Matematics Form 1

In this assignment, you are to collect, display, analyse and interpret data regarding the
media channels used to promote a new product among teenagers. The media channels
include newspapers, television, radio, Internet, social media, magazines, catalogues,
pamphlets and others.
Write a report to suggest the media channels that can extensively promote the
new product among teenagers, and predict the media channels that would become
increasingly popular in the future. To support your suggestion, your report must
include questionnaires, frequency tables, suitable data representations using software,
data interpretations and conclusions.

Read the following article and discuss the questions posed. CHAPTER
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12
1. State the three categories that received the most number of complaints.
2. A statistician commented that “there is a possibility that the general consumer

products category received more complaints than the telecommunications
category”. Discuss why this comment may be true.
Note: You can think of how the data (number of complaints) have been arranged.
3. What can you tell about this data organisation?

291

Data Handling

13CHAPTER The Pythagoras’
Theorem

What will you learn?

• The Pythagoras’ Theorem
• The Converse of Pythagoras’ Theorem

 Why study this chapter?CHAPTER Right angles exist in many objects around us.
As the fundamental In the construction of buildings, how does a
knowledge for solving civil engineer ensure that the corners of the
problems involving right-angled walls of the building are built at right angles?
triangles. Discuss the f ields that
involve solving problems related to
right-angled triangles.

13

292

Chapter 13

Pythagoras

Pythagoras (569 B.C. – 475 B.C.) was a
mathematician and also a philosopher
who had contributed substantially to
the development of mathematics today.
He was the first person who proved
the Pythagoras’ theorem.

For more information:

http://goo.gl/r4JZ

Word Link

• converse of Pythagoras’ • akas teorem CHAPTER
  theorem Pythagoras

• hypotenuse • hipotenus

• Pythagoras’ theorem • teorem Pythagoras

13

Open the folder downloaded from page vii
for the audio of Word Link.

293

The Pythagoras’ Theorem

13.1  The Pythagoras’ Theorem LEARNING
STANDARDS
What is a hypotenuse?
Identify and define
the hypotenuse of a
right-angled triangle.

CHAPTER We are often told about the size of the monitor screen of a computer as 19 inches,
21 inches or 24 inches and so forth. The size is measured according to the length of the
monitor diagonally. What is the relationship between the size of the monitor screen and
the length and width of the monitor screen?

1 Group

Aim:  To identify the hypotenuse of a right-angled triangle.
Instruction: • Perform the activity in groups of four.
• Open the folder downloaded from page vii.
1. Open the f ile Hypotenuse.ggb using GeoGebra.

The screen displayed shows a right-angled triangle
with the length of each side.
2. Identify and record the longest side.
3. Click and drag points A, B or C to change the
shape of the triangle and repeat the exploration in
Step 2.
4. Discuss your findings with your friends.

From the results of Exploration Activity 1, it is found that the longest side of a right-
angled triangle is always the side opposite to the right angle.
The longest side opposite to the right angle is known as the hypotenuse of the right-
angled triangle.

13
a

95°
a

Discuss with your friends and explain why the side labelled as a is not the hypotenuse.

294

Chapter 13

1

For each of the following, identify the hypotenuse.
(a) A (b) S

PR

BC

Q

(a) AC is the hypotenuse. The side opposite to the right angle.

(b) PR is the hypotenuse of triangle PSR.
PQ is the hypotenuse of triangle PRQ.

Self Practice 13.1a

1. For each of the following, identify the hypotenuse. R
(a) B (b) b (c) S

C

ac Q
T

A

P

What is the relationship between the sides LEARNING
of a right-angled triangle? STANDARDS

2 Flip Determine the relationship
lassroo between the sides of a
right-angled triangle. Hence,
explain the Pythagoras’
theorem by referring to
the relationship.
CHAPTERm

Aim: To explore and explain the Pythagoras’ theorem. C

Instruction: • Explore by yourself before the lesson
begins and discuss in groups of four
during the lesson.
• Open the folder downloaded from page vii.

1. Open the f ile Pythagoras.ggb using GeoGebra. 13
The screen displayed shows a right-angled triangle
ABC with a square drawn on each side of the
triangle.

2. Click and drag the coloured shapes in the square
on sides AB and BC, and move them into the
square on side AC. Do all the coloured shapes f ill
up the square on side AC perfectly?

3. Click and drag the slider ‘Move all’ or click at
the checkboxes ‘Show line guides’ and ‘Move one
by one’ for help.

295

The Pythagoras’ Theorem

4. Click and drag points A, B and C to change the shape of the right-angled
triangle and repeat your exploration.

5. Discuss your f indings with your friends.
6. Consider the area of a square, state the relationship between sides AB, BC and

AC.

From the results of Exploration Activity 2, it is found that the area of the square on the
hypotenuse is equal to the total area of the squares on the other two sides.

A R Area of R = Area of P + Area of Q
P AC 2 = AB2 + BC2
This relationship is known as the
BQ C Pythagoras’ theorem.

2

For each of the following, state the relationship between the lengths of sides of the given
right-angled triangle.
(a) R (b) c

Pa

b
Q

(a) PR2 = QR2 + PQ2 (b) c2 = a2 + b2

Self Practice 13.1b

1. For each of the following, state the relationship between the lengths of sides of the
given right-angled triangle.
(a) B (b) L (c) r (d) x

C N p q yz
AM

How do you determine the length of theCHAPTER LEARNING
unknown side of a right-angled triangle? STANDARDS
13 Pythagoras’ theorem can be used to determine the length
of an unknown side in a right-angled triangle if the lengths Determine the lengths
of two other sides are given. of the unknown side of
(i) a right-angled

triangle.
(ii) combined

geometric shapes.

296

Chapter 13

3

For each of the following, calculate the value of x.
(a) (b)
x cm 24 cm
12 cm x cm

16 cm 26 cm Smart

(a) x2 = 122 + 162 (b) 262 = x2 + 242 The length of the
= 144 + 256 x2 = 262 – 242 hypotenuse can be
= 440000 == 676 – 576 calculated using the Pol
x = 20 x = 110000 function. For instance,
= Example 3(a),
= 10
press Pol( 1 2 ,

16 )=

4

Calculate the length of PQ in each of the following diagrams. id ou now
(a3) cm R 12 cm Q (b) S 8 cm 6Rcm
S Three numbers a, b and c
4 cm 15 cm Q that satisfy c2 = a2 + b2
are known as a
P Pythagorean triples (or
P triplets). For example,
(3, 4, 5), (5, 12, 13),
(7, 24, 25) and so forth.

(a) PR2 = 32 + 42 (b) QS2 = 62 + 82
= 9 + 16 = 36 + 64
= 25 = 100

PQ2 = PR2 + RQ2 PS 2 = QS2 + PQ2
= 25 + 122 PQ2 = PS 2 – QS2
= 116699 == 152 – 100
PQ = 125
= 13 cm PQ = 125
= 11.18 cm (2 decimal places)
CHAPTER
Self Practice 13.1c

1. For each of the following, calculate the value of x. Give your answer correct to two 13
decimal places if necessary.
(a) (b) x cm (c) 9 cm (d)
6 cm xm 1.0 m
x cm 7 cm x cm 1.2 m
8 cm 25 cm

14 cm

297

The Pythagoras’ Theorem

2. Calculate the length of QS in each of the following diagrams. Give your answer
correct to two decimal places if necessary.
(a) P (b)P (c) P 24 cm S
6 cm S
15 cm 12 cm 13 cm 26 cm Q 30 cm

Q RS Q 24 cm RR

How do you solve problems?

Help! LEARNING
Help! STANDARDS

Where is Pythagoras? Solve problems involving
We need his help! the Pythagoras’ theorem.

Can the ladder
reach the f loor
that is on f ire?

A f ireman climbs up a ladder to save a child who is trapped
on the third f loor as shown in the diagram. The third f loor
is 6 m high from the horizontal ground. The base of the
ladder is 4.5 m away from the wall of the building. How long
is the ladder?

Understanding Devising a plan Implementing Doing reflection

the problem • Draw a the strategy 7. 52 = 56. 25
• Distance of the right-angled 4. 52 + 62 = 56. 25
triangle PQR R
third f loor from to represent
the horizontal 6m

CHAPTER ground = 6 m the given P 4.5 m Q
• Distance of information.

13 the base of the • Use PR2 = PQ2 + QR2
ladder from the Pythagoras’ = 4. 52 + 62
building = 4. 5 m theorem. = 20.25 + 36
• Find the length = 56.25
of the ladder.
PR = 7.5 m
Thus, the length of
the ladder is 7.5 m.

298

Chapter 13

5

In the diagram, PVRQ and RUTS are squares. Calculate P V
the perimeter of the whole diagram. 12 cm R
S
VU 2 = VR2 + RU 2 Perimeter of the whole diagram Q U
= 20 + 16 + 16 + 16 + 12 + 12 + 12 16 cm
= 122 + 162 = 104 cm T
= 440000
VU =
= 20 cm

Self Practice 13.1d U 11 cm T S
1. In the diagram, PQSU is a rectangle. Calculate the 12 cm 4R cm
Q
perimeter of the shaded region. P 17 cm

2. Ship A is at 34 km north of ship B. Ship C is at 10 km west of ship A. Calculate the
distance between ship B and ship C, correct to two decimal places.

13.1 Open the folder downloaded from page vii for extra
  questions of Mastery Q 13.1.

1. In the diagram, PQR is a straight line. Calculate the 34 cm S
length of QR. 20 cm

2. In the diagram, PQR and RST are straight lines. P 30 cm Q R
Calculate the perimeter of the shaded region. 12 cm T
R
3 cm S

Q 5 cm 13 cm

3. One end of a rope of length 7.2 m is tied to the tip of P CHAPTER
a flag pole. The other end of the rope is tied to a spot 7.2 m
on the horizontal ground 4.5 m away from the base 13
of the flag pole. Calculate the height of the flag pole 4.5 m
and give your answer correct to two decimal places.

4. A ship departs from point O and sails towards southwest for a distance of 300 km
and then towards northwest for a distance of 450 km. Calculate the final distance of
the ship from point O and give your answer correct to two decimal places.

299

The Pythagoras’ Theorem

13.2 The Converse of Pythagoras’ Theorem LEARNING
STANDARDS

How do you determine whether a triangle is a Determine whether a
right-angled triangle? triangle is a right-angled
triangle and give
3 Flip justification based on
lassroo the converse of the
Pythagoras’ theorem.
Aim: To explore the converse of Pythagoras’ theorem.
C m

CHAPTERInstruction: • Explore by yourself before the lesson begins and discuss in groups
of four during the lesson.
• Open the folder downloaded from page vii.

