard dice. Since the outcomes 1, 2, …, 6 are equally likely (assuming that the dice is fair, that is, it is not ‘loaded’), the probability of getting a 1 is 1 , 6 the probability of getting a 2 is 1 , and so on. 6
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The total probability is 1 Question: If we toss a dice, what is the chance of getting 1 or 2 or 3 or 4 or 5 or 6? Answer: It is certain that we will get one of the numbers 1, 2, 3, 4, 5, 6, so we say that the total probability is 1.
Sample space Tossing a dice and recording the face that shows up is an example of doing an experiment. The numbers 1, 2, …, 6 are called the outcomes of this experiment. The complete set of possible outcomes for any experiment is called the sample space of that experiment. For example, we can write down the sample space for the experiment of tossing a dice as: 1
2
3
4
5
6
In this case, each outcome has probability 1 . 6
Events An event is something that happens. In everyday life, we speak of sporting events, or we may say that the school concert was a memorable event. We use the word ‘event’ in probability in a similar way. For example, suppose that we toss a dice and we are interested in getting a prime number. In this case ‘the number is prime’ is the event that interests us. Some of the outcomes will give rise to this event. For instance, if the outcome is 2, then the event ‘the number is prime’ takes place. We say that the outcome 2 is favourable to the event ‘the number is prime’. If the outcome is 4, then the event ‘the number is prime’ does not occur. The outcome 4 is not favourable to the event. Of the six possible outcomes, these are the ones that are favourable to the event ‘the number is prime’: 2
3
5
In many situations ‘success’ means favourable to the event and ‘failure’ means not favourable to the event.
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Example 2 Hassan tosses a dice. Hassan’s favourite numbers are perfect square numbers. a What is the sample space? b What are the outcomes that are favourable to the event ‘the result is a perfect square’?
Solution a The sample space is: 1
2
3
4
5
6
b The only perfect squares in the sample space are 1 and 4, so the outcomes favourable to the event are: 1
4
Probability of an event Suppose I throw a dice. Then a natural question to ask is: What is the probability that I get a 1 or a 6? To make sense of this, we have to decide how the question is to be interpreted. There are six possible outcomes: 1
2
3
4
5
6
All are equally likely. Of these, only two are favourable to the event ‘the result is 1 or 6’. They are: 1
6
We say that the probability of getting a 1 or a 6 is 2 = 1 because all of 6 3 the outcomes 1, 2, 3, 4, 5, 6 are equally likely. In general, the following rule is true.
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Probability of an event For an experiment in which all of the outcomes are equally likely: Probability of an event = number of outcomes favourable to that event total number of outcomes
Here is a good way of setting out the solution to a question about the probability of an event.
Example 3 I throw a dice. What is the probability that the result is a perfect square?
Solution The sample space of all possible outcomes is: 1
2
3
4
5
6
The outcomes that are favourable (to the event ‘the result is a perfect square’) can be circled, as shown. 1 2
3
4 5
6
Probability that the result number of favourable outcomes is a perfect square = total number of outcomes
=2
=
6 1 3
Notation From now on, we will often write ‘P’ instead of ‘probability of’. Thus we write the previous result as: P(perfect square) = 1 . 3
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Example 4 I toss a dice. What is the probability of getting a 2 or a 3 or a 4 or a 5?
Solution The sample space, with the favourable outcomes circled, is: 1
2 3 4 5 6
P(2 or 3 or 4 or 5) =
=
=
number of favourable outcomes total number of outcomes 4 6 2 3
Class discussion Can you see that this is equal to 1 – P(1 or 6)? Why is this so?
The word ‘not’ Sometimes we need to recognise the word ‘not’ hidden in the question. The next example gives a simple illustration of what we mean by this.
Example 5 Joe is playing a board game. If he tosses a 1 with the dice, he goes to gaol. What is the probability that he does not go to gaol?
Solution This is the same as the probability of NOT getting a 1, which means that Joe gets a 2, 3, 4, 5 or 6. So P(not a 1) = P(2, 3, 4, 5 or 6)
=5 6
Alternatively, we can consider the probability of getting anything except a 1.
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Then P(not a 1) = 1 – P(1)
=1– 1
5 , 6
=
6
as before.
The words ‘random’ and ‘randomly’ In probability, we frequently hear these words, as in the following situations. • A classroom contains 23 students. A teacher comes into the room and chooses a student at random to answer a question about history. What does this mean? It means that the teacher chose the student as if she knew nothing at all about the students. Another way of interpreting this is to imagine that the teacher had her eyes closed and had no idea who was in the class when she chose a student. • In a TV quiz show, a prize wheel is spun. The contestant is asked to guess which number the wheel will stop at. This is a random choice because the contestant really has no idea where the wheel will stop.
Example 6 In a pickabox show, there are five closed boxes, each containing a snooker ball. Two of the boxes contain yellow balls and the others contain a green ball, a black ball and a red ball. The contestant will win a prize if she chooses a yellow ball. She chooses a box at random and opens it. What is the probability that she wins a prize?
Solution Here is the sample space with the favourable outcomes circled. They are equally likely, because she chooses at random. Y
Y
G
B
R
P(prize) = 2 5
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Exercise 17B Example 2
1
A bag contains 10 marbles numbered 1 to 10. If a marble is taken out: a give the sample space for this experiment b write down the favourable outcomes of the event ‘a marble with an odd number is taken out’. A dice is thrown. Write down the outcomes favourable to each given event.
a An even number is thrown. b An odd number is thrown. c A number divisible by 3 is thrown. d A number greater than 1 is thrown.
3
a For the spinner shown opposite, write down the sample space of the experiment ‘spinning the spinner once’.
b Write down the outcomes favourable to the event ‘the number obtained is less than 4’.
4
A dice is thrown. Work out the probability that:
a an even number is thrown
3
4
2
Example 3,4
2
1
5
b an odd number is thrown
c a number divisible by 3 is thrown
d a number greater than 1 is thrown.
5
A bag contains 10 marbles numbered 1 to 10. If a marble is taken out, what is the probability of getting:
a a 3?
For the experiment of spinning the spinner shown opposite once, write down the probability of:
a obtaining a 4 b obtaining an odd number. ICEEM Mathematics Secondary 1B
1
4
6
5
e a number greater than 1?
2
182
d a number divisible by 3?
3
c a number greater than 2?
b an even number?
Ex
Example 5
7
The Scrabble pieces for the word MELBOURNE are placed in a box.
M3 E1 L1 B3 O1 U1 R1 N1 E1
One piece is withdrawn. What is the probability of obtaining:
a an M?
8
A dice is thrown. What is the probability of:
a obtaining a 3?
c obtaining a number divisible by 3?
d obtaining a number not divisible by 3?
9
The first five letters of the alphabet are written on separate cards and the cards are put in a hat.
b an E?
c a vowel?
d a consonant?
b not obtaining a 3?
A B C D E
One card is taken out of the hat. Find the probability of getting:
a the C
b a vowel
c a consonant.
10 A bag contains three red marbles numbered 1 to 3, five green marbles numbered 4 to 8, and two yellow marbles numbered 9 and 10. A single marble is withdrawn. Find the probability that:
a the marble is numbered 3
c the marble is yellow
e the number is greater than 6
b the marble is green d the number is odd f the number is green and even.
11 The letters of the alphabet are written on flashcards and put into a bag. A card is drawn out at random. Find the probability that:
a it is D
b it is not W, X, Y or Z
c it is a vowel d it is in the word ‘PERTH’
e it is in the word ‘CANBERRA’. Chapter 17 Probability
183
Example 6
12 There are three green apples and two red apples in a bowl. Giorgia is blindfolded and randomly chooses one of the apples. What is the probability that she chooses a green apple? 13 A bag of sweets contains 10 red ones, nine green ones, six yellow ones and five blue ones. They are all the same size and shape. You reach in a take one out at random. Find the probability that the sweet you pick is:
a blue d purple
b green or yellow e not yellow or blue.
c not red
14 Cards with numbers 1 – 100 written on them are placed in a box. One card is randomly chosen. What is the probability of obtaining a number divisible by 5?
17C Empirical probability In our discussion of probability so far, we have been concerned with measuring ‘how likely’ an event is to occur. We have approached this by thinking about equally likely events. When we perform practical experiments, we do not always find that things work out exactly. For example, if we toss a dice six times, we do not usually get each face appearing exactly once. The word ‘empirical’ refers to the results of practical experimentation. We will now look at what this means in terms of probability. A dice with six faces is thrown 600 times and the outcomes are recorded as follows: Outcome
1
2
3
4
5
6
Number of appearances
85
107
110
109
96
93
Although the six possible outcomes are equally likely, we do not, in practice, get each face appearing exactly 100 times when we throw a dice 600 times. Nevertheless, in the above experiment, each outcome appears approximately 100 times.
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Example 7 If we toss a dice 6000 times, how many times approximately would we expect each face to appear?
Solution Since each of the six outcomes is equally likely, in 6000 tosses of the dice we would expect each face to appear approximately 1000 times.
Further experiments We now look at two further experiments involving different aspects of empirical probability.
Example 8 A pile of cards has six cards marked with a 1, six cards marked with a 2, six cards marked with a 3 and six cards marked with a 4. The cards are shuffled and one card is chosen. The number on it is noted and the card is returned to the pile. The cards are then reshuffled. If this experiment is performed 72 times, roughly how many times would we expect to see a 3?
Solution Since each type of card is equally likely to be chosen, the probability of getting a 3 is 6 = 1 , so we would expect to see a card 24 4 with the number 3 in 1 of the experiments. We would expect a 4 card with the number 3 to be chosen approximately 72 ÷ 4 = 18 times.
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Example 9 A school has 1200 students. The table below gives information about whether or not each student plays a musical instrument. Boys
Girls
Plays a musical instrument
325
450
Does not play an instrument
225
200
Each student has a school number between 1 and 1200. Each student’s number is written on a card and the 1200 cards are placed in a hat. One card is pulled out of the hat. What is the probability that the number pulled out belongs to: a a boy?
b a girl?
c a student who plays a musical instrument? d a student who does not play an instrument? e a boy who plays a musical instrument?
Solution a Number of students = 1200
Number of boys = 325 + 225
= 550
P(boy’s number) = 550
=
1200 11 24
b Number of students = 1200
Number of girls = 450 + 200
= 650
P(girl’s number) = 650
=
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ICEEM Mathematics Secondary 1B
Ex 7,
c
Number of students = 1200
Number who play a musical instrument = 325 + 450 = 775
P(plays a musical instrument) = 775
= d
1200 31 48
Number of students = 1200
Number who do not play a musical instrument = 225 + 200 = 425 P(not plays a musical instrument) = 425
= e
1200 17 48
Number of students = 1200
Number of boys who play a musical instrument = 325
P(boy who plays a musical instrument) = 325
=
1200 13 48
Exercise 17C Example 7,8
1
In a light bulb factory, it is found that on a particular production run the probability of a bulb being defective is 0.1. Estimate how many defective bulbs there would be in:
a 1000 bulbs
2
The probability that a particular type of battery is defective is 1 . 200 Estimate how many defective batteries there would be in:
a 800 batteries
b 8000 bulbs
c 100 000 bulbs.
b 16 000 batteries.
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3
At home, toss a coin 100 times and record the number of heads.
a Write the fraction total number of tosses .
number of heads
b Put your results together with those of other students who completed the same experiment. Do this by adding the total number of heads recorded by everyone in the group. Record also the total number of tosses of everyone in the group. Write the fraction number of heads recorded by the group total number of tosses made by the group .
(Discuss these results with your teacher.) 4
Over a period of 10 years, the number of rainy days in July in Melbourne was recorded. The results are as shown below. Rainy days in July in 16 Melbourne over 10 years
18
14
20
12
17
15
16
19
13
a Find the total number of rainy July days in Melbourne over the 10 years. b Using these observations, estimate the probability of rain falling on a particular day in the month of July. Example 9
5
A bowl contains green and red normal jelly beans and green and red doubleflavoured jelly beans. The numbers of the different colours and types are given in the table below. Green
Red
Normal jelly beans
250
400
Doubleflavoured jelly beans
175
75
A jelly bean is randomly taken out of the bowl. Find the probability that: a it is a doubleflavoured jelly bean b it is a green jelly bean c it is a green normal jelly bean d it is a red doubleflavoured jelly bean e it is a red jelly bean.
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6
In a traffic survey, during a 30minute period, the number of people in each passing car was noted, and the results tabulated as follows. Number of people in car
1
2
3
4
5
Number of cars
60
50
40
10
5
a What is the probability that there was only one person in a car during this period? b What is the probability that there was more than one person in a car during this period? c What is the probability that there were less than four people in a car during this period? d What is the probability that there were five people in a car during this period? 7
A door prize is to be awarded at the end of a concert. It is announced that the winner will be chosen randomly. The numbers of people at the concert in different age groups are given in the table below. Age group
0−5 6−11 12−18 19−30 30−40 40+
Number of people in age group
10
150
350
420
125
85
What is the probability of the prize winner being:
a in the 0–5 age group?
b in the 19–30 age group? d aged between 12 and 30?
c aged 40 or less
f older than 40?
e older than 5?
8
A school has 1500 students. The table below gives information about whether or not a student plays a sport for the school.
Boys
Girls
Plays sport
500
400
Does not play sport
300
300
A student is chosen at random. What is the probability that the student is a boy who does not play sport? Chapter 17 Probability
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Review exercise 1
There are eight marbles, numbered 1 to 8, in a bowl. A marble is randomly taken out. What is the probability of getting:
a an 8?
c a number less than 6?
b an even number? d a number divisible by 3?
2
Twelve cards are put in a hat. Five of the cards are red and are numbered 1 to 5, and the other seven cards are blue and are numbered 6 to 12. A card is taken randomly from the hat. Find the probability of taking out:
a a blue card
c a card with a prime number
b a card with an even number d a red card.
3
From data collected over 100 years for the rainfall in a particular town, it is found that the probability of rain on any day in September is 1 . 6 How many rainy days would you expect in this town next September?
4
The scrabble pieces for the letters of the word ‘PROBABILITY’ are put in a box. One piece is chosen at random. What is the probability of choosing:
a the P?
5
Seven red marbles numbered 1 to 7 and six blue marbles numbered 8 to 13 are placed in a box. A marble is taken randomly from the box. Find the probability of taking out:
a a red marble
b a B?
c a vowel?
b a blue marble
c a marble with a prime number
190
d a marble with a multiple of 3
e a red marble with an even number
f a blue marble with an odd number
g a red marble with a number less than 6 on it. ICEEM Mathematics Secondary 1B
d a consonant?
6
A large bowl contains two types of lollies: chocolates and toffees. The chocolates and toffees are manufactured by the Yummy Sweet Company and the ACE Confectionary Company. The number of chocolates and toffees in the bowl of the two different brands are given in the table below. Yummy Sweet Company
ACE Confectionary Company
Chocolates
250
170
Toffees
300
140
A lolly is randomly taken out of the bowl. Find the probability that: a it is a chocolate manufactured by the Yummy Sweet Company b it is a toffee manufactured by the Ace Confectionary Company c it is a chocolate d it is a toffee.
7
a A dice is tossed 1002 times. How many times would you expect a 6 to be obtained? b A dice is tossed 3000 times. How many times would you expect a number less than 3 to occur?
8
The Scrabble pieces for the letters of the word ‘PROSPECTIVE’ are put in a box. One piece is chosen at random. What is the probability of choosing:
a the I?
9
Each number from 1 to 500 is written on a card and put in a box. A card is withdrawn at random. What is the probability of obtaining a card with:
a a number less than 100?
b a number divisible by 10?
b a P?
c an E?
d a vowel?
c a number divisible by 5?
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Challenge exercise 1
In a raffle, 50 tickets are sold. The tickets are numbered from 1 to 50. They are placed in a hat and one is drawn out at random to win a prize. Find the probability that the number on the winning ticket:
a is 43
c is even
b is not 28 d is less than or equal to 20
e is an even number less than 20 f contains the digit 7
g does not contain the digit 9
h contains the digit 2 at least once
i contains the digit 2 only once j does not contain the digit 2. 2
Suppose the raffle tickets in Question 1 are coloured: those numbered 1 to 25 are green, the tickets numbered 26 to 40 are yellow and the rest are blue. Find the probability that the ticket that wins the prize:
a is blue
c is green or blue
b is not green d is both green and blue f is oddnumbered and yellow
e is evennumbered and blue
3
Thomas, Leslie, Anthony, Tracie and Kim play a game in which there are three equal prizes. No player can win more than one prize. a List the ten ways the prizes can be allocated. b What is the probability of Leslie winning a prize? c What is the probability that Leslie does not win a prize?
192
4
Find the probability that a randomly chosen threedigit number is divisible by:
a three
b seven
ICEEM Mathematics Secondary 1B
Chapter 18 Transformations and symmetry Consider the triangle T drawn below. Each of the triangles T1, T2, T3 is obtained from the triangle T by moving it.
T
T1 Diagram 1
T
T2
Diagram 2
T
T3
Diagram 3
When the figure is moved, we call this movement a transformation. The resulting figure is called the image of the original figure. We will be dealing with three types of transformation in this chapter: •
translation (Diagram 1)
•
rotation (Diagram 2)
•
reflection (Diagram 3)
Note that none of these transformations changes the size or shape of an object. Transformations have many applications including computer imaging.
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All of these transformations satisfy two important properties. Properties of translation, rotation and reflection When a figure is translated, rotated or reflected: • intervals move to intervals of the same length • angles move to angles of the same size. Many of these exercises will require graph paper. The work on rotation will require some work on polar graph paper, also called circular graph paper. You can download samples of this from the ICEEM website at www.iceem.org.au, or photocopy the samples at the end of this chapter. Carefully drawn diagrams are essential to gain the most from this chapter.
18A Translation Shifting a figure in the plane without turning it is called translation. To describe a translation, it is enough to say how far left or right and how far up or down the figure is moved. In the diagram opposite, triangle ABC has been translated to a new position on the page, and the new triangle has been labelled AʹBʹCʹ. You can see that triangle ABC has been shifted 2 units right and 1 unit up. Point A has moved to Aʹ, point B to Bʹ and point C to Cʹ.
Aʹ A Bʹ B
C
The image of a point is usually written with a dash attached. Thus the image of A is written Aʹ (read as ‘A prime’ or ‘A dash’). The image of ABC is written AʹBʹCʹ. You can see that our translation has not changed the side lengths nor the angles of the triangle, in agreement with the properties listed in the box on the above box.
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Cʹ
Example 1 Describe how the triangle in the diagram opposite has been translated.
A
Aʹ
Bʹ
C
B Cʹ
Solution The figure has been translated 3 units left and 1 unit down. (Also note that 1 unit down followed by three units left is the same.) We note that triangle ABC is shown by a broken line while the image triangle AʹBʹCʹ is shown with a solid line.
Example 2 Translate XYZ in the diagram opposite 2 units down and 3 units left.
X Y Z
Solution We only need to shift the vertices of this triangle. We then join them up to give the translated figure and label it XʹYʹZʹ.
X Xʹ
Y Z Yʹ
Zʹ
Notice that in each of the examples above, every point in the triangle has moved. This is a special property of translation.
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A special property of translation When a translation is applied: • every point moves, that is, there are no fixed points. The translation which translates a figure 0 units right and 0 units up is called the identity translation. The identity translation leaves all point fixed.
Exercise 18A Example 1
1
Describe the translation shown in each diagram below. You need not copy the diagrams.
a
Aʹ
b
Aʹ
c
A
A Aʹ Bʹ
C
B
B
A
d
e
A
f
A
Aʹ
A
Aʹ Aʹ
C
B B
Bʹ C
Cʹ Bʹ
Example 2
Cʹ
2
In each part below, a figure is shown and a translation described. Copy each figure and draw its image after the translation. Remember to label your images using dashes.
a 4 units up, 3 units right
b 3 units right, 1 unit up
A
196
Cʹ
Bʹ
A
ICEEM Mathematics Secondary 1B
B
d 2 units right, 1 unit down
c 2 units up, 1 unit left
A A C
B
e 3 units left, 1 unit down
3
A
B
f 1 unit down, 2 units right A
B
C
C
In the diagram below, XʹYʹZʹ is the image under translation of XYZ. Copy the diagram onto your graph paper.
X Xʹ
Y
Z Yʹ
Zʹ
a Measure angles XYZ and XʹYʹZʹ. What do you notice? b Write down the angle in the image that is the same size as: i ∠XZY
ii ∠ZXY.
(Check with a protractor.) c Measure the lengths of the intervals XY and XʹYʹ. Comment. d Calculate the area, in square units, of Comment.
XYZ and
XʹYʹZʹ.
Chapter 18 Transformations and symmetry
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4
In the diagram below, AʹBʹCʹ is the image under translation of ABC. Copy the diagram onto your graph paper.
A Aʹ
B
L
C Cʹ
Bʹ
a Mark in the image of L under the same translation; label it Lʹ. b Describe the translation in words. c Are the intervals AB and AʹBʹ parallel? d Write down the interval in the image that is parallel to: i AC
ii AL.
e Write down the interval in the image that is parallel to and the same length as: i BL
ii CA.
f Are there any points in the original figure that have not moved under the translation? 5
In the diagram below, AʹBʹCʹ is the image under translation of ABC. Copy the diagram onto your graph paper.
A
B
Aʹ
Bʹ C
Cʹ
a Describe the translation in words. b Are the intervals AB and AʹBʹ parallel?
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ICEEM Mathematics Secondary 1B
c Write down the side of the triangle that is parallel to AC. d Are there any points in the original figure that have not moved under the translation?
18B Rotation A rotation turns a figure about a fixed point, called the centre of rotation. The centre of rotation is usually labelled by the letter O. A rotation is specified by: • the centre of rotation O • the angle of rotation • the direction of rotation (clockwise or anticlockwise). In the first diagram below, the point A is rotated through 120˚ clockwise about O. In the second diagram, it is rotated through 60˚ anticlockwise about O. Aʹ Aʹ
O
O
60˚
120˚ A
A
The examples above demonstrate the following special properties of any rotation.
Special properties of rotation When a rotation is applied: • there is only one fixed point – the centre of rotation • the distance of a point from the centre of rotation does not change. The first special property does not apply to the identity rotation, which rotates a figure through 0˚ and leaves every point fixed.
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To help us in working with rotations, we may use special graph paper called polar or circular graph paper. This is available at the end of the chapter. We will use two types of such paper, examples of which can be seen in the next two diagrams. The first is marked with radial lines 30˚ apart, while in the second the lines are 45˚ apart. The centre of rotation O is always the centre of the graph paper (the origin). O
O
In this graph paper, lines radiate out from O at 30˚ angles. The diagram is a bit like a clock face, with the lines pointing towards the hours.
The second diagram is like a compass, with the lines 45˚ apart pointing to N, NE, E, SE, S, SW, W and NW.
The next example shows how the image of an interval under rotation may be found by rotating its endpoints and joining their images.
Example 3 Rotate the interval ED in the diagram by 120˚ in a clockwise direction about O.
E D
O
Solution If we want to rotate the points D and E by 4 × 30˚ = 120˚ clockwise, then we must shift each point four ‘hours’ around the ‘clock’. Each point stays the same distance from the centre of rotation, that is, it stays on the same circle. The points D and E are rotated and then joined up to give DʹEʹ.
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ICEEM Mathematics Secondary 1B
E D
O Eʹ Dʹ
Example 4 Rotate XYZ in the diagram by 90˚ anticlockwise about O.
X O Z
Y
Solution A rotation of 90˚ anticlockwise moves every point though two divisions on the ‘compass’. To move the triangle XYZ, we rotate the three corners and join them up to see how the triangle has rotated.
X O Z
Yʹ
Y
Xʹ
Zʹ
You may like to try question 1 of Exercise 18B before considering the next example, which is related to question 2. What do we do if the picture is not on polar graph paper? One way is to construct your own polar grid about the centre of rotation – lines radiating out to help you rotate a point. In the following example, we will be concerned with rotating figures 90˚, 180˚ or 270˚, so a set square is ideal to construct these angles.
Example 5 Rotate the line AB by 90˚ clockwise about O.
B A
O
(continued on next page)
Chapter 18 Transformations and symmetry
201
Solution We only need to rotate the endpoints A and B. The image of the segment AB is obtained by joining Aʹ and Bʹ. This is explained in the following diagrams.
B
B
A
A Aʹ
Aʹ O
O
Bʹ
To rotate A we have drawn lines at 90˚, starting with OA. Compasses then help us rotate A around O to Aʹ, keeping it always the same distance from O.
The same process is repeated to rotate B around O to Bʹ.
Exercise 18B Use a polar graph paper and mark the points and figures carefully. Example 3,4
1
Rotate each figure about the centre O through the given angle. Remember to label your images with dashed letters.
a Rotate 135˚ anticlockwise
N O
A O
M
B
C
c Rotate 90˚ anticlockwise
d Rotate 30˚ clockwise
O X
202
b Rotate 90˚ clockwise
P
Z Y
ICEEM Mathematics Secondary 1B
O
Q
R
Ex
e Rotate 120˚ anticlockwise
f Rotate 30˚ anticlockwise
A O
O
Example 5
2
Rotate each figure through the given angle about the given centre O. Remember to label your images with dashed letters.
a Rotate 90˚ anticlockwise
b Rotate 90˚ anticlockwise
O O
A
A
c Rotate 90˚ clockwise
d Rotate 90˚ clockwise
A
A O
O
B
B
e Rotate 270˚ clockwise
f Rotate 270˚ anticlockwise
O
O
Chapter 18 Transformations and symmetry
203
g Rotate 180˚ clockwise
A
B O
h Rotate 180˚ clockwise A
C
O
i Rotate 180˚ anticlockwise B
B
A
j Rotate 180˚ anticlockwise
B
A O
O
C
C
k Rotate 90˚ clockwise
D
l Rotate 90˚ clockwise
A O
A O
3
Consider each of the parts of question 2 above. Were any points fixed under the rotations – or did they all move?
4
Consider the following diagram of a house. a Draw the image of the house after a rotation of 90˚ anticlockwise about O. Label it AʹBʹCʹDʹEʹ. (Hint: Remember EʹCʹ ⊥ EʹAʹ etc.) b AE and BC are parallel. What do you notice about AʹEʹ and BʹCʹ?
D
O
E
C
A
B
c Measure AB and AʹBʹ. Comment. d Find the area of ABCE and AʹBʹCʹEʹ in square units. Comment.
