Format 190mm X 260mm Extent : 136pg (6.94mm) (All 2C/ 60gsm) Status CRC Date 28/2
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CC033233
FORM
3 KSSM FOCUS
Mathematics
FOCUS KSSM Form 3 – a complete and precise series of reference books with Mathematics FORM
special features to enhance students’ learning as a whole. This series covers the 3 KSSM
latest Kurikulum Standard Sekolah Menengah (KSSM) and integrates Ujian Akhir
Dual Language Programme
Sesi Akademik (UASA) requirements. A great resource for every student indeed! Mathematics
REVISION REINFORCEMENT EXTRA • Ng Seng How
› Comprehensive Notes & ASSESSMENT FEATURES • Ooi Soo Huat
• Yong Kuan Yeoh
› Example and Solution › Formative Practices › Maths Info • Samantha Neo
› Tips › Summative Practices › HOTS Challenge
› Common Mistakes
› UPSA Model Paper › Daily Application
› UASA Model Paper › TIMSS Challenge
› Answers › Digital Resources QR Code
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DIGITAL RESOURCES
› › Full Solutions for Summative Practice,
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UPSA & UASA Model Papers
› › Info › › Video › › Infographic
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CC033233
ISBN: 978-629-7537-77-1
Fulfil UASA
Assessment Format
PELANGI
Format: 190mm X 260mm TP Focus Tg123 2023 Maths F3_pgi CRC
Mathematics FORM
Dual Language Programme 3 KSSM
• Ng Seng How
• Ooi Soo Huat
• Yong Kuan Yeoh
• Samantha Neo
Format: 190mm X 260mm TP Focus Tg123 2023 Maths F3_pgii CRC
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Mathematics PT3 Mathematical Formulae
Exclusive Features of this Book
Learning Area: Number and Operations
Infographic
visually highlights Chapter 1 Indices Mathematics Form 3 Chapter 7 Plans and Elevations
4
the key concepts The diagram on the right shows a solid with E
a rectangular base ABCD. Rectangle HIJL is a
horizontal plane. Edges AF, DE, BI, CM, HG and LK
of each chapter to KEYWORDS are vertical and HG = LK. Draw to full scale, F 2 cm G K M
(a) the plan of the solid,
as viewed from X,
(b) the elevation on a vertical plane parallel to AB
enhance students’ • Index (c) the elevation on a vertical plane parallel to BC 5 cm A 1 cm D H 3 cm L I J 3 cm C
• Base
• Repeated
as viewed from Y.
multiplication
learning. • Index notation Solution: X 7 cm B 3 cm Y Example
• Law of indices
• Fractional index
• Exponent
• Power
Side elevation provides
(c)
Front elevation
(b)
F 3 cm E F/E solution
Chapter Introduction INFOGRAPHIC 2 cm 1 cm G K G/K for example of
Access to
contains picture and I/H 2 cm J/L M 1 cm H/L questions in the
question to stimulate 3 cm 5 cm 3 cm I/J/M subtopics.
interest and thinking B/A C/D A/D 3 cm
about the chapter’s 45° E/D 7 cm 7 cm M/C B/C
will become a molecule that is identical to its original. These molecules will keep splitting and
each splitting causes the number of DNA molecules doubled and this situation can be written
content. A DNA molecule can split in half to form a copy of itself. Each of the copies 7 CHAPTER K/L J 1 cm
How can we represent this multiplication in the simplest form?
as 2 × 2 × 2 × 2 × 2 × ... .
1 2 cm
F/A
G/H
3 cm
I/B
Plan
(a)
Mathematics Form 3 Chapter 6 Angles and Tangents of Circles
Mathematics Form 3 Chapter 6 Angles and Tangents of Circles Try questions 3 – 5 in Formative Practice 7.2
2. Find the values of x and y in each of the Section A 6 Full
Summative Practice
solution
HOTS Challenge HOTS Challenge E D following diagrams. (b) x y 82 1. (b) The diagram below shows a cyclic quadrilateral
Praktis Formatif 1
(a)
inscribed in a circle with centre O.
provide questions A 71º 103º C 68º 110º y x (d) 74º 102º x O 74° p s q O t
that stimulate The diagram above shows a pentagon ABCDE (c) 2y 94º 4x 63º 3y 2x A 37° C 45° Mark (3) in the empty box for the correct
Find the value of x.
The diagram above shows a circle with centre O.
r
B
inscribed in a circle. Given BC = CD, find ∠ADC.
B 40°
D 53°
relationship.
student’s higher Solution: 3. Find the values of m and n in each of the 2. S R (i) p = 2q [2 marks] Summative Practice
relationship and (7) for the incorrect
3x
∠ABD = 180° – 103°
= 77°
(ii) p + t = 180°
order thinking ∠ADB = 180° – 71° – 77° following diagrams. (b) m 117º In the diagram above, PQRS is a cyclic Section C provides ample
m
88°
114°
P
Q
(a)
= 32°
m
skills. ∠BCD = 180° – 71° 2 120º 98º n n 80º 3. A 114° D 92° 1. (a) In diagram below, KLM is tangent to the circle questions to test
quadrilateral. Determine the value of m.
= 109°
at L.
C 66°
B 88°
∠BDC = 180° – 109°
b
= 71° 2 (c) (d) 116º 80° a 20º c d students’ mastery
= 35.5° 96º 51º 33º
n – 10º K 32º
∠ADC = ∠ADB + ∠BDC 2m 3n m + 30º x L M
= 32° + 35.5° 114º 71º CHAPTER The diagram above shows a circle. Find the value Complete the following empty boxes. of the chapter.
= 67.5° of x. (i) a = [4 marks]
A 80°
Try questions 4 – 9 in Formative Practice 6.2 6 CHAPTER B 140° C 120° (ii) b =
4. E D 25º C Section B D 100° (iii) c =
Formative Practice 6.2 95º B 1. (a) 6 (iv) d =
1. Explain whether the quadrilateral in each of the A a b c d (b) A
following diagrams is a cyclic quadrilateral.
Formative Practice (a) (b) In the diagram above, ACE is a triangle and f e B
ABDE is a cyclic quadrilateral. Find ∠AEC.
provides questions 5. 96º A x B angles which are equal in size. d [2 marks] parallelogram. E D C
Based on the diagram above, match the
a
In the diagram above, ABCE is a
to test students’ (c) (d) D 114º 68º C b e f (i) Show that AD = AE. [2 marks]
[1 mark]
(ii) If ∠DAE = 16°, find ∠ABC.
understanding of The diagram above shows a cyclic quadrilateral 74 Analysing
ABCD. Find the value of x.
the subtopics and 65
reinforce learning. Mathematics Form 3 Chapter 5 Trigonometric Ratios Mathematics Form 3 Answers ANSWERS Answers
(b) PQ is the opposite side. Chapter 1 Indices 8. (a) 9 –4
Sine, Cosine and Tangent of QR is the adjacent side. (b) 14 –8 Section B help students
(c) q –12
5.1 Acute Angles in Right-angled Try questions 1 and 2 in Formative Practice 5.1 1. (a) 5 7 9. (a) (i) 5 1. (a) h –6 ; h 4 × h 2 ;
Formative Practice 1.1
(d) (2y) –7
(b) (–3) 4
(b) a 8 ÷ a 2 ; (a 2 ) 3
Triangles Define trigonometric ratios (c) (–1.2) 5 (d) 1 4 2 9 3 (ii) 10 2 p 3 2. (a) (i) 2 5 (ii) 0 to check and
(iii) 1 q 2
5.1.2
5.1.1 Identify the opposite side and 1. The ratio between two sides of a right- 2. (a) 3 6 10 4 (b) (–2) 5 (b) (i) (ii) 1 8 2 1 6 –3 –1 Section C (b) 12 5 evaluate their
adjacent side
(c) 1 6 2
ratio. This ratio remains constant if the
1. In a right-angled triangle, the longest side angled triangle is known as trigonometric 3. (a) 243 10. (a) 4 (iii) 1 y 2 x 3 –6 1. (a) (i) p 6 (ii) x 7
(d) 0.5 4
size of the angle in the triangle changes
which is opposite to the right angle is
(b) –1
(c) 3
(b) 4 096
(c) 125
2. (a) m 3 n 4
proportionally.
216
(b) 16
Tips called hypotenuse. Hypotenuse 2. For an acute angle, A, in a right-angled 5. –2 (d) –0.0000128 11. (a) 25 2 4 15 h 1 1 Chapter (b) 108(3 n ) 2 5 performance.
(c) x = 3 , y = 4
(c)
4. a = 9, n = 4
Opposite
triangle ABC:
side
2
A
6. 5
θ
(b) 64 6
Adjacent side
points out 2. The side which is opposite to the angle q Adjacent Hypotenuse C 28a 8 TIMSS 12. (a) 81 5 or ( 81 ) 5 4 1. 3 significant figures. The length of the
Standard Form
1
(c) k 8
4
Formative Practice 2.1
side
is called opposite side and the side which
Mathematics Form 3 Chapter 5 Trigonometric Ratios
(b) 3 125 2 or ( 3 125) 2
Opposite
B
important tips for is next to the angle q is called adjacent Ratio of the opposite side to the 1. (a) 3 11 Determine the values of (c) 7 x 3 or ( 7 x ) 3 2. (a) 3 significant figures
Formative Practice 1.2
rod can be read as 5.21 or 5.22. The
side
2
(d) y 9 or ( y ) 9
5.1.4
third digit in both readings is uncertain.
(b) 18 7
side.
2
(c) 5 15 trigonometric ratios of acute
13. (a) 27 3 5
(d) k 22 angles
(b) 0.04 2
(b) 4 significant figures
students to take For a right-angled triangle, hypotenuse always hypotenuse is named as sine. x y CHAPTER 3. (a) 7 2 (b) m 10 n 6 14. (a) 16 3 7 (c) 5 significant figures
2. (a) 12x 6
TIPS
z
(d) 2 significant figures
(c) p 5
sin A = opposite side
(e) 1 significant figure
1. The values of sine, cosine or tangent of an
(d) q 2
(c) 28a 10 b 5
remains the same whereas the opposite side
(f) 6 significant figures
(d) p 7 q 15 r 14
hypotenuse θ
acute angle in a right-angled triangle can
(g) 4 significant figures
(b) 243
State the following trigonometric ratios based
(h) 6 significant figures
position of the given angle.
of the triangle are given.