1. Open the file Converse of Pythagoras.ggb using GeoGebra. The screen
displayed shows a triangle ABC with angles at vertex B and vertex C.

2. Click and drag point A towards left or right and observe the change in the
information displayed in red colour. Copy and record your observations in the
following table for a few sets of values. Click and drag point B or C to change
the shape of the triangle if necessary.

Comparative value Size of angle
(red) (red)

AB2  AC2 + BC2

AB2  AC2 + BC2

AB2 = AC2 + BC2

3. Repeat Step 2 for the information displayed in blue colour.

4. Discuss your findings with your friends. • An acute angle is an
5. What are the conclusions that can be made? angle less than 90°.

From the results of Exploration Activity 3, it is found that • An obtuse angle is an
angle more than 90°
but less than 180°.

AA A

bc c bc
b
C aB C aB
C aB
13 If c2 , a2 + b2, then the If c2 = a2 + b2, then the
angle opposite to side c If c2 . a2 + b2, then the angle opposite to side c
is an acute angle. angle opposite to side c is a right angle.
is an obtuse angle.

The converse of Pythagoras’ theorem states that:
If c2 = a2 + b2, then the angle opposite to side c is a right angle.

300

Chapter 13

6

Determine whether each of the following triangles is a right-angled triangle.
(a) D (b) S 20 cm
24 cm E 12 cm
25 cm 7 cm U
F
T 18 cm

(a) The longest side = 25 cm (b) The longest side = 20 cm
Thus, 252 = 625 Thus, 202 = 400
242 + 72 = 576 + 49 182 + 122 = 324 + 144
= 625 = 468
Hence, DEF is a right-angled Hence, STU is not a right-angled
triangle. triangle.

Self Practice 13.2a

1. Determine whether each of the following triangles is a right-angled triangle.
(a) 15 cm (b) (c) 32 cm
34 cm
17 cm 18 cm 30 cm 40 cm

8 cm 24 cm

How do you solve problems? LEARNING
STANDARDS
7 S
In the diagram, calculate Solve problems involving
the value of x. 28° the converse of the
29 cm 20 cm Pythagoras’ theorem.

Q x CHAPTER
P 21 cm 155° R

20 cm, 21 cm and 29 cm satisfy c2 = a2 + b2. A housing contractor and 13
Thus, ∠PQS = 90° a civil engineer will use
∠SQR = 360° – 90° – 155° the Pythagoras’ theorem
= 115° to solve problems
x = 180° – 115° – 28° involving right angles
= 37° in the construction
of buildings.

301

The Pythagoras’ Theorem

8
Sheila is given three straws to form a frame in the shape of a right-angled triangle. The
straws are 15 cm, 20 cm and 25 cm long respectively. Will she be able to form the frame
in the shape of a right-angled triangle?

152 + 202 = 225 + 400
= 625
252 = 625
152 + 202 = 252
Therefore, Sheila will be able to form the frame in the shape of a right-angled triangle.

Self Practice 13.2b

1. A 2.5 m long ladder leans against the wall of a 2 m 2.5 m
building. The base of the ladder is 1.5 m away
from the wall. Explain how you would determine 1.5 m
whether the wall is vertical.
6m Q
2. In the diagram, f ind ∠PQR. P 65°

10 m 8 m 45° R

S

13.2 Open the folder downloaded from page vii for extra
  questions of Mastery Q 13.2.

1. Explain whether the lengths of sides in each of the following can form a right-angled
triangle:
(a) 9 cm, 40 cm, 41 cm (b) 27 m, 45 m, 35 m
(c) 2.5 cm, 6 cm, 6.5 cm (d) 13 m, 84 m, 85 m

2. A carpenter wishes to f ix a triangular piece of wood
measuring 12 cm, 16 cm and 20 cm onto a L-shaped
CHAPTER wooden structure, as shown in the diagram. Explain
whether the triangular piece of wood can be fixed
perfectly onto the L-shaped structure.
20 cm
13

3. Kanang draws a quadrilateral with measurements as shown
in the diagram. What is the name of the quadrilateral he
has drawn? Explain. 25 cm 15 cm
15 cm

20 cm

302

Chapter 13

Book Website
Teacher
ac • c is the hypotenuse.
b • c is the longest side

opposite to the right
angle.
• c2 = a2 + b2

Pythagoras’
theorem

B
ac

C bA

If c2 = a2 + b2,
then, ∠ACB = 90°

Discussion

Very Work
good harder

identify and define the hypotenuse of a right-angled triangle. CHAPTER

determine the relationship between the sides of a right-angled triangle. Hence, 13
explain the Pythagoras’ theorem by referring to the relationship.

determine the lengths of the unknown side of
(i) a right-angled triangle.
(ii) combined geometric shapes.

solve problems involving the Pythagoras’ theorem.

determine whether a triangle is a right-angled triangle and give justif ication based
on the converse of the Pythagoras’ theorem.

solve problems involving the converse of the Pythagoras’ theorem.

303

The Pythagoras’ Theorem

S R
8 cm
1. In the diagram, calculate the length of 26 cm Q
(a) PR (b) SR 150 m 6 cm

2. A car moves up a slope from P to Q. When the car P
reaches Q, the horizontal distance and the vertical
distance it has covered are 150 m and 2 m respectively. P Q
Explain how you would calculate the actual distance 2m
the car has moved, correct to two decimal places.

3. Based on the diagram, P 8 cm Q 9 cm
(a) calculate the length of QS. 17 cm R
(b) explain whether PQS is a right-angled triangle.
12 cm
4. In the diagram, PQST is a rhombus and TSR is a
straight line. Calculate the area of the whole diagram. S
PQ

8 cm

T 10 cm S R

5. The width of a river is 18 m. Imran swims across the QR
river from point P to point Q, as shown in the diagram.
Due to strong water currents, Imran eventually lands at P
point R which is 6 m away from Q. Explain how you
would calculate the actual distance Imran has swum,
correct to two decimal places.

CHAPTER 6. In the diagram, PQRS is a rhombus. PR and SQ are 16 cm P Q
and 30 cm long respectively. Explain how you would
calculate the length of side SR. S R
P Q
13 7. In the diagram, PQRS and UVST are squares. Given 18 cm
TQ = 30 cm, find the area of UVST . UV
W R
304
TS
Chapter 13

 8. The diagram shows a ladder PQ leaning against the wall. P
The ladder is 2.5 m long and the base of the ladder is hm
0.7 m away from the wall. When the top of the ladder
slides down a distance of h m, the base of the ladder Q
becomes 1.5 m away from the wall. Find the value 0.7 m
of h.
1.5 m
 9. The distance between two poles, PQ and RS, on a
horizontal ground is 12 m. A piece of 15 m long string 15 m S
is tied to both tips of the poles. If the height of pole Q 20 m
RS is 20 m, explain how you would f ind the height of
pole PQ. P 12 m R

10. Ali wishes to saw a piece of wood into the shape of 35 cm
a right-angled triangle. The hypotenuse of the wood
must be 35 cm long and the lengths of the two other
sides of the wood must be in the ratio 3 : 4. Explain
how you would f ind the lengths of the other two sides
and hence help Ali saw the wood.

11. A loop of thread is marked with 12 points so that
the adjacent points are of the same distance from one
another. Explain how you would form a right-angled
triangle using the loop of thread.

Making a wooden stool 51 cm Platform
A carpenter wishes to make a wooden stool
according to the design shown in the diagram. Frame 15 cm 15 cm CHAPTER

The wooden stool he wants to make sWupopoodretns 30 cm
consists of three parts. Part 1 is the frame; part
2 is the platform top for sitting on while part 3 10 cm 13
are the supports for the corners of the frame.
Each end of the supports is fixed at 15 cm 45 cm

from the corners of the frame, and the other end is fixed at 10 cm from the legs of the
wooden stool. The platform is 51 cm long. Explain how you would use the Pythagoras’
theorem to help the carpenter make the wooden stool.

305

The Pythagoras’ Theorem

CHAPTER A The lengths of the three sides of a right-angled triangle such as (3, 4, 5) and
(8, 15, 17) are known as Pythagorean triples.

You can explore the Pythagorean triples through the following activity.

Open the folder downloaded from page vii for this activity.

1. Open the file Triples.ggb using GeoGebra.
2. Is the triangle shown a right-angled triangle?
3. Click and drag the slider m and the slider

n and observe the changes on the screen
displayed.
4. Do the lengths of sides shown constitute a
set of triples?
5. Click and drag the slider m and the slider n
for other combinations.
6. Explain what you have observed.
7. Present your findings in class.
B Constructing a Pythagorean Tree
A Pythagorean tree is a pattern constructed based on right-angled triangles and
squares related to Pythagoras’ theorem. The Pythagorean tree was invented by a Dutch
mathematics teacher in 1942.

13 Construct your own Pythagorean tree beginning from a right-angled triangle with
squares drawn on each side of the triangle. Then similar triangles are drawn on the
sides of the squares. Then squares are drawn on the sides of the new triangles and
so on.