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18C Reflection A reflection is a transformation that flips a figure about a line. This line is called the axis of reflection. A good way to understand this is to suppose that you have a book with clear plastic pages and a figure drawn, as in the first diagram below. If the page is turned, the triangle is flipped over. We say it has been reflected; in this case the axis of reflection is the binding of the book.
Hello
olleH
Our later examples will look more like the diagram opposite. An axis of reflection has also been drawn (notice it does not have to be vertical, as in our first example).
A Aʹ
The image of each point can be determined in the following way. To reflect A, a line has been drawn at right angles to the axis of reflection and a ruler or compasses used to mark Aʹ at the same distance from the axis of reflection as A, but on the other side. This is best done as a construction using straight edge and compasses.
A
A Aʹ
As was the case for rotation, we can find the image of a triangle under reflection by reflecting the vertices and joining the image vertices.
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205
Example 6 Reflect ABC in the axis of reflection shown.
Ex
A B
C
Solution Aʹ
A B
C
Aʹ
A
Bʹ
B
Cʹ
Reflect each of the vertices.
Bʹ
C
Cʹ
Join them up to get the image triangle.
Here is a list of special properties of reflections. The second property can be seen in the previous example.
Special properties of reflection When a reflection is applied: • all points on the axes of reflection are fixed points • if the points A, B, C, … are in a clockwise order, then the points Aʹ, Bʹ, Cʹ, … will be in anticlockwise order, and vice versa.
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Exercise 18C Example 6
1
Reflect each figure in the given axis of reflection. (In each diagram, the axis of reflection is marked with an arrow at each end.) Remember to label your images in the proper way.
a
b
A
A B B C
c
d
X A
Z
B
Y
e
f
A
X Z
B
Y C
Chapter 18 Transformations and symmetry
207
g
h A
A
B B C
i
j
l
n
X Z Y
k
m
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ICEEM Mathematics Secondary 1B
o
2
This question involves two reflections. First reflect the triangle in the axis of reflection labelled m, and then in the axis of reflection labelled n. m n
3
a Carefully perform, on a sheet of graph paper, the construction outlined below to construct the reflection of ABC. The diagrams will help you if you get stuck, but try and do it on your own first. This time you will need to use a set square, a protractor or a pair of compasses to construct the lines perpendicular to the reflection.
A
B
A
B
A
B
Bʹ Aʹ C
C
C Cʹ
b Fold the page along the axis of reflection and note that onto AʹBʹCʹ.
ABC falls
c Check that ∠AʹBʹCʹ = ∠ABC. d Check that the intervals AB and AʹBʹ are the same length. e Are there any fixed points? Chapter 18 Transformations and symmetry
209
f Check that the lines AC and AʹCʹ make the same angle with the axis of reflection. g What do you notice about the order of the vertices ABC and the order of the image vertices AʹBʹCʹ?
18D The three transformations Here is a summary of what we have discovered about translation, rotation and reflection. • Intervals move to intervals of the same length. • Angles move to angles of the same size. • Pairs of parallel lines move to pairs of parallel lines. Transformation
Defined by
Fixed points
Translation
Vertical and horizontal shift
None
Rotation
Centre, angle and direction
The centre of rotation
Reflection
Axis of reflection
Points on the axis of reflection
Reflection is the only one of the three transformations that reverses the order of vertices. Thus if a triangle is labelled ABC in clockwise order, then the vertices of AʹBʹCʹ will appear in anticlockwise order after reflection. Aʹ
A
C
B
Bʹ
Cʹ
This is different from the case when a translation or rotation is applied to ABC with vertices in clockwise order. The vertices of AʹBʹCʹ will also be in clockwise order as shown below.
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ICEEM Mathematics Secondary 1B
Aʹ
A
Aʺ
Cʺ
C
B Cʹ
Bʹ
Bʺ
The following exercise involves no new transformations, but uses combinations of the three transformations we have learned about. You will be asked to transform points or figures in the number plane. If you get stuck on any step, look back at the earlier exercise dealing with the relevant kind of transformation.
Exercise 18D You will need graph paper to do these exercises. Each question requires that you draw a number plane with labelled x and yaxes. A single number plane for each question is fine (for example, one number plane for question 1 parts a–f), if points are neatly labelled. The coordinates of the image point(s) should also be given. Translation 1
Mark the following points on your graph paper and translate each one 2 units right and 3 units up. Be careful to label each point and its image. Also write down the coordinates of each image point.
a A(0, 0)
b B(2, 1)
c C(–3, 1)
d D(–5, –5)
e E(–3, 2)
f F(0, 2)
2
Given the following points and their images under a translation, plot the points and describe the translation.
a A(0, 0) and Aʹ(2, 1)
c C(0, 4) and Cʹ(0,–4)
e E(4, 4) and Eʹ(0, 0)
b B(1,–2) and Bʹ(3, 3) d D(2, 3) and Dʹ(–2,–3) f F(3, 3) and Fʹ(3,–1)
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Rotation 3
Mark the given points on a number plane and rotate them 90˚ anticlockwise about the origin O. Write down the coordinates of each image point.
a A(2, –1)
b B(2, 1)
c C(–3, 1)
d D(–5, –5)
e E(0, 0)
f F(0, 2)
4
Mark the given points on a number plane. Find their images under a rotation of 180˚ anticlockwise about the origin O. Draw each image point and write down its coordinates.
a A(2, –1)
b B(2, 1)
c C(–3, 1)
d D(–5, –5)
e E(0, 0)
f F(0, 2)
Reflection 5
Copy the given points onto a number plane. Find their images when reflected in the xaxis.
a A(2, –1)
b B(2, 2)
c C(–3, 3)
d D(–5, 0)
e E(0, 4)
f F(–1, –1)
6
Mark the given points on a number plane. Find their images when reflected in the xaxis.
a i A(2, 1)
ii B(3, 3)
b Join points A, B and C in part a to form ABC under reflection in the xaxis. 7
ABC. Find the image of
Mark the given points on a number plane. Find their images under a reflection in the yaxis. a i A(3, –1)
ii B(–3, 3)
b Join points A, B and C in part a to form ABC under reflection in the yaxis.
212
iii C(5, 0)
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iii C(0, 4) ABC. Find the image of
18E Symmetry Symmetry is an important idea in mathematics. We also see it in art and architecture. Here we are going to consider two types of symmetry, namely rotation symmetry and reflection symmetry.
The figure shown opposite is symmetric (reflection symmetry) because it falls back exactly on itself when it is reflected in the axis of reflection marked on the diagram. We call the axis of reflection an axis of symmetry.
Think of rotating the figure shown opposite by 90˚ anticlockwise. The figure falls back on itself, and we say that it has rotation symmetry. We look at the situation after repeated anticlockwise rotations of 90˚. For example, the original point A in diagram 1 moves to the point A in diagram 2. It then moves to the point A in diagram 3, and so on.
O
C
A D
Diagram 1
A B
D
O
A
O
D
C
C
anticlockwise rotation of 90˚ Diagram 2
B
2 anticlockwise rotations of 90˚ Diagram 3
C D
B
B
O
C B
O A
A D 3 anticlockwise rotations of 90˚ 4 anticlockwise rotations of 90˚ Diagram 4 Diagram 5 Chapter 18 Transformations and symmetry
213
It takes four rotations of 360 ˚ = 90˚ anticlockwise for the point A to 4 return to its original position. Try repeated rotations of angles smaller than 90° such as 360 ˚ and 360 ˚. 5 6 You should be able to convince yourself that there is no angle of rotation less than 90° that can be applied repeatedly to take the figure back on itself. What about angles larger than 90° such as 360 ˚ = 180°? While it can be 2 seen that a rotation of 180˚ does in fact take the figure back on itself, each 180° rotation achieves the same result as two rotations of 90°. We say that our figure has rotation symmetry of order 4 because: • repeated rotations of amount 360 ˚ carry the figure back on itself 4
• none of the rotations by the smaller angles 360 ˚, 360 ˚, 360 ˚, ... does 5 6 7 the same thing. Can you see that this propeller
has rotation symmetry of order 3?
The rotation by 360° is called the identity, because any point in a figure returns to where it started after a rotation of 360°.
Example 7 a Draw any axes of reflection symmetry for the figure shown opposite.
b Write down the order of rotation symmetry.
Solution a The axes of symmetry are shown in the solution diagram opposite. There are 4 of them. The axes of symmetry are shown as dotted lines. b The order of rotation symmetry is 4, because the cross falls back on itself after rotations of 90˚, 180˚, 270˚ and 360˚ about the centre O.
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O
Ex
This example shows that figures may have more than one axis of symmetry, and figures may have both rotation and reflections symmetry. A figure has rotation symmetry of order 1 if there is no suitable angle of rotation smaller than 360˚.
Exercise 18E Example 7
1
In the table below, the letters of the alphabet and several common symbols are drawn. Copy the table into your book. a For each of these letters and symbols: mark any axes of reflection symmetry. b Does each figure have rotation symmetry? If so, what is the centre of these rotations? (Mark it on the diagram.) Give the order of rotation symmetry.
Note: The symmetry of some letters may vary, depending on how you draw them. For example, O is not drawn here as a perfect circle. Copy the letters accurately and comment on any issues of this type.
A B C D E
F
Order:
G H
I
J
K L
Order:
M N O P Q R Order:
S
T U V W X
Order:
Y Z = + $ % Order: Chapter 18 Transformations and symmetry
215
2
In the following table, a number of common mathematical figures are drawn. Copy the table into your book. a For each of these shapes: mark any axes of reflection symmetry. b Does each figure have rotation symmetry? If so, mark the centre of rotation on the diagram and give the order of rotation symmetry.
Assume that the rhombus is not a square, the kite is not a rhombus, the isosceles triangle is not equilateral, the trapezium has unequal vertical sides and the parallelogram is not a rhombus.
square
rectangle
rhombus
kite
equilateral triangle
isosceles triangle
right triangle
trapezium
circle
ellipse
Order:
Order:
Order:
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parallelogram
18F Regular polygons A regular polygon is a plane figure in which every side is an interval of the same length and all internal angles are equal. A polygon is named after its number of angles; this is the same as the number of sides. For example, the figure on the right is called a regular pentagon, from the Greek word pente, meaning five and gonos meaning angled. The equal angles are marked. The names of some common regular polygons are given in the table below. Number of sides
Name
3
equilateral triangle
4
square
5
regular pentagon
6
regular hexagon
7
regular heptagon
8
regular octagon
9
regular nonagon
10
regular decagon
12
regular dodecagon
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217
Drawing a regular pentagon Use a protractor in the following to form angles of 72˚. B
B
C O
C
A
72˚
A
D
D E
Draw a circle of radius 5 cm. Mark a radius OA.
A
O
O
Draw rays 360˚ ÷ 5 = 72˚ apart, starting at OA. Let these rays intersect the circle at B, C, D, E.
E
Join A, B, C, D, E to form the regular pentagon.
Symmetry of regular polygons Regular polygons show a lot of symmetry. For example, the regular pentagon has five axes of reflection symmetry, as shown in the diagram opposite. As we turn the pentagon about the centre point O, every 72˚ it will ‘fall back on itself’, looking identical to the original diagram even though the points have moved. The five rotations 72˚, 144˚, 216˚, 288˚, 360˚ about the centre O are symmetries. The points of the figure do not return to their original positions until the complete 360˚ rotation is done. For this reason, we say that the regular pentagon has rotation symmetry of order 5.
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ICEEM Mathematics Secondary 1B
3
4 5
2 O
1
Exercise 18F 1
a Draw a ‘small’ regular pentagon, using a circle of radius 4 cm and the method above. b Mark the axes of reflection symmetry.
2
This question requires you to draw a regular nonagon (also called a 9gon). a Explain why the rays in your construction have to be 40˚ apart. b Using a protractor, construct the regular nonagon in a circle of radius 5 cm. c Draw the axes of symmetry. Write down the order of rotation symmetry of a regular nonagon.
3
a How many sides does a regular octagon have? b Construct a regular octagon using only a straight edge and compasses and by following the steps below. A
B
A
E
B
A
B
H D
C
D
C
F
D
C G
Construct a square of side length 4 cm. Mark the diagonals of the square to locate its centre.
Place the point of your compasses on the centre of the square, and draw a circle around the square, touching its four vertices. Mark the midpoints of the sides of the square.
Connect the opposite midpoints with a line that extends each end to meet the circle. Join the points A, E, B, F, C, G, D and H to form a regular octagon.
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219
c What is the angle between the rays in your construction? d Draw the axes of symmetry and write down the order of rotation symmetry.
e Use a different colour to join every second vertex and so construct a regular 4gon (a square!). Which of the axes of symmetry of the octagon will be axes of symmetry of the square? What is the order of rotation symmetry of the square?
4
a Follow the steps below to construct a regular hexagon without using a protractor. B
C O
A
E
Draw a circle of radius 5 cm. Leave your compasses set at 5 cm and mark a point A on the circle.
A
O
D
F
Place the point of your compasses on A and mark an arc intersecting the circle at B. Move the point of your compasses to B and mark an arc intersecting the circle at C. Continue around the circle.
B
C
A
O
D E
F
To check your accuracy, draw a final arc centred at F. It should pass through A. As before, join the points A, B, C, D, E and F to form a regular hexagon.
b Explain why this construction works. c Draw in the axes of reflection symmetry and write down the order of rotation symmetry.
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ICEEM Mathematics Secondary 1B
5
a Construct a regular dodecagon in a circle of radius 5 cm, using the following steps. A
B
A
G
B
L C
F
E
D
Start by constructing a hexagon using the method in question 4. Mark the midpoints of each side of the hexagon.
A H C
F I
K E
J
D
Connect the opposite midpoints with a line that extends each end to meet the circle.
G
B
L
H C
F I
K E
J
D
Join the points A, G, B, H, C, I, D, J, E, K, F and L to form a regular dodecagon.
b Label the vertices 1, 2, 3, …, 12 clockwise (with 12 as the topmost vertex). What have you constructed? c How many axes of symmetry will this polygon have, and what is its order of rotation symmetry? 6
What are the orders of rotation symmetry of a regular 36gon and a regular 37gon? Do not attempt to draw them!
Chapter 18 Transformations and symmetry
221
18G Combined transformations We can combine two or more transformations by applying them one after another, as in the examples below.
Example 8 Translate ABC 2 units right and 1 unit up, then translate the image 3 units left and 2 units up. Give a single translation that has the same effect.
C
A
B
Solution The transformation is shown in the diagram opposite. The combined translation is the same as a translation 1 unit left and 3 units up.
Cʺ Cʹ Aʺ A
C Bʺ
Aʹ
B
Bʹ
Example 9 Rotate the point A by 45˚ and then by a further 90˚, both rotations being anticlockwise about O. Give a single rotation that has the same effect as the combination of these two rotations.
O
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ICEEM Mathematics Secondary 1B
A
Ex
Solution The first rotation moves A to Aʹ (see the diagram opposite). The second rotation then moves Aʹ to Aʺ. This combined transformation is the same as a single transformation involving a rotation of 135˚ anticlockwise about O.
135˚
Aʹ
45˚
Aʺ A 0˚
O
Example 10 Describe two successive transformations that will have the combined effect of mapping ABC onto AʹBʹCʹ as shown opposite.
Aʹ
C
A
B
Bʹ Cʹ
Solution There are many possible answers. One possible answer is: 1 first translate ABC so that A moves to Aʹ, that is, 4 units right and 3 units up 2 then reflect the triangle in AʹBʹ.
Exercise 18G Example 8
1
Perform each pair of translations on the given figure. Then find a single translation that has the same effect.
b shift 3 down, then 2 right a shift 1 right, 2 up, then 3 right, 1 up
A
Chapter 18 Transformations and symmetry
223
c shift 3 right, 2 down, d shift 1 right, 3 up, then 2 down, 2 right then 2 up
Example 9
2
Carry out the first rotation given for each figure, draw the image, and then carry out the second rotation. Find a single rotation that will have the same effect.
a 60˚ anticlockwise, then 30˚ anticlockwise
b 30˚ anticlockwise, then 60˚ clockwise
c 120˚ anticlockwise, then 90˚ anticlockwise
a
3
a
Copy the first diagram below.
b
Reflect the triangle in m, then n.
c
Now reflect the original triangle in n, then m.
d
Are your answers to parts a and b the same?
e–h Repeat the exercise for the second diagram.
b
c
n
n
m
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ICEEM Mathematics Secondary 1B
m
Ex
Example 10
4
In each part, describe two successive transformations that will have the combined effect of mapping ABC onto AʺBʺCʺ.
a
A
Bʺ
b A
Aʺ
Cʺ
Cʺ C
B
Bʺ
B
c
Aʺ
C
d Aʺ
Bʺ Bʺ Aʺ A
C
Cʺ C
B
A
B
e
Cʺ
f
Cʺ B C
Aʺ
Bʺ C, Aʺ
A A
B
Bʺ
Cʺ
Chapter 18 Transformations and symmetry
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Review exercise 1
Describe the translation shown in each of these diagrams.
a
b
c
B
B
Bʹ Aʹ
Bʹ
C
A
A
A Aʹ
Aʹ
d
e
f
A
Cʹ B
Aʹ
Bʹ
Dʹ
C
D
B
Bʹ A
B A
Aʹ
Cʹ
Bʹ
2
Cʹ
Aʹ
Rotate the interval AB by 60˚ in a clockwise direction about the point O. O A B
3
Rotate the triangle ABC by 90˚ clockwise about O.
C O B A
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ICEEM Mathematics Secondary 1B
4
Rotate the triangle ABC by 90˚ anticlockwise about O.
B O
5
C
A
Reflect triangle ABC in the axis of reflection. B
A
C
6
Mark the points given below on graph paper and translate them 3 units right and 1 unit up. Be careful to label each point and its image. Also write down the coordinates of each image point.
a A(0, 0)
7
Given a point and its image under translation in each part, plot the points and describe each translation.
a A(0, 0) and Aʹ(3, 1)
b B(2, 1)
c C(–3, 1)
b B(1,–3) and Bʹ(6, 2)
c C(0, 3) and Cʹ(0,–3) 8
Mark the points given below on a number plane and rotate each one 90˚ anticlockwise about the origin O. Write down the coordinates of each image point.
a A(1, –2)
9
Mark the points given below on a number plane. Find the image of each under a rotation of 180˚ anticlockwise about the origin O. Draw each image point and write down its coordinates.
a A(3, –2)
b B(1, 4)
b B(3, 2)
c C(–2, 1)
c C(–2, 1)
10 Mark these points on a number plane. Find their images when reflected in the xaxis.
a A(3, –2)
b B(3, 2)
c C(–2, 1) Chapter 18 Transformations and symmetry
227
11 State the order of rotation symmetry of each of the diagrams.
a
b
c
e
f
d
12 Perform each pair of translations on the given figure. Then find a single translation that has the same effect.
b Shift 2 left, 3 up Shift 1 right, 2 down a then 2 left, 1 down. then 1 left, 1 up.
B
C
A
B A
C
13 Perform each pair of rotations on the given figure. Then find a single rotation that has the same effect. All rotations are about O.
a rotation of 30˚ anticlockwise b rotation of 30˚ anticlockwise then 60˚ anticlockwise. then 60˚ clockwise.
B
C
A O
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ICEEM Mathematics Secondary 1B
B
C A
O
14 Mark the given points on the number plane. Reflect them in the xaxis and then reflect the image in the yaxis. State the coordinates of the final image.
b B(2, –1)
a A(5, –1)
c C(–2, –3)
15 Mark the following points on your graph paper. First translate 1 unit up and 3 units to the right and then translate the image 2 units up and 2 units to the left. State the coordinates of the final image.
b B(–5, –3)
a A(1, –2)
c C(–1, 4)
16 In each part, describe two successive transformations that will have the combined effect of mapping ABC onto AʺBʺCʺ.
a
A
B
Aʺ
b Aʺ C
C Cʺ
Bʺ
B
Cʺ
Bʺ
c
A
d A
B B A Aʺ
Cʺ C C Cʺ
Bʺ
Aʺ
Bʺ
e
f
C Bʺ Cʺ
B
A Cʺ, A
Aʺ Aʺ
Bʺ
B
C Chapter 18 Transformations and symmetry
229
Challenge exercise Answer these questions on square grid paper. 1
a Reflect triangle ABC in line l1 and then reflect the image in l2. Label the first image AʹCʹBʹ and the second AʺBʺCʺ.
A l1
C
l2
B
b This time, reflect triangle ABC in line l3 and then reflect the image in l4. Label the first image AʹCʹBʹ and the second AʺBʺCʺ as before.
A l3
C
l4
B
c Describe the translation that takes triangle ABC to triangle AʺBʺCʺ in both parts a and b. What do you notice? 2
a Reflect triangle ABC in line l5 and then reflect the image in l6. Label the first image AʹCʹBʹ and the second AʺBʺCʺ.
A l5
C
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ICEEM Mathematics Secondary 1B
l6
b Describe the translation that takes triangle ABC to triangle AʺBʺCʺ in part a. c Describe the translation that takes triangle ABC to triangle AʺBʺCʺ if the lines of reflection are parallel 4 units apart. 3
The triangle ABC whose vertices have coordinates A(0, 0), B(2, 0), C(2, 4) is translated to the triangle AʹBʹCʹ with coordinates Aʹ(4, 0), Bʹ(6, 0), C ʹ(6, 4). Draw the two triangles on a number plane and draw in two axes of reflection, l1 and l2, such that when triangle ABC is reflected in l1 and its image is reflected in l2, the final result is triangle AʹBʹCʹ. Is there only one way to do this? What is the distance between lines l1 and l2?
4
The triangle ABC whose vertices have coordinates A(0, 0), B(2, 0), C(2, 4) is translated to the triangle AʹBʹCʹ with coordinates Aʹ(0, 4), Bʹ(2, 4), Cʹ(2, 8). Draw the two triangles on a number plane and draw in two axes of reflection, l1 and l2, such that when triangle ABC is reflected in l1 and its image is reflected in l2, the final result is triangle AʹBʹCʹ. Is there only one way to do this? What is the distance between lines l1 and l2?
5
Triangle ABC has been rotated through 90˚ in an anticlockwise direction about O to the triangle AʹBʹCʹ. Copy the diagram carefully into your book. a Draw two lines l1 and l2 through O such that, when triangle ABC is reflected in l1 and its image is reflected in l2, the final result is triangle AʹBʹCʹ.
Bʹ
Cʹ
Aʹ
O
C
A
B
b Measure the acute angle between lines l1 and l2. c Is there only one choice for the position of lines l1 and l2? Chapter 18 Transformations and symmetry
231
Circular graph paper (30˚)
Circular graph paper (45˚)
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ICEEM Mathematics Secondary 1B
Chapter 19 The five Platonic solids In this chapter, you will be building five famous symmetric threedimensional objects called the regular polyhedra or the Platonic solids. These solids are shown below.
Chapter 19 The five Platonic solids
233
The word polyhedra is the plural of the word polyhedron. A polyhedron is a threedimensional solid figure bounded by flat faces – poly  hedron is a Greek word meaning ‘many bases’. The boundary of a polyhedron is made up of: •
faces
•
edges
•
vertices.
Each face of a polyhedron is some kind of polygon. A regular polyhedron is a threedimensional solid whose faces are all identical regular polygons. For example, the cube shown on the previous page is a regular polyhedron, because all of its faces are identical squares. The ancient Greeks already knew that there are only five regular polyhedra – the ones shown in the diagram above. These five regular polyhedra are often called the Platonic solids, because they were very important in the ideas of the ancient Greek philosopher Plato.
In this chapter, you will build these five Platonic solids out of stiff A4 cardboard using a ruler, protractor, compasses, set square, scissors and sticky tape. Each construction starts by drawing a net of the polyhedron on A4 cardboard (except that nets will simply be provided for the two most difficult solids). A net is a drawing of the faces of the polyhedron, arranged so that it can be cut out and folded up to make the solid. All the regular polyhedra have their vertices lying on a sphere, just as the regular polygons have their vertices on a circle, as we saw in Chapter 18. Each time you build one of these solids, you should imagine it fitting neatly inside the sphere passing through all the vertices.
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ICEEM Mathematics Secondary 1B
The exercises will examine some of the properties of the Platonic solids. If you complete the work in the Challenge sections, you will examine the rotational symmetries about the various axes of each solid, and count its total number of rotation symmetries. This is a chapter in which it will be helpful to use the internet for further exploration. Just enter ‘Platonic solids’ into a search engine to find many interesting sites.
19A Building the regular tetrahedron A regular tetrahedron is a triangular pyramid whose four faces are all identical equilateral triangles. (The Greek prefix tetra means ‘four’.) You will need a piece of stiff A4 cardboard to build a regular tetrahedron following the steps below. The net that you will construct is an equilateral triangle dissected into four smaller identical equilateral triangles.
F
"
G %
$ $
B
B
%
$
B
%
A C
F
D
C
D
E
#
"
#
E
"
D
G C
#
Chapter 19 The five Platonic solids
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Be careful to add the tabs to your diagram. They will end up hidden away inside your solid, but they will help to keep it reasonably rigid once you have folded it up and stuck it together with sticky tape. Be careful also to crease the folds as sharply as possible with your fingernail before you fold the net up. The sharper the folds, the neater will be the final solid. Step 1: Draw a line right across your cardboard, about 6 cm up from the bottom. Step 2: Set your compasses to about 7 cm. Place the point about halfway along this line, and mark two equal lengths off the line. Step 3: Use a ruler and compasses to construct the four equilateral triangles shown on the diagram. Step 4: Draw tabs on the edges as shown in the diagram. Step 5: Add the names of all vertices, faces and edges exactly as on the diagram. Notice that: • the vertices are named with uppercase italic letters, like A, B and C • the edges are named with lowercase italic letters, like a, b and c • the faces are named with uppercase curly letters, like A, B and C. Step 6: Cut out the figure with scissors, being careful to cut around the tabs. Step 7: Crease all the edges and tabs sharply, leaving the letters on the outside. Step 8: When the solid is sufficiently creased to stay in place easily, stick it together with sticky tape.
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Exercise 19A Note: Remember that in this chapter, faces are named with curly capitals, like A, and edges are named with lowercase italic letters, like a. 1
a How many vertices does the regular tetrahedron have? b How many edges does the regular tetrahedron have? c How many faces does the regular tetrahedron have? d Check that the number of vertices plus the number of faces is two more than the number of edges. This result is called Euler’s formula.
2
a How many edges meet at each vertex?
b How many edges surround each face?
3
Take the face A and write down: a the edges on the boundary of the face A b the vertices lying on the face A c the vertex opposite the face A.