(c) 125
(b) 29 8
note of. and the adjacent side change according to the Ratio of the adjacent side to the 5 4. (a) 6h 6 be determined when the lengths of sides (b) k 5 3. (a) 90
(c) 12 5
on the length of sides of the right-angled
hypotenuse is named as cosine.
(d) 128
(b) 40 000
(d) y 9
4
triangle above.
(c) 40 000
15. (a) 6 4
(d) 430
(a) sin q
(c) 23 4
(b) a 3 b 14
hypotenuse
(c) k 7 h 2
16. (a) 27
MATHS INFO INFO cos A = adjacent side Determine the value of trigonometric ratio (d) x 14 (e) 4 100
(b) cos q
(f) 98 000
(d) 4x 8 y 6
(b) 2
of the acute angle given in each of the right-
Besides alphabetical letters, the angle labelled in Ratio of the opposite side to the adjacent 5. (a) 9 18 17. (a) 64p 7 q 2 4. (a) 4 (h) 181 000
(c) tan q
(g) 7 050
angled triangles below.
(c) 1
a triangle can be indicated by using Greek letters
3
Solution:
(i) 745 600
(b) 0.4 40
(a)
(d) 2
(c) d 14 sin x
side is named as tangent.
such as q (theta), a (alpha) and b (beta).
(d) y 12
(a) y
(b) 20
(c) 20
(b) p 8 r 36 3 cm
(b) x
tan A = z z opposite side 6. (a) 64x 15 4 cm x (b) 4 (c) a 7 t 2 b 4 (f) 410
adjacent side
(d) 5.8
(e) 8.0
(c) 3 8
1
y 14
Identify the opposite side and the adjacent side (c) y x 7. (a) 1 s 21 tan y 5 cm 18. x = 3, y = 6 (g) 1.94
(d) x 12
(d) p 7 q 14
with respect to the shaded angle for each of the
(b)
y 15
(h) 96.0
(i) 770.0
4 5
right-angled triangles below. MATHS INFO (b) 50 2 6 cm y 10 cm 19. 36 hours 27 minutes 5. (a) 0.003 Digital Resources
1
INFO Try question 3 in Formative Practice 5.1
20. 2 17
(b)
(b) 0.2
(a) C Q Abbreviation for sine, cosine and tangant are sin, (c) 1 p INFO (c) 0.9
cos and tan respectively. (d) 1 x 6 8 cm Section A q = sin q 1 (e) 0.035 help learners
(d) 0.00037
The impact of changing the size
Summative Practice
5.1.3
(f) 0.87
tan
Maths Info A B P R CHAPTER of the angles on the values of 120 (c) cos z 13 cm 5 cm 1. B 2. A cos q 4. D (h) 0.0830 better comprehend
(g) 0.0479
trigonometric ratios
3. C
Mnemonic for memorizing
(i) 0.01005
trigonometric ratios
(a) BC is the opposite side.
consists of Solution: 5 Value INFO When the size When the size Solution: z 12 cm concepts and
of the angle
of the angle
AB is the adjacent side.
increases
decreases
49
decreases
sine
increases
extra info and cosine decreases increases (a) sin x = 3 5 opposite side deepen learning.
hypotenuse
decreases
tangent
increases
knowledge related Each of the following shows the sine, cosine Opposite side Hypotenuse 4 cm x *Video/Info
3
3 cm
5 cm
to the chapter. and tangent of an angle. Arrange the values of (b) tan y = 8 opposite side
the sine, cosine and tangent in ascending order.
(a) sin 62°, sin 29°, sin 90°, sin 34°
adjacent side
(b) cos 44°, cos 76°, cos 6°, cos 15°
(c) tan 26°, tan 0°, tan 85°, tan 57° = 4 6
Solution: 3
(a) sin 29°, sin 34°, sin 62°, sin 90° Adjacent side y 10 cm
(b) cos 76°, cos 44°, cos 15°, cos 6° 6 cm
(c) tan 0°, tan 26°, tan 57°, tan 85° 8 cm
Opposite side
Try questions 4 and 5 in Formative Practice 5.1
50
iii
F3 Exclusive Features.indd 3 28/02/2023 12:20 PM
Mathematical Formulae
NUMBER AND OPERATIONS
a × a = a m + n
n
m
a ÷ a = a m – n
m
n
(a ) = a m × n
m n
Simple interest, I = Prt
Total savings = P(1 + rt)
r
Matured value, MV = P(1 + n )
nt
Value of return of investment, ROI = Total return × 100%
Value of initial investment
Investment amount
Average cost =
Number of share units owned
Total repayment, A = P + Prt
MEASUREMENT AND GEOMETRY
measurement of the drawing
Scale of a drawing =
measurement of the actual object
sin q = opposite side
hypotenuse
adjacent side Opposite Hypotenuse
cos q = side
hypotenuse
θ
opposite side Adjacent side
tan q =
adjacent side
iv
F3 Formulae Mathematics.indd 4 28/02/2023 11:59 AM
CONTENTS
Mathematical Formulae iv Measurement and
LEARNING AREA
Geometry
LEARNING AREA Number and Operations
Chapter
4 Scale Drawings 37
Chapter
1 Indices 1 4.1 Scale Drawings 38
Summative Practice 4 46
1.1 Index Notation 2
1.2 Laws of Indices 3
Chapter
Summative Practice 1 9 5 Trigonometric Ratios 48
Chapter 5.1 Sine, Cosine and Tangent of Acute
2 Standard Form 10 Angles in Right-angled Triangles 49
Summative Practice 5 57
2.1 Significant Figures 11
2.2 Standard Form 14 Chapter
Summative Practice 2 18 6 Angles and Tangents of 59
Circles
6.1 Angle at the Circumference and
Chapter Consumer Mathematics: Central Angle Subtended by
3 Savings and Investments, an Arc 60
Credit and Debt
20
6.2 Cyclic Quadrilaterals 63
3.1 Savings and Investments 21
6.3 Tangents to Circles 66
3.2 Credit and Debt Management 28
6.4 Angles and Tangents of Circles 72
Summative Practice 3 35
Summative Practice 6 74
v
0b F3 Content.indd 5 28/02/2023 12:00 PM
Chapter LEARNING AREA Relationship and Algebra
7 Plans and Elevations 76
7.1 Orthogonal Projections 77 Chapter
9 Straight Lines 106
7.2 Plans and Elevations 80
Summative Practice 7 91
9.1 Straight Lines 107
Summative Practice 9 117
Chapter
8 Loci in Two Dimensions 95 Answers 120
8.1 Loci 96
8.2 Loci in Two Dimensions 97
Summative Practice 8 103
UPSA Model Paper
(Ujian Pertengahan Sesi Akademik)
https://qr.pelangibooks.com/?u=FocusMatF3UPSA
UASA Model Paper
(Ujian Akhir Sesi Akademik)
https://qr.pelangibooks.com/?u=FocusMatF3UASA
Answers for UPSA and UASA Model Papers
https://qr.pelangibooks.com/?u=FocusMatF3AnsUjian
vi
0b F3 Content.indd 6 28/02/2023 12:00 PM
Learning Area: Number and Operations
Chapter
1 Indices
KEYWORDS
• Index
• Base
• Repeated
multiplication
• Index notation
• Law of indices
• Fractional index
• Exponent
• Power
Access to
INFOGRAPHIC
A DNA molecule can split in half to form a copy of itself. Each of the copies
will become a molecule that is identical to its original. These molecules will keep splitting and
each splitting causes the number of DNA molecules doubled and this situation can be written
as 2 × 2 × 2 × 2 × 2 × ... .
How can we represent this multiplication in the simplest form?
1
F3 Chapt 1.indd 1 28/02/2023 12:05 PM
Mathematics Form 3 Chapter 1 Indices
1.1 Index Notation 1.1.2 Rewrite a number in index form
and vice versa
1.1.1 Represent repeated 1. We can express a number in index form by
multiplication in index form the following steps:
1. The repeated multiplication of a number Write the number in Write the product
a can be written in index form by using the form of repeated using base and
the index notation, a , as shown in the multiplication. index.
n
CHAPTER
following.
1
For example, 32 = 2 × 2 × 2 × 2 × 2 = 2 5
a × a × a × ... × a = a n Index
2 as the base that is
Repeated multiplication multiplied by 5 times.
of a by n times Base
n
In index form, a , a is known as the base 2. We can find the value of a number in index
and n is power that is known as index. n form by the following steps.
also is known as exponent.
Write the number in Find the
the form of repeated product.
multiplication.
MATHS INFO
INFO
For example, 7 = 7 × 7 × 7 × 7 = 2 401
4
• Index notation a means the value of a is 7 is multiplied by 4 times
n
multiplied repeatedly by n times.
• a is read as a raised to the nth power. TIPS
n
Using scientific calculator, press
1 7 4 =
^
Express each of the following repeated
multiplication in index form.
(a) 6 × 6 × 6 × 6 2
(b) 0.8 × 0.8 × 0.8 × 0.8 × 0.8 × 0.8 Express each of the following numbers in index
1
× –
(c) – 2 2 1 2 2 1 2 2 form.
× –
7
7
7
(a) 81 (b) –100 000 (c) 64
Solution: 125
Solution:
(a) 6 × 6 × 6 × 6 = 6 4
(a) 81 = 9 × 9 or 81 = 3 × 3 × 3 × 3
6
(b) 0.8 × 0.8 × 0.8 × 0.8 × 0.8 × 0.8 = 0.8 2 4
= 9 = 3
1
× –
(c) – 2 2 1 2 2 1 2 2 1 2 2 3 (b) –100 000
× –
= –
7
7
7
7
= (–10) × (–10) × (–10) × (–10) × (–10)
= (–10) 5
64 4 4 4
Explanation for the (c) = × ×
solutions 125 5 5 5
INFO 4 3
= 1 2
5
Try question 1 in Formative Practice 1.1 Try question 2 in Formative Practice 1.1
2
F3 Chapt 1.indd 2 28/02/2023 12:05 PM
Mathematics Form 3 Chapter 1 Indices
3 1.2 Laws of Indices
Find the value of each of the following.
(a) 2 6
(b) (–8) 4 1.2.1 Relate the multiplication of
1
(c) – 3 2 3 numbers in index form with CHAPTER
the same base
7
Solution: 1. Multiplication of numbers in index form
(a) 2 = 2 × 2 × 2 × 2 × 2 × 2 with the same base can be related by 1
6
= 64 repeated multiplication as follows.
4
(b) (–8) = (–8) × (–8) × (–8) × (–8) Repeated multiplication Repeated multiplication
= 4 096 of 9 by 3 times. of 9 by 5 times.