306

Chapter 13

ANSWERS

Chapter 1 Rational Numbers Self Practice 1.2a Self Practice 1.3a
1. (a)
Self Practice 1.1a 1. (a) 8 (b) –7
(c) 1 (d) 2 –1 – 3 – 1 0110 1 1
1. (a) +1 000 m, –250 m (e) –10 (f) 3 5 5 2 1
(b) +RM2 000, –RM500 (g) 13 (h) –2
(b)
Self Practice 1.1b Self Practice 1.2b
–2–123 –1 – 31 0 16 12
1. 7 3 3 7 3 3 7 3 7 3 1. (a) 18 (b) –14
(c) –32 (d) – 48
2. (e) – 4 (f) 3 Self Practice 1.3b
48 (g) –3 (h) –5
1. (a) –   5  , –   5  , –   1  , 7  , 3
–12 – 6 Self Practice 1.2c 6 12 4 24 8
Integer
(b) –   5 , –   13 , –   2 , –   15 , 1 , 5
0 458 1. (a) –18 (b) –24 6 18 3 24 3 8
59 (c) 1
(e) 39 (d) –3 2. (a) 5 , 3 , –   1 , –   7 , –   5
1 6 5 8 20 12
Self Practice (f) 3 11

Self Practice 1.1c 1.2d (b) 1118 , 2 , –   7 , –   1 , –   5 ,
1. (a) 9 18 2 9
1. (a) 426 (b) 56 700
–5 –3 1 5 (c) 452 120 (d) 1 380 – 172 
(b) (e) 6 (f) 4 495
Self Practice 1.3c
–10 –8 02 Self Practice 1.2e
13 13
2. (a) 1. A loss of RM9 200 1. (a) 1 120 (b) – 4 18
2. (a) 26°C (b) 32°C
–12 –8 –4 0 4 8 (c) 172 (d) 7 2
(b) 7
 Mastery 1.2
(e) 1 7 (f) –2 71
1. –12 + (–2); 8 240
–56 –48 –40 –32 –24 –16 6 × (–2) – 2;
Self Practice 1.3d
5 – 11 – 8;
Self Practice 1.1d –2 × (–3) – 15 – 5; 1. 22 1 cups
2
1. – 6, – 4, –2, 0, 1, 3, 5 8 × (–2) + 2;
2. 4, 3, 2, –1, – 3, –4, – 5 or other valid answers 2. RM1 080

 Mastery 1.1 2. (a) ×  , –  Mastery 1.3
(b) ÷  , –
1. (a) 20 m below sea level 1. 1 – 4 + 1 ;
(b) a movement of 90 m to the 5 5 10
3. (a) –5, –1, 3
south (b) –8, 64, –128 3 ÷ 2 × 1– 2 2;
(c) – 800 10 5 3
(d) –1 000 4. (a) –23°C
(b) –11°C 3 ÷ 6 – 1 3 ;
2. (a) – 80 2 5 4
(b) +76
5. Below sea level = –50 m or other valid answers
3. (a) – 8, –7, – 6, –5, – 4, –3, –2,
–1, 0, 1, 2, 3, 4 After 2 minutes 2. (a) 2 (b) 2516
120 3
(b) –12, –11, –10, – 9, –8, –7, = –50 + 5 ×2 1
– 6, –5, – 4, –3, –2 = –2 m 3
3. (a) –2 (b) – 8
4. Integer: –14, 12, –26, 85, 0, –2 The diver had not reached the
Not integers: 3.9 7
sea surface after 2 minutes. 4. 3 30 m
5. – 4 °C, –3 °C, 1 °C, 2 °C, 4 °C
6. (a) Cheque valued at RM1 730 5. 128 ml
(b) RM1 382

307

Answers

Self Practice 1.4a (c) 3 409 (d) –3 5 11.
1. (a) 648 24 –10

–1 –0.7 –0.3 0 0.2 0.6 1 Self Practice 1.5c –30 –60
(b)
1. RM10.05 million 36
–2 –1.3 –0.4 0 0.3 0.7 1 2. 0.925 m
24 3
2. (a)
–2 –1.84 –1.62 –1.46 –1.20 –1  Mastery 1.5 1
(b) 2
1. (a) –20 5 8 4
–0.5 –0.39 –0.25 –0.17 –0.08 0 6


(b) 17 11   or other valid answers
12
Chapter 2 Factors and
Self Practice 1.4b 2. (a) –5.7, –6.8 Multiples

1. (a) –1.48, –1.23, –0.034, 0.34, (b) – 1 Self Practice 2.1a
1.034 8
1. (a) No (b) Yes
(b) –1.654, –1.546, –1.456, 3. (a) 1 238 (c) Yes (d) Yes
1.456, 1.564 (b) 17070 (e) Yes (f) Yes
(g) Yes (h) No
2. (a) 2.522, 2.452, –2.005, 4. Jia Kang is at a level 2.2 m
–2.052, –2.505 2. (a) 1, 3, 5, 15
lower than Ishak. 17 (b) 1, 2, 4, 8, 16, 32
(b) 0.621, 0.065, –0.068, 60 (c) 1, 2, 4, 5, 8, 10, 20, 40
– 0.639, – 0.647 Suresh is at a level 1 m (d) 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
lower than Ishak. (e) 1, 3, 17, 51
Self Practice 1.4c (f) 1, 3, 29, 87
Let’s Practise (g) 1, 2, 7, 14, 49, 98
1. (a) 2.36 (b) – 43.75 (h) 1, 2, 4, 31, 62, 124
(c) 1.68 (d) –27.72 1. B 1
(e) 1.77 (f) 2.23 100
2. (a) Self Practice 2.1b

Self Practice 1.4d (b) – 4.3 1. (a) 3 and 5 are prime factors of 30.
7 is not a prime factor of 30.
1. RM19.85 (c) 2.5 (b) 3 is a prime factor of 54.
2. 30.2°C 5 and 9 are not the prime
3. –2
factors of 54.
 Mastery 1.4 4. 348 years

1. 1.2 + 1.5 – 5.2; 5. Mass of alms for each victim 2. (a) 2, 3 (b) 2, 3
0.4 – 2.1 + (– 0.8); 1 (c) 2, 29 (d) 3, 11
– 0.2 × 4.5 ÷ 0.36; =2+ 2 + 0.4
or other valid answers 3. (a) 42 = 2 × 3 × 7
= 29 kg (b) 96 = 2 × 2 × 2 × 2 × 2 × 3
10 (c) 120 = 2 × 2 × 2 × 3 × 5
2. (a) – 0.6, 0 (b) 4.2, –33.6 (d) 135 = 3 × 3 × 3 × 5
Mass of alms in three vans
3. (a) –1.84 (b) – 6.2 29
= 80 × 10
4. (a) RM0.70 (b) RM4.10 Self Practice 2.1c

5. 15.64 m = 232 kg 1. (a) Yes (b) Yes
(c) No (d) Yes
Self Practice 1.5a Mass of alms in a van (e) Yes (f) No
= 232 ÷ 3
1. –2 , 8 , – 12 ,   153 , 12 , –21 = 77.33 kg 2. (a) 1, 2, 3, 6
4 7 15 20 5 5 (b) 1, 2, 3, 4, 6, 8, 12, 24
6. (a) –, + (c) 1, 5, 7, 35
Thus, –2 , 8 , –1.2 , 7.65, 2 2 , (b) +, – (d) 1, 2
4 7 1.5 5 (e) 1, 5
– 4.2 are rational numbers. 7. –7°C (f) 1, 2, 3, 4, 6, 12
(g) 1, 2
Self Practice 1.5b 8. 0.75 m towards right of O (h) 1, 2, 3, 4, 6, 12
(i) 1, 2, 5, 10
1. (a) –1 4 (b) 116 9. Moves 5.6 m towards west
5
10. 14.7 m below level H

308

Answers

Self Practice 2.1d 4. 48 Self Practice 3.1g
5. 27
1. (a) 24 (b) 18 6. These two numbers do not have 1. (a) 92 cm to 110 cm
(c) 12 (d) 6 (b) Enough. It is because the
(e) 18 (f) 6 common factors except 1. For maximum length of the white
(g) 4 (h) 3 instance, 2 and 3. lace needed is 4.4 m
7. 6 and 60
Self Practice 2.1e 8. 5:15 p.m.  Mastery 3.1
9. 30 cm × 30 cm
1. 6 boxes 10. The second Saturday 1. (a) No
2. 15 plates 11. (a) 6 pages (b) No
(b) 4 photographs and (c) Yes
 Mastery 2.1
7 newspaper cuttings. 1024: 2 × 2 × 2 × 2 × 2
1. 15 × 2 × 2 × 2 × 2 × 2
2. 1 968 = 2 × 2 × 2 × 2 × 3 × 41 Chapter 3 Squares, Square
4 968 = 2 × 2 × 2 × 3 × 3 × 3 × 23 Roots, Cubes and 2. 100 = 2 × 5 × 2 × 5
HCF = 24 Cube Roots 100 = 10

3. 4, 8, 16, 20, 28, 32 Self Practice 3.1a 3. 6 10 14 19 22

4. 2 segments 4
49
5. 20 cm

Self Practice 2.2a 1. (a) No (b) Yes 4. (a) 36 (b)
(c) No (d) Yes
1. (a) Yes (b) No (c) 1699 (d) 65.61
(c) Yes (d) Yes
(e) Yes (f) Yes Self Practice 3.1b (e) 19 (f) 3
8 7
1. (a) 5 × 5  = 5 (g) 5 (h) 1.1
2. (a) 10, 20, 30, 40, 50 (b) 8 × 8  = 8
(b) 15, 30, 45, 60, 75 (c) 24  2  = 24 5. (a) 16 129 (b) 1 197.16
(c) 198, 396, 594, 792, 990
(d) 120, 240, 360, 480, 600 Self Practice 3.1c (c) 0.009409 (d) 46441
(e) 24, 48, 72, 96, 120
(f) 120, 240, 360, 480, 600 1. (a) 64 25 (e) 8.72 (f) 10.41
(g) 120, 240, 360, 480, 600 (c) 1.96 36
(h) 180, 360, 540, 720, 900 2. (a) 841 (b) (g) 0.63 (h) 1.61
(c) 234.09
Self Practice 2.2b (b) 18211 6. Length of side of the base of the
pyramid
1. (a) 144 (b) 30 Self Practice 3.1d
(c) 24 (d) 180 = 52 900
(e) 90 (f) 224 = 230 m
(g) 252 (h) 60
1. (a) 9 (b) 7 7. (a) 90 000 (b) 2 500
Self Practice 2.2c (c) 11 (d) 30 (c) 0.0016 (d) 64
(e) 4 (f ) 15
1. 36 seconds (e) 7 (f ) 2 2 (g) 3 (h) 0.7
2. 30 pieces 9 3
(g) 85 8. (a) 10 (b) 6, 8
(h) 1.5
Self Practice 3.2a
 Mastery 2.2 Self Practice 3.1e
1. (a) 27 = 3 × 3 × 3
1. 40 1. (a) 6.56 (b) 6.15 27 is a perfect cube.
2. 72 (c) 0.68 (d) 3.58
3. 24 days (b) 45 = 3 × 3 × 5
4. 150 cm Self Practice 3.1f 45 is not a perfect cube.

Let’s Practise 1. (a) 3 600 (b) 400 (c) 215 = 5 × 43
(c) 81 (d) 0.04 215 is not a perfect cube.
1. 60 (e) 6 (f) 4
2. 2 (g) 11 (h) 0.9 (d) 343 = 7 × 7 × 7
3. 60 343 is a perfect cube.