4
Take the edge a and write down: a the two vertices that are joined by a b the two faces on each side of a c all the edges that intersect with a d all the edges that are parallel to a e all the edges that are skew to a. (Recall from Chapter 5 that two lines in threedimensional space are called skew if they do not meet and are not parallel.)
5
a Name the edge where the faces A and D meet.
b Check that any two faces intersect in an edge.
c Name the edge that joins the vertices B and D.
d Check that any two vertices are joined by an edge.
Chapter 19 The five Platonic solids
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6
Name all three pairs of skew edges.
7
a Move your finger along a path that starts at vertex A and travels along edges until it arrives back at vertex A. The path you follow must visit every other vertex, but never travel along any edge more than once. b Try to move your finger along a path that starts and finishes at vertex A, and that travels along every edge once and no more than once. (You will not be able to do it.)
8
a Imagine an axis that passes through the vertex A and through the midpoint of the opposite face A. The diagram opposite shows the tetrahedron being held so that it can rotate about this axis.
b Hold the tetrahedron as shown above, between finger and thumb, with the face B facing towards you (see above). Now rotate it very slowly about this axis towards the left. After a onethird turn, the tetrahedron will be in an identical position, except that the face C will be facing towards you, as in the diagram opposite. c After another onethird turn, the tetrahedron will again be in an identical position, except that face D will be facing towards you, as in the next diagram.
d After a final onethird turn, the original face B will be facing towards you again. This means that the tetrahedron has rotation symmetry of order 3 about the axis A .
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9
a Now imagine an axis that passes through the midpoints of the opposite edges a and d. The diagram opposite show the tetrahedron being held so that it can rotate about this axis.
b Hold the tetrahedron as shown above, between finger and thumb, with the edge a at the top, the edge d at the bottom, and the face A facing towards you. Now rotate it very slowly about this axis towards the left. After a halfturn, the tetrahedron will be in an identical position, except that the face B will be facing towards you, as in the diagram opposite. c After a second halfturn, the original face A will be facing towards you again. This means that the tetrahedron has rotation symmetry of order 2 about the axis ad. Challenge: Finding all the rotation symmetries of the regular tetrahedron Questions 1 and 2 in the Challenge exercise at the end of this chapter describe how to find all the rotation symmetries of the regular tetrahedron. This would be a good time to try these two questions.
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19B Building the regular hexahedron or cube A cube has six faces, all identical squares, and so is sometimes called a regular hexahedron. (The Greek prefix hexa means ‘six’.)
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The diagram on the left below shows a net that folds up easily into a cube. Use a ruler and a set square or protractor to construct a large copy of this net on stiff A4 cardboard. Make all the edge lengths 6 cm. Be careful to mark out the tabs.
Step 1: Copy the names of the vertices, edges and faces exactly as on the diagram. Step 2: Cut around the figure, making sure that you do not cut off the tabs. Step 3: Crease the edges sharply, including the edges of the tabs, until the solid stands up by itself. Step 4: When the solid is sufficiently creased to stay in place easily, stick it together with sticky tape.
Exercise 19B Use your model to answer the questions. 1
a How many vertices does the cube have? b How many edges does the cube have? c How many faces does the cube have? d Check that the number of vertices plus the number of faces is two more than the number of edges. (This is Euler’s formula.)
2
a How many edges meet at each vertex? b How many edges surround each face?
3
Take the edge a and write down: a the edges that intersect with a b the edges that are parallel to a c the edges that are skew to a.
4
a Name the three pairs of parallel faces. b Name the four pairs of opposite vertices. (By opposite we mean diametrically opposite across the sphere that contains the solid.) c Name the six pairs of opposite edges.
5
a Move your finger along a path that starts and finishes at vertex A, that visits every other vertex, but never travels along any edge more than once. Chapter 19 The five Platonic solids
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b Can you find a path that starts and finishes at vertex A, and that travels along every edge once and no more than once? 6
a Imagine an axis that passes through the midpoints of the opposite faces A and B. The diagram below shows the cube being held so that it can rotate about this axis.
b Hold the cube as shown above, with the face C facing towards you. Now rotate it very slowly about this axis towards the left. After a quarterturn, the cube will be in an identical position, except that the face F will be facing towards you. c Keep rotating the cube quarterturns about this axis, noting how the cube is in an identical position after each quarterturn. d It takes four quarterturns to bring the original face C back facing towards you again. This means that the cube has rotation symmetry . of order 4 about the axis 7
a Now imagine an axis that passes through the opposite vertices A and G, as shown in the diagram opposite.
b Hold the cube as shown above, with the face A facing towards you. Now rotate it very slowly about this axis towards the left. After a onethird turn, the cube will be in an identical position, except that the face C will be facing towards you. c After another onethird turn, the cube will again be in an identical position, except that the face F will be facing towards you.
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d It takes three onethird turns to bring the original face A back facing towards you again. This means that the cube has rotation symmetry of order 3 about the axis AG. 8
a Now imagine an axis that passes through the midpoints of the opposite edges a and g, as shown in the diagram below.
b Hold the cube as shown above, with the edge a at the top, the edge g at the bottom, and the edge c facing towards you. Now rotate it very slowly about this axis towards the left. After a halfturn, the cube will be in an identical position, except that the edge e will be facing towards you. c After a second halfturn, the original edge c will be facing towards you again. This means that the cube has rotation symmetry of order 2 about the axis ag. Challenge: Finding all the rotation symmetries of the cube Questions 3 and 4 in the Challenge exercise describe how to find all the rotation symmetries of the cube. This would be a good time to try these two questions.
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19C Building the regular octahedron A regular octahedron has eight faces, all identical equilateral triangles. (The Greek prefix octa means ‘eight’.) The most straightforward net for the octahedron is the twopiece net drawn below. The two pieces are identical except for the letters used to name them. %
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Exercise 19C Use your model to answer the questions. 1
a How many vertices does the octahedron have? b How many edges does the octahedron have? c How many faces does the octahedron have? d Check that the number of vertices plus the number of faces is two more than the number of edges. (This is Euler’s formula.)
2
a How many edges meet at each vertex? b How many edges surround each face?
3
Take the edge a and write down:
a the edges that intersect with a b the edges that are parallel to a
c the edges that are skew to a. 4
a Name the four pairs of parallel faces. b Name the three pairs of opposite vertices. c Name the six pairs of opposite edges.
5
a Find a path along the edges that starts and finishes at vertex A, visiting every other vertex, but never travelling along any edge more than once. b Find a path that starts and finishes at vertex A, travelling along every edge once and no more than once.
6
a Imagine an axis that passes through the opposite vertices A and B. The diagram opposite shows the octahedron being held so that it can rotate about this axis. b Hold the octahedron as shown opposite, with the vertex C facing towards you. Now rotate it very slowly about this axis towards the left. After a quarterturn, the octahedron will be in an identical position, except that the vertex D will be facing towards you. Chapter 19 The five Platonic solids
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c Keep rotating the octahedron quarterturns about this axis, noting how the octahedron is in an identical position after each quarterturn. d It takes four quarterturns to bring the original vertex C back facing towards you again. This means that the octahedron has rotation symmetry of order 4 about the axis AB. 7
a Now imagine an axis that passes through the midpoints of the opposite faces A and G, as shown in the diagram opposite. b Hold the octahedron as shown opposite, with the vertex A facing towards you. Now rotate it very slowly about this axis towards the left. After a onethird turn, the octahedron will be in an identical position, except that the vertex F will be facing towards you. c After another onethird turn, the octahedron will again be in an identical position, except that the vertex C will be facing towards you. d It takes three onethird turns to bring the original vertex A back facing towards you again. This means that the octahedron has . rotation symmetry of order 3 about the axis
8
a Now imagine an axis that passes through the midpoints of the opposite edges a and g, as shown in the diagram opposite. b Hold the octahedron as shown opposite, with the edge a at the top, the edge g at the bottom, and the edge c facing towards you. Now rotate it very slowly about this axis towards the left. After a halfturn, the octahedron will be in an identical position, except that the edge e will be facing towards you. c After a second halfturn, the original edge c will be facing towards you again. This means that the octahedron has rotation symmetry of order 2 about the axis ag.
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19D Building the regular dodecahedron A regular dodecahedron has 12 faces, all identical regular pentagons. (The Greek prefix dodeca means ‘twelve’.) The net for a dodecahedron is shown below.
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Chapter 19 The five Platonic solids
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a How many vertices, how many edges and how many faces does the regular dodecahedron have? b Check that the number of vertices plus the number of faces is two more than the number of edges. (This is Euler’s formula.)
2
a How many edges meet at each vertex?
b How many edges surround each face?
3
Take the edge a and write down:
a the edges that intersect with a (or the edge extended meets a extended)
b the edges that are parallel to a
c the edges that are skew to a.
4
a Name the six pairs of parallel faces.
b Name the 10 pairs of opposite vertices.
c Name the 15 pairs of opposite edges.
5
a Can you find a path along the edges that starts and finishes at vertex A, visiting every other vertex, but never travelling along any edge more than once? b Can you find a path that starts and finishes at vertex A, travelling along every edge once and no more than once?
6
a Imagine an axis that passes through the midpoints of the opposite faces A and B. Hold the dodecahedron so that you can rotate it slowly about this axis, and find the order of rotation symmetry about the axis. b Imagine an axis that passes through the opposite vertices A and P. Hold the dodecahedron so that you can rotate it slowly about this axis, and find the order of rotation symmetry about the axis. c Imagine an axis that passes through the midpoints of the opposite edges a and h. Hold the dodecahedron so that you can rotate it slowly about this axis, and find the order of rotation symmetry about the axis.
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19E Building the regular icosahedron A regular icosahedron has 20 faces, all identical equilateral triangles. (The Greek prefix icosa means ‘twenty’.) The net for an icosahedron is shown below. ,
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Chapter 19 The five Platonic solids
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Exercise 19E Use your model to answer the questions. 1
a How many vertices, how many edges and how many faces does the regular icosahedron have? b Check that the number of vertices plus the number of faces is two more than the number of edges. (This is Euler’s formula.)
2
a How many edges meet at each vertex?
b How many edges surround each face?
3
Take the edge a and write down:
a the edges that intersect with a
b the edges that are parallel to a
c the edges that are skew to a.
4
a Name the 10 pairs of parallel faces.
b Name the six pairs of opposite vertices.
c Name the 15 pairs of opposite edges.
5
a Can you find a path along the edges that starts and finishes at vertex A, visiting every other vertex, but never travelling along any edge more than once? b Can you find a path that starts and finishes at vertex A, travelling along every edge once and no more than once?
6
a Imagine an axis that passes through the opposite vertices A and B. Hold the icosahedron so that you can rotate it slowly about this axis, and find the order of rotation symmetry about the axis. b Imagine an axis that passes through the midpoints of the opposite faces A and P. Hold the icosahedron so that you can rotate it slowly about this axis, and find the order of rotation symmetry about the axis. c Imagine an axis that passes through the midpoints of the opposite edges a and p. Hold the icosahedron so that you can rotate it slowly about this axis, and find the order of rotation symmetry about the axis.
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Challenge exercise In this exercise we shall find all the rotation symmetries of the regular tetrahedron and the cube. The treatment of the regular dodecahedron, regular octahedron and the regular icosahedron is similar. See the website at www.icemaths.org.au for the details and for the exercises on these three solids. Finding all the rotation symmetries of the regular tetrahedron In question 1 we will find 12 different rotation symmetries of the regular tetrahedron. In question 2 we will show that there cannot be more than 12 rotation symmetries of the regular tetrahedron. Thus the regular tetrahedron has exactly 12 rotation symmetries. 1
The regular tetrahedron has seven axes of symmetry: • Four of these axes join a vertex and the midpoint of the opposite face, as in question 8 of Exercise 19A. • The other three axes join the midpoints of opposite edges, as in question 9 of Exercise 19A. a Copy the table below and list these axes in the top row. b Write the order of rotation symmetry of each axis in the bottom row. c To find the number of rotation symmetries, find the sum of all the orders – the sum should be 18.
Axis
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Chapter 19 The five Platonic solids
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d But by adding up these orders, we have counted the ‘do nothing’ rotation seven times, when it should have been counted only once, so we subtract 7 – 1 = 6 from the total of the orders. Thus there are 12 distinct rotation symmetries of the tetrahedron. e Check that these 12 rotation symmetries are different since they fix different points on the surface of the tetrahedron. 2
In this question we show that there are exactly 12 rotation symmetries of the tetrahedron. Place the tetrahedron flat on the table with one of its edges lined up with the front edge of the table. A e C
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From your model you can see that the position of all vertices, edges and faces is fixed by placing the tetrahedron in this way.
Note that under rotation a face is taken to a face, a vertex to a vertex and an edge to an edge.
To find the maximum possible number of rotation symmetries we count the number of ways that faces can be placed on the table with an edge lined up with the front edge of the table. In summary, we have found 12 different rotation symmetries and we have shown that here are at most 12 of them. Hence there are exactly 12 rotation symmetries.
a How many ways can you choose which face to place flat on the table? b How many ways can you choose which edge to then line up with the edge of the table? c Deduce that there are exactly 12 ways to line the tetrahedron up with an edge of the table.
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Finding all the rotation symmetries of the cube In question 3 we will find 24 different rotation symmetries of the cube. In question 4 we will show that there cannot be more than 24 rotation symmetries of the cube. Thus the cube has exactly 24 rotation symmetries. 3
The cube has 13 axes of rotation symmetry: • Three of these axes join the midpoints of opposite faces, as in question 6 of Exercise 19B. • Four of these axes join opposite vertices, as in question 7 of Exercise 19B. • The other six axes join the midpoints of opposite edges, as in question 8 of Exercise 19B. a Copy the table below and list the axes in the top row. b Write the order of each axis of rotation symmetry in the bottom row. c To find the number of rotation symmetries, find the sum of all the orders – the sum should be 36. d But by adding up these orders, we have counted the ‘do nothing’ rotation 13 times, when it should have been counted only once, so we subtract 13 – 1 = 12 from the total of the orders. Thus there are 24 distinct rotation symmetries of the cube. Axis
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e Check that these 24 rotation symmetries are different since they fix different points on the surface of the cube.
Chapter 19 The five Platonic solids
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In this question we show that there are exactly 24 rotation symmetries of the cube. Place the cube flat on the table with one of its edges lined up with the front edge of the table. F j B
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From your model you can see that the position of all vertices, edges and faces is fixed by placing the cube in this way.
Note that under rotation a face is taken to a face, a vertex to a vertex and an edge to an edge.
To find the maximum possible number of rotation symmetries we count the number of ways that faces can be placed on the table with an edge lined up with the front edge of the table. a How many ways can you choose which face to place flat on the table? b How many ways can you choose which edge to then line up with the edge of the table? c Deduce that there are exactly 24 ways to line the cube up with an edge of the table.
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In summary we have found 24 different rotation symmetries and we have shown that there are at most 24 of them. Hence there are exactly 24 rotation symmetries.
The same procedures can be used to find all the rotation symmetries of the regular octahedron, the regular dodecahedron and the regular icosahedron. The first of these has 24 rotation symmetries and the other two have 60.
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b List the integers greater than –10 and less than –1.
2 The sequence –60, –57, –54, … is ‘going up by threes’. Give the next four terms. 3 Simplify:
a –(–5)
b –(–34)
c –(–(–7))
d –(–(–13))
4 Write the answers to these additions.
a –3 + 11
b –5 + 7
c –5 + 12
d –29 + 18
e –35 + (–7)
f –12 + 18
g –3 + (–2)
h –6 + (–11)
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Write the answers to these subtractions.
a 3 – 6
b 6 – 18
c –3 – 11
d –15 – (–8)
e –8 – 26
f 17 – (–3)
g –16 – 19
h –11 – (–22)
i –36 – (–32)
j –23 – (–9)
k –20 – (–11)
l 20 – (–30)
6 The temperature in Montreal on a winter’s day went from a minimum of –22°C to a maximum of –7°C. By how much did the temperature rise? 7
Write the answers to these multiplications.
a 5 × (–7)
b 6 × (–3)
c 10 × (–11)
d 12 × (–7)
e 10 × (–16)
f –5 × (–10)
g –11 × 4
h –12 × (–4)
Chapter 20 Review and problem solving
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Complete these divisions.
a –18 ÷ (–3)
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Complete these divisions.
a 18 –6
b –36 ÷ 4
c –35 ÷ (–7)
d –27 ÷ (–3)
b –63
c –85
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e –3 × (–8) + 24 – 15
11 In an indoor cricket match, a team has made 30 runs and lost 8 wickets. What is the score of the team? (A run adds 1 to the score and a wicket subtracts 5.) 12 Evaluate:
a –6 × (6 – 7)
b 8 × (12 – 20)
c –3 × (25 + 15)
d –6 × (–14 – 6)
e –12 × (–6 + 40)
f –(–5)2
g (2 – 11) × (11 – 20) h (10 – 3) × (–3 + 10) i (–5 – 10) × (10 – 4)
Chapter 12: Algebra and the number plane
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1
Evaluate each expression for x = – 4.
a 2x
b –x
c x + 2
d x – 3
g –x3
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b m + p
e np
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Given that m = –12, n = 4 and p = – 3, evaluate:
a m + n
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Evaluate each expression for x = –5.
a 5x + 4
b –5x + 4
c 5(x + 4)
d –5(x + 4)
e –5x2
f (–5x)2
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b m + p
c m – p
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i x = 20
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Angel has $2000 in a bank account. She takes $x from the bank account every day.
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a How much money does she have in the account after:
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b Find the value of her bank account after 5 days if:
i x = 100
7
Evaluate each expression for x = –10.
a 5x + 5
b –4x + 5
c 5(x + 5)
d –4(x + 10)
e –5x2
f 5(–x)2
8
Give the coordinates of each of the points marked on the number plane below.
ii x = 200
iii x = 450
y A
4
G
3 2
B
1 –4 –3 –2 –1
0
1
2
3
4
x
–1
D
–2
E
F C
–3 –4
9
Do each part of this question on graph paper. a Plot the points O(0, 0), A(5, 0), C(5, 5) and D(0, 5), and join points to form OA, AC, CD and DO. Describe the shape formed. b Plot the points A(–5, 0), B(5, 0) and C(0, 10), and join points to form AB, BC and CA. Describe the shape formed.
Chapter 20 Review and problem solving
259
c Plot the points A(–3, 0), B(3, 0), C(3, 3) and D(–3, 3), and join points to form AB, BC, CD and DA. Describe the shape formed. d Plot the points O(0, 0) and A(4, 4), and draw the line passing through them. Now plot the points B(2, –3) and C(5, 0) and draw the line passing through these points. Describe the relationship between the lines. 10 For each given rule, complete the table, list the coordinates of the points, and plot the points on a number plane. (After you have completed the table, decide on the numbering of your axes.) a y = 4x
x
–3
–2
–1
0
1
2
3
–3
–2
–1
0
1
2
3
–2
–1
0
1
2
3
y
b y = –x
x y
c y = x – 4
–3
x y
Check the points are collinear.
Chapter 13: Triangles and constructions 1
Find a, b, c or i in each diagram below.
a
B
C
82˚
D
260
B
55˚
a D
A
b
c
A
b
ICEEM Mathematics Secondary 1B
b
c
C 92˚
a C 85˚
a B
35˚ A E
E
D
d
e
B
S B
c
70˚ A
g V P c
R
24˚ i V 21˚ B c b Q O 92˚ a
C 88˚ a D
i
a 88˚ b B R
P
52˚
h
B
49˚
2
a C
Q Q
A
A
c
b 70˚ A
30˚ b a C
f
T
55˚
68˚
S C
A
Use a ruler and compasses to construct an equilateral triangle of side length 6 cm.
Chapter 14: Negative fractions 1
Arrange the numbers –3 1 , 2 1 , –1 3 , 0, 2, –2 1 , 3 1 from smallest to largest.
2
Calculate:
a – 3 + 4
b –1 5 + 4
e –1 – 1
1 4
f –2 +
3
Calculate:
a 3 + 3
b 3 – 3
c 1 – 7
d 3 – 3
e
f – 3 5
g
h
4
Calculate:
a 2 × – 3
b 7 × – 16
c –1 1 × – 4
d – 3 × – 5
e – 5 × – 2
f 1 1 × – 35
g – 2 ÷ – 7
h 1 1 ÷ – 6
i –2 1 ÷ – 2
j – 3 ÷ 2
k 1 2 ÷ –1 4
l – 2 × 3 1
4
4
5 2 3
+
3
6
1 4
4 7 8
4
8
3
2
3 5
5
4
8
2
5
+
7 10
21
5 4
48
5
4
2
2
c – 2 + 3
d – 5 + 3
5
5 2 g –5 + 2 5
2 1 2
–
2
3 3
8
8
1 2
h –11 – 13
8 7 20
3
12
5
1 2
11 10 3 5 – 7 8
4 3 3
8
11
5
Chapter 20 Review and problem solving
261
5
Evaluate each expression for x = – 3 .
a 3x
b –x
c x – 3
d 4x
e (–x)3
f –x 3
g 3 – x
h 4 – 4x
6
Given that m = –2 1 , n = 5 and p = –5, evaluate:
a m + n
b m + p
c mn
d np
e mnp
f 2m + p
g 4m
h –8n
4
4
8
Chapter 15: Percentages 1
Write the following fractions as percentages.
a 9
2
Write the following decimals as percentages.
a 0.78
3
a Express 30 cents as a percentage of $1.
b 17
10
20
b 0.095
c 6
d 12
5
25
c 0.97
d 1.35
b Express 1 kg as a percentage of 800 grams. c Express four months as a percentage of one year. 4
Arrange the following from smallest to largest.
a 0.39, 12 , 4 1 %
5
Fill in the table with the equivalent fraction, decimal and percentage for each number.
32
Fraction
b 64%, 0.6, 2
2
3
Decimal
Percentage
0.375 1 3 75% 2
3 4 300%
262
6
Write each percentage as a fraction.
a 62 1 % 2
b 87 1 % 2
ICEEM Mathematics Secondary 1B
c 1 1 % 4
d 5 1 % 4
7
Express $7.50 as a percentage of $25.
8
Phuong gave 20% of her week’s wages to her mother. She still had $450 to spend. What was Phuong’s weekly wage?
Chapter 16: Solving equations 1
Solve:
a x + 3 = 5
b x – 2 = 9
c x – 4 = 8
d 2x = 10
e 3x = 7
f z – 10 = 12
g 5x = 20
h x = 4
i z + 8 = 16
2
Solve:
m 3
j x – 11 = 13
k
a 2x + 3 = 7
b 3x – 6 = 9
c 2x – 4 = 8
e 3x – 5 = 7
f 6z – 10 = 12 g 5x + 10 = 20 h x + 3 = 4
i 5z + 10 = 45 j 2x – 11 = 13 k
3
Solve each equation for x.
a x + 3 = 2 x 5
m 3
l
3 z 5
= 10
d 2x + 8 = 16
+ 6 = 10
l
3 z 5
–4=6
b 2x + 4 = –2
c 2 – x = 8
e 2x – 5 = –6x + 7
f 2(x – 2) = 11x
d
4
In each case below, write an equation and solve it.
+ 4 = 11
=6
a Three is added to a number and the result is 6. b Five is subtracted from a number and the result is 10. c A number is multiplied by 7 and the result is 84. d A number is divided by 4 and the result is 15. e A number is multiplied by 8 and the result is 15. f A number is multiplied by 4, and 3 is subtracted from the result. The result of this is 17. g Six is added to a number and the result is multiplied by 3. The result of this is 28. h Seven is subtracted from a number and the result is multiplied by 6. The result of this is 15.
Chapter 20 Review and problem solving
263
5
A rectangle is five times as long as it is wide. The perimeter of the rectangle is 120 cm. Find the length and the width of the rectangle.
6
The difference between two numbers is 18. The sum of the two numbers is 20. Find the two numbers.
7
Solve:
a x + 5 = –12
b x – 7 = –14
c x + 7 = 8
d 5x = –12
e –6x = –24
f –3x = 9
g x = –20
h –3x = –6
i 4x + 1 = –11
j 5x – 3 = –18
k –2x + 10 = –12
l 4x – 6 = –8
8
Solve each equation for x.
a 5x + 4 = 2x + 5
b 6x + 7 = 2x + 10
c 5x + 2 = x + 1
d 4x – 2 = 5x + 1
e x + 14 = 3x + 10
f 10x + 4 = 2x + 7
9
Solve these equations for x.
a 2x – 3 = 3 – 4x
4
2
b 2x – 6 = 3 – 4x
c 2x – 12 = 6 + 8x
10 When a number is multiplied by 5 and divided by 11, the result is 2. Find the number. 11 If you add 13 to a number, you get the same result as when you subtract half the number from 4. What is the number?
Chapter 17: Probability 1
A jar contains 20 blue marbles and 30 red marbles. A marble is chosen at random. What is the probability that it is:
a red?
2
A coin is tossed four times. List the 16 outcomes in the form (H, H, H, H), (H, H, H, T), ….
What is the probability of:
a getting two heads and two tails? b getting three heads and a tail?
b blue?
c getting four heads?
d not getting a head?
e getting a head on the first throw?
f getting a head on the last throw? g getting two or more heads?
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ICEEM Mathematics Secondary 1B
3
A box contains 24 balls. There are eight red balls numbered 1 to 8, six blue balls numbered 9 to 14, and ten green balls numbered 15 to 24. One ball is taken out of the box. Find the probability of getting:
a a red ball
d a green ball with an odd number
e a green or a red ball
f a number greater than 10
g a number divisible by 5
h a prime number
i a ball that is not red.
4
A dart is thrown at the board shown opposite. There is an equal chance of hitting any point on the board. What is the probability of:
b a green ball
c an odd number
Blue
Red
Green
Yellow
a hitting the blue square?
b not hitting the blue square?
c hitting one of the bottom two squares?
5
There are 10 boys and 3 girls in a room. One person is chosen at random. What is the probability that the person is a boy?
6
The scrabble tiles that make up the word MATHEMATICS are put in a sack together with the scrabble tiles for the word FOOTBALL. There are a total of 19 tiles. A tile is withdrawn. What is the probability of obtaining:
a an M?
e a consonant? f other than an M?