1
1
(c) – 3 2 3 = – 3 2 1 3 2 1 3 2 9 × 9 = (9 × 9 × 9) × (9 × 9 × 9 × 9 × 9) = 9 8
× –
× –
3
5
7
7
7
7
27
= – Total number of repeated multiplications
343
of 9 by 8 times (which is 3 + 5 = 8 times).
TIPS
• (–a) is a negative value if n is an odd number. 2. The product of two numbers in index form
n
• (–a) is a positive value if n is an even number. with the same base can be obtained by
n
adding the indices of both numbers.
n
m
Try questions 3 – 6 in Formative Practice 1.1 a × a = a m + n
Formative Practice 1.1
4
1. Express each of the following repeated
multiplication in index form. Simplify each of the following.
5
6
3
9
(a) 5 × 5 × 5 × 5 × 5 × 5 × 5 (a) 4 × 4 (b) 13 × 13 × 13
2
3
7
(b) (–3) × (–3) × (–3) × (–3) (c) 2p × 5p 4 (d) xy × x y 5
(c) (–1.2) × (–1.2) × (–1.2) × (–1.2) × (–1.2) Solution:
4
(d) × 4 × 4
9 9 9 (a) 4 × 4 = 4 5 + 3 Retain the bases and
3
5
8
2. Express each of the following numbers in index = 4 add the indices.
form. 6 9 6 + 9 + 1 1
(a) 729 (b) –32 (b) 13 × 13 × 13 = 13 16 13 = 13
1 296 = 13
(c) (d) 0.0625
10 000 (c) 2p × 5p = 2 × 5 × p × p 4
7
7
4
3. Find the value of each of the following. = 10 × p 7 + 4
(a) 3 (b) (–4) = 10p 11
6
5
5
(c) 1 2 3 (d) (–0.2) 7 (d) xy × x y = x × x × y × y
2
3 5
5
2
3
6
= x 1 + 3 × y 2 + 5
4. Given that a × a × a × a = 9 , where a and n = x y
n
4 7
are positive integers. State the value of a and
of n.
Try questions 1 and 2 in Formative Practice 1.2
5. Given k = –128, find the value of k.
7
6. The product of the repeated multiplication of TIMSS Challenge
a number by n times is 7 776. Determine the
5
value of n. Simplify: 7a × 4a 3
3
F3 Chapt 1.indd 3 28/02/2023 12:05 PM
Mathematics Form 3 Chapter 1 Indices
1.2.2 Relate the division of numbers 1.2.3 Relate the numbers in index
in index form with the same form raised to a power
base
1. Numbers in index form raised to a power
1. Division of numbers in index form with can be related by repeated multiplication
the same base can be related by repeated as follows.
multiplication as follows.
Repeated multiplication
5 8 of 4 by 3 times.
2
6
5 ÷ 5 =
8
CHAPTER
5 6
1
2
2
2
5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 (4 ) = 4 × 4 × 4
2 3
5 ÷ 5 =
6
8
2 3
5 × 5 × 5 × 5 × 5 × 5 (4 ) = (4 × 4) × (4 × 4) × (4 × 4)
(4 ) = 4 × 4 × 4 × 4 × 4 × 4
2 3
5 ÷ 5 = 5 Indices are subtracted, (4 ) = 4 Indices are multiplied, 2 × 3 = 6
8
2
6
6
2 3
2 3
5 ÷ 5 = 5 8 – 6 8 – 6 = 2 (4 ) = 4 2 × 3
8
6
2. Numbers in index form raised to a power
2. The quotient of two numbers in index form can be simplified by multiplying its indices.
with the same base can be obtained by
subtracting the indices of both numbers. (a ) = a m × n
m n
m
n
a ÷ a = a m – n where m and n are positive integers.
where m and n are positive integers, a ≠ 0
6
5 Simplify each of the following.
7 4
3 5
(a) (6 ) (b) (p )
Simplify each of the following. m 8 2
(c) (3x ) (d) 1 2
2 4
5
30 8 n
(a) 12 ÷ 12 5 (b)
11
30 2 Solution:
(a) (6 ) = 6 7 × 4
7 4
a b 8
7
(c) 48r ÷ 6r 14 (d) = 6
21
28
4
a b
3 5
Solution: (b) (p ) = p 3 × 5
Retain the bases and = p 15
11
(a) 12 ÷ 12 = 12 11 – 5 subtract the indices. Common
5
2 4
4
= 12 6 (c) (3x ) = 3 × x 2 × 4 mistakes
= 81x 8 INFO
30 8
(b) = 30 8 – 2 8 × 2
m
30 2 (d) 1 2 = m
8 2
= 30 6 n 5 n 5 × 2
m 16
48 =
21
14
(c) 48r ÷ 6r = r 21 – 14 n 10
6
= 8r 7
TIPS
a b 8
7
(d) = a 7 – 4 × b 8 – 1 • (a × b ) = a × b qn
q n
pn
p
a b
4
a
p n
= a b • (a ÷ b ) = 1 2 = a pn
3 7
p
q n
b q b qn
Try questions 3 and 4 in Formative Practice 1.2 Try questions 5 and 6 in Formative Practice 1.2
4
F3 Chapt 1.indd 4 28/02/2023 12:05 PM
Mathematics Form 3 Chapter 1 Indices
1.2.4 Verify that a = 1 and Solution:
0
1
a = (a) 1 = 6 –9
–n
a n 6 9
1
1. We can simplify a ÷ a as follows: (b) r 3 = r –3
n
n
1
a n
n
n
n
a ÷ a = a n – n or a ÷ a = a n CHAPTER
n
= a 0 1 TIPS
= 1 1
Therefore, it is verified that • a 1 –n = a n
a
b
• 1 2 –n = 1 2 n
a = 1 where a ≠ 0 b a
0
7
For example, 9 = 1, 1 2 0 = 1, (–0.6) = 1 Try question 8 in Formative Practice 1.2
0
0
8
n
0
2. For a ÷ a , we can simplify as follows: 9
a 0 a = 1
0
0
a ÷ a = a 0 – n or a ÷ a = a n Express
0
n
n
= a –n 2 –1
9
= 1 (a) 1 2 in the form of positive index.
a n
x
Therefore, it is verified that (b) 1 2 4 in the form of negative index.
y
1
–n
a = where a ≠ 0 Solution:
a n
2
–n
1 1 (a) 1 2 –1 = 1 a = a 1 n
2
For example, 3 = , z = 9
–1
–5
3 z 5 9
2
7 = 1 ÷ 9
1 = 9
State each of the following in . 2
a n
–3
(a) 5
(b) 0.7 –6 x 4 x 4
(b) 1 2 =
Solution: y y 4
1 y 4
(a) 5 = = 1 ÷ x 4
–3
5 3
1 y 4
–6
(b) 0.7 = = 1 ÷ 1 2
0.7 6 x
1
Try question 7 in Formative Practice 1.2 =
y
1 2 4
x
8 y –4
= 1 2 1 n = a –n
x
State each of the following in a . a
–n
1 1
(a) (b)
6 9 r 3 Try question 9 in Formative Practice 1.2
5
F3 Chapt 1.indd 5 28/02/2023 12:05 PM
Mathematics Form 3 Chapter 1 Indices
1.2.5 Determine and state the Solution:
relationship between fractional 8 3 — 8 3 —
8
3
3
8
indices and roots and powers (a) 27 = √27 or 27 = ( √27 )
3 — 3 —
2
2
3
3
1. The relationship between the fractional (b) 100 = √100 or 100 = (√100)
indices and the roots can be written as:
Try question 12 in Formative Practice 1.2
1
n —
n
a = √a
CHAPTER
1 — 1 —
1
2
For example, 25 = √25 and 128 = √128 13
7
7
— 1 — 1 m
2
Conversely, √25 = 25 and √128 = 128 7 Express each of the following in the form of a .
7
n
m — —
4
5
2. For the number in fractional index a : (a) √243 (b) (√0.04 ) 5
n
m m × 1 1 — Solution:
• a = a n = (a ) = √a m — 4
n
m n
n
5
5
4
m 1 × m 1 — (a) √243 = 243
n
n m
n
• a = a n = (a ) = ( √a ) m 5
—
(b) (√0.04 ) = 0.04 2
5
10 Try question 13 in Formative Practice 1.2
—
State each of the following in the form of √a.
n
1 1
3
(a) 8 (b) 256 14
8
Solution: Without using calculator, find the value of
1 — 7 3
3
3
(a) 8 = √8 (a) 32 (b) 625 4
5
1 —
8
(b) 256 = √256 Solution:
8
7
Try question 10 in Formative Practice 1.2 5 5 — 7
(a) 32 = ( √32 )
= 2 7
11 =128
1
n
Write each of the following in the form of a . 3 —
4
4
— — (b) 625 = ( √625 ) 3
(a) √121 (b) √729
6
= 5 3
Solution: = 125
— 1
2
(a) √121 = 121 TIPS
— 1
6
(b) √729 = 729 6 Check answer by using scientific calculator.
(a) Press
Try question 11 in Formative Practice 1.2
b
3 2 ^ ( 7 a /c 5 ) =
12 (b) Press
— 6 2 5 ( 3 b 4 ) =
n
State each of the following in the form of √a m ^ a /c
—
or ( √a ) .
m
n
8 3 Try question 14 in Formative Practice 1.2
(a) 27 (b) 100 2
3
6
F3 Chapt 1.indd 6 28/02/2023 12:05 PM
Mathematics Form 3 Chapter 1 Indices
Solution: 1.2.6 Perform operations involving Solution:
8 — 8 — laws of indices 54y 3 54y 3
3
3
3
3
8
(a) 27 = √27 or 27 = ( √27 ) (a) — =
8
6y × ( √y ) 2 3 6
2 6
3
3 — 3 — 15 6y × (y )
(b) 100 = √100 or 100 = (√100) 54y 3
2
2
3
3
Simplify each of the following. = 6y × y 4
Try question 12 in Formative Practice 1.2 7 × (7 ) CHAPTER
2 6
4
(a) 3 × 3 ÷ 3 7 (b) = 54 × y 3 – 1 – 4
5
8
3
7 × 7 8 6
Solution: = 9y –2 1
13
8
(a) 3 × 3 ÷ 3 = 3 5 + 8 – 7 = 9
7
5
2
m = 3 6 y
n
Express each of the following in the form of a .