309

Answers

Self Practice 3.2b Self Practice 3.2h 3.
2
1. (a) 3 8 × 8 × 8 1. (a) 16 (b) 0.75
4 (d) 0.4 4
= 8 (c) – 125
(f) 0.018
(b) 3 0.3 × 0.3 ×0.3 (e) –1 (h) 536
9
= 0.3 (g) – 4 Area of the square

1 3 (i) –3 =2× 1 ×4×2
2 2
(c) 3 1– 2 = 8 unit2
 Mastery 3.2

= – 1 1. (a) No =Len8gtuhnoitf side of the square
2 (b) Yes
343 = 7 × 7 × 7
Self Practice 3.2c (c) Yes 4. 400 = 2 × 2 × 5 × 2 × 2 × 5
1 000 = 2 × 5 × 2 × 5 × 2 × 5 Length of side = 2 × 2 × 5
1. (a) 216 (b) –343 = 20 m
8 (d) – 0.027
(c) – 729 2. 3 375 = 3 × 5 × 3 × 5 × 3 × 5
33 375 = 15 5. (a) 512 = 2 × 2 × 2 × 2 × 2 × 2
(e) 2112957 64
3. (a) –125 (b) 125 × 2 × 2 × 2

2. (a) 17 576 (b) –132.651 (c) – 343 (d) –32.768 Prime factors can be grouped
5 832 216 into three identical groups.
(c) 0.027 (d) – 1 331 (e) 5 (f) – 8
(e) 13128524 (b) 512
(g) 9 (h) –30 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
(i) 52 4 Prime factors cannot be
(k) – 0.8 (j) – 7
grouped into two identical
Self Practice 3.2d (l) 1.1 groups.

1. 21 (b) –5 4. (a) 8 242 408
2. 14 (d) –10
3. (a) 3 (b) –5 451.776 6. 6 m
(c) 7
(c) 0.000068921 7. (a) 10 cm
4 913 (b) 100 cm2
(d) – 343

4. (a) 2 (b) – 1 (e) 3.26 (f) 6.00 8. (a) (i) RM3.20
5 3
(c) 32 (d) 45 (g) – 0.98 (h) – 0.86
(ii) RM20.00
5. (a) 27 (b) –1 000
(e) 0.1 (f) – 0.4 (c) 3 375 (d) –1 (b) Number of 20 sen coins with
(e) 4 (f) 7
(g) – 0.6 (h) 0.07 (g) –9 (h) 2 an amount of RM60
60
Self Practice 3.2e = 0.2

1. (a) 2.47 (b) – 4.20 6. (a) 50 mm (b) 503­ = 300
(c) 5.48 (d) 0.92
(e) –1.12 7. (a) 2.09 (b) – 80 300 is between 289 and 324,
(c) 2254 th2a8t9isa,nd30032i4s .between
Self Practice 3.2f (e) –225 (d) 12 300 is between 17 and 18.

1. (a) 8 (f) – 19
(b) – 1 000 21
(c) 8 000 Therefore, the measurement
(d) – 64 000 Let’s Practise
of the largest square that can
2. (a) 2 1. 4, 81, 49 be arranged is 17 × 17.
(c) –5
(b) 4 2. 36 – (– 0.1)3 9. (a) 112
(d) –3 25
(b) No.
Self Practice 3.2g 6 112 is not a perfect square.
5
1. Not enough. = – ( – 0.001  ) 10. 10, 13, 17
The length of the wire needed is
= 1 201
156 cm. 1 000

310

Answers

Chapter 4 Ratios, Rates and (b) Rate = 20 litres  ;  Mastery 4.3
Proportions 1 time
175 ml olive oil
volume (litre) and number of 1. (a) 50 ml vinegar

Self Practice 4.1a times the water is pumped. 300 ml olive oil
x ml vinegar
1. (a) 14 : 16 : 7 (c) Rate = RM240  ; =
(b) 2 : 1 : 5 4 subjects
(c) 2 : 1 : 48 amount of money (RM) and (b) 73.8homumrs = 11.7 mm
(d) 2 : 3 : 4 t hours
number of subjects.
2. 56 : 12 : 3 2. 12 times
(d) Rate = RM500  ; 3. RM60
Self Practice 4.1b 10 hectares

1. 36 : 48 : 90, amount of money (RM) and Self Practice 4.4a

0.6 : 0.8 : 1.5, area (hectare). 1. 2 : 9 : 7
2. 8 : 12 : 3
2 : 8 :1 (e) Rate = 600 revolutions  ;
5 15 3 seconds
Self Practice 4.4b
2. (a) 1 : 3 and 2 : 6 number of revolutions and
(b) 3 : 8, 6 : 16 and 9 : 24 1. RM48
time (second). 2. 63 kg
3. RM200
3. 2 : 4 : 6, 1 : 2 : 3 2. Object A 4. 35 questions
or other valid answers. 3. 30 g per 100 cm2 5. 3 : 1 : 2
4. 35 g per m2
Self Practice 4.1c
 Mastery 4.2 Self Practice 4.4c
1. (a) 3 : 20 (b) 5 : 7 : 8
(c) 9 : 20 (d) 3 : 4 : 8 1. 58 calories
2. RM49.00
 Mastery 4.1 1. 2.7 g per cm3 3. 2 cups
4. USD50
1. (a) (i) 2 : 3 : 4 2. (a) (i) 1 : 21; 1 : 11
(ii) 8 : 12 : 16 (ii) 606 kg 7 kg
(iii) 4 : 9 : 16 days  ;  60 days
(b) 2 : 3 : 4 is equivalent to
(b) mass and time
8 : 12 : 16.
(c) The numbers in the ratio 3. (a) RM2.25  , RM4.00 , Self Practice 4.4d
250 ml 500 ml
of areas are the squares of 1. 300 fish
the numbers in the ratio of RM7.50 2. (a) 12 times
lengths. 1000 ml (b) 4 : 3
(c) 15 times
2. 25 : 1 (b) RM8.00

3. (a) 1 Amanah and 1 Cekap (c) 1 l carton of milk. It has  Mastery 4.4
(b) 16 : 20 : 12 the lowest price compared
(c) 12 : 7 to 250 ml carton of milk 1. 8 teachers
and 500 ml carton of milk
4. as as as as priced at RM9 per litre and 2. 3 : 4 : 7
2:10 12:27:6 1:2 12:2 3:12:5 RM8 per litre respectively.
or other valid answers. 3. Hardtail scad and one f inlet
Self Practice 4.3a scad or hardtail scad and Indian
5. (a) 5 : 9 is equivalent to 15 : 27 mackerel
1. (a) RM3 5 = RM20
(b) 5 : 12 is equivalent to 10 : 24 12 Pomfret: RM15/kg
and 15 : 36 Hardtail scad: RM6.40/kg
(b) 244cm = 78 cm One finlet scad: RM9.00/kg
6. 6 : 4 : 2,  18 : 12 : 6 13 Indian mackerel: RM13.50/kg
or other valid answers.
(c) 1135 boys = 65 boys 4. (a) Bacteria A, 40 million per
girls 75 girls minute
Self Practice 4.2a
(b) 20 million
1. (a) Rate = 2 RM154  ; Self Practice 4.3b (c) 3.3 minutes
passengers
1. 50 players 5. (a) 3 : 4 : 2
amount of money (RM) 2. 100 times (b) Nitrogen: 4.5 kg
3. 510 chilli plants Phosphorus: 6 kg
and number of passengers Potassium: 3 kg

(person).

311

Answers

Self Practice 4.5a New house: Self Practice 5.1c
Total cost of monthly rental
1. 75% 1. (a) 12 (b) 30
2. 7 : 13 and petrol (c) 45 (d) 9
= RM368.80
Self Practice 4.5b Lai Huat should not 2. (a) 8x + 4y
consider moving to the new (b) 80 kg
1. (a) 25%
(b) 40% house because he can save 3. (a) p(m – n) or p (n – m)
(c) 7 500 000% RM30. 40 while staying in (b) RM3
2. RM11.25 his current house.
Self Practice 5.1d
8. (a) RM3.60
Self Practice 4.5c 1. (a) 6k and 2k
(b) Mr Ong should go back to
1. (a) 20% his off ice to get his parking (b) x2 and 9xy
(b) 18 : 50 or 9 : 25 ticket. The parking charge is (c) a3b , 2a and 5b
RM10.60 compared to the 7x
lost ticket penalty of RM20. (d) 4pq, 2 , 8p2q and 1

 Mastery 4.5 2. (a) –8 (b) – y2

1. (a) 83  ;  0.375 Chapter 5 Algebraic (c) –8x (d) 8y2
(b) 37.5% Expressions
Self Practice 5.1e
2. (a) 3 : 2 Self Practice 5.1a
(b) Same 1. (a) Like terms
(c) The length of each side of 1. (a) k represents the mass of (b) Unlike terms
each student in the class. (c) Like terms
P9Q9R9 is 1. 5 times the (d) Unlike terms
length of the corresponding k has a varied value because
side of PQR. the mass of each student is  Mastery 5.1
different.
3. Buying online is a better choice. 1. k is the amount of money
The online price is 50 sen (b) x represents Zaini’s marks invested and d is the dividend
cheaper than the price offered in a Mathematics test. given.
by the bookshop.
x has a f ixed value because k has a fixed value because it is
Let’s Practise 1 : 15 it is Zaini’s score for the the initial amount invested and
1. 6 : 9 Mathematics test. d has a varied value because the
dividend rate will vary based on
6 : 90 2 : 3 (c) h represents the distance economic conditions every year.
between Arman’s house and
his school. 2. 1 m – 5 2 kg
2
9 : 60 3 : 20 h has a fixed value because
the distance between 3. (xy – 225) m2
2. (a) 4 : 5 Arman’s house and his (b) 142
(b) For aquarium A, add 4 f ish. school is the same for every 4. (a) 4 21
For aquarium B, add 5 f ish. trip. 5. (a) n – 4x

3. (a) 36 (d) c represents the temperature (b) RM22.20
(b) 9 : 8 at the peak of Gunung
(c) Nurin, 84 Kinabalu in a day. 6. (a) p – 3x
(b) 35
4. 1.8 l bottle c has a varied value because
the temperature at the peak 7. Algebraic Coeff icient Variable
5. 0900 of Gunung Kinabalu is term
always changing within a
6. 10 m day. –10ab c

7. (a) 0.804 litre Self Practice 5.1b –10abc –10ac b
(b) Current house: –10b ac
Total cost of monthly rental 1. ((ac)) x4x– 7 ((db)) myp+9+znq
–10 abc
and petrol
= RM338.40 (e) h – 2k or other possible answers.