7
Amanda, Barbara, Carol and Dianne are in one room at a party, and Anthony, Barry, Carl and David are in another. For a party game, one person from each room is chosen randomly. List all the possible pairs of boys and girls to help find the probability of :
a Barry being chosen?
b Amanda and Barry being chosen?
c Carol not being chosen?
d A boy and a girl being chosen?
e Anthony or Barry being chosen?
f Barbara and Carol not being chosen?
g Barbara and David not being chosen?
b an A?
c an F?
d a vowel? g an M or an A?
Chapter 20 Review and problem solving
265
Chapter 18: Transformations and symmetry
266
1
Copy the points listed below onto your graph paper and translate them 3 units right and 1 unit up. Be careful to label each point and its image. Also write down the coordinates of each image point.
a A(0, 0) d D(–5, –5)
2
Given the following points and their images under a translation, plot each point and describe the translation.
a A(0, 0) and A′(1, 3) c C(0, 5) and C′(0, –5) e E(5, 5) and E′(0, 0)
3
Plot the points listed below onto a number plane and rotate them 90˚ anticlockwise about (0, 0). Write down the coordinates of each image point.
a A(2, –1) d D(–4, –4)
4
Plot the points listed below onto a number plane. Find their images under a rotation of 180˚ anticlockwise about (0, 0). Draw each image and write down its coordinates.
a A(2, –1) d D(–4, –4)
5
Copy these points onto a number plane and find their images under reflection in the xaxis.
a A(2, –1) d D(–5, 0)
6
a Copy these points onto a number plane and find their images under reflection in the xaxis.
i A(3, 2)
b Join points A, B and C in part a to form ABC under reflection in the xaxis.
b B(1, 2) e E(–3, 2)
c C(–2, 1) f F(0, 2)
b B(2, –2) and B′(3, 4) d D(3, 4) and D′(–3, –4) f F(2, 2) and F′(2, –1)
b B(2, –2) e E(0, 0)
b B(2, –2) e E(0, 0)
b B(2, 2) e E(0, 4)
ii B(4, 4)
ICEEM Mathematics Secondary 1B
c C(–3, 4) f F(0, 5)
c C(–3, 4) f F(0, 5)
c C(–3, 3) f F(–1, –1)
iii C(6, 1) ABC. Find the image of
20B Tessellations A tiling pattern with no gaps or spaces between the tiles is called a tessellation. The tessellations we are going to look at are made up of tiles that are polygons and completely cover a plane. You will need a ruler, compasses and a protractor. A single tile is said to tessellate if the tessellation is made up of copies of the one tile. We will begin with tessellations that can be created using just one polygon. If we limit ourselves to regular polygons, there are only three possibilities: • equilateral triangles will tessellate • squares will tessellate • regular hexagons will tessellate. The resulting patterns are called the regular tessellations.
Activity 1 (Why are these the only regular polygons that tessellate by themselves?) Use a template to show what happens when you try to use a different regular polygon – say a pentagon, an octagon or a nonagon – to tessellate the plane. Copy and complete this table of the sizes of interior angles. Then explain convincingly why the equilateral triangle, square and hexagon are the only regular polygons that can be used by themselves to tessellate. Polygon
Number of sides
Interior angle sum
Size of each angle
Triangle
3
180˚
60˚
Quadrilateral
4
360˚
Pentagon
540˚ 6
(table continued over page) Chapter 20 Review and problem solving
267
Heptagon
7 135˚ 9
1260˚
10 Dodecagon
144˚
12
Activity 2 (Tessellations with nonregular polygons) If we consider polygons that are not regular, we can find many other singlepolygon tessellations. But even a rectangle can be used more inventively to tessellate.
It can be proved that any triangle can tessellate the plane by itself. Try doing this using a triangle like this: and setting it out along a line like this:
B
A
C
Use a similar construction to explain why any quadrilateral can be used to tessellate the plane. Activity 3 (The Cairo tessellation) The Cairo tessellation is so named because tiles such as these were used for many years on the streets of Cairo. Each tile is a pentagon with all sides of equal length.
268
ICEEM Mathematics Secondary 1B
Note that this pentagon is not regular because it does not have all its interior angles of the same size.
D E
C
45˚ A
D B A A B D
45˚ M
B
You can draw one such tile using your protractor ruler and compass. Step 1: Draw the interval AB. Step 2: Find the midpoint M of AB. Step 3: Draw rays from M at 45˚ to AB. Step 4: Use your compasses set at the length of AB to complete the pentagon. The angles of the pentagon at E and C are 90˚. Draw one of these pentagons on card. Use it to form the Cairo tessellation. Why does this pentagon tessellate? Another way of drawing such a pentagon is suggested by the diagram shown. Draw two identical rightangled isosceles triangles on card. Cut them out, join them at D as shown, and rotate one around D so that AB is the same length as the equal sides of the isosceles triangles.
D
C
E A
B
Activity 4 Here is a heptagon that can be used to tessellate the plane. Draw it on card and use it to cover a rectangular piece of paper at least 15 cm × 10 cm to show how the tessellation works. The angles of the heptagon are all either 45˚, 90˚ or 270˚.
Chapter 20 Review and problem solving
269
Activity 5 After Activity 1, you probably noticed that, because the angles about a point add up to 360˚, there are some combinations of regular polygons that can be used to tessellate the plane. These are called semiregular tessellations and one way of describing them is to look at the set of regular polygons that meet at a point and write down the numbers of sides these polygons have, ‘in cyclic order’ (that is, clockwise starting from the highest number), in square brackets. For example, the code for the second tessellation shown below is [6, 4, 3, 4]. Write the codes for the other seven semiregular tessellations. 1
2
3
4
5
270
ICEEM Mathematics Secondary 1B
6
7
8
Activity 6 The pentagon in the tessellation shown below has 5 equal sides and angles of 60˚, 140˚, 100˚, 80˚ and 160˚. It can be used to tessellate the plane but this has been done wrongly here – the arrows show places where no tile will fit.
a Find the sizes of these two angles without using a protractor. b Draw one of these pentagons on card, using a ruler and protractor. Use it to show that a repeating tessellating pattern can be drawn with it.
Chapter 20 Review and problem solving
271
20C Sets and Venn diagrams A set is just a collection of objects, but we need some new words and symbols and diagrams to be able to talk sensibly about sets. Some tricky counting problems can be solved by using a type of diagram called a Venn diagram to illustrate the sets involved.
A Describing and naming sets In daytoday life, we try to make sense of the world we live in by classifying collections of things. English has many words for such collections. For example, we speak of ‘a flock of birds’, ‘a herd of cattle’, ‘a swarm of bees’ and ‘a colony of ants’. We do a similar thing in mathematics, and classify numbers, geometrical figures and other things into collections that we call sets. The objects in these sets are called the elements of the set. Describing a set A set can be described by listing all of its elements. For example: S = {1, 3, 5, 7, 9} We read this as ‘S is the set with elements 1, 3, 5, 7 and 9’. Notice how the five elements of the set are separated by commas and the list is enclosed between curly brackets. A set can also be described by writing a description of its elements between curly brackets. Thus the set S can also be written as S = {odd whole numbers less than 10}, which we read as ‘S is the set of odd whole numbers less than 10’. A set must be well defined It is important that our description of the elements of a set is clear and unambiguous. For example, {tall people} is not a set, because people tend to disagree about what tall means. A set must be well defined, like the following set: {letters in the English alphabet}.
272
ICEEM Mathematics Secondary 1B
Equal sets Two sets are called equal if they have exactly the same elements. Thus {a, e, i, o, u} = {vowels in the English alphabet}. On the other hand, the sets {1, 3, 5} and{1, 2, 3} are not equal, because they have different elements. This is written as {1, 3, 5} ≠ {1, 2, 3}. The order in which the elements are written between the curly brackets does not matter at all. For example: {1, 3, 5, 7, 9} = {3, 9, 7, 5, 1} = {5, 9, 1, 3, 7} If an element is listed more than once, it is only counted once. For example, {a, a, b} = {a, b} The set {a, a, b} has only the two elements ‘a’ and ‘b’. The second mention of ‘a’ is an unnecessary repetition and can be ignored. However, it is normally considered poor notation to list an element more than once. The symbols ∈ and ∉ The phrase ‘is an element of’ occurs so often in discussing sets that the special symbol ∈ is used for it. For example, if A = {3, 4, 5, 6}, then 3 ∈ A. (We read ‘3 ∈ A’ as ‘3 is an element of the set A’.) The symbol ∉ means ‘is not an element of’. For example, if A = {3, 4, 5, 6}, then 8 ∉ A. (We read ‘8 ∉ A’ as ‘8 is not an element of the set A’.)
Describing and naming sets • A set is a collection of objects, called the elements of the set. • A set must be well defined, meaning that its elements can be described or listed without ambiguity. For example: {1, 3, 5} and {letters of the English alphabet}. • Two sets are called equal if they have exactly the same elements. • If a is an element of a set S, we write a ∈ S. • If b is not an element of a set S, we write b ∉ S.
Chapter 20 Review and problem solving
273
Exercise A
274
1
List the elements of each of these sets.
a {odd whole numbers between 0 and 12}
b {two digit numbers ending with 3}
c the set of months ending in ‘y’
d {continents}
e {multiples of 6 between 10 and 40}
f {multiples of 7 between 10 and 40}
g {perfect squares less than 50}
h {prime numbers less than 30}
2
Give a verbal description of each set.
a {a, b, c}
3
State whether or not the following sets are well defined, and give a reason in each case if not.
a {best films of 1955}
c {whole numbers less than 1 000 000}
d {brightly coloured birds}
e {small numbers}
f {people now in this room}
g {people who vote Labor}
4
Which of the following pairs of sets are equal?
A {2, 3, 5} and {3, 5, 2}
C {64, 7 + 2, 11} and {2, 9, 11}
D {letters of the alphabet following v} and {w, x, w, y, z, x, w}
5
Replace
a {a, b, c}
{c, b, a}
b {1, 2, 4}
c {a, b, c}
{x, y, z}
d {7 + 2, 15 – 11}
e {citizens of Australia}
b {1, 3, 5}
c {2, 3, 5, 7}
b {States of Australia}
B {7, 11, 14} and {11, 9, 14}
in each statement by either = or ≠.
ICEEM Mathematics Secondary 1B
{1, 4, 2, 1, 4, 4} {4, 9}
{men, women and children in Australia}
6
Let S = {8, 9, 10, 11, 12}. Rewrite each sentence in symbols, using ∈ or ∉.
a 9 is an element of S.
b 12 is an element of S.
c 5 is not an element of S.
d 0 is not an element of S.
7
Replace
a 3
{0, 1, 2, 3, 8}
b ‘dog’
c 5
{0, 1, 2, 4}
d Sydney
e g
{vowels}
f 20
g 11
in each statement by either ∈ or ∉.
{prime numbers} h 6
{fiveletter words} {cities north of the equator}
{perfect squares} {even whole numbers}
B Finite and infinite sets All the sets we have seen so far have been finite sets, meaning that we can list all their elements. Here are two more examples: {whole numbers between 2000 and 2005} = {2001, 2002, 2003, 2004} {whole numbers between 2000 and 3000} = {2001, 2002, 2003, …, 2999} The three dots ‘…’ in the second example stand for the other 995 numbers in the set. We could have listed them all, but to save space we used dots instead. This notation can only be used if it is clear what it means, as in this case. Infinite sets A set can also be infinite – all that matters is that it is well defined. Here are two examples of infinite sets: {even whole numbers} = {0, 2, 4, 6, 8, 10, …} {whole numbers greater than 2000} = {2001, 2002, 2003, 2004, …} Both these sets are infinite because no matter how many elements we list, there are always more elements in the set that are not in our list. This time the dots ‘…’ have a slightly different meaning, because they stand for infinitely many elements that we could not possibly list, no matter how long we tried.
Chapter 20 Review and problem solving
275
The symbol S for the number of elements of a finite set If S is a finite set, the symbol S stands for the number of elements of S. For example: If S = {1, 3, 5, 7, 9}, then S = 5. If A = {1001, 1002, 1003, …, 3000}, then A = 2000. If T = {letters in the English alphabet}, then T = 26. Oneelement sets A set may have only one element, like the set S defined by S = {5}, ‘S is the set whose only element is 5.’ It is important to distinguish between the number 5 and the set S = {5}: 5 ∈ S but 5 ≠ S. In this case S = 1. A tin containing a single biscuit is quite a different object from the biscuit itself. The biscuit tin can be thought of as a set with one element, whereas the biscuit is a biscuit. The empty set The symbol ∅ represents the empty set, which is the set that has no elements at all. The empty set is like a biscuit tin after all the biscuits have been eaten: there is nothing in the whole universe that is an element of ∅: ∅ = 0 and x ∉ ∅, no matter what x may be. There is only one empty set, because any two empty sets have exactly the same elements (that is, no elements) and so they must be equal to one another.
Finite and infinite sets • A set is called finite if we can list all of its elements. • An infinite set has the property that no matter how many elements we list, there are always more elements in the set that are not in our list. • If S is a finite set, the symbol S stands for the number of elements of S. • The set with no elements is called the empty set, and is written as ∅. Thus ∅ = 0.
276
ICEEM Mathematics Secondary 1B
Exercise B 1
Is each of these a well defined set? If it is well defined, is it finite or infinite?
a {large whole numbers}
b {adults in Australia who like music}
c {fractions between 1 and 2}
d {working mobile phones in Australia}
e {points on a particular line }
f {past and present Australian senators} g {whole numbers between 1 million and 1 trillion}
2
Find the size A of each set.
a A = {5, 6, 8}
b A = {0, 1, 4, 5, 8, 9}
c A = {10, 20, 30, …, 200}
d A = {letters of the alphabet}
e A = {states of Australia}
f A = {primes less than 20}
g A = {perfect squares less than 20} h A = {50, 51, 52, …, 70}
i A = ∅
j A = {p}
k A = {0, 1, 1}
l A = {vowels}
3
List, or begin to list, the elements of each of the following sets (you may need to use dots). Then state whether it is a finite or an infinite set.
a {whole numbers less than 10}
b {odd whole numbers between 0 and 20}
c {all perfect squares}
d {whole numbers greater than 50}
e {even whole numbers between 1 and 7}
f {multiples of 7 greater than 50}
g {prime numbers less than 30}
Chapter 20 Review and problem solving
277
4
State whether each set is the empty set or is a nonempty set.
a {dogs 50 m high}
b {whole numbers greater than 4 but less than 6}
c {English speakers on Mars}
d {whole numbers greater than 7 but less than 8}
e {English words with three Es}
f {letters in the alphabet that are not vowels or consonants}
5
Explain whether each statement is true or false. a The set of points on a line is a finite set. b {whole numbers less than 1000 000} is a finite set. c {human beings who have ever lived} is an infinite set. d If two finite sets are equal, then they have the same number of elements. e If two finite sets have the same number of elements, then they are equal. f If A and B are empty sets, then A = B.
6
Should the symbol
in each statement be replaced by = or by ≠?
a If A = {7, 10, 38} and B = {Peter, Paul, Mary}, then A b If A = {7, 10, 38} and B = {Peter, Paul, Mary}, then A c {1, 3, 5, 7}
{7, 5, 1, 3}
d {8, 3, 5, 7}
{7, 5, 7, 8, 3, 8, 5}
e If T = {primes less than 10}, then T 
h If L = {0}, then L
0.
i If A = {0, 0}, then A j If A = ∅, then A
278
1. 0.
ICEEM Mathematics Secondary 1B
10.
B.
5.
f If H = {positive multiples of 3 less than 20}, then H g If K = {0, 1, 2, …, 10}, then K
B.
6.
C Subsets of a set Sets of things are often further subdivided. For example, owls are a particular type of bird, so every owl is also a bird. We express this by saying that the set of owls is a subset of the set of birds. A set S is called a subset of another set T if all the elements of S are elements of T. This is written as S ⊆ T. (We read this as ‘S is a subset of T’.) The new symbol ⊆ means ‘is a subset of ’. Thus {owls } ⊆ {birds} because every owl is a bird. Similarly, if A = {2, 4, 6} and B = {0, 1, 2, 3, 4, 5, 6}, then A ⊆ B, because every element of A is an element of B. The sentence ‘S is not a subset of T’ is written as S ⊆ T. This means that at least one element of S is not an element of T. For example, {birds} ⊆ {ﬂying creatures} because an ostrich is a bird, but it does not fly. Similarly, if A = {0, 1, 2, 3, 4} and B = {2, 3, 4, 5, 6}, then A ⊆ B, because 0 ∈ A, but 0 ∉ B. The set itself and the empty set are always subsets Any set S is always a subset of itself, because every element of S is an element of S. For example: {birds} ⊆ {birds} If A = {1, 2, 3, 4, 5, 6}, then A ⊆ A. Furthermore, the empty set ∅ is a subset of every set S, because every element of the empty set is an element of S, as there is no element in ∅ that lies outside S. For example: ∅ ⊆ {birds} If A = {1, 2, 3, 4, 5, 6}, then ∅ ⊆ A. Subsets and the adjective ‘all’ Statements about subsets can be rewritten as sentences using the word ‘all’. For example: {owls} ⊆ {birds} means ‘All owls are birds’. {birds} ⊆ {flying creatures} means ‘Not all birds are ﬂying creatures’. Chapter 20 Review and problem solving
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Venn diagrams Diagrams make mathematics easier because they help us to see the whole situation at a glance. The English mathematician John Venn (1834 –1923) began using diagrams to represent sets. His diagrams are now called Venn diagrams. The universal set In most problems involving sets, it is convenient to choose a larger set that contains all of the elements in all of the sets being considered. This larger set is called the universal set, and is usually given the symbol E. In a Venn diagram, the universal set is generally drawn as a large rectangle, and then other sets are represented by circles within this rectangle. For example, If V = {vowels}, we could choose the universal set to be E = {letters of the alphabet} and all the letters of the alphabet would then need to be placed somewhere within the rectangle, as shown below. E
bcdf ghjklm npqrstvwxyz
V aei ou
In the Venn diagram below, the universal set is E = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, and each of these numbers has been placed somewhere within the rectangle. E 0
A
2
5
1
9
7 6
10
4
3
8
The region inside the circle represents the set A of odd whole numbers between 0 and 10. Thus we place the numbers 1, 3, 5, 7 and 9 inside the circle, because A = {1, 3, 5, 7, 9}. Outside the circle we place the other numbers 0, 2, 4, 6, 8 and 10 that are in E but not in A.
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Representing subsets on a Venn diagram When we know that S is a subset of T, we place the circle representing S inside the circle representing T. For example, let S = {0, 1, 2}, and T = {0, 1, 2, 3, 4}. Then S is a subset of T, as illustrated in the Venn diagram below. E 3
4
T
0 S 2 1
Subsets of a set • If all the elements of a set S are elements of another set T, then S is called a subset of T. This is written as S ⊆ T. • If at least one element of S is not an element of T, then S is not a subset of T. This is written as S ⊆ T. • For any set S, ∅ ⊆ S and S ⊆ S. • Statements about subsets can be rewritten using the words ‘all’ or ‘not all’. • Subsets can be represented using a Venn diagram.
Exercise C 1
Choose whether ⊆ or ⊆ should replace
a {cats}
c {primes}
e {boys}
2
Let A = {1, 2, 3, 4, …, 10}, B = {1, 2, 3, 8, 9} and C = {2, 3, 5, 7, 9}. with either ⊆ or ⊆. Copy these statements, replacing
a B
e {3, 5}
i ∅
{mammals}
A
∅
in these sentences.
b {animals}
{mammals}
{odd numbers} d {brick buildings} f {adults}
{males}
{homes}
{people}
b B
C
c C
A
d {1, 4, 9, 16}
B f ∅
A
g A
A
h B
j C
{prime numbers}
A
∅
k {2, 4, 6}
A
Chapter 20 Review and problem solving
281
3
Explain why each statement below is true.
a If S is a set, then ∅ ⊆ S.
b If A ⊆ B and B ⊆ C, then A ⊆ C.
c If S is a set, then S ⊆ S.
d If A ⊆ B and B ⊆ A, then A = B.
e If x ∈ S, then {x} ⊆ S.
f If x ∈ S and y ∈ S, then {x, y} ⊆ S.
g If A ⊆ B, where A and B are finite sets, then A ≤ B.
4
Are the statements below true or false? Explain your answers.
a If A ⊆ B and A and B are finite sets, then A < B.
b If A ⊆ B and A and B are finite sets, then A < B.
c If A = B and A and B are finite sets, then A = B .
d If A ⊆ B ⊆ C and A, B and C are finite sets, then A ≤ C .
5
a List all subsets of each set listed below. (Recall that the set itself and the empty set are subsets.)
i {a, b, c}
ii {a, b}
iii {a}
b How many subsets did each set have? What pattern do those numbers form? 6
Rewrite each of the following statements using set notation and the symbol ⊆ or ⊆.
a All fish are animals.
b Numbers ending in 6 are all even.
c Not all children with blonde hair are naughty.
d Not all numbers ending in 4 are multiples of 4.
e Not all adults are sensible.
7
a Construct three particular sets A, B and C such that A ⊆ B, A ⊆ C and B ⊆ C. b Construct three particular sets A, B and C for which A ⊆ B, A ⊆ C and B ⊆ C.
8
a Construct two particular sets A and B such that A ⊆ B and B ⊆ A. b Construct two particular sets A and B such that A ⊆ B and B ⊆ A.
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D Sets and the number line Whole numbers The whole numbers are the numbers 0, 1, 2, 3, …. These are often called the ‘counting numbers’, because they are the numbers we use when counting things. In particular, we have been using these numbers to count the number of elements of finite sets. Zero is a whole number, because zero is the number of elements of the empty set. The set of all whole numbers can be represented by dots on the number line. 0
1
2
3
4
5
6
7
Any set of whole numbers can be represented on the number line. For example, here is the set {0, 1, 4}: 0
1
2
3
4
5
6
7
Sets and the number line • The set of whole numbers is infinite. • The set of whole numbers can be represented on the number line.
Exercise D 1
Write down, using set notation, the set represented by the blue dots.
a
b
c
d
2
Represent each set on a number line.
a {1, 2, 5}
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
b {0, 1, 5, 6}
c {2, 3}
d {4}
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e {whole numbers less than 5}
f {even whole numbers less than 5 units from 6}
g {odd whole numbers between 0 and 8}
h the set of prime numbers less than 12
i the set of numbers that are 2 units from 3
j {even whole numbers between 3 and 9}
k {whole numbers between 1 and 4 inclusive}
l {odd whole numbers between 3 1 and 4 1 }
3
List the elements of the set A such that A ⊆ {whole numbers}, A = 7, and the largest number in A is 6.
2
2
E Union and intersection We often need to talk about the overlap between sets. We use some new words and new notation to describe the situation. The intersection A ∩ B of two sets The intersection of two sets A and B consists of all elements belonging both to A and to B. This is written as A ∩ B. For example, some musicians are singers and some play an instrument. If A = {singers} and B = {instrumentalists}, then A ∩ B = {singers who play an instrument}. Here is an example using small sets of letters. If V = {vowels} and F = {letters in ‘dingo’}, then V ∩ F = {i, o}. This can be represented on a Venn diagram as follows. E V
F a e u
284
i o
d n g
ICEEM Mathematics Secondary 1B
We return to a discussion of intersection and Venn diagrams in the next section. Intersection and the word ‘and’ The word ‘and’ tells us that there is an intersection of two sets. For example: {singers } ∩ {instrumentalists} = {people who sing and play an instrument} {vowels} ∩ {letters of ‘dingo’} = {letters that are vowels and are in ‘dingo’} The union A ∪ B of two sets The union of two sets A and B consists of all elements belonging to A or to B. This is written as A ∪ B. Note that the elements may belong to both. Continuing with the example of singers and instrumentalists: If A = {singers} and B = {instrumentalists}, then A ∪ B = {musical performers}. In the case of the sets of letters: If V = {vowels} and F = {letters in ‘dingo’}, then V ∪ F = {a, e, i, o, u, d, n, g}. This can be represented on a Venn diagram as follows. E
V
F a e u
i o
d n g
We return to a discussion of union and Venn diagrams in the next section. Union and the word ‘or’ The word ‘or’ tells us that there is a union of two sets. For example: {singers } ∪ {instrumentalists} = {people who sing or play an instrument} {vowels} ∪ {letters in ‘dingo’} = {letters that are vowels or are in ‘dingo’}
Chapter 20 Review and problem solving
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Note that the word ‘or’ in mathematics always means ‘and/or’, so there is no need to add ‘or both’ to these descriptions of the unions. For example, let A = {even whole numbers that are less than 15} and B = {whole numbers divisible by 3 that are less than 15}. Then: A = {0, 2, 4, 6, 8, 10, 12, 14} and B = {0, 3, 6, 9, 12} A ∪ B = {0, 2, 3, 4, 6, 8, 9, 10, 12, 14} Note: 6 and 12 are in both A and B. Disjoint sets Two sets are called disjoint sets if they have no elements in common. For example: The sets M = {men} and W = {women} are disjoint. The sets S = {2, 4, 6, 8} and T = {1, 3, 5, 7} are disjoint. E
T
S
6
2
1
8
3
4
5
7
Another way to define disjoint sets is to say that their intersection is the empty set: Two sets A and B are called disjoint if A ∩ B = ∅. In the two examples above, M ∩ W = ∅ because no person is both a man and a woman, and S ∩ T = ∅ because no number lies in both sets.
Union and intersection Let A and B be two sets. • The intersection A ∩ B is the set of all elements belonging to A and to B. • The union A ∪ B is the set of all elements belonging to A or to B. In mathematics, the word ‘or’ always means ‘and/or’. • The two sets are called disjoint if they have no elements in common, that is, if A ∩ B = ∅.
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Exercise E 1
Use the words ‘and’ and ‘or’ to describe the sets P ∩ Q and P ∪ Q. a P = {boys who play cricket}, Q = {boys who play tennis} b P = {girls who study mathematics}, Q = {girls who study history} c P = {triangles with a right angle}, Q = {triangles with two equal sides} d P = {perfect square}, Q = {even whole numbers} e P = {animals with fur}, Q = {animals with webbed feet}
2
In each part, describe the sets A ∩ B and A ∪ B. Then state whether or not the sets are disjoint. a A = {girls who play winter sport}, B = {girls who play summer sport} b A = {popular singers}, B = {classical singers} c A = {hardboiled eggs}, B = {softboiled eggs} d A = {people born in Australia}, B = {people born overseas}
3
In each part, list the sets X ∩ Y and X ∪ Y, then state whether or not the sets are disjoint.
a X = {0, 1, 5, 8}, Y = {0, 5, 8, 10, 12}
b X = {0, 1, 4, 9, 16}, Y = {2, 3, 5, 7, 11, 13}
c X = {a, b, c, d, e, f}, Y = {b, c, f, g, k}
d X = {2, 4, 6, 8}, Y = {1, 3, 5}
e X = {1, 2, 3, …, 20}, Y = {15, 16, 17, …, 50}
f X = {2, 5, 10, 12, 14}, Y = {5, 10, 14}
4
In each part, list the elements of A ∩ B and A ∪ B, then represent them on separate number lines. a A = {0, 3, 6}, B = {1, 2, 3} b A = {odd numbers between 0 and 10}, B = {whole numbers less than 6}.