4 4 × 9
— — 4 2 6 4 2 × 6 (m n ) m 2 × 9 n 3 Use (a × b )
2
3 9
p
q n
(a) √243 (b) (√0.04 ) 5 (b) 7 × (7 ) = 7 × 7 (b) = pn qn
4
5
7
7
7 × 7 8 7 × 7 8 m n 10 m n 10 = a × b
3
3
Solution: = 7 4 + 12 – 3 – 8 = m n 12
18
5
7
— 4 = 7 m n 10
5
m
4
5
(a) √243 = 243 = m 18 – 7 n 12 – 10 Use a ÷ a n
Try question 15 in Formative Practice 1.2 = a m – n
5
— = m n 2
11
(b) (√0.04 ) = 0.04 2
5
Try question 13 in Formative Practice 1.2 16 Try question 17 in Formative Practice 1.2
Find the value of each of the following. 1.2.7 Solve problems
7
—
14 (a) √6 ÷ 36 × 6
2
5
9 3 18
Without using calculator, find the value of (b) (4 × 5 ) ÷ (2 × 5 )
2 2
5
2
4
7 3 Given 8 x – 1 = 32, find the value of x.
5
(a) 32 (b) 625 4 Solution:
— 7 2 5 2 7 2 Solution:
Solution: (a) √6 ÷ 36 × 6 = 6 ÷ 6 × 6 x – 1
5
2
5 7 8 = 32
7 — 2 – 2 + 2 (2 ) = 2 5
3 x – 1
5
5
(a) 32 = ( √32 ) 7 = 6 3x – 3 5
= 2 7 = 6 4 2 = 2 Compare the indices
=128 = 1 296 3x – 3 = 5 on both sides.
9 3 x = 8
5
2 2
2
4
3 — (b) (4 × 5 ) ÷ (2 × 5 ) 3
4
(b) 625 = ( √625 ) 3 9 3 × 2
4
p
q n
4
= 5 3 = (2 2 × 2 × 5 ) ÷ (2 5 × 2 × 5 2 ) Use (a × b ) 19 Daily Application
qn
pn
= 125 = (2 × 5 ) ÷ (2 × 5 ) = a × b
10
4
3
9
= 2 9 – 10 × 5 4 – 3 Use a ÷ a = a m – n A Science quiz consists of two parts. There
n
m
10
TIPS = 2 × 5 are 2 ways to answer 10 true-false questions
–1
8
in part A and 2 ways to answer 8 true-false
5
Check answer by using scientific calculator. = 2 questions in part B. How many ways to answer
(a) Press all 18 questions in the quiz?
b
3 2 ^ ( 7 a /c 5 ) = Try question 16 in Formative Practice 1.2 Solution:
Number of ways to answer all questions
(b) Press 17 = 2 × 2 8
10
b
6 2 5 ^ ( 3 a /c 4 ) = Simplify each of the following. = 2 10 + 8
4 = 2 18
54y 3 (m n )
2
3 9
Try question 14 in Formative Practice 1.2 (a) — (b)
6y × ( √y ) m n 10 Try questions 18 – 20 in Formative Practice 1.2
7
2 6
3
7
F3 Chapt 1.indd 7 28/02/2023 12:05 PM
Mathematics Form 3 Chapter 1 Indices
Formative Practice 1.2 m n
13. State each of the following in the form of a .
1. Simplify each of the following. (a) √27 (b) √0.04 5
—
3 —
2
(a) 3 × 3 (b) 18 × 18 2 5 — 3 — 7
7
5
4
6
10
(c) 5 × 5 × 5 (d) k × k × k 3 (c) (√p ) (d) (√q )
9
8
2. Simplify each of the following. 14. Without using calculator, calculate the value of
5
2
(a) 6x × 2x (b) m n × m n (a) 64 (b) 81
3
7 5
4
2
4
3
(c) 4ab × 7a b (d) p q × p r × q r 3 7
9 10
2
3
6
5 4
9 2
2
CHAPTER
3. Simplify each of the following. (c) 25 (d) 8 3
(a) 7 ÷ 7 (b) 29 ÷ 29 15. Simplify each of the following.
6
5
13
4
1
12 8
(c) (d) y ÷ y 7 (a) 6 × 6 ÷ 6 (b) k × k ÷ k 9
16
10
4
8
7
5
12 3 5 3 4 7 3
(x )
4. Simplify each of the following. (c) 23 × (23 ) 3 (d) x × x 2
10
5
23 × 23
(a) 18h ÷ 3h (b) a b ÷ ab
10
4 17
3
4
20 7
9 5
(c) k h (d) 128x y 16. Find the value of each of the following.
5
2
k h 32x y (a) 27 × 9 ÷ 3 (b) √8 ÷ 16 × 2 1 2
12
2 3
—
2
3
4
3
5. Simplify each of the following. 2 5
2
8 5
(a) (9 ) (b) (0.4 ) (c) (25 × 27 ) ÷ (5 × 9 )
3 2
3 6
4
(c) (d ) (d) (y ) 2 3 5 3
2 7
6 2
6. Simplify each of the following. (d) 4 × 2
2
(a) (4x ) (b) (p r ) 32 5
2 9 4
5 3
3
2
4
(c) 1 2 2 (d) 1 pq 3 2 7 17. Simplify each of the following. 28t 5
y
s
2
7
3
3 –4
6
4 —
3 8
7. Write each of the following in the form of 1 . (a) p q × (2p q) (b) 7t × (√t )
1
a n (a b ) xy × (x y) 5
3
2 4
3
4
(a) 4 (b) 50 (c) a b (d) (√xy )
—
–5
–2
5 6
6 8
(c) p (d) x –6
–1
4 3
18. Given that (2a ) × 2 x = 32a , where x and y
y
6
8. State each of the following in the form of a . 2a
–n
1 1 are integers. Find the values of x and y.
(a) (b)
9 4 14 8 19. Water in a dam is channelled to a water
3
(c) 1 (d) 1 treatment plant at the rate of 243 m per minute.
q 12 (2y) 7 Find the time taken, in hours and minutes, to
channel 81 m of water to the water treatment
3
3
9. Express plant.
(a) in the form of positive index
p
(i) 1 (ii) 1 (iii) 1 2 –3 20. A type of bacteria reproduces by continuous
5 –1 10 –2 q cell division as shown in the diagram below.
9
(b) in the form of negative index In a certain condition, a number of 2 parent
x
(i) 6 (ii) 3 (iii) 1 2 6 cells have been reproduced at the same time.
3
8 y Find the number of daughter cells if all the
—
10. State each of the following in the form of √a. parent cells undergone 8 times of cell division
n
process.
Applying
1 1 1
4
2
(a) 4 (b) 16 (c) h 15
1
n
11. Write each of the following in the form of a .
38 —
6 —
—
(a) √25 (b) √64 (c) √k
n —
m
12. State each of the following in the form of √a
n —
or (√a) .
m
5 2
4
3
(a) 81 (b) 125
3 9
7
(c) x (d) y 2
8
F3 Chapt 1.indd 8 28/02/2023 12:05 PM
Mathematics Form 3 Chapter 1 Indices
Summative Practice 1 Full
solution
Section A (b)
a × a a ÷ a
8
2
2
3
1. Simplify. a 4 a 12
2 3
5 × (5 ) a –2 a 2 ( a ) CHAPTER
2 7
3
A 5
12
B 5 The diagram above shows the operations 1
17
C 5 21 involving indices. Based on the diagram,
27
D 5 choose and write the operations that give
the answer as shown in the following circle
2. Given √32 = 4. Find the value of m and n. map. i-THINK [2 marks]
m
n
A m = 2, n = 5
B m = 5, n = 2
C m = 3, n = 4 4
D m = 4, n = 3 a a 6
a –2
3. Which of the following values is not equal to
256?
8
A 2
B 4 2. (a) Fill in the blanks with the correct numbers.
4
C 8 —
3
2
D 16 (i) 5 √p = p [1 mark]
2
4. Given 6 × 6 x + 1 = 7 776. Find the value of x. (ii) (–8) = 1 [1 mark]
x
A 5 5 — 5 — 5 — 5 —
y
B 4 (b) Given √8 × √8 × √8 × √8 = 2 , find the value
C 3 of y. [2 marks]
D 2 Section C
Section B 1. (a) Simplify.
(i) (p ) ÷ p × 1 [2 marks]
7
3 5
1. (a) p 2
1 6 — h –6 1 3
√h
2
h 6 x × x 2
1 (ii) x –5 [2 marks]
4
2
h 6 h × h h 6
n
A B (b) Given that 3 × 27 = 81, find the value of n.
(3 )
n 2
The diagram above shows indices. Complete [3 marks]
the equivalent pairs below based on the x n – 1 y n – 5
given example. [2 marks] (c) Given 1 2 = 1 2 , find the value of n.
x
y
Analysing [3 marks]
A B
—
3 2
4
1 2. (a) Simplify √m n × (mn ) . [3 marks]
Example 6 — = h 6 5
√h
mn 2
1 = (b) Simplify 18 n + 1 × 3 1 – n . [3 marks]
h 6 2 n – 1
(c) Given 16(25 ) = 125(32 ), find the values of
y
x
h 6 = x and y. [4 marks]
Analysing
9
F3 Chapt 1.indd 9 28/02/2023 12:05 PM
Learning Area: Measurement and Geometry
Chapter
4 Scale Drawings
4
KEYWORDS CHAPTER
• Grid
• Proportion
• Scale drawing
• Ratio
• Scale
Access to
INFOGRAPHIC
In the automotive industry, the engineers together with the automotive designers produce
drawing showing how the outlook of a new car will be. What information is needed so that
their drawing will illustrate accurately the exact car that is going to be produced? What kind
of mathematical skills needed by an engineer or an automotive designer?
37
F3 Chapt 4.indd 37 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
The diagram shows three drawings of an
4.1 Scale Drawings aeroplane. The table below shows the
measurements of length and width for the three
4.1.1 Investigate and explain the drawings.
relationship between the actual Original Drawing I Drawing II
measurements of an object and
its drawing Length 7.2 10.8 3.6
(cm)
1. 13.8 cm Width
(cm) 3.6 5.4 1.6
4 cm
Based on the information given in the table,
explain the relationship between the original
measurement with the measurements of
Actual height of the car 1 400 mm drawing I and drawing II. Hence, determine
Actual length of the car 4 830 mm whether drawing I and drawing II are the scale
drawings of the original drawing.