312

Answers

Self Practice 5.2a 7. Let n represent a number.  Mastery 6.1

1. (a) 8x + 7y n + 7 = x 1. (a) x – 8 = 15
(b) 5ab – bc + 12 n2 = y (b) 14y = 42
(c) 17xy + 7k – 7 1 n 2 (c) p + 34 = 3p
(d) 6p – 3q – 3pq x + y = (n + 7) + 2
(e) fg – 6mn 2. (a) 16 (b) 4 (c) 2
= n + 7 + n (c) 24
2 3. (a) –   1 (b) – 4
Self Practice 5.2b 3 2 (c) 6
x + y = 2 n + 7
1. (a) (pq)3 4. (a) 130 (b) 7
(b) (6a – 1)2 8. RM60(3h + 4k – 2p – 3q)
(c) (8x + 3y)3 5. 20 boys
9. (6xy + 8x + 18) m
2. (a) (2 + 7x)(2 + 7x) 6. 1 hour and 30 minutes
(b) (h – 4k)(h – 4k)(h – 4k) 10. 27°F
(c) (5p+q)(5p+q)(5p+q)(5p+q) 7. Small table = 2.16 m2
Chapter 6 Linear Equations Large table = 4.32 m2

Self Practice 5.2c Self Practice 6.1a Self Practice 6.2a
(b) – 28m3n
1. (a) 15x4 1. (a) Yes, because the equation 1. (a) Yes, because the equation
(c) 4p5qr has one variable m and the has two variables, h and
power of m is 1. k, and the power of each
2. (a) 4x5y (b) 23ba2 variable is 1.
6p4r (b) Yes, because the equation
(c) – 5q has one variable p and the (b) No, because the equation
power of p is 1. has only one variable.
3. (a) 10m3 (b) – 3p5y2
3n2­ 2x (c) No, because the equation (c) Yes, because the equation
has two variables, x and y. has two variables, x and
 Mastery 5.2 y, and the power of each
(d) No, because the highest variable is 1.
1. (a) 43 x + 3 pq – 8y + 11 power of the variable k is 2.
2 (d) No, because the highest
Self Practice 6.1b power of variable p is 2.
(b) 8ab – 11mn

2. (2d + 7y) cm 1. (a) x = 12 Self Practice 6.2b
6
3. 4n + 4 1. (a) x + y = 258
(b) 5y = 40 (b) p – q = 15
4. n = 3, a = 9, b = – 2 (c) 8x + 5y = 265
5. 6(2 + 3p)2 cm2 (c) 4x + 1 000 = 1 400 (d) x + 2y = 40

6. (a) 12y4 (b) 2pq3 2. (a) Subtract 1 from a number, 2. (a) The total number of
z (b) 14x3y3z2 the result is 6. aluminium cans and glass
bottles collected during the
7. (a) 5pr2 (b) When Edri’s score is added
to 10, it will be 78. recycling campaign was 465.
8. 12a3b2 = 4a2b cm (b) The difference between
3ab (c) The total mass of four packs
of rice is 50 kg. the length and width of a
rectangle is 3 cm.
Let’s Practise Self Practice 6.1c

1. a = 8, b = 4, c = 4 1. (a) 13 (b) 2
(c) 12 (d) 10
2. RM(120 – 12x – 7y) (e) 3 (f) –15

3. –11 Self Practice 6.2c

4. RM4.80 1. (a) (7, 0), (8, 1), (9, 2)
(b) (0, 1), (2, –3), (4, –7)
5. 2k + 10 Self Practice 6.1d (c) (2, 0), (6, 24), (1, – 6)
or other valid answers
6. (a) mx + 2my 1. 53 marks
5 2. 266 cm2

(b) RM33.40

313

Answers

2. 1 gold medal and 3 bronze Self Practice 6.3a 3. First part of the wire = 24 cm
medals; Second part of the wire = 24 cm
1. (a) x + y = 6 Third part of the wire = 52 cm
2 gold medals and 2 bronze 20x + 40y = 160
medals; 4. Lai Yee: 45 stamps
y Khadijah: 15 stamps
3 gold medals and 1 bronze
medal. 6 x+y=6 5. Devaki's father: 30 years old
5 Devaki: 6 years old
Self Practice 6.2d 4
3 6. Sarah: RM35
1. Diagram (a) and Diagram (c). 2 20x + 40y = 160 Hui Chin: RM15
1
 Mastery 6.2 0 12345678 x Let’s Practise

1. (a) 30x – 20y = 8 (b) 2x + 2y = 12 1. 16 sweets
(b) x + 2y = 130
(c) x = 2y x – y = 2 2. 220 points

2. (a) (4, 3), (2, 5) y 3. Ella = RM600
(b) (1, 7), (2, 12) Zahida = RM960
or other valid answers 6 2x + 2y = 12
5 4. RM310
3. (a) y 4 5. 187 cm2
3 x–y=2
6. A cup of coffee: RM1.80
2 A piece of curry puff: RM0.80
1
3 x 7. 3 children
2 0 1 2 3 4 5 6
1 8. Asnita: 25 years old
–10 1 2 3 4 5 x Self Practice 6.3b Reslynna: 19 years old
–2
1. (a) x = 2, y = 7 9. Year 2013
(b) y (b) x = 9, y = – 4 Population of trees = 25
(c) x = 2, y = 0
8 (d) x = –26, y = 68 10. 12.7s
7
6 Self Practice 6.3c 11. Pineapple: RM2
5 Watermelon: RM1.50
4 1. 500 booklets of RM30 coupon
3 and 300 booklets of RM50 12. P : RM287 500
2 coupon Q : RM108 000
1
–2–1 0 1 2 x 2. p = 50 m,  q = 25 m

 Mastery 6.3 Chapter 7 Linear Inequalities

(c) y 1. x – y = 5 Self Practice 7.1a
2x + y = 7
4 1. (a) ,
3 y – 6 is less than 0.
2
1 7
–6–5–4 –3–2–1 0 6 (b) ,
12 x 5 2x + y = 7 1 1
7 4
4 is less than .
3
4. 1 shirt and 6 pairs of pants; 2 (c) .
2 shirts and 4 pairs of pants; 1
3 shirts and 2 pairs of pants. x 0.42 is greater than 0.072.
The maximum number: 3 shirts –10 1 2 3 4 5
–2 (d) . is greater than 4.5.
5. 10x + 20y = 500 –3 4.5

y –4 x–y=5 (e) .
–5 10 cm is greater than 50 mm.
25
20 Unique solution. (f) ,
15 1 200 g is less than 1.6 kg.
10 2. (a) x = 2, y = 3
5
(b) m = 5, n = –2 Self Practice 7.1b
0 10 20 30 40 50 x 1. 12 is greater than y.
(c) p = 9, q = 1 20 12 . y
10 3
(d) f = 3 , g =

314

Answers

2. 3 is less than b.  Mastery 7.1 (b) The area of an apartment is
3 , b more than 1 000 m2.
1. (a) , (b) , (c) ,
Self Practice 7.1c (d) . (e) . (f) . (c) The total expenditure of four
1. (a) m is less than or equal to 8. customers who patronize the
m < 8 2. (a) y is greater than x. restaurant is at least RM60.
(b) y . x
mഛ8 Self Practice 7.2b
3. (a) t = deposit
5 6 7 8 9 10 t is greater than or equal to 1. (a) x > 7
(b) x . –9
(b) t is greater than or equal to 100. (c) x , –21
21. (d) x > 4
(b) a ജ 100 2. At least 200 canned drinks in an
t > 21
t ജ 21 80 90 100 110 120 130 hour.
a > 100 3. 10 months
4. 5 hours
19 20 21 22 23 24 4. (a) xϾ3

Self Practice 7.1d 1 234 5 6 Self Practice 7.2c
(b) x Ͻ 15
1. (a) 14 . 5 1. (a) x . 5
(b) x < –3
(b) –8 , 8 3
(c) –32 , –23 12 13 14 15 16 17 (c) – 2 ,x<2

(d) 1.5 , 6.7 (c) x ജ –19 (d) –1 < x , 1

(e) 114 , 1 –21 –20 –19 –18 –17 –16 (e) x , 1 5
13 (d) x ഛ –5 7

(f) –11.8 , –2 (f) x . 3

2. (a) –2 , 10 –7 –6 –5 –4 –3 –2  Mastery 7.2
(b) –15 , 0 (e) y ഛ 8.3
(c) – 4.56 , 2.01
2 , 20 13 1. (a) x > 18 000
(d) –17 9 (b) t < 8
(c) h . 700
(e) 210 , 1 8.1 8.2 8.3 8.4 8.5
7
1 (f) p ജ –5.7 2. (a) The speed of the vehicle
(f) –8 , 9 should not exceed 30 km/h
when approaching the
Self Practice 7.1e –5.9 –5.8 –5.7 –5.6 –5.5 school area.

1. (a) (i) , (ii) , (iii) , (g) x Ͻ – 3 (b) The mass of a car is greater
5 than 1 100 kg.

(b) (i) . (ii) . (iii) . –1 4 3 2 1 (c) A student whose parents’
5 5 5 5 salary is less than RM900
(c) (i) . (ii) . (iii) . – – – – qualifies to apply for a
scholarship.
2. (a) (i) , (ii) , (iii) , (h) q Ͻ 7.8

(b) 8c , 16c
8 16
c , c 7.6 7.7 7.8 7.9 8.0 3. (a) x ജ 1 200

(c) 16c . 8c 5. (a) . (b) , (c) . 1000 1100 1200 1300 1400
16 8 (d) . (e) , (f) ,
c . c (b) x > 1 200
Self Practice 7.2a
3. (a) (i) , (ii) . (iii) . 1. (a) x > 450 000 4. 2 packets
(b) y , 50
(iv) . (v) . (c) k < 30 5. (a) m < 10.50
(d) q > 600 (b) m ഛ 10.5
(b) 6d . 12d 2. (a) The number of passengers
6 12
d . d in a taxi should not 8 9 10 11 12
exceed 4.
(c) 12d , 6d 6. 21 boys
12 6 7. 7 months
d , d

315

Answers

8. (a) x , – 5 (c) Appears much more than a (b) (i)
6 right angle.
(b) –5 < x , 8 Thus, estimated size is
(c) x < –1 about 130°. 78°
72° 78°
Let’s Practise 4. (a) 52° 30°
(b) 115°
1. (a) , (b) , (c) . (c) 146°
72 + 30° + 78° = 180°
2. (a) x . 18 Self Practice 8.1c
(b) x + 3 . 21 (ii)
(c) x – 5 . 13 1. (a) Reflex angle
(b) The angle on a straight line
3. 20 – 1.20x . 5 (c) The angle of one whole turn 78°
78°
4. (a) 170 + n 2. (a) (i) 72°
(b) 1, 2, 3, 4, 5, 6, 7 135° 30°
45°
5. (a) x , 4 (b) x , 2
88°
6. 14 g

7. 17 hours

8. (a) 15n + 200 . 290 72° 78°
(b) 7 135° + 45° = 180° 30° 78°

9. RM751 (ii)

10. (a) – 4 < x < 3 135°
45°
(b) x . 8
3 88°

(c) 3 < x < 7 72° 78°
(d) 52 < x , 3 30° 78°


Chapter 8 Lines and Angles 135°
45°
Self Practice 8.1a
88°
1. (a) Congruent; PQ = RS
(b) Not congruent; PQ ≠ RS 72° 78°
(c) Congruent; PQ = RS 30° 78°
2. (a) Congruent; ∠PQR = ∠ABC 135°
(b) Not congruent; 45°

∠PQR ≠ ∠ABC 88°
(c) Congruent; ∠PQR = ∠ABC

Self Practice 8.1b
135° 72° 78°
1. (a) 4 cm. Use a 4 cm long 45° 30° 78°
eraser.
88°
(b) 6 cm. Use a 3 cm long
paper clip.
A reflex angle is more A reflex angle is more
(or other objects of known lengths)
than 180° and less than than 180° and less than
2. (a) 4.3 cm 360°. 360°.
(b) 5.8 cm (iii) (iii)

3. (a) Appears slightly more than 135° 72° 78°
a right angle. 45° 30° 78°

Thus, estimated size is 88°
about 100°.