Chapter 20 Review and problem solving
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5
In each part, list the elements of A ∩ B ∩ C and A ∪ B ∪ C. a A = {0, 4, 8, 12, 16}, B = {0, 3, 6, 9, 12, 15}, C = {0, 10, 20} b A = {1, 3, 5, 7, 9, 11}, B = {1, 2, 3, 4, 5, 6}, C = {7, 8, 9, 10, 11, 12}
6
a Let A = {mammals}.
i Write down a subset S of A. ii Write down another subset T of A so that S and T are disjoint. b Let N = {whole numbers}.
i Write down an infinite subset S of N. ii Write down another infinite subset T of N so that S and T are disjoint.
288
7
Explain why the following statements are true, where A and B are any two sets.
a A ∩ B ⊆ A
b A ∩ B ⊆ B
c A ⊆ A ∪ B
d A ∩ B ⊆ A ∪ B
e ∅ ⊆ A ∩ B
f ∅ ⊆ A ∪ B
8
Copy and complete the following statements, where A and B can be any two sets.
a If A ⊆ B then A ∪ B =
b If A ⊆ B then A ∩ B =
c A ∪ ∅ =
d A ∩ ∅ =
e A ∪ A =
f A ∩ A =
9
Suppose A and B are finite sets. Are the following statements true or false? Explain your answers.
a A ∪ ∅ = A
b A ∩ ∅ = 0
c A ∪ B = A + B
d A ∩ B = A – B
ICEEM Mathematics Secondary 1B
F Set complements and Venn diagrams Complement of a set Suppose that a suitable universal set E has been chosen. The complement of a set S is the set of all elements of E that are not in S. The complement of S is written as S c. For example: If
E = {letters} and V = {vowels}, then V c = {consonants}
If E = {whole numbers} and O = {odd whole numbers}, then O c = {even whole numbers} Complement and the adverb ‘not’ The adverb ‘not’ always corresponds to the complement of a set. For example: If E = {letters} and V = {vowels}, then V c = {letters that are not vowels} = {consonants} If E = {whole numbers} and O = {odd whole numbers}, then O c = {whole numbers that are not odd} = {even whole numbers} Venn diagrams and set complements Note: In the next three Venn diagrams below, the universal set is E = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, and all these numbers are placed somewhere within the rectangle. In the first Venn diagram below, the region inside the circle represents the set A of odd whole numbers between 0 and 10. Thus we place the numbers 1, 3, 5, 7 and 9 inside the circle, because A = {1, 3, 5, 7, 9}. Outside the circle, we place the other numbers in E (0, 2, 4, 6, 8 and 10) that are not in A. Thus the region outside the circle represents the complement: Ac = {0, 2, 4, 6, 8, 10} E 0
5
1 7 6
10
4
A
2 9
3
8
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Venn diagrams of intersections and unions The Venn diagram below shows two sets A and B, where A = {1, 3, 5, 7, 9} and B = {1, 2, 3, 4, 5}. • The numbers 1, 3 and 5 lie in both sets, so we place them in the overlapping region of the two circles. • The remaining numbers in A are 7 and 9; these are placed inside A but outside B. • The remaining numbers in B are 2 and 4; these are placed inside B but outside A. E
A
10
B 9 7
1
5
2
3
6
4 8
0
Thus the overlapping region represents the intersection A ∩ B = {1, 3, 5}, and the two circles together represent the union A ∪ B, because A ∪ B = {1, 2, 3, 4, 5, 7, 9}. The four remaining numbers 0, 6, 8 and 10 are placed outside both circles. Representing disjoint sets on a Venn diagram When we know that two sets are disjoint, we represent them by circles that do not intersect. For example, let P = {0, 1, 2, 3} and Q = {8, 9, 10}. Then P and Q are disjoint, as illustrated in the Venn diagram below. E
Q
P
8
0 1 2 5
290
3 7
4
10
9
ICEEM Mathematics Secondary 1B
6
Set complements and Venn diagrams • We often choose a convenient universal set that contains all the elements under discussion. • The complement of a particular set consists of all elements of the universal set that are not in the set under discussion. • Sets are represented in a Venn diagram by circles inside the universal set. • The overlapping region of two circles represents the intersection of the two sets, and the two circles together represent their union. • When two sets are known to be disjoint, the circles are drawn without any overlap. • When one set is known to be a subset of another, its circle is drawn inside the circle of the other set.
Exercise F In all of the following, E stands for the universal set. 1
Write down the complement of each set S in the given universal set E. a S ={women}, where E = {adults} b S = {boys}, where E = {children} c S = {Australians living within 10 km of the coast}, where E = {Australians} d S = {weekdays}, where E = {days} e S = {whole numbers less than 100}, where E = {whole numbers} f S = {prime numbers}, where E = {whole numbers}
2
The diagram below represents the set A inside the universal set E.
E
A
2 5 0 1 6 4 3
List the elements of: a A c Ac e A ∩ Ac g A ∩ E i (Ac)c
b E d A ∪ Ac f E c h A ∪ E
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3
State whether each statement is true or false. Explain your answers.
a (Ac)c = A
b A ∪ Ac = E
c A ∩ Ac = E
d E c = E
e ∅c = E
f A = Ac
4
From the Venn diagram below, list all the elements of the following sets.
E
A
B
d b f a e g c
a E
b A
c B
d A ∩ B
e A ∪ B
f {elements of A that are not in B}
g {elements of B that are not in A} 5
From the Venn diagram below, list all the elements of the following sets.
E
a P
b Q
5 2
c P ∩ Q
d P ∪ Q
8 6 9
e P c
f Qc
11 10
g (P ∪ Q)c
h (P ∩ Q)c
i P ∩ Q c
j P ∪ Q c
k Q ∩ P c
l Q ∪ P c
P
Q
3
7
1
4
12
292
6
Draw a Venn diagram to illustrate each situation, where A and B are subsets of the universal set E.
a A ⊆ B
d A and B have some common elements but neither set is a subset of the other.
7
For each part, draw a Venn diagram, then list A ∪ B and A ∩ B.
a A = {1, 2, 4}, B = {3, 5}
b A = {a, e, i}, B = {e, i, o, s, t}
c A = {1, 2, 3, 5, 8, 10}, B = {3, 8, 10}
b B ⊆ A
ICEEM Mathematics Secondary 1B
c A and B are disjoint.
G Keeping count of elements of sets Before solving problems with Venn diagrams, we need to work out how to keep count of the elements of overlapping sets. E
A 6 9 1
3
E
B
8
10 5 4 2 7 12
A
B (3) (3)
11
(4) (2)
The lefthand diagram above shows two sets A and B inside the universal set E, where A = 6 and B = 7, with 3 elements in the intersection A ∩ B.
The righthand diagram shows only the number of elements in each of the four regions. These numbers are placed inside round brackets so that they don’t look like elements.
Notice carefully that A = 6 and B = 7, but A ∪ B ≠ 6 + 7. The reason for this is that the overlapping region A ∩ B should only be counted once, not twice. When we subtract the three elements of A ∩ B from the total, the calculation is then correct: A ∪ B = 6 + 7 – 3 = 10
Example 1 In the diagram shown below, A = 15, B = 25, A ∩ B = 5 and E  = 50. a Insert the number of elements into each of the four regions. E
A B
b Hence find A ∪ B and A ∩ B c. (continued on next page)
Chapter 20 Review and problem solving
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Solution E
A
B (10)
(5)
(20) (15)
a The intersection A ∩ B has 5 elements. Hence the region of A outside A ∩ B must have 10 elements, and the region of B outside A ∩ B must have 20 elements. This makes 35 elements so far, so the outer region has 15. b Hence A ∪ B = 35 and A ∩ B c = 10.
Example 2 In the diagram shown below, S = 15, T  = 20, S ∪ T  = 25 and E = 50. E
S
T
a Insert the number of elements into each of the four regions. b Hence find S ∩ T and S ∪ T c.
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ICEEM Mathematics Secondary 1B
Solution E
S
T (5)
(10)
(10) (25)
a The union S ∪ T has 25 elements, whereas S has 15 elements and T has 20 elements, so the overlap S ∩ T must have 10 elements. Hence the region of S outside S ∩ T must have 5 elements, and the region of T outside S ∩ T must have 10 elements. Finally, the region (S ∪ T )c inside E but outside S ∪ T must have 50 – 25 = 25 elements. b Hence S ∩ T  = 10 and S ∪ T c  = 40.
Number of elements in the union of two sets • The number of elements in the union of two sets A and B can be found by: Number of elements in A ∪ B = number of elements in A + number of elements in B – number of elements in A ∩ B
A ∪ B = A + B – A ∩ B
Exercise G 1
The sets A and B are subsets of the universal set E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. For each question part, draw a Venn diagram with two overlapping sets and insert in each region the number of elements, not the elements themselves. Then write down A ∪ B and A ∩ B.
a A = {1, 2, 3, 4}, B = {2, 3, 4, 5}
b A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8, 10}
c A = {5, 7, 8, 10}, B = {1, 3, 4}
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Example 1,2
2
Draw Venn diagrams to solve the following problems. a Find A ∪ B if A = 14, B = 11 and A ∩ B = 6. b Find A ∪ B if A = 25, B = 38 and A ∩ B = 13. c Find A ∪ B if A = 25, B = 21 and A ∩ B = 0. d Find A ∩ B if A = 7, B = 8 and A ∪ B = 10. e Find A ∩ B if A = 10, B = 15 and A ∪ B = 21. f Find A ∩ B if A = 16, B = 20 and A ∪ B = 36.
3
a A pair of dice was rolled 100 times. The sum of the numbers on the upturned faces was greater than 6 in 70 rolls, and less than 8 in 65 rolls. Draw a Venn diagram and find how many times this sum was 7. b One hundred boys all play cricket or soccer. Eighty play cricket and 67 play soccer. Draw a Venn diagram and find out how many play both sports. c One hundred girls each wrote down the value of the coins in their pockets. Sixty of them had more than $1 and 48 of them had less than $1.20. Draw a Venn diagram to find how many of them had $1.05, $1.10 or $1.15. d Among 100 concertgoers, 75 are singers, 65 are instrumentalists and 50 are both singers and instrumentalists. Draw a Venn diagram to find out how many are neither singers nor instrumentalists. e A coin collector buys a bag containing 100 coins. Sixtyfive are silver, 90 are round, and 61 are silver and round. Draw a Venn diagram to find out how many are neither silver nor round. f One hundred and sixty students all play the piano or the violin. One hundred and twenty play the piano and 80 play the violin. How many play both?
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H Problem solving using Venn diagrams This last section explains how to use Venn diagrams to solve counting problems. The approach is to identify the sets involved, and then construct a Venn diagram to keep track of the numbers in the different regions of the diagram.
Example 3 A travel agent surveyed 100 people to find out how many of them had visited the cities of Melbourne and Brisbane. Thirtyone people had visited Melbourne, 26 people had been to Brisbane, and 12 people had visited both cities. Draw a Venn diagram to find the number of people who had visited: a Melbourne or Brisbane
b Brisbane but not Melbourne
c only one of the two cities
d neither city.
Solution Let M represent the set of people who had visited Melbourne, let B be the set of people who had visited Brisbane, and let the universal set E be the set of people surveyed. Then the information given in the question can be rewritten as M = 31, B = 26, M ∩ B = 12, E = 100. Hence number in M only = 31 – 12 = 19 and number in B only = 26 – 12 = 14. E
M
B (19) (12)
(14) (55)
(continued on next page)
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a Number visiting Melbourne or Brisbane = 19 + 14 + 12 = 45 b Number visiting Brisbane only = 14 c Number visiting only one city = 19 + 14 = 33 d Number visiting neither city = 100 – 45 = 55
Problem solving using Venn diagrams • First identify the sets involved. • Then construct a Venn diagram to keep track of the numbers in the different regions of the diagram.
Exercise H Solve each of the following problems by drawing a Venn diagram. Example 3
1
Boys in a class play tennis or swim or both. Fifteen play tennis, 10 swim and 5 both play tennis and swim. a How many boys play tennis but do not swim? b How many boys swim but do not play tennis? c How many boys play tennis or swim but do not do both? d How many boys are in the class?
2
Of a group of 80 people on a railway station, 48 carry umbrellas and 52 wear raincoats, while 40 both carry an umbrella and wear a raincoat. a How many have an umbrella, but no raincoat? b How many have a raincoat, but no umbrella? c How many have an umbrella or raincoat or both? d How many have neither raincoat nor umbrella?
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ICEEM Mathematics Secondary 1B
3
The Royal Zoo has 500 animals, of which 400 can walk, 150 can swim, and 130 can both walk and swim.
a How many can walk, but not swim?
b How many can swim, but not walk?
c How many can walk or swim?
d How many can neither walk nor swim?
4
The Republican Zoo has a collection of 70 interesting snakes. Fiftyfive of them are poisonous, 45 are more than 1 m long, and 67 are either poisonous or more than 1 m long. a How many are poisonous and more than 1 m long? b How many are poisonous, but 1 m or less long? c How many are more than 1 m long, but not poisonous? d How many are neither poisonous nor more than 1 m long?
5
Two hundred and forty university students take part in a demonstration, all of them either carrying banners or sitting down on the road or both. Eightyfour carry banners and 204 sit on the road. Find how many students: a sit down on the road and carry banners b do only one of these things.
6
In a factory, 1000 light bulbs were subjected to two quality control tests A and B. Five hundred and sixty passed test A and 410 passed test B, while 150 passed both tests. How many light bulbs:
a failed both tests?
b passed test A but not B?
c passed test B but not A?
d passed only one of the tests?
e failed at least one of the tests?
f failed in only one of the tests?
Chapter 20 Review and problem solving
299
7
A 14thcentury Book of Saints contains the names of 300 men and 200 women. There are 250 martyrs amongst the men and 180 amongst the women; 210 of the men and 105 of the women healed the sick; and 200 of the men and 100 of the women both healed the sick and were martyrs. a How many saints were martyrs, but didn’t heal the sick? b How many saints healed the sick, but were not martyrs? c How many saints neither healed the sick nor were martyrs? d How many of the women saints neither healed the sick nor were martyrs?
300
8
A group of 100 children was asked which of the three films Hook, Batman and Dracula they had seen. Twentyseven had seen Hook, 38 had seen Batman and 16 had seen Dracula. Eleven children had seen both Hook and Batman, 8 had seen both Hook and Dracula, and 6 had seen Batman and Dracula, while 3 had seen all three films. Find how many of the children had seen:
a at least one of the three films
b Hook only
c only one of the films
d none of the films
e exactly two of the three films.
9
In a certain school, there are 180 pupils in Year 7. One hundred and ten pupils study French, 88 study German and 65 study Indonesian. Forty pupils study both French and German, 38 study German and Indonesian, and 26 study both French and Indonesian, while 19 study German only. Find the number of pupils who study:
a all three languages
b Indonesian only
c none of the languages
d at least one language
e either one or two of the three languages.
ICEEM Mathematics Secondary 1B
Answers to exercises
Chapter 11 answers Exercise 11A 1 a –4, –3, –2, –1, 0, 1, 2 b –7, –6, –5, –4, –3, –2 c –9, –8, –7, –6, –5 d –131, –130, –129, –128, –127, –126, –125, –124 2 a –1000, –100, –10, –6, 0, 5, 10, 100 b 55, 45, 26, –26, –30, –31, –45, –550 3 a
–8
–7
–6
–5
–4
–3
–2
–1
0
–8
–7
–6
–5
–4
–3
–2
–1
0
–8
–7
–6
–5
–4
–3
–2
–1
0
–3
–2
–1
0
1
2
3
b c d 4
–9, –7, –5
5
–15
–14
–13
–12
–11
–10
–9
–8
–7
–6
–5
–4
–6
–5
–4
–3
–2
–1
0
1
2
3
4
–3, –5, –7
6
–7
–35, –30, –25
–50
7 a –5
–45
b 4
–40
c 10
–35
–30
d 12
–25
e –7
f 8
8 a 2
b 7
c 20
d –10
9 a 3 < 5
b 3 > –5
c –7 < –4
d 2 < –(–3)
10 a –15°C
b –30°C
c –18°C
d –40°C
e –30
g 4
h 3 f 40
Exercise 11B 1 a 2 i –7
b 1 j –8
c 5 k 18
d 3 l –39
e 4 m –50
2 a –1 i –15
b –6 j –17
c –15 d –111 e –23 k –100 l –1990 m –15
f –3 n 34
g 6 o –65
h 5 p 5
f –7 n –100
g –8 h –5 o –1200 p –33
Answers to exercises
301
3 a 1 g –28
b –11 h 15
c –29 i –45
d –111 j 3
e –43 k –60
f –7 l –6
4 a –34
b –14
c 0
d 1
e –20
5 a $1600
b $1214
c –280 m
d –9°C
e $240
1 a 3 g –8
b –4 h –15
c 5 i 7
d –15 j –8
e 4 k –50
f –7 l –10
2 a 11
b 18 h 15
c 5 i –15
d 15 j –8
e 4 k 5
f –3 l 70
b –14 h 71
c 14 i 16
d –1 j 44
e 2 k 200
f 9 l 1550
4 a 434
b –229
c 135
d –6401
5 a –5
b 35 h –17
c 65 i –19
d 12 j 41
e 22 k 116
f –23 l –25
Exercise 11C
g –4 3 a –36 g –27
g 42 6 a i 8
ii 8
7
17°C rise
9
Minimum (°C)
8 10°C rise Maximum (°C)
–10
10
15
–15
5
20
–25
–3
22
–20
–15
5
–7
–5
2
–11
2
13
–13
5
18
11 –14˚C
6 minutes
12
Increase (°C)
5
Temperature inside (°C)
Temperature outside (°C)
Inside – outside (°C)
M
20
25
–5
T
13
18
–5
W
24
20
4
T
10
–5
15
F
–5
–10
5
S
3
–6
9
A negative number means it is warmer outside than inside. 13
302
Jane will have $70 left.
ICEEM Mathematics Secondary 1B
Exercise 11D 1 a f k p
–10 –10 120 –60
b g l q
–12 –12 –48 153
c h m r
–5 –50 126 102
d i n s
–44 –44 169 –280
e j o t
–192 –48 –152 180
2 a f k p
–5 –5 17 24
b g l q
–13 –18 6 3
c h m r
–5 –9 2 50
d i n s
–7 –3 –11 –27
e j o t
–40 –456 16 –36
3 a –5 g 50 4 a 4 g –15 m –24 5 a 36 g 2700 6 a –15 g –104
b 5 h –2
c –3 i –5
d –2 j –4
e 1 k 3
f –1 l –1
b –4 h –73 n –64
c –5 i –8 o 14
d 14 j –29 p 18
e –33 k –14
f –120 l 10
b –168 h –204
c 2400 i 40
d 6300 j 10
e –150 k 50
f –12 l –48
b –13 h –8
c –6 i –3
d –7 j –17
e –5 k –14
f –5 l 169
Exercise 11E 1 a 64 b –121 g –125 h 16
c 32 i –32
2 a 128
b –96
c –256 d 20
e 48
3 a –1 f 149
b 26 g –42
c –9 d 22 h –120 i 65
e –16
4 a 56
b 48
c –78
d 30
5 a 14 g 126
b 27 h –140
c 0 i 49
6 a 24
b –10
7 a –1
b –24
8 a 80
b –100 000
c –1900 d 2000
9 a –8
b –2
d –1
10
d –81 j 64
e –900 f 144 k –1 l 1 f –2
g –108 h –7
e –45
f –18
g 34
h –110
d –48 j –532
e –18 k –350
f 93
c –25
d –60
e –40
f –189
g 9
h –170
c 10
d –12 f –29
g 7
h 15
c –8
e 23
$3200, –$200, a loss of $200
11 a $15 000 b –$5000 12a $340 13 a $140 loss
b $430
b $220 profit
c $20 loss c $40 profit Answers to exercises
303
Review exercise 1 a 23 h –60 o –6 2
b –14 i 70 p –85
c 15 d –95 j –80 k 32 q –190 r –65
e –16 f –68 l –50 m –25 s 70 t –82
g –180 n –135 u –547
–10
3a 25°C
b –25°C
4a 25°C
5 a –250
b –396
c –1750
d 2040
e –48
f 1150
h 500
i –440
j –2400
k 100
l 100
b –4 h 150
c –7 3 i –2 4
d 17 j –6
e 4 k 14
f 5 l –25
b –63 h 49
c –60 d 60 i –90
e –168 8 –56
f –16 9 –6
g 3200 6 a –25 g 8 7 a 4 g 16
b –25°C
1
10 a 64
b –64
c –12
d –22
e –26
f –1000 g –1 2
11 a 55
b –93
c –13
d –64
e –37
f 30
h 8 g –119 h 51
Challenge exercise 1 a –32 3
2a 1
a
1
–1
–2
b –1
c
0
–7
8 0
–6 –4
–26
–6
–6
d
–18
–34
–7
2
–18
–10
–12
4 a 4 + (–7) – (–2)
b 4 + (–2) – (–7)
c –7 + (–2) – 4
5 a 5 + (–4) – (–5)
b –5 + (–4) – 5
c 5 + (–5) – (–4)
6
–2
7 a (5 + (–3)) × (3 + 4) = 14 c (5 – 5) × 6 + 7 × 6 – 5 = 37
ii –6
10 100
11 –6
12a –10
b –52
2 –3 –2
–5 –1 3
0
b 5 + (–3) × 3 + 4 × 2 = 4
9a i –6
8
304
–1
–6
–9
–8
b –1
1
–4
ICEEM Mathematics Secondary 1B
b –50
Chapter 12 answers Exercise 12A 1 a –4 g 8
b 2 h 4
c 0 i 5
d –5 j 7
e –1 k 1
f –8 l 12
2 a –57
b 36
c –64
d 35
e 66
f 125
g 15
3 a –1
b –28
c 20
d 96
e –72
f 6
g 288
h –17 h –8 1 3
4 a –9
b –20
c –10
d 75
e –30
f
5 a –6
b 14
c 10
d –10
e –20
f 100
6 a –1
b 15 g 36
c –3 h 216
d 7 i –25
e 27
b 56
c –20
d 20
e –500
f 27 7 a –44 8 a 7
b 5
c –1
d –1
e –1
f 4
g –2
9 a –7
b 0
c –7
d –10
e –25
10 a 9
b 81
c 11
d –216
e 27
11 a –30
b 30
c –60
d 200
e –1000 f
f 60 h 7 f –1 f –18
2 –5
g
5 –2
h 10 000
12 a –400
b –10 000
c –10
d –8000
e –900
f 0
13 a 40
b –16
c 14
d 64
e 12
f –48
14 a i $(1000 – x)
ii $(1000 – 5x)
b i $500 ii $0 iii –$250
15 a i (25 – x)°C
ii (25 – 6x)°C
b –5°C
Exercise 12B 1
A(4, 4), B(3, –1), C(0, 4), D(–5, 0), E(–3, –3), F(–5, 5), G(5, –4)
2
y
3
y
5
B
E
4
E
3
2
A
1 –5 –4 –3 –2 –1 0
1
–5
4
5
x
–5 –4 –3 –2 –1 0
A
F
1
2
3
4
5
x
D
–1
C
–2
B
–3 –4
3
1
F
–2
C
2
D
–1
H
4 3
G
2
G
5
–3 –4
H
–5
Answers to exercises
305
4
a
y
b
y
3
6
2
5
1 0 –3 –2 –1
4 1
2
3
x
–1
2
–2
1 –6 –5 –4 –3 –2 –1 0
–3
1
2
3
4
5
6
x
–1
–2
The letter E c
–3
y
–4
4
–5
3
–6
2
An eight pointed star 1
5a A(3, 2), B(4, 0), C(3, –3), D(–1, –1), E(–3, 4), –3 –2 –1 1 2 3 4 x 0 F(–2, 3), G(0, 2) –1 b A(1, 2), B(3, 3), C(3, –1), D(2, –1), E(0, 0), –2 F(–3, 0), G(–2, 3) –3 c A(1, 1), B(3, 3), C(3, –1), D(3, –2), E(2, –3), –4 F(–4, –3), G(0, –2) Ned Kelly d A(2, 3), B(1, 0), C(1, –1), D(0, –3), E(–1, –2), F(–3, –2), G(–2, 0) 6 a
b
y 5
D
y 5
C
4 3
3
2
2
1 –5 –4 –3 –2 –1 0
A
B 1
2
3
A 4
5
x
1
–5 –4 –3 –2 –1 0
–1
–1
–2
–2
square; area = 9 cm 2
c
2
3
4
5
2
C
1
0 –6 –5 –4 –3 –2 –1
B 1
triangle; area = 12 cm2
y D
C
4
1
2
3
4
5
6
7
x
–1 –2 –3
rectangle; area = 55 cm2 –4 A B
306
ICEEM Mathematics Secondary 1B
x
d
e
y
y 6
6
C
5
B
A
4
4
2
2
A
1
1
–6 –5 –4 –3 –2 –1 0
D B
3
3
C
5
1
x
–6 –5 –4 –3 –2 –1 0
1
2
3
4
5
6
7
x
–1
rightangled triangle; area = 7.5 cm2
–2 –3 –4
The lines meet at right angles.