Note that
(a) the ratio of the height of the car on Solution:
drawing to the actual height of the car, Drawing I
height on drawing : actual height length of drawing I : original length
= 40 mm : 1 400 mm = 10.8 cm : 7.2 cm
= 1 : 35 2
= 1 :
CHAPTER
(b) the ratio of the length of the car on 3
4
drawing to the actual length of the car, width of drawing I : original width
length on drawing : actual length = 5.4 cm : 3.6 cm
= 138 mm : 4 830 mm 2
= 1 :
= 1 : 35 3
2. It is found that the ratio of the height of Drawing II
the car and the ratio of the length of the length of drawing II : original length
car are the same, that is = 3.6 cm : 7.2 cm
= 1 : 2
height of the length of the car
car on drawing on drawing
= . width of drawing II : original width
actual height actual length = 1.6 cm : 3.6 cm
of the car of the car = 1 : 2.25
The equalisation of these two ratios is a
proportion. For drawing I,
10.8 cm : 7.2 cm = 5.4 cm : 3.6 cm.
3. A drawing which is congruent to the actual
object and follows an equal proportion is Since the two ratios are proportional, therefore
known as scale drawing. drawing I is the scale drawing of the original
drawing.
1 For drawing II,
3.6 cm : 7.2 cm ≠ 1.6 cm : 3.6 cm.
Since the two ratios are not proportional,
therefor drawing II is not the scale drawing of
the original drawing.
Original Drawing I Drawing II
38
F3 Chapt 4.indd 38 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
Alternative Method Scale Meaning
drawing
× 1 1 : 2 Every 2 units of length on the
2
actual object is represented
Original Drawing I Drawing II by 1 unit of length on the
Length 7.2 × 1.5 10.8 3.6 scale drawing.
(cm) Conclusion:
Width The scale drawing produced
(cm) 3.6 × 1.5 5.4 1.6 is smaller than the actual
object, that is 1 times the
4 2
× actual object.
9
Drawing I is the scale drawing of the original 1 : 1 Every 1 unit of length on the
drawing whereas drawing II is not the scale 2 2
drawing of the original drawing. actual object is represented
by 1 unit of length on the
scale drawing.
TIPS
Conclusion:
Scale drawing is the representation of the The scale drawing produced
actual object or the original object which is is bigger than the actual
either enlarged or diminished following a certain object, that is 2 times of the
proportion. CHAPTER
actual object.
1 : 1 Every 1 unit of length on the
Try question 1 in Formative Practice 4.1 4
actual object is represented
by 1 unit of length on the
4.1.2 Interpret the scale of a scale scale drawing.
drawing
Conclusion:
The scale drawing produced
is the same size as the actual
object.
TIPS
Generally, given a scale drawing 1 : n
• If n . 1, the scale drawing produced is
smaller than the actual object.
In the lower part of the map, there is a ratio • If 0 , n , 1, the scale drawing produced is
such as 1 : 25 000. What is the meaning of bigger than the actual object.
1 : 25 000?
1. In scale drawing, the ratio of the length of 2
drawing to the actual length of object is
known as scale drawing. Which of the following scales will produce the
Scale drawing = measurement of drawing largest scale drawing of the object as compared
to the object itself? Explain your answer.
measurement of actual
object 1
2. Scale drawing can be stated in the form of 1 cm : 25 cm 1 cm : 1 m 1 cm : 10 cm
1
1 : n or 1 : such that n = 1, 2, 3, …
n
39
F3 Chapt 4.indd 39 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
Solution:
Praktis Formatif 1
1 cm : 25 cm means the scale drawing produced HOTS Challenge
1
is smaller than the actual object, that is times Scale of a scale drawing is 10 cm : 1 mm. Which
25
the actual object. is bigger, the actual object or the scale drawing?
Give justification for your answer.
1 cm : 1 m = 1 cm : 100 cm Solution:
When defining a scale, convert the units 10 cm : 1 mm
of the measurement to similar units. = 100 mm : 1 mm
= 100 : 1
1 cm : 100 cm means the scale drawing = 1 : 1
produced is smaller than the actual object, that 100
1 Since 0 , 1 , 1, the scale drawing is bigger
is times the actual object. 100
100
than the actual object.
1
1 cm : cm means the scale drawing produced
10
is bigger than the actual object, that is 10 times 4.1.3 Determine the scales,
the actual object. measurements of objects or
measurements of scale drawings
Therefore, the scale that produces the biggest
1
drawing is 1 cm : cm.
10
Try question 2 in Formative Practice 4.1
CHAPTER
4
MATHS INFO
INFO
Scale enables comparison between the actual size
of an object with the size of its drawing. A scale
can be expressed as ratio (for example 1 : 15 000) 3
or fraction (for example 1 ). In a map of Malaysia, the distance between
15 000 Kuala Lumpur and Kangar is 9.6 cm. The scale
of the map is 3 cm : 125 km. What is the actual
distance between the two cities, in km?
HOTS Challenge Solution:
Praktis Formatif 1
Explain how would you know whether a drawing Assuming the actual distance between Kuala
drawn to scale 5 : 2 is bigger or smaller than the Lumpur and Kangar is d km.
actual object. distance on the map
Scale drawing =
Solution: actual distance
2 3 9.6
Scale 5 : 2 = 1 : = 1 : 0.4 =
5 125 d
Since 0 , 0.4 , 1, the scale drawing produced 3d = 125 × 9.6
is bigger than the actual object.
d = 125 × 9.6
3
= 400
Therefore, the actual distance between Kuala
Interpreting scale
Lumpur and Kangar is 400 km.
VIDEO
Try question 3 in Formative Practice 4.1
40
F3 Chapt 4.indd 40 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
4
The actual diameter of a piece of 10 cents
coin is 18.8 mm. Determine the scale used to
produce a diagram of the 10 cents coin which
has a diameter of 4.7 mm.
Solution:
diameter of the coin
on diagram
Scale drawing = Solution:
actual diameter of the
coin Length of the mural, p
1 4.7 = scale drawing
n = 18.8 Length of the picture
p = 1
By using the method of proportion, 15 1
÷ 4.7 20
1
1 4.7 p = 15 × 1
= 20
4 18.8 p = 20 × 15
= 300 cm
÷ 4.7
The scale drawing used is 1 : 4. Width of the mural, l
= scale drawing
Width of the picture
Alternative Method l 1 CHAPTER
10.5 = 1
By using scale ratio, 20 4
Scale drawing 1 l = 10.5 × 1
diameter of the coin actual diameter 20
= :
in the diagram of the coin l = 20 × 10.5
= 4.7 : 18.8 = 210 cm
= 1 : 4 Therefore, the mural painting has length of 3 m
and width of 2.1 m.
The scale drawing used is 1 : 4.
Try questions 7 and 8 in Formative Practice 4.1
Try questions 4 – 6 in Formative Practice 4.1
TIMSS Challenge Common mistakes
The actual length of an antenna is 18 m. INFO
Determine the scale for a drawing which shows
the length of the antenna as 1 cm.
4.1.4 Draw the scale drawings of
objects and vice versa
5
Ramli and his classmates prepared a mural 6
painting on the wall of the school hall by
referring to the picture given. If the scale A rectangular shape swimming pool has
1 dimension of 50 m × 25 m. Draw a scale drawing
drawing used is 1 : , determine the length
20 of the swimming pool on the square grid with
and width, in m, of the mural painting. side of 0.5 cm by using the scale of 1 : 2 500.
41
F3 Chapt 4.indd 41 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
Solution: Step 2: Draw a grid with side of 0.5 cm on a
piece of blank paper.
0.5 cm
Step 3: Copy the drawing of the logo from each
0.5 cm
plot on the grid with side of 1 cm onto
the respective plot on the grid with side
of 0.5 cm.
Step 4: Repeat step 2 until step 3 by using the
grid with side of 1.5 cm.
1 : 2 500 means that the scale drawing
produced is 1 times the actual object. (a)
2 500 0.5 cm
Length of drawing = 1 × 50 m
2 500 0.5 cm
= 0.02 m
= 2 cm
1
Width of drawing = × 25 m
2 500
= 0.01 m
= 1 cm (b) 1.5 cm
Try question 9 in Formative Practice 4.1
1.5 cm
7
CHAPTER
4
Ambok is a member of art club. During one of Try question 10 in Formative Practice 4.1
the weekly meeting, he intended to draw the
logo of the club as shown above, with size that
is
(a) smaller on the square grid with side of 8
0.5 cm,
(b) bigger on the square grid with side of 0.5 cm
1.5 cm. 0.5 cm
Explain briefly on how Ambok do it. Then,
show the drawings made by Ambok.
Solution:
Step 1: Draw a grid with side of 1 cm onto the
logo.
1 cm
The diagram above shows a trapezium redrawn
1 cm by Ghanesh. The drawing was drawn by using
1
a scale of 1 : on the square grid with side
2
of 0.5 cm. Draw the trapezium with actual size
on a piece of blank paper.
42
F3 Chapt 4.indd 42 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
Solution: paint the wall except the hung blackboard. If a
can of paint contains one litre of paint which
1 cm
is able to cover a surface area of 12 m , find
2
the minimum number of cans of paint needed.
1.5 cm
Solution:
Length (m) × Surface
0.5 cm 1 cm 0.5 cm
Height (m) area (m )
2
Try question 11 in Formative Practice 4.1
Measurement of 8 × 2.8 22.4
the wall
4.1.5 Solve problems involving scale Measurement of
drawings the blackboard 5 × 1.5 7.5
9 Area of the wall needed to be painted
= 22.4 – 7.5
= 14.9 m 2
Let the volume of paint needed to paint the
Scale
1 : 20 area of the wall is x litre.
The scale drawing of a square shape photo 12 m 2 = 14.9 m 2
frame is shown in the diagram above. The 1 litre x CHAPTER
scale drawing has side of 5 cm. Find the actual 12 × x = 1 × 14.9
perimeter, in m, of the photo frame. 14.9 4
x =
Solution: 12
Length of drawing = 1.24 litre
= Scale drawing
Length of photo frame The volume of paint needed is 1.24 litre.
Therefore, minimum of 2 cans of paint is
5 1
= needed.
Length of photo frame 20
Length of photo frame = 100 cm 11
= 1 m
Therefore, actual perimeter = 4 × 1
= 4 m
4 cm
10
2.8 cm 1.5 cm Blackboard 6 cm
Wall The diagram above shows the plan of a living
5 cm room with length of 6 cm and width of 4 cm.
8 cm
The actual length of the living room is 7.5 m.