(b) Appears slightly less than a The angle of one whole The angle of one whole
right angle.
turn is 360°. turn is 360°.
Thus, estimated size is
about 80°.

316

Answers

Self Practice 8.1d (b) 7. (a)

1. (a) True (b) False PM C Q
(c) True (d) True 7 cm
(e) True P

Self Practice 8.1e Q 30°
Self Practice 8.1g AB
1. a = 44°, b = 136°, c = 77°,
d = 57° 1. (a) (b) 47°

2. p = 116°, q = 64° Self Practice 8.2a

Self Practice 8.1f 1. (a) (i) Vertically opposite
angles: ∠p, ∠q
The following diagrams are not
drawn to actual scale. (ii) Adjacent angles at
intersecting lines:
1. (a)
∠q and ∠r; ∠p and ∠r
A 6 cm B (b)
(b) 5.4 cm Q (b) (i) Vertically opposite
angles: ∠a, ∠b
P
(c) (ii) Adjacent angles at
intersecting lines:
R 7.3 cm S 45°
2. (a) (c) ∠a and ∠c; ∠b and ∠c

(c) (i) Vertically opposite
angles: ∠t, ∠r

(ii) Adjacent angles at
intersecting lines:

∠t and ∠s; ∠r and ∠s

PQ 75° 2. (a) z
yx
(d)
z

(b) (b)
zx
A B 105°
3. (a) y
z



 Mastery 8.1 (c) y

zz

1. (a) True (b) False x
(c) True (d) True
P Q (e) True
(b) M Self Practice 8.2b
2. p = 45°, q = 45°
1. (a) x = 140°, y = 100°
3. 7 cm (b) x = 24°, y = 148°

4. p = 105°, q = 75° Self Practice 8.2c
1. x = 66°, y = 58°
5. p = 72°, q = 288° 2. x = 90°, y = 25°

M Q 6. (a) R

P Q 6 cm  Mastery 8.2
M Q
60° 1. x = 34°, y = 62°
4. (a) P 2. 15°
P (b) 5 cm 3. 28°



317

Answers

Self Practice 8.3a (b) Corresponding angles; 3. x = 125°, y = 100°
1. (a) ∠a = ∠b 4. (a) ∠QOR
(b) ∠TOU
c 5. (a) x = 70°, y = 48°
(b) d (b) 250°
(c) 70°
a 6. x = 33°, y = 58°
(c) b
Let’s Practise
(c) Alternate angles; ∠a = ∠b
1. x = 65°, y = 118°
b c 2. x = 60°, y = 84°
da 3. C

(d) Corresponding angles; 5 cm
∠a = ∠b
60°
a
c A 8 cm B
b
d
4. x = 139°, y = 94°
2. (a)

2. (a) ca q 5. x = 72°, y = 108°
rp
A b 6. (a) x = 75°, y = 45°
P (b)
B R
(b) P A Q S
A
B (b) a bq 6 cm

pr 105° 120°
p cr

P 4 cm Q

Self Practice 8.3c 7. x = 64°, y = 32°
1. (a) Parallel
B (b) Parallel 8. x = 72°, y = 120°, z = 60°
Q (c) Not parallel
9. x = 90°, y = 52°
(c) P Self Practice 8.3d
A 10. (a) 46° (b) 42°
1. a = 76°, b = 70°, c = 70°, d = 70°
2. a = 67°, b = 42° Chapter 9 Basic Polygons

A Self Practice 8.3e Self Practice 9.1a
1. Aeroplane
Q B B 1. (a) Number of vertices: 6
Number of diagonals: 9
(b) Number of vertices: 9
Number of diagonals: 27
3. (a) Yes Jasni a (c) Number of vertices: 12
(b) Yes b Number of diagonals: 54
(c) No (d) Number of vertices: 20
Number of diagonals: 170
Self Practice 8.3b Rock

1. (a) Interior angles; Self Practice 8.3f Self Practice 9.1b
∠a + ∠b = 180°
1. (a) x = 117°, y = 88° 1. (a) A
(b) 88°

abcd  Mastery 8.3 EB

1. x = 45°, y = 93° D C
2. x = 106°, y = 58° Pentagon ABCDE

318

Answers

(b) A B Self Practice 9.3a 2. (a)

HC 1. All sides have the same length;
opposite sides are parallel;
GD diagonals are perpendicular
bisectors of each other.
FE (b)

Octagon ABCDEFGH 2. Rectangle Parallelogram

(c) B C Opposite sides Opposite sides
AD
are of the same are of the same
JE 3. x = 164°, y = 74°
length and length and 4. x = 34°, y = 66°
I F
HG parallel. parallel. Let’s Practise

Decagon ABCDEFGHIJ All of its Opposite 1. (a) 3
(b) 3
 Mastery 9.1 interior angles angles are (c) 7
(b) True (d) 3
1. (a) True are 90°. equal. 2. (a) Rectangle, rhombus
2. A B (b) Scalene triangle
Diagonals Diagonals are (c) Square, rhombus
are of equal bisectors of (d) Rectangle, square
length and are each other. 3. (a) 85°
bisectors of (b) 24°
HC each other. (c) 50.5°
(d) 15°
GD It has two axes It does not 4. x = 53°, y = 53°
of symmetry. have any axes 5. x = 15°, y = 60°
FE 6. The largest number in the ratio
Name of polygon: octagon of symmetry.
Number of diagonals: 20 is 6; 120°
Self Practice 9.3b 7. 75°
Self Practice 9.2a 1. 121° 8. x = 112°, y = 44°
2. x = 84°, y = 90° 9. x = 30°, y = 74°
1. (a) Obtuse-angled triangle; or 10. x = 56°, y = 62°
scalene triangle Self Practice 9.3c 11. x = 53°, y = 127°
1. x = 152°, y = 121°
(b) Equilateral triangle 2. x = 132°, y = 138°
(c) Right-angled triangle
(d) Acute-angled triangle; or Self Practice 9.3d
1. x = 72°, y = 108°
scalene triangle 2. x = 30°, y = 72°

Self Practice 9.2b  Mastery 9.3

1. (a) 35° (b) 47° 1. Similarities: Chapter 10 Perimeter and Area
(c) 27° (d) 42° • Opposite sides are of equal
2. (a) 123° (b) 54° length and parallel. Self Practice 10.1a
(c) 82° (d) 76° • Opposite angles are equal.
1. (a) 74 cm
Self Practice 9.2c Differences: (b) 71 cm
Parallelogram (c) 13 cm
1. x = 56°, y = 32° • Opposite sides are of equal
2. x = 60°, y = 89°
length.
 Mastery 9.2 • Diagonals are bisectors of Self Practice 10.1b

1. (a) 1 (b) 3 each other. The estimated answers depend on
(c) 1 (d) 1 • It does not have any axes of the students.

2. (a) Equilateral triangle symmetry. 1. (a) Estimation: 10 cm
Rhombus Perimeter: 9.9 cm
(b) Isosceles triangle • All sides have the same length.
• Diagonals are perpendicular The difference between the
(c) Obtuse-angled triangle; or estimated value and the actual
bisectors of each other. value is small, therefore the
scalene triangle • It has 2 axes of symmetry. estimated value is accurate.

(d) Right-angled triangle

3. x = 56°, y = 118°

4. x = 108°, y = 144°

5. x = 160°, y = 62°

319

Answers

(b) Estimation: 11.5 cm Self Practice 10.3b (ii) Q = {3, 6, 9, 12, 15, 18,
Perimeter: 12.3 cm 21, 24}
The difference between the 1. 420.25 m2
2. 93.75 cm2 (iii) Q = {x : x is a multiple
estimated value and the actual 3. 48 cm2 of 3 and x , 25}
value is small, therefore the
estimated value is accurate.  Mastery 10.3 2. (a) True
(b) False
Self Practice 10.1c 1. 128 cm (c) True
(d) False
1. 76 m 2. Draw a square with length of
side of 12 cm. The area is Self Practice 11.1c
2. RM1 720 144 cm2. 1. (a)  (b) 
(c)  (d) 
3. 22 cm 3. 432 cm2 2. (a)  (b) 
(c)  (d) 
 Mastery 10.1 4. 875 m2 (e)  (f) 

1. 9 m Let’s Practise Self Practice 11.1d

2. 42 cm 1. 41 cm 1. (a) n(A) = 5 (b) n(B) = 5
2. 30 cm (c) n(C) = 3 (d) n(D) = 9
3. Perimeter of shape A 3. 81 cm2
= x + y + 6 + (y – 5) + (x – 6) + 5 2. (a) 6 (b) 7
= (2x + 2y) cm 4. (a) 24 cm2
Perimeter of shape B (b) 4.8 cm Self Practice 11.1e
= x + y + 3 + (y – 2) + (x – 3) + 2 5. 111.5 cm2
= (2x + 2y) cm 6. 42 cm 1. (a) Yes (b) No
Thus, perimeters of both the (c) No (d) Yes
7. (a) 4 cm
shapes A and B are equal. (b) 32 cm  Mastery 11.1
8. 25 cm2
4. RM24 000 9. Form a square with length of