f
y
parallelogram; area = 12 cm 2
6 5 4
B
C
3 2
A D
1 –3 –2 –1 0
1
2
3
4
5
6
7
8
9 10
x
–1 –2 –3 –4
7
a, b, c
y
b (0, 0), (3, –1), (4, –2), (4, –4), (2, –4), (1, –3), (0, 0)
6 5
d (0, 0), (–3, –1), (–4, –2), (–4, –4), ( –2, –4), (–1, –3), 3 (0, 0)
4
2
(0, 0), (–3, 1), (–4, 2), 1 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 (–4, 4), (–2, 4), (–1, 3), 0 x –1 (0, 0) –2 –3 –4 –5 –6
Answers to exercises
307
Exercise 12C 1 a y = 3x
b y = –2x
x
−3 −2 −1
0
1
2
3
y
−9 −6 −3
0
3
6
9
y
x
−3 −2 −1
0
y
6
0
(3, 9)
4
2
5
(–2, 4)
4
7
(2, 6)
6
3
(–1, 2)
2
5 4
(1, 3)
3
1
(0, 0)
0
1
–5 –4 –3 –2 –1
1
(0, 0)
0
1
2
2
3
4
5
x
(1, –2)
–1
2
(–1, –3)
3
6
8
–5 –4 –3 –2 –1
2
y
(–3, 6)
9
1
−2 −4 −6
–2 3
4
5
x
(2, –4)
–3
–1
–4
–2
–5
–3
–6
(3, –6)
–4
(–2, –6)
–5 –6 –7
(–3, –9)
–8 –9
d y = x + 2
c y = x – 2
x
−3 −2 −1
y
−5 −4 −3 −2 −1
0
3
0
1
−3 −2 −1
0
1
2
3
y
−1
2
3
4
5
0
1
y 5
4
4
3
3
1 0 –1
(–1, –3) (–2, –4) (–3, –5)
x
5
2
–5 –4 –3 –2 –1
308
2
y
1
–2
(–1, 1) (–2, 0)
(3, 1) (2, 0) 1
2
3
(1, –1) (0, –2)
4
5
x
2 1
–5 –4 –3 –2 –1 0
(–3, –1)
–1 –2
–3
–3
–4
–4
–5
–5
ICEEM Mathematics Secondary 1B
(3, 5) (2, 4) (1, 3) (0, 2) 1
2
3
4
5
x
e y = 2x + 1 f y = 1 – x
x
−3 −2 −1
0
1
2
3
y
−5 −3 −1
1
3
5
7
y
x
−3 −2 −1
0
1
y
4
1
0
3
2
(3, 7)
5
6
(–3, 4) (–2, 3) (–1, 2)
(2, 5)
5 4
(1, 3)
(–1, –1) (–2, –3)
3 2
–5 –4 –3 –2 –1
(0, 1)
(0, 1) (1, 0)
0
1
2
0
1
2
3
4
5
–2
x
–1
–3
–2
–4
–3
–5
3
4
5
x
(2, –1) (3, –2)
–1
1 –5 –4 –3 –2 –1
4
1
3 2
3
y
7
2
−1 −2
–4
(–3, –5)
–5
2a y = x +
g y = 3 – 2x
x
−3 −2 −1
0
1
y
9
3
1
7
5
2
3
−1 −3
y
y
−3
−2
−1
0
1
2
3
1 1 1 –2 2 –1 2 – 2
1 2
1 12
1 22
1 32
9
(–3, 9)
x
1 2
y 5
8
7 (–2, 7) (–1, 5)
6
3
5
1 –2) 2 1
4
(–1, (0, 3)
–5 –4 –3 –2 –1
3
(1, 1)
2 1 –5 –4 –3 –2 –1
4
0 –1
1
2
3
4
5
(2, –1)
0
1
2
3
4
5
x
–1 –2 –3 –4 –5
–2 –3
x
1 (–2, –1 2 ) 1 (–3, –2 2 )
1
(3, 3 2 ) 1 (2, 2 2 ) 1 (1, 1 2 ) 1 (0, 2 )
(3, –3)
–4
Answers to exercises
309
b y = x –
1 2
x
−1
0
1
1 1 1 y –3 2 –2 2 –1 2
1 –2
1 2
−3
1 2
c y = 2x +
−2
2
3
1 2
1 22
1
y
5
4
2 1
1 (–1, –1 2 ) 1 (–2, –2 2 ) 1 (–3, –3 2 )
−3
−2
−1
0
1
2
3
1 1 1 y –5 2 –3 2 –1 2
1 2
1 22
1 42
1 62
y
0
1
–1
(3, 2 2 ) 1 (2, 1 2 ) 1 (1, 2 ) 2
3
4
5
(3, 6 2 )
6
1
5
(2, 4 2 )
4
1
3
(1, 2 2 )
2
x
1
1
(0, – 2 )
–2
1
7
1
3
–5 –4 –3 –2 –1
x
–5 –4 –3 –2 –1
–3
(–1,
1 –1 2 )
0
1
(0, 2 ) 1
2
3
4
5
x
–1 –2
–4
(–2,
–5
1 –3 2 )
–3 –4
d y = –x +
1
(–3, –5 2 )
1 2
x
−3
−2 −1 0 1 2 3
y
1 32
1 22
1
1 2
1 2
1 –2
–5 –6
1 1 –12 –2 2
y 5 1 32)
4
(–3, 1 (–2, 2 2 ) 1 (–1, 12 )
3 2
1
(0, 2 )
1
–5 –4 –3 –2 –1
0
1
–1 –2 –3
2 13 4 5 x (1, – 2 ) 1 (2, –1 2 ) 1 (3, –2 2 )
–4 –5
3 a
310
x
0
1
2
3
4
5
6
y
–7
–2
3
8
13
18
23
ICEEM Mathematics Secondary 1B
b
x
–4
–3
–2
–1
0
1
2
3
4
y
25
21
17
13
9
5
1
–3
–7
4 y = x2
y
x
−3 −2 −1
0
1
2
3
y
9
0
1
4
9
4
1
10
(–3, 9)
(3, 9)
9 8
7 6 5
(–2, 4)
(2, 4)
4 3
(–1, 1)
2 1
–5 –4 –3 –2 –1
(1, 1) (0, 0)
0
1
2
3
4
5
x
Exercise 12D 1 a y = 1 × x + 1 b d = 3 × t + 6 c y = 2 × x + 2 = x + 1 = 3t + 6 = 2x + 2 e n = 3 × m – 3 f y = –2 × x – 1 d d = –3 × t + 6 = –3t + 6 = 3m – 3 = –2x – 1 2 a 2 c
4
5
6
9
11
13
b 2
3a
4
5
6
10
12
14
b 2
3 c m
c
14
14
13
13
12
12
11
11
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1 0
1 1
2
3
4
d m = 2t + 1
5
6
t
0
1
2
3
4
5
6
t
d c = 2t + 2
Answers to exercises
311
4 a 40 b
c
4
5
6
28
24
20
c 40 36
d c = –4n + 44
32 28
e 6
24 20 16 12 8 4 0
5
a
1
2
b
y 4
4
3
3
2
2
1
1
0
1
2
3
4
5
–5 –4 –3 –2 –1
x
y=x
–3
–3 –4
–5
–5
y
d
1
2
3
4
5
x
5
x
y=x+1
y
5
5
4
4
3
3
2
2
1
1
0
n
6
–2
–4
–1
312
0 –1
–2
–5 –4 –3 –2 –1
5
y 5
–1
c
4
5
–5 –4 –3 –2 –1
3
1
2
3
y = 2x
4
5
x
–5 –4 –3 –2 –1
0
2
3
4
–1
–2
–2
–3
–3
–4
–4
–5
–5
ICEEM Mathematics Secondary 1B
1
y=x–2
6 a
2
3
4
960 940 920 7 a
4
5
5
6
900
880
d 50 months
b h = 5w
c 60
d 9
8a y = x + 3
b y = –2x
c y = –2x + 1
d y = x – 4
e y = 3x
f y = –2x – 1
6
b d = 1000 – 20m c $800
20 25 30
Review exercise 1 a 1 d 6
b –7 e –12
c 5 f 6
y
2
C
6
F
5 4 3 2
A
1
D
–5 –4 –3 –2 –1 0
1
2
3
4
5
x
–1
E
–2
B
–3 –4
3
A(4, 4), B(–4, 3), C(–3, –2), D(3, –3), E(1, 1), F(0, –4), G(–2, 0)
4 a y = 3 – x
b y = 2x – 3
x
−3 −2 −1
0
1
2
3
x
−3 −2 −1
0
1
2
3
y
6
3
2
1
0
y
–9
–3
–1
1
3
5
4
y
(–3, 6) (–2, 5) (–1, 4)
6
6
5
4
4
2
1 –5 –4 –3 –2 –1 0
(0, 3) (1, 2) (2, 1) (3, 0) 1
2
3
4
5
–1
6 a i $(600 – x)
y
8
2
b 7
–5
7
3
5 a 13
–7
–5 –4 –3 –2 –1 0 –2
x
(–1, –5) (–2, –7) (–3, –9)
–12
f 225
(0, –3)
3
4
5
x
(1, –1)
–8 –10
e – 9 ii $(600 – 5x)
2
–6
–3
d –27
1
–4
–2
c 9
(3, 3) (2, 1)
g –225 h 20
b i $100
ii –$400 Answers to exercises
313
7 a i (20 – 2x)°C c 0°C
ii (20 – 10x)°C d 11:00 pm
8 D(6, 0)
9 D(7, –6)
10 a y = x + 2
b d = 2t + 8 e d = –3t + 15
d q = 2p – 1
b 12°C
c n = 4m – 4 f y = –3x – 7
Challenge exercise 1
a
CDs for David
0
1
2
3
4
5
6
7
8
9
10
CDs for Angela
10
9
8
7
6
5
4
3
2
1
0
b (0, 10), (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1), (10, 0)
c
CDs for Angela
d y = 10 – x
y (0, 10) 10 (1, 9) 9 (2, 8) 8 (3, 7) 7 (4, 6) 6 (5, 5) 5 (6, 4) 4 (7, 3) 3 (8, 2) 2 (9, 1) 1 (10, 0) 1 2 3 4 5 6 7 8 9 10 x 0 CDs for David
2
a
Pairs of CDs for David and Andrew
5
4
3
2
1
Single CDs for Angela
0
2
4
6
8
b (5, 0), (4, 2), (3, 4), (2, 6), (1, 8) c
y 10 9
(1, 8)
CDs for Angela
8 7
(2, 6)
6 5
(3, 4)
4 3
(4, 2)
2 1
0
(5, 0) 1
2
3
4
5
6
7
8
9 10
x
Pairs of CDs for David and Andrew
314
ICEEM Mathematics Secondary 1B
d y = 10 – 2x
3 a B(8, 9) b Y(5, 5), Z(5, 0) or Y(–5, 5), Z(–5, 0) c C(0, 8), D(–4, 4) or C(8, 0), D(4, –4) 4
C can be any point with ycoordinate 8 or –8.
5
C(6, 10), D(6, 4) or C(–6, 10), D(–6, 4)
6
C(7, 10), D(7, 4) or C(–7, 10), D(–7, 4)
7 a
Stacks of mangoes
Stacks of pineapples
3
0
2
2
1
4
0
6
Stacks of pineapples
c
b (0, 6), (1, 4), (2, 2), (3, 0)
y (0, 6)
d 2x + y = 6
6 5
(1, 4)
4 3
(2, 2)
2
(0, 3)
1
0
1
2
3
4
5
6
x
Stacks of mangoes 8 a (0, 45), (2, 36), (4, 27), (6, 18), (8, 9), (10, 0) c i 2 adults and 36 children
b
y (0, 45) 50 (2, 36) 40 (4, 27) 30 (6, 18) 20 (8, 9) 10 (10, 0)
0
1
9 a S = a 2
2
3
4
b
5
6
7
8
9 10
ii 4 adults and 27 children
x
a
1 2
1
1 12
2
1 22
S
1 4
1
1 24
4
1 64
c S
1
1
(2 2 , 6 4 )
6 5 4 3 2 1 0
(2, 4) 1
1
(12 , 2 4 ) 1 1
( 2 , 4 ) (1, 1) 1 2
1
1
1
1 2 2 2 2
a
Answers to exercises
315
Chapter 13 answers Exercise 13A 1 a i = 70˚ (corresponding angles, AB  PQ) b a = 110˚ (alternate angles, AB  PQ) c b = 95˚ (cointerior angles, AB  PQ) 2 a i = 110˚ (straight angle at B) b a = 33˚ (adjacent and vertically opposite angles at O) c b = 125˚ (revolution at M) d a = 108˚ (cointerior angles, RS  UT), b = 72˚ (cointerior angles, TS  UR), c = 108˚ (cointerior angles, RS  UT) e i = 118˚ (corresponding angles, LS  AG), c = 62˚ (cointerior angles, AL  GS) f a = 82˚ (vertically opposite angles at G), b = 62˚ (alternate angles, IE  TH) g a = 35˚ (corresponding angles, AB  NO), b = 55˚ (adjacent angles at N) h c = 75˚ (cointerior angles, LA  SR), i = 45˚ (corresponding angles, EP  LA; adjacent angles at P) i a = 140˚ (cointerior angles, MJ  NK), b = 150˚ (cointerior angles, OL  NK), c = 70˚ (revolution at N) 3 a BC  DA (alternate angles are equal); ∠CAT = 90˚ (cointerior angles) b AC  TK (alternate angles are equal); ∠CAT = 38˚ (corresponding angles) c OG  CA (corresponding angles are equal); ∠CAT = 107˚ (corresponding angles) d AC  TD (cointerior angles supplementary); ∠CAT = 110˚ (cointerior angles)
Exercise 13B 2 a a = 80˚ e a = 73˚
b b = 140˚ f b = 43˚
c c = 55˚ g c = 60˚
d i = 14˚ h i = 90˚
c c = 31˚ g c = 117˚
d i = 27˚ h i = 123˚
i a = 60˚ (alternate angles, RS  PQ), b = 70˚ (alternate angles, PQ  RS), c = 50˚ (straight angle at B) 3 a a = 120˚ e a = 53˚
316
b b = 44˚ f b = 152˚
ICEEM Mathematics Secondary 1B
4 a No, because two obtuse angles add to greater than 180˚. b No, because two right angles add to 180˚. c 2, because of parts a and b. d Yes, if the interior angle is obtuse. e No, because 2 obtuse interior angles would be required, which is impossible. 5 a a = 72˚ (straight angle at N), a + b = 112˚ (exterior angle of
), b = 40˚
b i = 115˚ (straight angle at H), c = 105˚ (exterior angle of ) c a = 45˚ (straight angle at M), b = 135˚ (exterior angle of ) d c = 110˚ (straight angles at W and A; exterior angle of e i = 90˚ (straight angles at A and B; exterior angle of f a = 93˚ (exterior angle of b = 117˚ (exterior angle of
)
FLS), ALS)
g a = 85˚ (straight angle at L; exterior angle of b = 115˚ (exterior angle of
)
GIL),
AGI)
h 125˚ = a + 20˚ (exterior angle of EGH), a = 105˚ a = b + 32˚ (vertically opposite angles at E; exterior angle of 6 a a = 30˚ e a = 40˚
b b = 30˚ f b = 70˚
c c = 60˚ g i = 30˚
EFG), b = 73˚
d i = 50˚ h c = 12˚
7 a i = 35˚ (alternate angles, HO  ER), a = 75˚ (angle sum of ORS) b 115˚ = 75˚ + i (exterior angles of
), i = 40˚, b = 40˚ (alternate angles, UL  RI)
c a = 55˚ (alternate angles, RA  DC), b = 25˚ (alternate angles, CA  DR), c = 100˚ (angle sum of ADR) d a = 50˚ (alternate angles, PL  YA), a + b = 110˚ (cointerior angles, PY  LA), b = 60˚, c = 60˚ (alternate angles, PY  LA) e a = 32˚ (alternate angles, PA  TR), b = 58˚ (angle sum of TRP), c = 100˚ (angle sum of
ARP)
Answers to exercises
317
f b = 45˚ (angle sum of FGH), a = 45˚ (alternate angles, IF  HG), c = 20˚ (angle sum of FHI) g 95˚ = a + 40˚ (exterior angle of
FGW), a = 55˚, b = 55˚ (alternate angles, FG  TI), c = 40˚ (alternate angles, FG  TI)
h a = 150˚ (straight angle at A), b = 65˚ (alternate angles, MS  TR), a = b + c (exterior angle of ART), c = 85˚ 8 a ∠AVB = 25˚ (angle sum of
AVB; corresponding angles, AB  PQ)
b ∠AVB = 86˚ (angle sum of
AVB; corresponding angles, VB  PQ)
c ∠VAP = 90˚ (angle sum of
RAP), VAB)
∠AVB = 63˚ (angle sum of
d ∠APV = 72˚ (exterior angle of ∠AVB = 97˚ (exterior angle of
BOP), APV)
e ∠APV = 102˚ (cointerior angles, AP  QB), ∠AVB = 130˚ (exterior angle of APV ) f ∠VAB = 115˚ (cointerior angles, AV  PQ), ∠AVB = 20˚ (exterior angle of ABV ) g AP  BR (alternate angles at P and Q are equal), ∠AVB = 70˚ (alternate angles, AP  BR) h PA  QV (cointerior angles are supplementary), ∠VAP = 45˚ (cointerior angles, PA  QV), 45˚ = 25˚ + ∠AVB (exterior angle of ABV), ∠AVB = 20˚ i ∠BCV = 70˚ (corresponding angles, AB  VC), 110˚ = 40˚ + ∠AVB (cointerior angles, PQ  RS), ∠AVB = 70˚
Exercise 13C 1 a centre b radius
c diameter d chord
2 a O b RT, SU c RS, ST e SRT, RUS, UTR, TSU
e minor arc f major arc
d OR, OU, OT, OS f the curves RS, RU, UT, TS
5 e ∠A ≈ 31.5˚, ∠B ≈ 39˚, ∠C ≈ 109.5˚ 6 b BC ≈ 9.2 cm 7 b Angle sum of a triangle is 180˚.
318
ICEEM Mathematics Secondary 1B
c AC = 4.3 cm, BC = 5.5 cm
Exercise 13D 1 a P b PQ and PR, 3 cm d ∠PQR = ∠PRQ ≈ 67˚
c QR e Base angles are equal.
2 a Yes, base angles are equal.
b Yes, ∠B = ∠C. d No, all angles are different. f No, all angles are different.
c No, all angles are different. e Yes, ∠B = ∠A. 3 a ∠ABC = 70˚
b ∠OPQ = 62˚
4 a a = 43˚
b a = 45˚, x = 5 mm d b = 45˚, y = 6 mm e a = 30˚, x = 5 mm f a = c = 60˚, b = 120˚, x = 6 mm h a = 45˚, x = 4 mm
5 a a = 75˚
g a = 90˚, b = 60˚
i a = 60˚, x = 5 mm b b = 36˚ e a = 46˚, b = 67˚
d b = 31˚
c a = 75˚, b = 90˚
c a = 110˚ f a = 90˚
Exercise 13E 4 a b c d
Construct an equilateral triangle on AB and then bisect angle at A twice. Construct a right angle at A and bisect angle twice. Construct 30˚ as in part a and take the supplement. Construct 45˚ as in part b and construct the supplement.
Exercise 13F 1 a a = 85˚ b b = 60˚
c a = 114˚ d a = 65˚
2 a 60˚
c 85˚
b 90˚
e a = 70˚
f a = 99˚
d 135˚
3 a a = 90˚, b = 115˚ c a = c = 118˚, b = 62˚
b a = 112.5˚ d a = 90˚
4 a a = 160˚ c a = 110˚, b = c = 35˚ e a = 101˚
b a = 50˚, b = 130˚, c = 120˚, i = 40˚ d a = 52˚, b = 48˚ f b = 46.5˚
5 a a = 13˚ d a = 72˚
b a = 135˚ e a = 40˚
6 a 90˚, 45˚ and 45˚
b 135˚, 22.5˚ and 22.5˚
c b = 62˚ f a = 78˚, b = 143˚
Answers to exercises
319
Review exercise 1 a a = 30˚ (vertically opposite angles at A), b = 90˚ (alternate angles, CB  QP), c = 60˚ (angle sum of ABC) b a = c = 65˚ (cointerior angles, AB  DC and AD  BC), b = 115˚ (cointerior angles) c b = 90˚ (cointerior angles, FI  GJ), a = 110˚ (corresponding angles) d i = 70˚ (construct MX parallel to KL and ON; alternate angles) 2 a i = 110˚ (exterior angle of ∆RST) b a = 40˚ (angle sum of UVW, vertically opposite angles at W) c d e f 3
b = 22.5˚ (angle sum of BRT) 125˚ = c + 90˚ (exterior angle of DEF), c = 35˚ i = 70˚ (angle sum of ABC; corresponding angles, BC  PQ) 5i = 140˚ (exterior angle of JKM), i = 28˚
∠XAB = b and ∠YAC = c (alternate angles, XY  BC), so a + b + c = 180˚ (straight angle at A).
4 a b c d e f
a = 25˚, b = 130˚ (base angles of isosceles triangle are equal) c = 78˚ (base angles of isosceles triangle are equal) i = 35˚ (base angles of isosceles triangle are equal) a = b = 20˚ (base angles of isosceles triangle are equal; alternate angles, PQ  RS) i = 100˚ (angle sum in quadrilateral) i = 65˚ (angle sum in quadrilateral; vertically opposite angles at A, B, C and D)
5 a 10 cm (equal angles)
b 8 cm (∠S = ∠R) c 5 cm (∠M = 40˚ due to alternate angles, PQ  GF, so ∠P = 70˚ = ∠F)
6 a b c
AB  DC (cointerior angles are supplementary) MN  PQ (alternate angles ∠N and ∠P are equal) PQ  AC, AP  BQ (alternate angles ∠APB and ∠PBQ are equal), BP  CQ (alternate angles ∠PBQ and ∠BQC are equal)
8 b All sides and hence all angles are equal, so each angle is a third of 180˚. 9 a Construct an equilateral triangle. b Bisect the 60˚ angle. c Bisect the 30˚ angle. e Bisect the 90˚ angle. f Bisect the 45˚ angle. 12 a a = 32˚, b = 68˚ d a = 90˚ f a = 79˚
320
d Construct a right angle.
b a = 108˚ c a = 94˚ e b = 65˚, a = 45˚, c = 110˚ g a = 90˚ h a = 64˚
ICEEM Mathematics Secondary 1B
Challenge exercise 1 a b = a + ∠DCB, so b > a. b ∠ACD = a (Base angles of
ABC are equal.)
c It is impossible that b > a and b < a at the same time, so AB and AC must have equal lengths. 2 b Draw in radius OA, and construct two 30˚ angles starting from A on either side of OA, cutting the circle at B and C. 3 c ∠POB = 2∠APO and ∠POA = 2∠BPO (exterior angles of so ∠APB = ∠APO + ∠BPO =
1 2
1 2
(∠POA + ∠POB) =
BOP and × ∠AOB = 90˚.
AOP),
5 d Proof follows from the result in question 3. 6
b Draw in two perpendicular diameters, AOB and COD.
7
b Construct right angles at A and B, and their bisectors. Let the bisectors meet at C.
8
c Let the base angles of the four isosceles triangles be a, b, c and i. Then the angle sum of ABCD is 2(a + b + c + i) = 360˚, so a + b + c + i = 180˚. Observing that ∠A + ∠C = a + b + c + i = ∠B + ∠D completes the proof.
d Follow the same procedure as in part c, except now let the flattest triangle have base angle i. Then the angle sum of ABCD is 2(a + b + c – i) = 360˚, so a + b + c – i = 180˚. In this case, ∠A + ∠C = a + b + c – i = ∠B + ∠D, and the result is proved.
Chapter 14 answers Exercise 14A 7
1 1
11
1 a –2, – 5 , –1, 4 , 2 , 1 11
4
2 12
11
c – 12, – 5 , – 3 , 13 2
–312
–4
3 a 4 a g 5 a
g
1 2 1 4 1 – 15 3 –4 3 – 20
–5 1
b 2 2 b h b h
1 10 1 –1 45 13 – 10 27 – 40
15
d –2, –113, – 13, –1
– 4 34 –6
7 1 7 2
b – 5 , – 4 , 4 ,
–314
–112
–3
–2
1
d 7
c – 12
7
d
i – 10
1
j
c – 24
11
d
i
1 –1 10
j
c – 5
1 12
–1
0
4
e – 2
4 21 7 – 12 10 – 21 13 – 25
1
1
f –2 2
e – 10
3
f –1 21
k
l
e k
1 15 3 10 4 15
1
2 1
g – 3
h
2 5
5
f l
1 15 1 – 15 11 – 15
Answers to exercises
321
1
1
1
2
i 2 4
j –3 5
k –2 18 l –2 14 m – 2 n 1 2 o – 15 p – 12 9 11 17 17 31 37 s – 20 t – 35 u 2 36 v –3 35 w –4 91 x 56
7
5
5
q –1 20 r – 12 19
1
19
c –4 20 d –5 40
14
k –5 20 l
b – 35
2
c – 20
2 – 21
j – 45
7 a 2 44
b 4 20
7
j 3 15
i –510
19
31
13
13 40
f 15
3
c –1
3
e –1 2
1
b 2 2
1
d – 3
1
6 a 2 2
1
g – 2
1
3
11
3
e – 4
f –510
g
17 30
o
1
m –6 10 n
5 12 11 – 15
h – 5 5
10
h –6 21 14
p 115
Exercise 14B 1 a
– 101 35
9
d – 77
15
4
k
7
3 10
1
m 3
3
e 10
f 19
3
e 4 2
i
2 a –12
b –11 3 c – 10
d – 7
3 a –4
b –3
d – 4
g 38
h –7 7
c 9
3
3
4
5
4 a – 4
b 24
c
g 4 44
h –2 15
i
31
5 a –9
14
21
f 44
10
b 1 11 1
g –1 5
1
i –6 5
c h
6
g – 55
1
n
1
1
f – 3 3
d – 6
e –1 2
f –1 9
j 1 27
k –5 24
l –2 40
5
8
2 d 125 i
3 10
3
1
j 65 7
55 108 5 –1 13 83 –1192 10 –1 11
1
f – 6
l 5
h – 78
2
2
e – 7
k –17 4 l –12 7 1
2
5
4
e 3 5
7
78 161
Exercise 14C 1 a 2.6
b 5.25
c 1.2
2 a –0.7
b –8.2
e 3.5
f 0.7
c 6.12 d –0.4
e –5.8
f –10.3 g 4.93 h 9.47
3 a –0.24 b 0.28
c 0.24 d 0.82
e –0.75 f 0.002 g 0.024 h 0.63
4 a –5
c 41
e –0.3
b 41
d –1.2
d –25
g –2.5 h –1.55
f 2.1
g 1.1
5
a –0.027
b –0.064 c –0.091
d 0.009
e –0.00801
6
a –5.0 9
b –30.4
c –16.35
d –7.76
e –3.03
f –14.67
g –26.26
h –6. 0 9
i 4.5
7
a 0.021
b 0.0075
c 0.21
d –0.02625
..