Scale 1 cm : 1 m Determine
(a) the scale drawing of the plan for the living
The diagram above shows the scale drawing of room in the form of 1 : n,
a front wall of a class. Encik Syahir intends to (b) the actual area, in m , of the living room,
2
43
F3 Chapt 4.indd 43 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
(c) the minimum number of tiles measured State the relationship between the measurement
20 cm × 20 cm each, needed to cover the of the original drawing with the measurement
floor of the living room. of drawing A and the measurement of drawing
B. Hence, determine whether drawing A and
Solution: drawing B are the scale drawing of the original
(a) Scale = 6 cm : 7.5 m drawing.
= 6 cm : 750 cm
= 6 : 750 2. Which of the following scales, will produce
= 1 : 125 the smallest scale drawing of a the object as
compared to the object? Explain your answer.
Scale drawing
1
= drawing measurement : actual object measurement 1 cm : 4 cm 1 cm : 0.5 m 1 cm : cm
4
(b) 1 : 125 means 1 cm on the drawing
represents 125 cm on the actual object, that 3.
is the actual measurement of the object is
125 times the measurement of the drawing.
Actual width of the living room = 125 × 4
= 500 cm
= 5 m
Actual area of the living room = 7.5 × 5 The diagram above shows the scale drawing
= 37.5 m of a congkak which is drawn to the scale of
2
1 : 14. Determine the original length and width
(c) Area of a piece of tile = (20 × 20) cm of the congkak, in cm.
2
CHAPTER
= (0.2 × 0.2) m
2
4
= 0.04 m 4.
2
37.5
Number of tiles needed = Kota Kinabalu Sandakan
0.04
= 937.5 Sabah
Therefore, the minimum number of tiles
needed is 938 pieces.
The flight distance from Kota Kinabalu to
Try questions 12 and 13 in Formative Practice 4.1 Sandakan is 228 km. This distance on the map
is 7.6 cm. Determine the map scale in the form
of 1 : n.
Formative Practice 4.1
5. Hasnah plans to explore by cycling a distance
1. The diagram below shows three letters I drawn of 45 km. The distance is 30 cm on the map.
on square grid with side of 1 unit. Determine the scale of the map.
6. Virus is smaller than a bacterium. A certain
virus has a diameter as small as 0.0001 mm.
A drawing of this virus drawn by a scientist
shows a diameter of 5 mm. Determine the scale
drawing used.
Original
7. The length of a mackerel is 73.8 cm. Calculate
the length of the fish on a drawing which is
drawn to a scale of 1 : 9, in cm.
A
B
44
F3 Chapt 4.indd 44 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
8. The scale of a map is 1 : 50 000. The runway 11. 0.5 cm
of an airport on the map is represented by a
straight line. If the actual length of the runway 0.5 cm
is 4 km, determine the length, in cm, of the
runway on the map.
9. 3 m
Aiman drew a scale drawing of a plate as
2.5 m shown above. His drawing was drawn using a
2 m scale 1 : 6 on square grid with side of 0.5 cm.
Draw the plate with actual measurement on a
5 m piece of blank paper.
Diagram above shows the sketch of a bedroom. 12.
Draw a scale drawing of the bedroom on a 0.5 cm
square grid with side of 1 cm by using a scale
of 1 : 50. 1 cm
2 cm
10. Scale 1 : 100
B
The scale drawing of a cuboid shape tank is
shown in the diagram above. Find the actual
A C
3
E volume, in m , of the tank. CHAPTER
13. 4 cm 4
1 cm Living room
Second
bedroom
2 cm Kitchen Master
D Bathroom bedroom 1.5 cm
2.7 cm
The diagram above shows a replica of a kite,
ABCD is drawn on a square grid with side of The diagram above shows the plan of a two
0.5 cm. AC and BD are two bamboo sticks bedroom flat. In the plan, the length of the flat
which are perpendicular to each other tied with is 4 cm and the length of the master bedroom
a wire at the intersecting point E. The points of is 1.5 cm. Given the actual length of the flat
intersection A, B, C and D are also tied with is 8 m. Determine
wires to form the kite. Draw kite ABCD with (a) the scale drawing in the form of 1 : n for
the size that is the plan of the flat,
(a) smaller than the replica on square grid with (b) the actual length, in m, of the master
side of 0.4 cm, bedroom,
(b) bigger than the replica on square grid with (c) the actual area, in m , of the kitchen of the
2
side of 2 cm. flat.
45
F3 Chapt 4.indd 45 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
Summative Practice 4 Full
solution
Section A 2. (a) The diagram below shows a circle with radius
1. of 0.6 cm.
P A B C D
The circle is redrawn by using a scale of
1 : n. For each of the following diagrams,
The diagram above is drawn on a square grid. write the scale used. [2 marks]
Which of the following shape A, B, C and D is
the scale drawing of object P? Scale
Drawing Radius
2. Given that the scale drawing 1 : n, if n . 1, then 1 : n
the size of the scale drawing is (i)
A bigger than the object.
B smaller than the object. 0.3 cm
C the same size as the object
D twice the size of the object.
(ii)
3. Which of the following objects does not require
a scale when drawn on a piece of paper?
A Map C House
CHAPTER
B Elephant D Apple 1.2 cm
4
4. Surjit draws a map of a river that is 10 cm long.
The actual length of the river is 2 km. Find the
scale used by Surjit.
A 1 : 2 000 C 1 : 200
B 1 : 20 D 1 : 20 000
(b) Match. [2 marks]
Section B
1. (a) A scale drawing is a drawing The size of the scale Scale
with the actual object and is drawn according drawing compare to
to a . [2 marks] actual size 1 : n
(b) The diagrams below are drawn on square Smaller n , 1
grids of equal size.
Bigger n . 1
Section C
P
1. (a) (i) The diameter of a Rafflesia on a scale
III
drawing is 9.5 cm. If the actual diameter of
the Rafflesia is 0.76 cm, find the scale use.
I II [2 marks]
(ii) The actual distance between recreational
park X and recreational park Y is 8 km.
What is the distance on the map, in cm,
between both recreational park if the
Mark (3) for the diagram that represents the scale used is 1 : 20 000? [2 marks]
scale drawing of P. [2 marks]
46
F3 Chapt 4.indd 46 28/02/2023 12:11 PM
Mathematics Form 3 Chapter 4 Scale Drawings
(b) The diagram below shows polygon S which 2. (a) Qʹ
is drawn on square grid with side of 0.5 cm.
Q
S P R Pʹ Rʹ
The diagram above shows object PQR and
the scale drawing P’Q’R’ drawn on a square
grid. State the scale used in the form of
1 : n. [2 marks]
On the square grid with side of 0.5 cm, (b) The diagram below shows a polygon
redraw polygon S with a scale of 1 : 1 . ABCDEF.
3
[2 marks] A 48 m B
(c) The diagram below shows the plan of two
rooms with rectangular shape floor. The plan 24 m D C
is drawn using a scale of 1 : 200. 12 m
F E
If the polygon is redrawn by using a scale of
2 cm 1 : 600, calculate, in cm, the length of AB
and AF. [3 marks] CHAPTER
(c) In conjunction with the science and maths
1.3 cm week, Shaiena plans to build a model of the 4
telecommunication tower. The actual height
4 cm 2 cm of the tower is 320 m. The model built by
Shaiena is to be loaded at the hallway of the
The cost of covering the floor with tiles is entrance of the hall which has a height of
RM20 for every meter square. Calculate the 3.8 m from the floor to the ceiling. If Shaiena
cost of covering the floor with tiles, in RM, uses a scale of 1 : 80 to build the model,
for the whole floor of the two bedrooms. can the model be loaded at the hallway of
Applying [4 marks] the entrance of the school hall? Explain your
answer and show your calculation.
Evaluating [5 marks]
47
F3 Chapt 4.indd 47 28/02/2023 12:11 PM
Mathematics Form 3 Answers
ANSWERS
–4
8. (a) 9 Section B
Chapter
1 Indices (b) 14 –8 1. (a) h ;
–6
(c) q h × h ;
–12
4
2
(d) (2y) –7 (b) a ÷ a ; (a )
2 3
2
8
Formative Practice 1.1 9. (a) (i) 5 2. (a) (i) 2 (ii) 0
(ii) 10 2 5
1. (a) 5 (b) (–3) 4 (iii) 1 2 3 (b) 12
7
q
4
(c) (–1.2) 5 (d) 1 2 3 p 5
9
(b) (i) 1
2. (a) 3 6 (b) (–2) 5 6 –3 Section C
6
8
6
(c) 1 2 4 (ii) 1 2 –1 1. (a) (i) p (ii) x 7
(b) –1
3
10
y
(d) 0.5 4 (iii) 1 2 –6 (c) 3 3 4
3. (a) 243 x 2. (a) m n n
(b) 108(3 )
(b) 4 096 10. (a) 4 (c) x = 3 , y = 4
4
(c) 125 (b) 16 2 5
216 (c) 15 h
(d) –0.0000128 1 Chapter
2
4. a = 9, n = 4 11. (a) 25 2 Standard Form
1
6
5. –2 (b) 64
1
6. 5 (c) k 8 Formative Practice 2.1
4
4
5
TIMSS 12. (a) 81 or ( 81 ) 5 1. 3 significant figures. The length of the
rod can be read as 5.21 or 5.22. The
3
3
2
28a 8 (b) 125 or ( 125) 2 third digit in both readings is uncertain.