5. 6 cm side of 7.5 cm.
10. Fence up a region with a length
Self Practice 10.2b 1. (a) X is the set of vowels.
of 22.5 m and a width of 20 m;
1. (a) 1  mn (b) tk (b) Y is the set of perfect
2 85 m squares which are less than
50.
(c) 12  (p + q)r (d) 21 st
2. (a) P = {Mercury, Venus,
Self Practice 10.2c Chapter 11 Introduction of Set Earth, Mars, Jupiter, Saturn,
Uranus, Neptune}
1. (a) 15 cm2 (b) 24.5 cm2 Self Practice 11.1a
(c) 16.8 cm2 (d) 24 cm2 (b) Q = {2, 3, 5}
1. Land transport: car, lorry, van,
2. 146 m2 bus 3. (a) G = {x : x is a month which
begins with the letter ‘M’}
 Mastery 10.2 Water transport: sampan, boat,
ship, ferry (b) H = {x : x is a multiple of 7
2. 1 550 m2 and 1 < x < 100}
Air transport: rocket, aeroplane,
3. 122.5 cm2 helicopter, hot-air balloon 4. (a) Yes
(b) No
4. 75.25 cm2 Self Practice 11.1b (c) No
(d) Yes
Self Practice 10.3a 1. (a) (i) P is the set of colours
of the rainbow 5. n(A) = 3, n(B) = 13, n(C) = 5
1. T, S, Q, R, P
The bigger the difference (ii) P = {red, orange, 6. 2
yellow, green, blue,
between the length and the indigo, violet} Self Practice 11.2a
width of the rectangle, the
smaller the area. (iii) P = {x : x is a colour of 1. (a) Universal set
the rainbow} (b) Not a universal set
2. P, R, T, S, Q (c) Universal set
The bigger the difference (b) (i) Q is the set of multiples 2. (a) P9 = {0, 1, 2, 4, 5, 7, 8}
of 3 which are less than (b) Q9 = {0, 1, 4, 6, 8, 9}
between the length and the 25.
width of the rectangle, the
bigger the perimeter.

320

Answers

Self Practice 11.2b  Mastery 11.2 6. (a) R , Q (b) Q9

1. (a) ξ 1 A 5 1. ξ B0 7. (a) Set P is a universal set. It
3 A1 2 45 is because set P consists
36 78 of all the elements under
79 11 9 discussion.

(b) ξ B 13 18 \ A9 = B (b) R is the set of odd numbers.
12 11 19
14 17 16 2. (a) P , Q 8. 8
(b) R , Q
15 9. Not true. The complement of
set A is the girls who are not
3. (a) { }, {1}, {4}, {9}, {16}, prefects and the boys in Class 1
{1, 4}, {1, 9}, {1, 16}, Bakti.
{4, 9}, {4, 16}, {9, 16},
2. (a) ξ P 27 Chapter 12 Data Handling
6 12 {1, 4, 9}, {1, 4, 16}, {1, 9, 16},
3 24 21 {4, 9, 16}, {1, 4, 9, 16}

9 18 15 Self Practice 12.1a

(b) ξ Q P N (b) 56 7 8 10 1. (a) Numerical data
L3 4 11 (b) Numerical data
E G L 2K 1 12 (c) Categorical data
S A 20 9 16 13 (d) Categorical data
19 15 14 (e) Numerical data
18 17 (f) Numerical data
(g) Categorical data
Self Practice 11.2c K,L (h) Numerical data

1. (a) , (b) , 4. ξ P R Q 2. T-shirt Tally Frequency
(c)  (d)  size

2. (a) ø, {p}, {q}, {p, q} S| 1

(b) ø, {2}, {3}, {5}, {7}, {2, 3}, M |||| |||| || 12
{2, 5}, {2, 7}, {3, 5}, {3, 7},
{5, 7}, {2, 3, 5}, {2, 3, 7}, L |||| |||| 9
{2, 5, 7}, {3, 5, 7}, {2, 3, 5, 7}
5. C , B , A , j XL ||||  5

Self Practice 11.2d Total 27

Let’s Practise Self Practice 12.1b
1. Ways of Travelling to School
1. (a) A 1. P is the set of quadrilaterals.
1030B2600 405070 10
2. {0} 8
6
3. (a) P = {2, 4, 6, 8, …} 4
Q = {2, 4, 6, 8, …} 2
P = Q 0
(b) M N (b) Set A has an element ‘0’, Number of students
Car
B A K therefore set A is not an School bus
I J empty set. Public bus
A ≠ B Bicycle
(c) E = {1, 3, 5, 15} Walk
F = {15, 30, 45, 60, …}
E ≠ F

Self Practice 11.2e 4. 6

1. ξ e P 5. ξ BA
g Q C
R d b a Way of travelling
f h c
The bar chart is suitable for
comparing the number of
students with how they travel
to school.

321

Answers

2. Accommodation Around 5. Number of Text 2. Blood Groups of Donors
Historical City of Melaka Messages Sent
10
113223005505500000000 3 4 5 6 7 8 9 10 11
Price (RM) The dot plot is suitable for 8
Hotel
Homestay displaying the number of text 6
Budget messages sent by a group of
hotel students. 4
Hostel 6. Ages of Magazine Readers
Stem Leaf 2
Number of donors
2 25 0 A B AB O
Type of accommodation 3 011234689 Blood Group
4 24477
Normal Holiday 5 113788 The bar chart is suitable as the
rate season 6 05 data of the blood groups of
Key: 2 | 2 means 22 years old donors is a categorical data.
The dual bar chart is suitable
for comparing different types of The stem-and-leaf plot is Self Practice 12.1d
accommodation with the normal suitable for displaying the age
and holiday season rates. of each reader of the magazine 1. (a) 8 m
in the survey. (b) 7.5 m
3. (c) Year 2013
Self Practice 12.1c (d) The tree has a nearly
Children’s Favourite Songs 1. Average Monthly
uniform growth rate from
Dayung Rainfall of Impian City 2010 to 2015 or there is
Lompat Sampan no increase in height from
Si Katak 400 2015 to 2016.
Lompat (e) 16 m
Geylang 30° 15° Rasa 300
Sayang
Si Paku 45° Ikan 200 2. (a) 30
Geylang 60° 120° Kekek
Bangau (b) Largest diameter = 28.6 mm
Oh Bangau Smallest diameter = 24.5 mm

(c) 20%

The pie chart is suitable for Average rainfall (mm) (d) The distribution of the
comparing the favourite songs diameters of the axles is
of a group of children. quite symmetric since most
of the diameters range from
4. Kamil’s Height 26.0 mm to 26.9 mm.

170 3. (a) 120

(b) 12

Height (cm) 160 100 (c) Number of tested samples

with the mass of 241 g to

150 0 260 g

July = 28 + 32 + 24 + 12
August
140 September = 96
October
130 November Percentage of tested samples
December
Year with the mass of 241 g to
The line graph is suitable Month
2011 260 g
for displaying the changes in2012 The line graph is suitable for
Kamil’s height over a period of2013 displaying the changes in the = 96 × 100%
6 years. 2014 average monthly rainfall of 120
2015 Impian City over a six-month
2016 period.

= 80%

Thus, this batch of biscuits

produced satisfies the

required specifications.

322

Answers

Self Practice 12.1e 3. (a) 35 bookings. Bar chart. (b) In Diagram (a), most of the
The difference in room taller bars lie in the middle
1. No, because not all the results bookings can be obtained of the histogram. It shows
of the survey are shown in the by calculating the difference that most students obtained
bar chart, that is: in length of the bars for the average marks in the
Internet and counter. Science test.
16 + 6 + 12 + 4 , 40
(b) A pie chart because the pie In Diagram (b), most of the
2. No, because the line graph chart shows 48% of the taller bars lie on the
seems to show a rapid change rooms are booked via the right-hand side of the
in the temperature every 2 Internet and that is almost histogram. It shows that
hours in the day. The scale on half of the total room most students obtained
the vertical axis should start bookings. higher marks in the
at zero so that the information Mathematics test.
displayed will not be confusing. (c) A bar chart because the
number of rooms booked (c) More students obtained
 Mastery 12.1 via the Internet can be read higher marks in the
directly from the length of Mathematics test than in the
1. (a) – 3°C the bar for the Internet. For Science test.
(b) – 4°C a pie chart, the number of
(c) 1100 and 1700 room bookings can only be 6. (a) Shop A and shop B sold
known by calculation. the same number of lamps,
2. (a) (i) which was 120.
(d) No, it is not suitable as
Maximum Speeds Recorded the data does not show (b) Not valid because shop E
120 the changes in the number sold 30 more lamps than
100 of room bookings over a shop C.
80 certain period.
60
40 (e) Pictogram
20
4. (a)
0 Daily Allowances of
Speed (km/h) Factory Workers Let’s Practise
Cheetah
20 21 22 23 24 25 26 27 28 29 30 1. (a) House A
Zebra Daily allowance (RM) (b) 5th year
Lion
Horse (b) (i) The distribution of the 2. (a) Dramas
Deer data ranges from 20 to
Ostrich 30. (b)

(ii) The data clusters around Students’ Favourite
22. Television Programmes

Animal (iii) 30 deviates greatly Sports Documentaries
from other values on
(ii) Maximum Speed the distribution of data, Cartoons 72°
Recorded thus this data has one 54° 144°
extreme value, that is, Dramas
Stem Leaf 30.
5 3. (a) Between the first week
6 05 5. (a) Science test and the second week, and
7 0 = 45 students between the second week
8 5 and the third week.
9 Mathematics test
10 0 = 75 students Between the fourth week
11 and the fifth week, and
between the fifth week and
Key: 6 | 5 means 65 km/h the sixth week.

(b) A bar chart is more suitable (b) Between the third week and
as the data displayed is a the fourth week.
categorical data. The stem-
and-leaf plot displays only (c) The line graph can display
the maximum speeds but the changes in height of
does not display the types the plant over the six-week
of animals. period.

323

Answers

4. (a) False Number of 39, 41, 42, 43, 45, 47, 48, Self Practice 13.2b
(b) True Laptops Sold 52, 56
(c) True 1. 1.52 + 2.02 = 2.52
5. (a) (b) 12 minutes Thus, the wall is vertical.