..
h 310 f –0.0064
e 0.3125
f –0.00575
Exercise 14D 1 a –1 1
g 8
c 1 2
1
i 3 2
h 4 2
2 a –3 3
322
1
b 2 1
b –4 2
1 1
1
c –3 2
d 2
ICEEM Mathematics Secondary 1B
1
1
d –3 2
e 2
f – 8
j 4
k 4
l 2
1
e – 6
f
1 8
2
g 3
1
h –1 2
11
1
4
1
1
3 a – 30
b –6 5
c 3 5
d 6
e –4 6
f 4 6
4 a 3.5
b 4.5
c 19.5
d –19.5
e –0.05
f 0.25
5 a 4
b 8
c 28
d –28
3
1
6 a 6 4
27
b 5 4
7 a –1 1
2
2
b –5 3
9
d – 1024 e – 16 c 4.1 d 4.5
b –5.1
8 a 1 6
243
c – 64
4
c –2 3
1
d 6 4
e 0.5
e –8
f
9 a –0.31
b –4.2
c 3.89
d 0.44
10 a 11.1
b 8.9 f 19.36
c –1.331 g –6.05
d –1.61051 h 12.2
e –1.21 19
9
11 a – 20
9
b –3 20
c 3 20
d –25
1
e – 5 f –12 2 1 1 1 f 2 4 g –1 8 h 6 2 f 0.02
9
1
1
e 44
g 33 3
9
19
11 100
h – 32 f 0.05
25 e 0.8
f 911
g –1 25 h
12 a –4.55
b 3.75
c –3.45
d –0.24
e –0.015
f –1.5
13 a 0.159
b –0.161
c 0.007
d 2.8
e 7000
f –20.14
Challenge exercise 1
–3
–2
2 3
2 3
5
9
2 –12 , –24
.
.
7
3a –3.63, –3.63, –3 11 , –3. 6
..
2 1
2 –5 –1 3 3
. .
.
4
b –1.435, –1.435, –1.435, –1 9 2
2 1 4 54 3 5 There are 16 solutions. Find them. –4 – – 3 3 11 –
–7 5
10
–9 5
– 19 10 – 13 10
–8 5 –3 2
– 17 10
–6 5
Chapter 15 answers Exercise 15A 1 a 100% 2 a g
2 25 1 15
3 a 0.07 g 1.27
b 66% b h
3 20 13 120
b 0.25 h 1.68
c 40% c i
3 4 1 2 10
c 0.75 i 4.1
d 50% d j
3 5 1 42
d 0.06 j 4.6
e 37.5% e k
3 10 1 14
e 0.35 k 1.7
f
4 5 12
l 425 f 0.9 l 4.96 Answers to exercises
323
4 a 27% g 90%
b 97% h 0.7%
c 13% i 5.3%
d 127% j 52.7%
e 229% k 98.7%
f 70% l 32.8%
5 a 60%
b 85%
c 44% i 106%
d 35% j 208%
e 46% k 435%
f 76% l 152%
c 70% i 129%
d 30% j 12.5%
e 73% k 37.5%
f 130% l 87.5%
h 340%
g 95% 6 a 35%
b 27% h 789%
g 560% 7 a 8
1 40
b
11 200
c
1 16
d
7 8
Decimal
0.5
0.25
0.75
0.4
1
0.457
0.23
0.403
0.625
Fraction
1 2
1 4
3 4
2 5
1
457 1000
23 100
403 1000
5 8
50%
25%
75%
40%
100%
45.7%
23%
40.3%
62.5%
Percentage
Exercise 15B 1 a 80% g 125%
b 62.5% h 93.75%
c 28% i 200%
d 85% j 87.5%
e 30% k 20%
f 37.5% l 312.5%
2 a 60%
b 75%
g 24%
h 45%
c 8.5% i 62.5%
d 24% j 0.4%
e 46% k 23.5%
f 15% l 57%
3 a 72%
b 80%
c 12%
d 67.5%
e 62.5%
b 40
c 200 i 45
d 150 j 200
e 15 k 236
f 1.5 l 7
c 41 h 560
d 207 i 205
e 150 4 $26
5 $64.80
Exercise 15C 1
$222
2 a 20
h 750
g 150 3 a 184
b 36
f 102
g 48
6 a 3.36 hours
b 8.4 hours
c 35.5%
Review exercise 1
1 a 4
b
14 25
2 a 0.34 b 0.99
c
b 45% f 1000% g 180% k 3% l 5% g 73.2
324
2
d 1 5
c 0.23 d 0.02
3 a 20%
4 a 33
3 100
b 89 h 111.15
c 98% h 80% m 25% c 12 i 29.58
ICEEM Mathematics Secondary 1B
2
e 1 25 e 0.09
99
f 9 100 f 1.23
g
1 40
h
g 2.5
d 20% i 370% n 225%
e 230% j 40% o 0.04%
d 18 j 236
e 18 k 30
49 400
h 3.83
f 9 l 1500
5
5 a 48% 6
92%
b 5 9 %
c 11.839%
7 129
8 26 3 %
d 8.625%
2
Challenge exercise 1
$800
2 68%
3a 6
4
$144
5 $824.32
b 5
6 625 000
c 3
1
7 9 11 mL
Chapter 16 answers Exercise 16A 1 a 3x + 4 = 31
b x = 9
2 a x = 7 g x = 4
c x = 15 1 i x = 4 2
b x = 15 h x = 7
3 a a = 8
d x = 5 j x = 9
b a = 30 f a = 3
e a = 35
e x = 21 k x = 24
c a = 15 g a = 7
f x = 21 l x = 100
d a = 6 h a = 10
4 a 23a + 17 = 707 b a = 30
5a 10x + 23 = 523 b x = 50
6 a 6n = 72
b n = 12
7a 6c + 7 = 127
8 a 19 – x = 12
b x = 7
9
10 $85.80
$34.65
11 x = 11
b c = 20
12 x = 5
13 x = 24
Exercise 16B 1 a m = 3 b m = 11 f m = 16 g m = 9
c m = 2 h m = 1
d m = 16 i m = 16
e m = 17
2 a n = 3
c n = 5
d n = 5 3
1
e n = 8 3
b n = 3
5
f n = 2 9 g n = 3 5
2
h n = 8 2
i n = 5 3
3 a x = 36 b x = 30
c x = 80
d x = 20
e x = 18
f x = 32
2 55
c x = 20
d x = 9
e x = 12
f x = 4
i x = 10
j x = 14
k x = 7
l x = 2 3
4 a x = 4
b x =
g x = 9
h x = 6
1
1
1
1
2
5
m x = 2 4 n x = 2 7
o x = 2 6
5 a x + 5 = 21, x = 16
b 7x = 35, x = 5
d 5x = 50, x = 10 g x – 15 = 37, x = 52
e
6 a z + 7 = 12, z = 5
b z + 12 = 19, z = 7
d 9 – z = 6, z = 3
e
g 5z = 45, z = 9
h
x = 6
10, x = 60
2 3z = 5, z = 1 3 6 7z = 20, z = 2 7
1
2
c 5x = 37, x = 7 5 f
x = 3
23, x = 69
c z – 6 = 14, z = 20 2
f 7z = 37, z = 5 7 i
z = 6
7, z = 42 Answers to exercises
325
Exercise 16C 2
1 a x = 3 f x =
3 44
b x = 2 5
c x = 2
g x =
h x =
1 45
2 a z = 24 b z = 45 g z = 27 h z = 175 2
3 a x = 4 4 a x =
b x = 5 5
1 42
b a = 14
1 85
d x = 4 i x =
c z = 60 i z = 132
d z = 187 j z = 165
c x = 5
d x = 1 5
c z = 8
7 b = 10 11 1 z = 22
h a = 11 3
2
i a = 7 2
j
m a = 2 5 n x = 204
o x = 84
p x = 240
g b = 9
1
3
d
3
5 a 7x + 6 = 20, x = 2 d
x – 3 z + 4
1 97
13 = 14, z = 4
h
a 11
e z = 72 k z = 78 1
f x = 45
e x = 3
f x = 70
10
1
k k = 1 11
l n = 2 2
q z = 84
r p = 252
f
+ 6 = 30, a = 264
f z = 160 l z = 1700
e x = 4 2
c
3 = 7, x = 24
e 4x + 6 = 30, x = 6
4 = 23, x = 81
g 7x + 4 = 68, x = j
b
x + 6
e x = 4
7 7 10
i
x + 11 = 20, x = 36 4 x – 4 = 50, x = 378 7 m – 11 = 20, m = 217 7
k 6m – 14 = 10, m = 4
Exercise 16D 1 a x = –9 b x = –6 g x = –4 h x = 0 m x = –3 n x = –15 2 a x = –3 e x = –12 i x = 3 m z = –100 3 a x = –7
b f j n
c x = –17 d x = 5 i x = –6 j x = –7 o m = –45 p z = –14
b x = –3
3
f x = 1 5 k x = 7 1
c g k o
x = –5 x = –50 m = 9 a = –72 c x = 7
1
h x = –9 5 m x = 5
1
g x = –2 2 l m = –4
4
e x = 0 k x = –9 d h l p
x = 3 x = –28 m = –14 p = –99
f x = 5 l m = 0 x = –1 x = –110 x = –5 b = –90 1
d x = 1
e x = –1 2
i x = –27 n x = –2
j x = 22 2 o x = 5
d x = –224 i m = –192
e x = –35 j m = –18
1
1
p m = –1 5
q p = –2 3
r z = 25 2
4 a x = –12 f x = –36 k n = 40
b x = –12 g m = –180 l p = –240
c x = –3 h m = –30
Exercise 16E 1 a 4x + 20 e 14a – 28 2
b 3x – 12 f 24 + 18x 1
2 a z = 1 3 b z = 10 5
326
c 12x – 24 g 54x – 99 1
c z = 7 6
ICEEM Mathematics Secondary 1B
3
d z = 3 4
d 15x – 25 h x2 – 3x 1
e z = –4 7
1
f z = 2 6
13
7
6
3 a x = 5 10
b m = 1 18
c m = 2 35
d x = 9
f n = 2 28
g m = 7 14
h n =
i m =
25
9
43 49
2
e m =
8 55
25 36
2
2
4 a 3(x + 4) = 32, x = 6 3 b 5(x – 2) = 42, x = 10 5 c 3(x + 7) = 50, x = 9 3 1
1
1
d 2(x – 3) = 15, x = 10 2 e 4(x + 11) = 23, x = –5 4 f 3(x – 6) = 16, x = 11 3 1
1
g 5(m – 10) = 26, m = 15 5 h 6(z + 5) = 42, z = 2 2 j 6(z – 8) = 100, z = 24 3
i 12(n + 11) = 150, n = 1 2
Exercise 16F 1 a 14x b 16x c 13x g 10x + 2 h 14x + 7y i 7x + 2y
d 7x j 8x + 6y
e 16x f 7 + 9x k 7x + 7 + 10y l 2x + 2y + 11
2 a 7x + 6 b 24x – 6 f 11x – 12 g 13x – 8
c 27x + 22 d 30 e 46x + 14 h 19x + 34 i 10x – 14 j 19x – 36
3 a x = 6
c x = 5 7 h x = 2
d x = 2 i x = 6
4 a z = 1 6 b z = 3 3
c z = 5
d z = 1 5
5 a m = 3
b m = 1 9
c x = –1 19 d m =
1 a x = 3
b x = 3
c x = 4
g x = 1
h x = 3
i x = 20
2 a m = –5 b z = 1
c n = – 2
4 d m = – 13 e z = – 12
f m = 1 15
3 a x = 16 23 b x = 5 34
c x = 20
d x = 2
f x = 2 12
g x = 5
h x = 8
i x = 5
j x = 45
e x = 2 67
b 34
c 98
d 12
e 4 7
f 6
j 7
k 11
l 6 3
f x = 2
2
b x = 3 g x = 3
1
1
4
2
6
4
e x = –4 5 1 j x = 1 2 1
1
e z = 3 5
f z = 1 7
e x = 9
f n = 1 23
d x = 5
e x = 2
f x = 1 4
j x = 8
k x = 2
l x = 2 2
20 23
15
Exercise 16G 3
1
8 k x = 1 17
1 1
l x = 7 15
Exercise 16H 1 a 8 g
1 47
h
2 18 3
i
2 12 7
3 12 cm; 36 cm
3
2
21 cm
6
5 and 4 4
8
27x + 21 = 2181, x = 80. There are 80 bananas in each case.
1
4 15 and 18
1
5 54 and 56
7 32 cm
9 a 4x – 3 = 6x – 7, x = 2 b 11x – 14 = 9x + 6, x = 10 c 3x – 8 = 12 – 2x, x = 4 10
x = 5
1
11 x = 3 3
12 x = 140
Answers to exercises
327
Review exercise 1 a 12x + 3 = 363
b x = 30 2
1
2 a x = 12 b a = 5 5 g n = 14 m x = 3 s x =
1 52
c z = 41 3
h p = 3 n x = t x =
i y =
1 47 4 25
u x =
e x = 9 5
f x = 70
j x = 108
k x = –20
l x = 160
q x = 15
r m = 14
1 42
o x = 6
p x =
4 17
3
d b = 9
v x =
3 15 1 2 18
w m =
b 7x = 25, x = 3 7
c 6x + 2 = 44, x = 7
d
x – 7
3 = 6, x = 63
4
Length is 48 cm, width is 12 cm.
5 28, 16
6
a x = –16 b x = –5
d x = –4
g x = –28 h x =
7 a x = 1
c x = 0
1 53
b x = 2
4
x p = 11
4
3 a x + 11 = 23, x = 12
3 44
9
8 a x = 2 13 b x = 3 11
1 i x = –2 4 1 c x = – 4
d x= –1
c x = 2
d x = 11 2
j x =
1 –2 5 1
e x = 3
f x = –1
k x = 11
l x = – 2
1
e x = 1 3
1
f x =
e x = 7 2
1
f x =
9
x 5
11
x + 163 = 390; 227 pages
12 3x + 8 = 44; 12 years old
13
2x 3
14 2x + 18 = 40; length = 11 cm
15
2n – 1 = 81; n = 41
17
Let x be Brett’s amount x + x + 16 + x + 24 = 400 3x + 40 = 400; Brett: $120, Jennifer: $136, David: $144
18
3 4x =
= 4; x = 20
= 34; 51 cakes
69; 92
20 Suppose Anne runs x km, x – 12 =
3 11 9 8 20
10 x + 29 = 65; the man is 36 years old.
16 6x – 6 = 3x – 4; x =
2 3
19 2x – 5 = 55; 30 cm 1 2 x;
Anne: 24 km, Jennifer: 12 km, Carl: 12 km
21 6(x – 7) = 48; x = 15 3
22 a 8x + 6 = 3x – 2, x = –1 5 c 11x + 7 = 10 – x, x =
b 6x – 10 = 7x + 6, x = –16
1 4
23 x = 10 24 x = 50 25 Let Jason receive x dollars; x + 3x + 2x = 50; Jason: $75 David: $225, Isabell: $150.
Challenge exercise 1 a 10x cm
2
b x = 6 5
2 a 2(20 –10(x + 6)) = –80 – 20x 3 a 5:12 pm, 192 km from Tantown 4
328
student: 11; sister: 14; father: 36
ICEEM Mathematics Secondary 1B
7
b x = –3 11
17
c x = 43 21
Chapter 17 answers Exercise 17A 1 a 1
1 1
b 0
c 2 , 2
d 1
e 0
Exercise 17B 1 a 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
b 1, 3, 5, 7, 9
2 a 2, 4, 6
c 3, 6
b 1, 3, 5
d 2, 3, 4, 5, 6
3 a 1, 2, 3, 4, 5 b 1, 2, 3 1
b 2
1
c 3
1
d
1 10 1 5 1 6 1 10 1 26
b 2
1
c 5
4
d
b 5
3
7a 9
1
b
b 6
5
c 3
1
d
1
c 5
1
d
11 13 1 6
c
5 26 1 2
d
4 a 2 5 a 6 a 8 a 10 a 11 a 12
3 5
b 2 b 13a
b
c
5 6 3 10 2 9 2 3 1 2 5 26 2 3
e c 9a e e
9 10 4 9 1 5 2 5 3 13
f
5 9 2 5 3 10
e
19 30
d b
d 0
c
3 5
1
14 5
Exercise 17C 1 a 100
b 800
4 a 160
b
4 11 1 114
b
6 a 7 a
b
16 31 7 11 7 19
c 10 000 5a c c
5 18 10 11 211 228
b
5
d 4 c d
2a 4 c
5 18
d
1 12
e
19 36
e
113 114
f
17 228
8
1 5
1
2a
1
c
5 12
d
3 13
g
5 13
d
4 11 4 13 22 43
7 12 7 11 3 13
7a 167
d
4 11
9a
c
1 5
d d
17 36 1 33 77 114
Review exercise 1
1 a 8
1
c 8 b
c
2 11 6 13 21 43
c
2 11
b 2
b
6 a
7 13 25 86
b
1 11 6 13 7 43
8 a
1 11
b
2 11
3
5
5
a
4a
c
Challenge exercise 1 a f
1 50 1 10
b g
49 50 9 10
1
c 2 h
b 80
7 25
e
99 500
b 2
f
d
5 12
b 1000 b
1 10
2
e
13 50
j
d 5 i
9 50 18 25
Answers to exercises
329
1
1
2 a 5
b 2
c
7 10
d 0
3 a
Thomas Leslie Anthony Thomas Leslie Tracie Thomas Leslie Kim Thomas Anthony Tracie Thomas Anthony Kim Thomas Tracie Kim Leslie Anthony Tracie Leslie Anthony Kim Leslie Tracey Kim Anthony Tracie Kim
e
3
c 5
1 3
b
f
7 50
2
b 5
4a
1 10
32 225
Chapter 18 answers Exercise 18A 1 a 3 units right and 5 units up c 3 units right and 2 units down e 3 units right 2
b 2 units right and 1 unit up d 3 units right and 2 units down f 2 units right and 2 units down
b
a
Aʹ
c
Aʹ
Aʹ
A
Bʹ
A
Bʹ
B
C
B
e
d
A Cʹ
f A
A
Aʹ
Aʹ A Aʹ
B Bʹ
3 a They are the same. c They are the same.
d i A′C′ ii A′L′
330
B
C Bʹ
Cʹ
b i ∠X′Z′Y′ ii ∠Z′X′Y′ d They are both 12.5 square units.
4 a L′ is the midpoint of B′C′. 5 a 4 units right
C
Cʹ
b yes
ICEEM Mathematics Secondary 1B
b 2 down, 10 right e i B′L′ ii C′A′ c A′C′
d no
c yes f no
Exercise 18B 1
b
a Mʹ
N
c
Zʹ
Yʹ
A O Cʹ
M
O
Nʹ
O
Xʹ Z
B
C
X
Aʹ
Y
Bʹ
d
e
f Aʹ
P Q Pʹ Qʹ
A
R
Rʹ 2
b
a
Aʹ
c Bʹ A
Aʹ
Aʹ O
O O
A
A
B
e
d
f
A Aʹ
Bʹ
B
O h
g B Cʹ
O
O
O
Aʹ
Bʹ
A
B
A
O
C Aʹ
Bʹ
i
B
A
C
O Aʹ
Cʹ
Bʹ
Answers to exercises
331
j
B
k
A Cʹ
Dʹ
O C
D
Aʹ
Bʹ
l
A
O
A
O Aʹ
Aʹ 3
Only the centre of rotation is fixed.
4 a
b Parallel
c They have the same length.
Cʹ Bʹ
d They are both 12 square units. D
Dʹ Eʹ
Aʹ O
E
C
A
B
Exercise 18C 1
a
b
Aʹ
A B
A
Aʹ
Bʹ B
Bʹ C
c
Cʹ
d
X
Xʹ A
Z Zʹ Y
Yʹ
332
ICEEM Mathematics Secondary 1B
Aʹ
B Bʹ
e
f
X
A Cʹ
Z
Bʹ B
Y Yʹ C
Zʹ
Aʹ
Xʹ
g
h
A B
A
Cʹ
Bʹ
Aʹ
B Bʹ
C
Aʹ i
j
l
Xʹ Yʹ Z, Zʹ
X
Y
k
Answers to exercises
333
m
n
2
o
m n 3 e C is fixed. g Vertices A, B, C appear in clockwise order; A′, B′, C′ are in anticlockwise (that is, opposite) order.
Exercise 18D 1
Eʹ y Cʹ E C Dʹ
D
334
Bʹ
4 3
Aʹ F
2 1
–5 –4 –3 –2 –1 0
Fʹ
5
2 y 5 C E 4 D F 3 Bʹ 2
B A
1
2
Aʹ
1
3
4
5
x
–5 –4 –3 –2
Eʹ –1 0 A 1
–1
–1
–2
–2
Dʹ
–3
–3
–4
–4
–5
–5
A′(2, 3), B′(4, 4), C′(–1, 4), D′(–3, –2 ), E′(–1, 5), F′(2, 5)
ICEEM Mathematics Secondary 1B
2
3
4
5
Fʹ B Cʹ
a 1 up and 2 right b 2 right and 5 up c 8 down d 4 left and 6 down e 4 left and 4 down f 4 down
x
3
4
y
y
5
4
4
Bʹ C Fʹ
–5 –4 –3 –2
3
3
F Aʹ
2
B
1
Eʹ –1 0 E 1
2
3
4
Aʹ
–5 –4 –3 –2
x
E –1 0 Eʹ1
Bʹ
–2 –3
D
4
5
x
A Cʹ Fʹ
–4
A′(1, 2), B′(–1, 2), C′(–1, –3), D′(5, –5), A′(–2, 1), B′(–2, –1), C′(3, –1), D′(5, 5), E′(0, 0), F′(–2, 0) E′(0, 0), F′(0, –2)
5
6
y
5
C
Fʹ D Dʹ F
E
4
3
3
B
2
1
2
3
4
5
–5 –4 –3 –2 –1 0
x
A Bʹ
–1
–3
A
1
Aʹ
1
B
2
–1
1
2
Cʹ 3
4
5Cx
Aʹ
–2 –3
Eʹ
–4
–4
–5
–5
A′(2, 1), B′(2, –2), C′(–3, –3), D′(–5, 0), a i A′(2, –1) E′(0, –4), F′(–1, 1) iii C′(5, 0) y
7
5
–2
Cʹ
y
4
–5 –4 –3 –2 –1 0
3
–5
–5
2
–1
–3
Dʹ
B
1
–2
–4
D
5
F
2
C
A
–1
Cʹ
Dʹ
5
Bʹ
ii B′(3, –3) b A′B′C′
5
Cʹ C B
4
Bʹ
3 2 1
–5 –4 –3 –2 –1 0
Aʹ
–1
1
2
3
4
5
x
A
–2 –3
a i
A′(–3, –1)
iii C′(0, 4)
ii B′(3, 3) b A′B′C′ Answers to exercises
335
Exercise 18E 1
A B Order:
1
1
G H Order:
1
2
C D E
F
1
1
1
1
I
J
K
L
2
1
1
1
M N O P Q R Order:
1
S Order:
Order:
336
2
2
1
1
1
T U V W X
2
1
1
1
Y
Z
=
+
1
2
2
4
ICEEM Mathematics Secondary 1B
1
2
$ % 2
2
2
Order:
Order:
square
rectangle
rhombus
kite
4
2
2
1
equilateral triangle
isosceles triangle
right triangle
trapezium
3
1
1
1
any diameter is an axis of reflection
ellipse
parallelogram
2
2
circle Order:
infinite
Exercise 18F 1
a,b 2 3 4 9 8 5 2 7 1
360˚
a 9 = 40˚ c 9
6 5 4 3
3 a 8
2
1
c 45˚
d The diameters shown are the axes of symmetry, together with the lines joining opposite m idpoints; 8. e the diameters shown, 4
Answers to exercises
337
4 b
All acute triangles thus constructed c are equilateral, and hence have angles of 60˚, which is the angle required for 360˚ the hexagon 6 = 60 .
5
b a clock face
6
36 and 37, respectively
order = 6
6 5 4 3
c 12, 12 1
2
Exercise 18G 1 a
Aʺ
4 right, 3 up
b
3 down, 2 right
3 right
d
3 right, 1 up
Aʹ A
c
2 a
b
c
90˚ anticlockwise 3
a–c
30˚ clockwise 210˚ anticlockwise d yes
n
m
338
ICEEM Mathematics Secondary 1B
e–g
h no n
m
4
Many answers are possible. For example:
a Rotate 90˚ clockwise about B; translate 3 right, 4 up. b c d e f
Reflect in AB; translate 5 right, 3 up. Reflect in AC; translate 1 up. Reflect in AB; translate 3 right, 6 up. Reflect in BC; translate 2 right, 3 up. Rotate 90˚ clockwise about C; translate 1 down.