7
7
3
(c) x or ( x ) 3
Formative Practice 1.2 (d) y or ( y ) 9 2. (a) 3 significant figures
9
1. (a) 3 2 3 (b) 4 significant figures
11
(c) 5 significant figures
(b) 18 7 13. (a) 27 5 2 (d) 2 significant figures
(c) 5 (b) 0.04 (e) 1 significant figure
15
3
(d) k 22 (c) p (f) 6 significant figures
5
2. (a) 12x 6 (d) q 7 2 (g) 4 significant figures
(b) m n 6 (h) 6 significant figures
10
(c) 28a b 14. (a) 16
10 5
(d) p q r (b) 243 3. (a) 90
7 15 14
3. (a) 7 2 (c) 125 (b) 40 000
(c) 40 000
(b) 29 8 (d) 128
(c) 12 5 4 5 (d) 430
(d) y 9 15. (a) 6 (b) k (e) 4 100
(c) 23 4 (d) x 14 (f) 98 000
4. (a) 6h 6 (g) 7 050
(b) a b 16. (a) 27 (h) 181 000
3 14
(c) k h 2 (b) 2 (i) 745 600
7
(d) 4x y (c) 1
8 6
5. (a) 9 (d) 2 3 4. (a) 4
18
(b) 0.4 40 (b) 20
(c) d 17. (a) 64p q (c) 20
14
7 2
(d) y 12 (b) 4 (d) 5.8
6. (a) 64x t 2 (e) 8.0
15
(b) p r 8 36 (c) a b 7 4 (f) 410
(g) 1.94
(c) 3 8 x 12
y 14 (d) y 15 (h) 96.0
7 14
(d) p q (i) 770.0
s 21 18. x = 3, y = 6 5. (a) 0.003
1 (b) 0.2
7. (a) 19. 36 hours 27 minutes
4 5 20. 2 (c) 0.9
17
(b) 1 (d) 0.00037
50 2 (e) 0.035
(c) 1 Summative Practice 1 (f) 0.87
p (g) 0.0479
1 Section A (h) 0.0830
(d)
x 6 1. B 2. A 3. C 4. D (i) 0.01005
120
F3 Answers .indd 120 28/02/2023 12:04 PM
Mathematics Form 3 Answers
(b) (i) 3 significant figures 9. (a) RM0.57
Formative Practice 2.2 (ii) 4 significant figures (b) Could lower the investment risk
3. (a) (i) Azilah and the monthly purchase did not
1. Standard form Reason (ii) Balqis: A = 34 . 10 give significant pressure to the
(A × 10 , 1 < A , 10) cash flow compared to purchasing
n
7.3 × 10 5 A = 7.3 and n = 5, all at once.
1 < A , 10 and n Chandra: n = 4, the obtained
value is 34 000.
is an integer 10. (a) 3.03%
(b) (i) –3 (b) RM0.81
2 × 10 –4 A = 2 and n = –4, (ii) 4.5 (c) 16.5%
1 < A , 10 and n (d) Invest in unit trust is more
is an integer Section C profitable because it has higher
1. (a) 3.75 × 10 g return of investment.
2
Not standard (b) (i) 6.83 × 10 12
form Reason (ii) 4.7 × 10 –8 Formative Practice 3.2
–3
0.241 × 10 A = 0.241 , 1 (c) 3.69 × 10 m
–2
6
45 × 10 7 A = 45 . 10 2. (a) 9.20 × 10 kg 1. (a) True
(b) False
(b) 1.50 × 10 seconds
4
0.089 × 10 1 A = 0.089 , 1 (c) 1.132 × 10 years 2. Advantage: provide cash withdrawal
2
service from ATM at
2. (a) 6.3 × 10 3 anytime
(b) 2.501 × 10 7 Consumer Mathematics: Disadvantage: impose initial fee for
(c) 7.38 × 10 Chapter Savings and Investments, cash withdrawal
3
–3
(d) 9.04 × 10 Credit and Debt Suggestion: use ATM card instead
–6
(e) 1.2 × 10 4 of credit card for cash
(f) 8.1 × 10 7 Formative Practice 3.1 withdrawal from ATM
–3
(g) 3.5 × 10 3. Finance charge = RM10.95
–1
(h) 1 × 10 1. Saving: saving account Late payment charge = RM10
7
3. (a) 5.8 × 10 km Investment: share, bond, foreign currency (minimum charge)
(b) 5.3 × 10 m 2. 4. RM3 116
–11
(c) 4 × 10 Annual Total Total 5. (a) Maisha can accumulate reward
11
(d) 1.7 × 10 g Principal interest Period interest savings points from credit card issuing
–24
4. (a) 4.3 × 10 7 (RM) rate (RM) (RM) company and redeem the reward
points for gifts or cash voucher.
(b) 8 × 10 –7 (b) RM2 907.62
(c) 2.45 × 10 (a) 2 000 1.5% 3 years 90 2 090 6. Total repayment = RM109 425
10
(d) 6.5 × 10 –1 Instalment = RM1 013.19
(b) 4 200 2% 15 months 105 4 305
5. (a) 5.68 × 10 5 7. Total repayment = RM63 750
(b) 2.2 × 10 –6 Instalment = RM1 062.50
13
(c) 4.13 × 10 (c) 3 000 1.8% 1 year 54 3 054 8. RM2 640.12
(d) 5 × 10 –4
(d) 5 000 1.2% 9 years 540 5 540 9. RM3 675.61
15
6. (a) 4.72 × 10 10. (a) RM1 237.50
(b) 3.92 × 10 15 (b) No. This is because the instalment
(c) 1.728 × 10 3. Simple interest = RM429 for 6 years (RM1 550) exceeds
30
(d) 1.08 × 10 1 Total savings = RM6 929 Ms Lily’s ability (RM1 300).
4. (a) RM19 764.11
7. 5 × 10 12
(b) RM19 546.74
8. 1.14 × 10 g cm –3
5. 16.4% Summative Practice 3
–2
9. 2.84 × 10 g
6. (a) –10%
4
10. (a) 1.2 × 10 km/h (b) Mr. Koh will face bigger loss because Section A
(b) 2 188 days he could not sell the polystyrene 1. D 2. C 3. B 4. A 5. D
food container after September. All 6. D 7. A
Summative Practice 2 excess stock on hand will become Section B
no value.
Section A 7. Investment in share. 1. (a) Investment
(b) Savings
1. C 2. B 3. A 4. D Risk: Investment in share will (c) Investment
Section B bear higher risk but it is (d) Investment
1. (a) (ii) 0.006508 → 0.00650 ; acceptable since Encik Hafiz 2. (a) (i) 3
is a risk taker.
0.00651 (ii) 7
Return: Investment in share could (b) Unit trust; Share
bring higher return as
(b) Not Section C
Significant compared to fixed deposit.
significant 1. (a) RM13 543.95
Liquidity: Investment in share is
(i) 3 slightly lower in liquidity as (b) RM97 372.80
(ii) 3 compared fixed deposit but (c) (i) RM1.22
it is acceptable for long term (ii) 3.45%
investment. 2. (a) RM4 961.62
2. (a) (i) mega (b) RM1 780.60
(ii) tera 8. RM0.77
121
F3 Answers .indd 121 28/02/2023 12:04 PM
Mathematics Form 3 Answers
(c) Total interest of fixed deposit for 9 × 4
years: RM6 555.06
Total interest reduced if RM20 000 is t = 1
used in loan: RM5 760 320 80
Mary shoud not use the money in × 4
fixed deposit account to reduce the t = 4 m
loan amount because the interest
of fixed deposit is higher than the 10. Draw the same kite ABCD on the grid The model built cannot be loaded at
interest of the car loan. with side of 0.4 cm and 2 cm. the hallway of the entrance of the hall
because the height of the model is 4 m
11.
whereas the height of the hallway of the
entrance is 3.8 m
Chapter
4 Scale Drawings
12 cm Chapter
5 Trigonometric Ratios
TIMSS
1 : 1 800 Formative Practice 5.1
12. 1 m 3
1. (a) KL is the opposite side.
13. (a) 1 : 200 LM is the adjacent side.
Formative Practice 4.1 (b) 3 m (b) RS is the opposite side.
(c) 10.4 m 2 ST is the adjacent side.
1. Height of A : Original height
= 3 units : 6 units 2. The opposite side is the side with length
= 1 : 2 Summative Practice 4 4 m; the adjacent side is the side with
Width of A : Original width length 3 m.
= 3 units : 6 units Section A 3. (a) q
= 1 : 2 1. C 2. B 3. D 4. D r
These two ratios are proportional. (b) p
Therefore, drawing A is the scale Section B r q
drawing of the original drawing. 1. (a) same, proportion (c) p
Height of B : Original height (b) II and III 4.
= 12 units : 6 units 2. (a) (i) 1 : 2
1 q 0° 15° 40° 65° 80°
= 1 : (ii) 1 : 1
2 2 sin q 0 0.2588 0.6428 0.9063 0.9848
Width of B : Original width (b)
= 9 units : 6 units Size of cos q 1 0.9659 0.7660 0.4226 0.1736
2 the scale
= 1 : Scale tan q 0 0.2679 0.8391 2.1445 5.6713
3 drawing 1 : n
These two ratios are not proportional. compare to 5. (a) (ii) 3
Therefore, drawing B is not the scale actual size (b) (i) 3
drawing of the original drawing. Smaller n , 1 (c) (i) 3
2. 1 cm : 4 cm means that the scale Bigger n . 1 8
drawing produced is smaller than the 6. (a) 15
actual object, that is 1 times than the Section C (b) 3
actual object. 4 2 5
1 cm : 0.5 m = 1 cm : 50 cm means that 1. (a) (i) 1 : 25 (c) 3
the scale drawing produced is smaller (ii) 40 cm 5
than the actual object, that is 1 times (b) (d) 21
the actual object. 50 29
1 7. (a) 24
1 cm : cm means that the scale 25
4
drawing produced is larger than the (b) 24
actual object, that is 4 times the actual 7
object. S 8. (a) 0.342
Therefore, the scale that will produce the (b) 0.530
smallest scale drawing is 1 cm : 0.5 m. (c) 0.613
3. Length = 79.8 cm, width = 18.2 cm 9. (a) 1 2
4. 1 : 3 000 000 √3
5. 1 : 150 000 (c) RM1 264 (b) 2
1 (c) 1
6. 1 : 1
50 000 2. (a) 1 : 3 (d) 1
7. 8.2 cm (b) AB = 8 cm; AF = 4 cm √2
1
8. 8 cm (e) 2
(c) Height of model, t = Drawing scale
9. Actual height (f) √3
122
F3 Answers .indd 122 28/02/2023 12:04 PM
Mathematics Form 3 Answers
10. (a) 13.79 4. 40° (c) x = 47°, y = 11°
(b) 7.23 5. x = 55°, y = 27.5° (d) x = 34°, y = 64°
(c) 28.48 6. 17° 8. (a) 81°
11. (a) x = 5, y = 5√2 (b) 109°
(b) x = 18√3, y = 36 Formative Practice 6.2 9. x = 37°, y = 66°
12. 5 cm 10. (a) 21.32° (b) 22.04 cm
13. 25 cm 1. (a) Cyclic quadrilateral because all four 11. (a) 128.68° (b) 20.47 cm 2
vertices lie on the circumference of
14. (a) 64° the circle. Formative Practice 6.4
(b) 18°
(c) 49° (b) Not a cyclic quadrilateral because 1. (a) 66°
there are two vertices out of the
15. (a) 26.6° circle. (b) 23°
(b) 13.9° (c) Cyclic quadrilateral because all four 2. (a) 9.0 cm
(c) 85.4° vertices lie on the circumference of (b) 41°
16. (a) 16.3° the circle. 3. (a) 188°
(b) 70.9° (d) Not a cyclic quadrilateral because (b) RM6 131 795.20
(c) 77.6° all four vertices do not lie on the
17. 483 circumference.