2013654 (c) Much of the time taken to 2. 160°
fix the leaking pipes ranges
from 30 minutes to 39  Mastery 13.2
minutes.
Number of 1. (a) Yes; 92 + 402 = 412
laptops 8. (a) Diagram (a) (b) No; 272 + 352 ≠ 452
(c) Yes; 2.52 + 62 = 6.52
January (b) Diagram (b). Not ethical. (d) Yes; 132 + 842 = 852
February The scale used on the
vertical axis of this line 2. 122 + 162 = 202
March graph is larger. \ The triangle is a right-angled
April
May triangle. Therefore, the triangular
June piece of wood can be f ixed
Month
(b) Chapter 13 The Pythagoras’ perfectly onto the L-shaped
5643201 Number of Theorem structure.
Laptops Sold
Self Practice 13.1a
Number of 3. 152 + 202 = 252
laptops 1. (a) AC All angles at the vertices of the
(b) b
January (c) PR is the hypotenuse of quadrilateral are right angles,
February thus the quadrilateral is a
∆PQR. rectangle.
March ST is the hypotenuse of
April
May ∆SPT.
June
Month Self Practice 13.1b Let’s Practise

The bar chart is more 1. (a) AC2 = AB2 + BC2 1. (a) 10 cm
suitable in representing this (b) LN2 = LM2 + MN2 (b) 24 cm
data because the number (c) r2 = p2 + q2 2. 22 + 1502 = 150.01 m
of laptops sold can be read (d) z2 = x2 + y2
from the height of the bars. 3. (a) 15 cm
Self Practice 13.1c
6. (a) Year 2010, RM13 million (b) Right-angled triangle;
(b) Year 2012, RM2 million 1. (a) 10 82 + 152 = 172
(c) Loss of RM8 million (b) 24
(d) (i) (c) 16.64 4. 104 cm2
(d) 0.66 5. 182 + 62 = 18.97 m
Profits of Cekap Company
2. (a) 14 cm 6. 82 + 152 = 172
14 (b) 8 cm SR = 17 cm
12 (c) 25.63 cm
10 7. 36 cm2
Profit (million RM)8 Self Practice 13.1d
20106 8. 0.4
20114 1. 54 cm 9. 20 – 152 – 122 = 11 m
20122 2. 35.44 km
2013–20 10. Using the concept of
2014–4  Mastery 13.1 Pythagorean triples, (3, 4, 5)
2015–6
–8 1. 12 cm 7 × 5 = 35 cm
–10 2. 28 cm 7 × 3 = 21 cm
3. 5.62 m 7 × 4 = 28 cm
Year 4. 540.83 km
11. Using the concept of
(ii) Profit of RM11 million Self Practice Pythagorean triples,
7. (a) 12, 15, 15, 16, 17, 18, 21, 3 + 4 + 5 = 12
1. (a) Yes
24, 25, 27, 28, 30, 30, 31, (b) No 13.2a
33, 34, 35, 36, 37, 37, 38, (c) Yes

324

Answers

GLOSSARY

adjacent angles (sudut bersebelahan) Two angles that are next to each other and whose
sum is 180°.

bar chart (carta palang) A type of data representation that uses bars to represent data.
common factor (faktor sepunya) A number that is a factor of a few other numbers.
complement of a set (pelengkap bagi suatu set) The elements in the universal set that are

not the elements of the set.
complementary angles (sudut pelengkap) Two angles with a sum of 90°.
congruent angles (sudut kongruen) Angles having the same size.
congruent line segments (tembereng garis kongruen) Line segments having the

same length.
conjugate angles (sudut konjugat) Two angles with a sum of 360°.
element (unsur) Each object in a set.
empty set (set kosong) A set that contains no elements.
equal sets (set sama) Sets in which every element of the sets are the same.
factor (faktor) A number that divides another number completely.
geometrical construction (pembinaan geometri) Method of using geometric tools or

geometry software to make drawings of accurate measurements.
hypotenuse (hipotenus) The longest side of a right-angled triangle which is opposite to

the right angle.
inequality (ketaksamaan) A relationship between two quantities of different values.
integer (integer) Positive and negative whole numbers including zero.
like terms (sebutan serupa) Terms that have the same variables raised to the same power.
line graph (graf garis) A type of data representation that displays changes of data over

a period of time.
linear inequality in one variable (ketaksamaan linear dalam satu pemboleh ubah)

An unequal relationship between a number and a variable to the power of one.

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negative number (nombor negatif ) A number having a value less than zero which is
written with a negative sign ‘–’. For example, –8, –45, –200.

pie chart  (carta pai)  A type of data representation that uses sectors of a circle to
represent data.

positive number (nombor positif ) A number having a value more than zero which is
written with a positive sign ‘+’ or without a sign. For example, +3, +56, +100 or 3,
56, 100.

prime factorisation (pemfaktoran perdana) Process of expressing a number as the
product of its prime factors.

proportion (kadaran) A relationship that states that two ratios or two rates are equal.

rate (kadar) A special case of ratio that involves two quantities of different units.

rational number (nombor nisbah) A number that can be written in fractional form p ,
where p and q are integers and q ≠ 0. q

reflex angle (sudut refleks) An angle whose size is more than 180° but less than 360°.

set (set) A group of objects which have the common characteristics and classified in the
same group.

simultaneous linear equations (persamaan linear serentak) Two linear equations which
have the same two variables.

statistics (statistik) The branch of mathematics that involves collecting, organising,
recording, representing, interpreting and analysing data.

subset of a set (subset bagi suatu set) A set whereby all of its elements are the elements
of another set.

supplementary angles (sudut penggenap) Two angles with a sum of 180°.

transversal (garis rentas lintang) A line that intersects two or more straight lines.

universal set (set semesta) A set that consists of all the elements under discussion.

unlike terms (sebutan tidak serupa) Terms that do not have the same variables raised to
the same power.

Venn diagram (gambar rajah Venn) A geometrical diagram used to represent sets.

vertically opposite angles (sudut bertentang bucu) Two angles opposite to each other at
an intersection of two lines and are equal in size.

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REFERENCES

Amanda Bearne, Sharon Bolger, Ian Boote, Greg Byrd, Meryl Carter, Gareth Cole,
Crawford Craig, Jackie Fairchild, Anna Grayson, June Hall, Mark Haslam, Fiona
Mapp, Phil Marshall, Avnee Morjaria, Keith Pledger, Robert Ward-Penny, Angela
Wheeler, 2008. Levels 3-5 Level Up Maths. England: Heinemann.

Chow Wai Keung, 2014. Discovering Mathematics 2A Normal (Academic) (2nd Edition).
Singapore: Star Publishing Pte Ltd.

Deborah Barton, 2012. Cambridge Checkpoint And Beyond Complete Mathematics for
Cambridge Secondary 1. Oxford: Oxford University Press.

Dr Joseph Yeo, Teh Keng Seng, Loh Cheng Yee, Ivy Chow, Neo Chai Meng, Jacinth
Liew, Ong Chan Hong, Jeffrey Phua, 2014. New Syllabus Mathematics Normal
(Academic). Singapore: Shinglee Publishers Pte Ltd.

Greg , Lynn Byrd, 2008. Levels 3-5 Level Up Maths Homework Book. England: Heinemann.
Istilah Matematik untuk Sekolah-sekolah Malaysia, 2003. Kuala Lumpur: Dewan Bahasa

dan Pustaka.
Kamus Dewan Edisi Keempat, 2007. Kuala Lumpur: Dewan Bahasa dan Pustaka.
M.J. Tipler, K.M. Vickers, 2002. New National Framework Mathematics 7+. United

Kingdom: Nelson Thornes Ltd.
Nicholas Goldberg, Neva Cameron-Edwards, 2010. Oxford Mathematics for the Caribbean

Fifth Edition. Oxford: Oxford University Press.
Peter Derych, Kevin Evans, Keith Gordon, Michael Kent, Trevor Senior, Brian Speed,

2014. Maths Frameworking 3rd edition Homework Book 3. London: HarperCollins
Publishers Limited.
Ray Allan, Nina Patel, Martin Williams, 2007. 7A Maths Links. Oxford: Oxford University
Press.
Sandra Burns, Shaun Procter-Green, Margaret Thornton, Tony Fisher, June Haighton,
Anne Haworth, Gill Hewlett, Andrew Manning, Ginette McManus, Howard Prior,
David Pritchard, Dave Ridgway, Paul Winters, 2010. AQA Mathematics Unit 3
Foundation. United Kingdom: Nelson Thornes Ltd.
Tay Choon Hung, Mark Riddington, Martin Grier, 2007. New Mathematics Counts
Secondary 1 Normal (Academic) 2nd Edition. Singapore: Marshall Cavendish
Education.

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INDEX

acute angle 207, 300 equivalent ratio 77 Pythagoras’ theorem 295,
additive inverse 156 exterior angle 209, 215 296
adjacent angles 186 factor 32
algebraic expression 108, 113 fraction 14 Pythagorean triples 297
algebraic term 110 frequency polygon 279 quadrilateral 212
alternate angles 190 frequency table 267 rate 81, 82
angle bisector 182 highest common factor 36 ratio 76
angle of depression 193 histogram 279 rational number 23
angle of elevation 193 hypotenuse 294 rectangle 213
area 231, 239, 240 Identity Law 11 reflex angle 174
Associative Law 11 integer 3 – 10 rhombus 213
backtracking 127, 128 interior angle 209, 215 right angle 207, 300
bar chart 268, 269 intersecting lines 185, 186 set 248
categorical data 267 kite 213, 235 set builder notation 249
coefficient 110 like terms 111 set notation 249
common factor 35 line graph 271, 272 simultaneous linear equations
common multiple 38 line segment 170, 176
Commutative Law 11 linear equation 125 138
complement of a set 254, 255 linear inequality in one
complementary angles 175 simultaneous linear
congruency 170 variable 160 inequalities 162
conjecture 208, 238 lowest common multiple 40
conjugate angles 175 multiple 38 square 49, 51, 213
converse of Pythagoras’ multiplicative inverse 157 square root 51, 56
numerical data 267 statistical question 266
theorem 300 numerical value 127 stem-and-leaf plot 273, 274
converse property 154 obtuse angle 207, 300 subset 256 – 258
corresponding angles 190 parallel lines 180, 188 substitution 140
cube 58, 61–63 parallelogram 213 – 215, 234 supplementary angles 175
cube root 61, 62 percentage 94 term 110
decimal 19 perfect cube 59, 60 transitive property 154
description 249 perfect square 49, 50 transversal 188
diagonal 202, 203 perimeter 226, 227, 239, 240 trapezium 213, 235
Distributive Law 11 perpendicular bisector 177 trial and improvement 127
dot plot 272, 273 perpendicular line 178 triangle 205, 233
element 249 – 251 pie chart 270, 271 universal set 254, 255
prime factor 33 unlike terms 111
elimination 140 prime factorisation 33, 36, variable 106
empty set 250 Venn diagram 255, 258, 259
equal sets 252 41, 50, 60 vertically opposite angles
equality 127, 128, 252 proportion 84, 85, 90, 95
186

vertices 202, 203

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