Review exercise 1 a 3 right, 1 up d 1 left, 2 down 2
b 2 down
c 2 right, 2 down
e 2 right
f 2 right, 1 up
3
Aʹ O
C
Cʹ
Aʹ O
B
A
A B
Bʹ 4
5
Aʹ B Bʹ
Cʹ O C
A
A
B
Bʹ
C
Cʹ
y
6
C
Bʹ
Aʹ 7
y
5
5
4
4
3
3
Cʹ
Bʹ
2 1
–3 –2 –1 0
1
2
1 3
4
C Bʹ
2
Aʹ
B
A
5
6
7
x
A
–3 –2 –1 0
–1
–1
–2
–2
–3
–3
–4
–4
–5
–5
a A′(3, 1) b B′(5, 2) c C′(0, 2)
Aʹ 1
Cʹ
a 3 right, 1 up c 6 down
2
3
4
5
6
7
x
B
b 5 right, 5 up
Answers to exercises
339
8
y
9
y
5
5 4
4
B
3
C
Bʹ
3
Aʹ
2
1
1 –4 –3 –2 –1 0
Cʹ
1
–1
2
3
B
2
C
Aʹ 4
5
6
–4 –3 –2 –1 0
x
–1
Bʹ
A
–2
–2
–3
–3
–4
–4
–5
–5
1
2
Cʹ
4
5
A
a A′(2, 1) b B′(–4, 1) c C′(–1, –2) a A′(–3, 2) c C′(2, –1)
b B′(–3, –2)
10
b B′(3, –2)
a A′(3, 2) y 5 c C′(–2, –1) 4 3
C
Aʹ, B
2 1
–4 –3 –2 –1 0
1
2
3
–1
Cʹ
4
5
6
x
A, Bʹ
–2 –3 –4 –5
11 a 2
b 4
d 2
e 2
12
c 6 f 2
a
b Bʹʹ
B
Bʹʹ A
Bʹ
Aʹʹ
Cʹʹ Bʹ
Aʹʹ Aʹ
C Aʹ
Cʹ B
Cʹ Cʹʹ
A
1 left and 3 down 3 left and 4 up
340
3
ICEEM Mathematics Secondary 1B
C
6
x
13 a CʹB
Bʹ
C
b B
AʹA Aʹʹ O
Cʹʹ
Bʹʹ
Cʹʹ
C Cʹ
Bʹ
Aʹʹ A Aʹ O
Bʹʹ 90˚ anticlockwise 30˚ clockwise 14
Aʺ(–5, 1), Bʺ(–2, 1), Cʺ(2, 3)
y
7 6 5
Cʹ(–2, 3)
4
Cʹʹ(2, 3)
3
Aʹʹ(–5, 1) Bʹʹ(–2, 1)
2
Bʹ(2, 1)
Aʹ(5, 1)
1 –7 –6 –5 –4 –3 –2 –1 0 –1
C(–2, –3)
1
2
3
4
5
6
7
x
B(2, –1) A(5, –1)
–2 –3
15
Aʺ(2, 1), Bʺ(–4, 0), Cʺ(0, 7)
y 7
Cʹʹ(0, 7)
6
C(–1, 4)
Cʹ(2, 5)
5 4 3 2
Bʹʹ(–4, 0)
–7 –6 –5 –4 –3 –2 –1 0
Bʹ(–2, –2)
16
1
2
3
4
5
6
7
x
Aʹ(4, –1)
–1 –2
B(–5, –3)
Aʹʹ(2, 1)
1
A(1, –2)
–3
Many answers are possible. For example:
a b c d e f
Rotate 90˚ anticlockwise about B; translate 3 left, 4 down. Reflect in AB; translate 5 left, 3 down. Reflect in AC; translate 1 down. Reflect in AB; translate 3 left, 6 down. Reflect in BC; translate 2 left, 3 down. Rotate 90˚ anticlockwise about C; translate 1 up. Answers to exercises
341
Challenge exercise 1
a
A
l1
B Cʺ
C
b
Aʺ Aʹ l2
A
Aʺ Aʹ l3
Bʺ Bʹ
Cʹ
l4 Bʺ Bʹ
B Cʺ
C
Cʹ
c Both are 4 units right. 2 a
A
b 6 units right
Aʺ Aʹ
c 8 units in a direction perpendicular l5 to the parallel line (can be left or right) B
C
l6 Bʺ Bʹ
Cʺ
Cʹ
y
3
5
y 10
Cʹ
C
4
4
9
3
8
2
7
1
B Aʹ
–3 –2 –1 0 A 1
2
–1
3
l1
4
5
6
6
Bʹ
7
Cʹ
x
5 4
l2
–2
3
–3
2
–4
1
Aʹ
l1
0 A 1
–5
l2
Bʹ C
B 2
3
4
5
6
7
8
9 10
x
No, there are many ways to do this; No, there are many ways to do this; t he distance between the lines is 2 units. the distance between the lines is 2 units.
5
a
l2
b 45˚
c No, but the angle between Bʹʹ Bʹ l1 and l2 must be 45˚. There are l1 many possibilities.
Cʹ
Aʹ Aʹʹ
Cʹʹ C
45˚ A O
342
ICEEM Mathematics Secondary 1B
B
Chapter 19 answers Exercise 19A 1 a 4
b 6
c 4
2a 3
3 a a, b, c
b B, C, D
c A
4 a C, D
b A, B
c b, c, e, f
5 a c
c b
6 a and d, b and e, c and f. There are many ways to do this.
b 12
c 6
d none
b 3
e d
Exercise 19B 1 a 8 3 a b, d, i, l
2a 3
b c, e, g
b 4 c f, h, j, k
4 a A and B, C and E, D and F b C and E, D and F, B and H, A and G c d and f, b and h, a and g, c and e, i and k, j and l 5 a There are many ways.
b no
Exercise 19C 1 a 6
b 12
3 a b, c, d, e, i, l
c 8 b g
2a 4
b 3
c f, h, j, k
4 a A and G, B and H, C and E, D and F b C and E, A and B, D and F c a and g, b and h, c and e, d and f, i and k, j and l 5 a There are many ways. b for example, A–C–F–B–E–F–A–E–D–B–C–D–A
Exercise 19D 1 a 20 vertices, 30 edges, 12 faces 3 a b, c, d, e, j, l, y, z
b h
2a 3
b 5
c the remaining 20 edges are skew to a
4 a A and B, C and H, D and I, G and L, F and K, E and J b A and P, B and Q, C and R, D and S, E and T, F and K, G and L, H and M, I and N, J and O c a and h, b and q, c and r, d and p, e and t, f and k, g and l, s and m, i and n, j and o, u and z, w and bʹ, v and aʹ, x and cʹ, y and dʹ 5 a yes
b no
6a 5
b 3
c 2
2a 5
b 3
Exercise 19E 1 a 12 vertices, 30 edges, 20 faces
Answers to exercises
343
3 a b, c, d, e, j, l, y, z
b p
c the remaining edges
4 a A and P, B and Q , C and R, D and S, E and T, F and K, G and L, H and M, I and N, J and O b A and B, C and H, D and I, E and J, F and K, G and L c a and p, b and q, c and r, d and s, e and t, f and k, g and l, h and m, i and n, j and o, u and z, v and aʹ, w and bʹ, x and cʹ, y and dʹ 5 a yes
b no
6a 5
b 3
c 2
Challenge exercise 1
Axis
AA
BB
CC
DD
ad
be
cf
Sum
3
3
3
3
2
2
2
18
Order
2 a 4 3
Axis Order
4 a 6
b 3 AB
CE
DF
CE
DF
BH
AG
ag
bh
df
ce
ik
jl
Sum
4
4
4
3
3
3
3
2
2
2
2
2
2
36
b 4
Chapter 20 answers 20A Review Chapter 11: Integers 1 a –5, –4, –3, –2, –1, 0, 1, 2, 3 2
b –9, –8, –7, –6, –5, –4, –3, –2
–51, –48, –45, –42
3 a 5
b 34
c –7
d –13
4 a 8
b 2
c 7
d –11
e –42
f 6
g –5
5 a –3 g –35
b –12 h 11
c –14 i –4
d –7 j –14
e –34 k –9
f 20 l 50
6 15ºC
7 a –35
b –18
c –110 d –84
e –160 f 50
g –44
8 a 6
b –9
c 5
d 9
10 a 81
b –64
c –4
d –65
12 a 6 f –25
b –64 g 81
c –120 d 120 h 49 i –90
9a –3
b –9
e 33
11 –10
e –408
Chapter 12: Algebra and the number plane
344
1 a –8
b 4
c –2
d –7
e –5
f –64
g 64
h 16
i 7
j 11
k –3
l –14
ICEEM Mathematics Secondary 1B
c
1 –7 12
h –17
h 48
2 a 1
b –65
c 55
d 300
e –360 f 12
3 a –8
b –15
c –9
d 36
e –12
4 a –21
b 29
c –5
d 5
e –125 f 625
5 a
x 5
2
b i 4 m
m
b i $1500 7 a –45 8
1
ii 8 5 m
6 a i $(2000 – x) b 45
f
g 1800 h –10
1 4
iii 19 5 m
ii $(2000 – 10x) ii $1000
iii –$250
c –25
e –500
d 0
f 500
A(1, 4), B(3, 1), C(4, –3), D(–3, –2), E(–2, –4), F(2, –2), G(–2, 2)
9 a square
b isosceles triangle y D
9
3
8
2
7
–5 –4 –3 –2 –1 0
O 1
2
3
4
5
6
A x
5
–1
4
–2
3
–3
2
–4
A
y
A
1
2
3
4
5
B x
d parallel lines
c rectangle
D
1
–5 –4 –3 –2 –1 0
–5
10
4
1
y C
C
5
y
5
5
4
4
C
3
3
2
2
1
–5 –4 –3 –2 –1 0
1
B 1
2
3
A
4
5
x
–5 –4 –3 –2 –1 0 O 1
–1
–1
–2
–2
–3
–3
–4
–4
–5
–5
C 2
3
4
5
x
B
Answers to exercises
345
10 a y = 4x
b y = –x
x
−3 −2 −1
0
1
2
3
y
–12 –8
0
4
8
12
–4
x
−3 −2 −1
0
1
2
3
y
3
0
–1
–2
–3
2
1
Points are: (–3, –12), (–2, –8), (–1, –4), Points are: (–3, 3), (–2, 2), (–1, 1), (0, 0), (0, 0), (1, 4), (2, 8), (3, 12) (1, –1), (2, –2), (3, –3) y
y
12
5
(3, 12)
10
4
(2, 8)
8
(–3, 3) (–2, 2) (–1, 1)
6
(1, 4)
4
–5 –4 –3 –2 –1
(0, 0) 2 –5 –4 –3 –2 –1 0
(–1, –4) (–2, –8) (–3, –12)
1
2
3
4
5
x
(0, 0)
0
1
–2
–4
–3
–6
–4
–8
–5
c y = x – 4 x
−3 −2 −1
0
1
2
3
y
–7
–4
–3
–2
–1
–5
Points are: (–3, –7), (–2, –6), (–1, –5), (0, –4), (1, –3), (2, –2), (3, –1)
y 3 2 1 –5 –4 –3 –2 –1
0 –1 –2 –3
(–1, –5) (–2, –6) (–3, –7)
–4
1
2
3
4
5
(3, –1) (2, –2) (1, –3) (0, –4)
–5 –6 –7
346
1
–2
–12
–6
2
–1
–10
3
ICEEM Mathematics Secondary 1B
x
2
3
4
5
(1, –1) (2, –2) (3, –3)
x
Chapter 13: Triangles and constructions 1 a c e g
b d f h
a = 98˚, b = 82˚, c = 98˚ a = 143˚ a = 52˚, b = 70˚, c = 58˚ a = 49˚, b = 43˚, c = 88˚, i = 43˚
a = 85˚, b = 40˚ a = 70˚, b = 80˚, c = 30˚ a = 67˚, b = 64˚, c = 43˚, i = 43˚ a = 149˚
Chapter 14: Negative fractions 1
1
1
3
1
1
–3 4 , –2 2 , –1 4 , 0, 2, 2 2 , 3 2 1
b 2 6
7
b – 20
4 a – 2
1
g
2 a 3 4 3 a 1 20
5 a
1 17 1 –2 4 5 –1 8
1
c 5
1
3
c – 8
d – 110
b – 3
2
c 2
d
h
i
3
4 –2 9 3 4 1 –7 4
b
c
b c Chapter 15: Percentages
6 a
1
1 64 3 –3 4 13 –1 32
3
e 1 24
13
f
5 12 25 – 27 27 64 1 7 32
f
15 32 7 –1 8
e k
d –3
e
1
e
d –3 8
1 a 90%
b 85%
c 120%
d 48%
2 a 78%
b 9.5%
c 97%
d 135%
3 a 30% 1
4 a 4 2 %, 5
3 8,
12 32 ,
1 33 3 %
b 125%
c
0.39
b 0.6, 64%,
.
1
2
e –2 2
j
1
d – 4
3
f –1 5
l f
f
1 10 7 –8 2 –2 15 27 64 1 –9 2
3
g –2 5 h –25 g
3 20
11
h – 56
3
g 3 4
g –9
h 7 h –5
2 3
37.5%; 0.3 , 33 3 %; 4 , 0.75; 2.75, 275%; 3, 3 5
7
6 a 8
b 8
c
1 80
d
21 400
7 30%
8 $562.50
Chapter 16: Solving equations e x = 2 3
1
f z = 22
k m = 30
l z = 30
d x = 4 j x = 12
e x = 4 k m = 12
f z = 3 3 l z = 50
d x = 35
e x = 1 2
1 a x = 2 g x = 4
b x = 11
c x = 12
d x = 5
h x = 12
i z = 8
j x = 24
2 a x = 2 g x = 2
b x = 5 h x = 3
c x = 6 i z = 7
3 a x = –1 b x = –3
c x = –6
4 a x + 3 = 6, x = 3
b x – 5 = 10, x = 15
d
x = 4
7
1
f x = –
4 9
c 7x = 84, x = 12
e 8x = 15, x = 1 8
15, x = 60
1
2
f 4x – 3 = 17, x = 5 1
g 3(x + 6) = 28, x = 3 3 h 6(x – 7) = 15, x = 9 2 5
50 cm, 10 cm
6 19, 1
Answers to exercises
347
2
7 a x = –17 b x = –7
c x = 1
d x = –2 5 e x = 4
f x = –3
g x = –80 h x = 4
i x = –3
j x = –3
k x = 11
l x = – 2
1
c x = – 4
1
d x = –3
e x = 2
f x =
3
8 a x = 3
b x = 4
9 a x = 1
b x = 1 2
1
2
c x = –3
10 4 5
1 16 5 24 2 3 10 13 7 19
e 2
1
3 8
11 –6
Chapter 17: Probability 1 a 5
3
b 5
3
b 4
1
c
3 a 3
1
b
c
f
7 12 1 4 2 19 1 4
g
5 12 1 6 3 4 3 19 1 16
2 a 8
4 a 6 a 7 a
2
b b b
1 16 1 2 3 8 1 2 1 19 3 4
h c c c
d d i 5 d
1
f 2
1
g
11 16
17 19 1 2
g
5 19 9 16
3
e 4
e
d 1
e
12 19 1 2
f f
g
Chapter 18: Transformations and symmetry 1
y C
2
y 4
C
2 1
–5 –4 –3 –2 –1 0
Dʹ D
Eʹ
4
Fʹ Bʹ
1
Aʹ
A 1
2
3
4
5
x
–5 –4 –3 –2 –1 0
–1
–1
–2
–2
Dʹ
–3
–4
–5
–5
ICEEM Mathematics Secondary 1B
a b c d e f
A, Eʹ 1
2
3
4
5
x
Fʹ B
–3
–4
a Aʹ(3, 1) b B′(4, 3) c C′(1, 2) d D′(–2, –4) e E′(0, 3) f F′(3, 3)
348
F
2
F B,Cʹ
Bʹ,D
Aʹ
3
3
E
E
5
5
Cʹ
1 unit right, 3 units up 1 unit right, 6 units up 10 units down 6 units left, 8 units down 5 units left, 5 units down 3 units down
3
4
y 5
C
y
F
4 3
1
1
2
–1
3
A B
4
5
D
Dʹ
–4
–5
–5
6
y E B
3
2
Aʹ
2
1
2
3
A Bʹ
4
5
x
–5 –4 –3 –2 –1 0
Fʹ
B
A C 1
2
3
4
–1
–2
–2
–3
–3
Eʹ
–4
–4
–5
–5
b B′(2, –2) a i
a A′(2, 1)
Cʹ
1
–1
Cʹ
x
b B′(–2, 2) d D′(4, 4) f F′(0, –5)
3
–5 –4 –3 –2 –1 0
F
5
4
1
D, Dʹ
4
5
4
Fʹ
A3 B
y
5
C
2
–3
–4
b B′(2, 2) a A′(–2, 1) d D′(4, –4) c C′(3, –4) f F′(–5, 0) e E′(0, 0)
5
1
–2
–3
a A′(1, 2) c C′(–4, –3) e E′(0, 0)
E,Eʹ
–1
–2
Cʹ D
2
–5 –4 –3 –2 –1 0
x
Dʹ
3
1
E,Eʹ
–5 –4 –3 –2 –1 0
F
4
Bʹ Aʹ
Aʹ Bʹ
2
Fʹ
5
C
A′(3, –2)
5x
Cʹ
Aʹ Bʹ
ii B′(4, –4)
d D′(–5, 0) iii C′(6, –1) f F′(–1, 1) b A′B′C′
c C′(–3, –3) e E′(0, –4)
20B Tessellations Activity 1 Polygon
Number of sides
Interior angle sum
Size of each angle 60˚
Triangle
3
180˚
Quadrilateral
4
360˚
90˚
Pentagon
5
540˚
108˚
Hexagon
6
720˚
120˚
Heptagon
7
900˚
128
4˚ 7
Answers to exercises
349
Octagon
8
1080˚
135˚
Nonagon
9
1260˚
140˚
Decagon
10
1440˚
144˚
Dodecagon
12
1800˚
150˚
The square, equilateral triangle and regular hexagon are the only polygons with angle sizes that divide 360˚, thus enabling several such shapes to meet at a vertex without leaving any spaces. Activity 3 The interior angles of a pentagon add up to 540˚. ∠A + ∠B + ∠D = 360˚ and 4 Cairo tiles fit together to form a hexagon which tessellates the plane. Activity 4 Activity 5 The other seven codes are: [6, 3, 6, 3], [8, 8, 4], [4, 3, 4, 3, 3], [12, 12, 3], [4, 4, 3, 3, 3], [12, 6, 4], [6, 3, 3, 3, 3] Activity 6 a
Both angles are 40˚.
b
20C Sets and Venn diagrams Exercise A 1 a {1, 3, 5, 7, 9, 11}
b {13, 23, 33, 43, 53, 63, 73, 83, 93}
c {January, February, May, July} d {Asia, Australia, Africa, Antarctica, Europe, North America, South America} e {12, 18, 24, 30, 36}
f {14, 21, 28, 35}
g {1, 4, 9, 16, 25, 36, 49}
h {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
b {odd numbers less than 6} 2 a {first 3 letters of the English alphabet} c {the first 4 prime numbers} Other answers are possible. 3 a no (‘best’ depends on people’s opinions)
b yes
c yes
d no (there is no consensus on ‘bright’ colours) e no (there is no consensus on ‘small’) f yes g no (some people change their votes with time) 4 A and D 5 a =
b =
c ≠
d =
6 a 9 ∈ S
b 12 ∈ S
c 5 ∉ S
d 0 ∉ S
7 a ∈
350
b ∉
c ∉
d ∉
ICEEM Mathematics Secondary 1B
e ∉
e ≠ (not everyone is a citizen) f ∉
g ∈
h ∈
Exercise B 1 a not defined e infinite 2 a 3 g 4
b finite f finite
b 6 h 21
c 20 i 0
c infinite g finite d 26 j 1
d finite e 6 k 2
f 8 l 5
3 a {0, 1, 2, …, 9}, finite b {1, 3, 5, …, 19}, finite d {51, 52, 53, …}, infinite e {2, 4, 6}, finite g {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}, finite
c {1, 4, 9, …}, infinite f {56, 63, 70, …}, infinite
4 a empty b nonempty c empty
d empty
e nonempty f empty
5 a false
b true
c false
d true
e false
6 a =
b ≠ g ≠
c = h ≠
d = i =
e ≠ j =
1 a ⊆
b ⊆
c ⊆
d ⊆
e ⊆
f ⊆
2 a ⊆
b ⊆ h ⊆
c ⊆ i ⊆
d ⊆ j ⊆
e ⊆ k ⊆
f ⊆
f =
f true
Exercise C
g ⊆
3 a Everything in ∅ belongs to S.
b Everything in A belongs to B, and everything in B belongs to C, so everything in A belongs to C.
c Everything in S is in S.
d Any element of A belongs to B and vice versa, so A and B must be equal.
e Everything in {x} is in S.
f Everything is {x, y} is in S.
g All elements of A are in B, so B must have at least as many elements as A.
4 a False; A could be equal to B. b False; they can have the same number of elements. c True; they have the same elements. d True; C has all the elements of A and maybe more. 5 a i {a, b, c}, {a, b}, {b, c}, {a, c}, {a}, {b}, {c}, ∅ ii {a, b}, {a}, {b}, ∅ iii {a}, ∅ b The number is 2 to the power of the number of elements. 6 a {fish} ⊆ {animals}
b {numbers ending in 6} ⊆ {even numbers} c {children with blonde hair} ⊆ {naughty people} d {numbers ending in 4} ⊆ {multiples of 4} e {adults} ⊆ {sensible people}
7
Answers may vary. Examples are:
a A = {1}, B = {1, 2}, C = {1, 2, 3}
b A = {1}, B = {1, 2, 3}, C = {1, 2}
Answers to exercises
351
8
Answers may vary. Examples are: b A = {1, 2}, B = {1, 3}
a A = {1, 2}, B = {1} Exercise D 1 a {1, 2, 4, 6} 2 a c e f g h i
j k 3
b {0, 2, 3, 5, 7} 4
5
c {2, 3, 4, 5} b
0
1
2
3
0
1
2
3
0
1
2
3
4
5
0
1
2
3
4
5
6
0
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
d {2}
0
1
2
3
4
0
1
2
3
4
7
8
9
10 11
6
7
8
6
7
8
9
10 11 12
7
8
9
d
5
6
6
l ∅
{0, 1, 2, 3, 4, 5, 6}
Exercise E 1 a P ∩ Q = {boys who play cricket and tennis}; P ∪ Q = {boys who play cricket or tennis}
b P ∩ Q = {girls who study mathematics and history}; P ∪ Q = {girls who study mathematics or history}
c P ∩ Q = {right isosceles triangles}; P ∪ Q = {triangles which are right or isosceles triangles}
d P ∩ Q = {even perfect square numbers}; P ∪ Q = {even numbers or perfect square numbers}
e P ∩ Q = {animals with fur and webbed feet}; P ∪ Q = {animals with fur or webbed feet}
2 a A ∩ B = {girls who play winter and summer sport}; A ∪ B = {girls who play winter or summer sport}; not disjoint
352
b A ∩ B = {singers who sing popular and classical music}; A ∪ B = {singers who sing popular or classical music}; not disjoint
ICEEM Mathematics Secondary 1B
c A ∩ B = ∅; A ∪ B = {hard or softboiled eggs}; disjoint
d A ∩ B = ∅; A ∪ B = {all people}; disjoint
3 a X ∩ Y = {0, 5, 8}; X ∪ Y = {0, 1, 5, 8, 10, 12}; not disjoint
b X ∩ Y = ∅; X ∪ Y = {0, 1, 2, 3, 4, 5, 7, 9, 11, 13, 16}; disjoint
c X ∩ Y = {b, c, f}; X ∪ Y = {a, b, c, d, e, f, g, k}; not disjoint
d X ∩ Y = ∅; X ∪ Y = {1, 2, 3, 4, 5, 6, 8}; disjoint
e X ∩ Y = {15, 16, 17, 18, 19, 20}; X ∪ Y = {1, 2, 3, …, 50}; not disjoint
f X ∩ Y = {5, 10, 14}; X ∪ Y = {2, 5, 10, 12, 14}; not disjoint
4 a A ∩ B = {3}
A ∪ B = {0, 1, 2, 3, 6} b A ∩ B = {1, 3, 5}
0
1
2
3
4
5
6
0
1
2
3
4
5
6
0
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
A ∪ B = {0, 1, 2, 3, 4, 5, 7, 9}
0
5 a A ∩ B ∩ C = {0}; A ∪ B ∪ C = {0, 3, 4, 6, 8, 9, 10, 12, 15, 16, 20} b A ∩ B ∩ C = ∅; A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} 6
Answers may vary. For example:
a i S = {rodents}
ii T = {primates}
b i S = {even whole numbers}
ii T = {odd whole numbers}
7 a b c d e
Anything in the intersection of A and B must be in A. Anything in the intersection of A and B must be in B. The union of A with anything must contain A. Elements common to A and B belong to the union of A and B. The empty set is a subset of any set. f As for e
8 a B
b A
9 a True; A ∪ ∅ = A.
c A
d ∅
e A
f A
b True; A ∩ ∅ = ∅.
c,d False; counterexamples are readily constructed.
Exercise F c {Australians living more than 10 km away from the coast} 1 a {men} b {girls} d {Saturday, Sunday} e {whole number greater than or equal to 100} f {0, 1, and composite numbers} Answers to exercises
353
2 a {5, 6} d {0, 1, 2, 3, 4, 5, 6} g {5, 6}
b {0, 1, 2, 3, 4, 5, 6} e ∅ h {0, 1, 2, 3, 4, 5, 6}
c {0, 1, 2, 3, 4} f ∅ i {5, 6}
3 a True; ‘not not in A’ is the same as ‘in A’. b True; something is either in A or not in A. c False; the right hand side should be ∅. d False; the right hand side should be ∅. e True; the universal set is everything. f False; counter examples are easily constructed, the size of Ac is the size of E – the size of A. 4 a {a, b, c, d, e, f, g} e {a, b, d, e, g}
d {b, e}
b {a, b, e} f {a}
5 a {1, 3, 5, 7, 8} d f i l 6 a
b {1, 2, 3, 6, 9, 11} {1, 2, 3, 5, 6, 7, 8, 9, 11} {4, 5, 7, 8, 10, 12} g {4, 10, 12} j {1, 3, 4, 5, 7, 8, 10, 12} {5, 7, 8} {1, 2, 3, 4, 6, 9, 10, 11, 12}
E
b
c {b, d, e, g} g {d, g} c e h k
{1, 3} {2, 4, 6, 9, 10, 11, 12} {2, 4, 5, 6, 7, 8, 9, 10, 11, 12} {2, 6, 9, 11}
E
B
A
A
B
c
E A
7 a
d
E A
B
E A 1
2 4
B 3 5
b
B
E A a
e i
o
B s
t
A ∪ B = {1, 2, 3, 4, 5}, A ∩ B = ∅
354
ICEEM Mathematics Secondary 1B
A ∪ B = {a, e, i, o, s, t}, A ∩ B = {e, i}
c
E 1 2 A 3 8 5 10 B
A ∪ B = {1, 2, 3, 5, 8, 10}, A ∩ B = {3, 8, 10}
Exercise G 1 a
E A
b
E A
B (1) (3) (1)
B (3) (2) (3)
(5)
(2)
A ∪ B = 5, A ∩ B = 3 A ∪ B = 8, A ∩ B = 2 c
E A
B (4) (0) (3) (3)
A ∪ B = 7, A ∩ B = 0 2 a 19
b 50
c 46
d 5
e 4
f 0
3 a 35
b 47
c 8
d 10
e 6
f 40
Exercise H 1 a 10
b 5
c 15
d 20
2 a 8
b 12
c 60
d 20
3 a 270
b 20
c 420
d 80
4 a 33
b 22
c 12
d 3
5 a 48
b 192
6 a 180
b 410
c 260
d 670
7 a 130
b 15
c 55
d 15
8 a 59
b 11
c 40
d 41
e 16
9 a 9
b 10
c 12
d 168
e 159
e 850
f 670
Answers to exercises
355