2. (a) x = 70°, y = 112° Summative Practice 6
18. 448 m (b) x = 78°, y = 106°
19. (a) 10.77 cm (c) x = 30°, y = 43° Section A
(b) 15.6° (d) x = 30°, y = 39° 1. A 2. B 3. D
3. (a) m = 98°, n = 120° Section B
Summative Practice 5 (b) m = 80°, n = 117° 1. (a) d
(c) m = 48°, n = 38°
(d) m = 86°, n = 81° a
Section A 4. 70° e
1. C 2. A 3. B b
5. 38° f
Section B 6. x = 49°, y = 100°
1. (a) z, w, x, y 7. (a) 112° (b) (i) 7
1 1 (b) 48° (ii) 3
(b) cos 45° = , cos 60° =
√2 2 8. 270° Section C
2. (a) tan; sin 9. ∠SQT = ∠SQR, ∠SQT = ∠SPT and 1. (a) (i) 65°
(b) 60; 18 (ii) 32°
∠SQR = ∠STP.
Therefore, ∠SPT = ∠STP and SP = (iii) 51°
Section C ST. (iv) 71°
1. (a) 26.6° (b) (ii) 82°
(b) 4.3 m (c) (i) 40°
(c) (5√3 – 5) m Formative Practice 6.3 (ii) 50°
2. (a) (i) 33 cm 2. (a) (i) 16.0 cm
(ii) a = 6 cm, c = 6√3 cm 1. (a) Yes, line AB touches the circle at (ii) 17.8 cm
(b) a = 3√3, b = 9, c = 9, d = 9 + 8√3 one point only. (b) (i) 80°
(c) 4 776.83 m (b) Not, when line AB is extended, it will (ii) 6.71 cm
cut through the circle at two points.
3. (a) 16 cm (c) Yes, line AB touches the circle at (c) 3.83 m
(b) 402.2 m one point only. 3. (a) h = 26°, k = 68°
(c) Length of wire A = 14.94 m, (d) Not, line AB cut through the circle at (b) (i) 16.6°
Length of wire B = 13.81 m; two points. (ii) 44.69 cm 2
Wire A is longer than wire B.
2. (a) 32°
(b) 63° Chapter
(c) 34° 7 Plans and Elevations
Chapter Angles and Tangents
6 (d) 25°
of Circles
3. (a) (i) 8 cm The following diagrams are not drawn in
(ii) 30° actual size.
Formative Practice 6.1 (b) (i) 15 cm
(ii) 113° Formative Practice 7.1
1. (a) 72° (c) (i) 20 cm
(b) 36° (ii) 34° 1. (a)
(c) 40° (d) (i) 17 cm 1 cm
(d) 51° (ii) 55° 3 cm
2. (a) 118° 4. 39°
(b) 240° 5. (a) 24 cm 6 cm
(c) 27° (b) 40° (b)
(d) 107° (c) 8.74 cm 2 cm
6. (a) 13.34 cm
3. (a) 30°
(b) 63° (b) 14° 3 cm
(c) 56° 7. (a) x = 60°, y = 70°
(d) 18° (b) x = 34°, y = 72° 3 cm
123
F3 Answers .indd 123 28/02/2023 12:04 PM
Mathematics Form 3 Answers
(c) 4 cm 2 cm Formative Practice 7.2 (c) (i) 2 cm
4 cm 1. (a) 4 cm
4 cm
3 cm
2 cm 5 cm
3 cm
(b)
2.5 cm
(d) (ii)
2 cm 2.5 cm 2 cm 2 cm
6 cm
3 cm 7 cm 4 cm
(c)
3 cm 7 cm
3 cm
2. (a) (i) 3 cm (d) (i)
2 cm
4 cm 5 cm
(d)
4 cm
2 cm
6 cm
5 cm 3 cm
(ii)
(ii) 7 cm
(e) 5 cm
5 cm 4 cm
6 cm
3 cm (f)
(e) (i)
5 cm 3 cm
4 cm
4 cm
(b) (i) 5 cm
2. (a) (i)
4 cm 5 cm
(ii)
3 cm 4 cm 4 cm
4 cm
(ii) 3 cm
5 cm
3 cm 4 cm
(ii) 6 cm
3.
4 cm
4 cm
L/E E 5 cm L
7 cm 1.5 cm
K/F F K
2.5 cm
3. (a) (i) Same length (b) (i) I/H 2 cm 8 cm H/G I/J
(ii) Same size 1 cm J/G
(iii) Congruent 1 cm 4 cm
(iv) Different 3 cm
(v) Different B/A 6 cm C/D A/D B/C
(b) (i) Different 5 cm 45°
(ii) Different E/D L/C
(iii) Congruent (ii)
(iv) Same length 1 cm 4 cm
(v) Different F/G K/J
5 cm 2 cm
4. (a) Same length 3 cm 4 cm
(b) Different H/A 5 cm I/B
(c) Different
6 cm
124
F3 Answers .indd 124 28/02/2023 12:04 PM
Mathematics Form 3 Answers
4. 9. 14. (a)
C B
E E F/I I F
G/H E/J P/N H/J G/E E 4 cm
P
5 cm 5 cm Q/K 4 cm 5 cm N 2 cm
Q
K
4 cm 4 cm 2 cm 4 cm 2 cm D 5 cm A
A/B D/L/C R/M B/C L/M A/D/R
A/D 4 cm B/C D/C 6 cm A/B
45° 45° N/M 4 cm P/R (b) E
D C
3 cm 6 cm
J/C K/L E/Q/D
E 2 cm
3 cm I F 2 cm D 2.5 cm 1.4 cm 2.5 cm B
A
C
A 4 cm B H/B G/A
10. 15. (a) A F E/D
5. L/F F 5 cm L
4 cm 3 cm 3 cm
M/E E M 1 cm
G 5 cm H G/H J/I/H 4 cm K/N/G H/G J/K B G 4 cm H/C
2 cm 4 cm 5 cm 4 cm 2 cm
J I J/I
K L K/L
1.5 cm F E F/E B/A 7 cm C/D A/D B/C (b) 2.2 cm
1 cm 1 cm F G E H
B/A C/D A/D 5 cm B/C 45° E/D 5 cm M / C
D
M/C
/
E
45° 3 cm 4 cm
E/D H/I L/C N 3 cm
F/G L/K
4 cm A B 4 cm D C
5 cm 1.35 cm 1.35 cm
H/A I J/B
F/A G/J K/B Summative Practice 7
2.5 cm 2.5 cm 11. 6 cm
N M N/M Section A
6. 4 cm 1. B 2. A 3. C 4. C
H/E 5 cm G E 6 cm H/G F/E H/I 7 cm E/I 4 cm F/H
3 cm G/J J G 2 cm Section B
3 cm 3 cm 6 cm 1 cm 1. (a) True
F F A/D/K B/C/L K/L D/C A/B
1 cm 1 cm (b) True
A/D B/C D/C A/B 45° (c) False
(d) True
45°
E/D F/C N/K M/L 6 cm 2. (a)
J
E/D I/C
6 cm 4 cm
(b) (3)
F/A G H/B ( )
H/A 5 cm G/B 3 cm 3 cm
(3)
12. (a) 3. (a) 3 (b) 3
7. (c) 3 (d) 7
2 cm 5 cm
Section C
3 cm
1.
6 cm 4 cm H 2 cm G
4 cm 3 cm I 1 cm I/F H/G
(b) 27.71 cm 3 F
3 cm 5 cm
6 cm 3.5 cm J J
13. (a) L 7 cm L/K
4 cm K 4 cm 3 cm
8. 5 cm B 4 cm C/A D/E 5 cm B/C/D
8 cm A/E
45°
F/E 5 cm G/D
3 cm
1 cm 12 cm 2 cm 2 cm
2 cm 3 cm I/J/A H/K/C
2 cm 4 cm
1 cm 4 cm (b) (i) 216 cm 3
(ii) 252 cm L/B
2
1 cm
125
F3 Answers .indd 125 28/02/2023 12:04 PM
Mathematics Form 3 Answers
2. 6. (a) (i) E/D 6 cm F/C (b) 8 cm
K H K/H 2.6 cm
L 1 cm G L/G 4 cm 2 cm 2 cm I/L
1 cm
J/M/N I/F/E N/E M/F J/I J/K 6 cm 4 cm 6 cm
5.5 cm 5.5 cm
3 cm 4 cm
B/A C/D A/D 6 cm B/C N/A H/M G/B
45° 2 cm 6 cm
E/D G/F H/I/C
2 cm 4 cm
4 cm
(ii) (c) 12 cm 2
N/A K/J/B 2.8 cm 2.8 cm
L/M E J I/F H G (d) 145.6 cm 2
3 cm
3. 6 cm
Chapter
P/Q Q 6 cm P K N/L M 8 Loci in Two Dimensions
2 cm 2 cm
N/M 1 cm M N D 4.2 cm A/C 4.2 cm B
E/H
F/E G/H 1 cm F/G Formative Practice 8.1
I/R R I
6 cm 5 cm 5 cm (b) (i) 1. (a) A locus, the motion of the cyclist
has a fixed path that is an inclined
2 cm 1 cm A/B 8 cm 5 cm line.
A/D/L T/S 4 cm B/C J/K L/S/K D/T/C/J 2 cm (b) A locus, the movement of the
3 cm satellite always follows its orbit.
45°
M/L Q/S 5 cm R/K (c) Not a locus, the direction of
2 cm 5.5 cm movement of the waves is
uncertain.
6 cm (d) A locus, the motion of the load has
(ii) 111.96 cm 3 a fixed path that is a vertical line.
H/C
N/E/D P/T I/J 2. The locus of the girl is an arc of a circle.
2 cm 2 cm 7. (a) (i)
F/A 6 cm G/B D/C
3 cm 4 cm
3.
4.
1 cm 1 cm E/A 4 cm F/B
G F J E H F/E G/H J
2 cm
(ii)
6 cm 6 cm E D F
Formative Practice 8.2
2 cm
2 cm A/D B/C 4 cm 1. (a) Locus X is the circumference of a
B A D C A C B circle with the radius of 2 cm and
45° 1.12 cm 2.88 cm centre O.
H/C
E/D
2 cm 4 cm J 2 cm
F/A (b) (i)
G/B 15 cm O
Locus X
5. 10 cm
M K/L L M/K 30 cm 18 cm (b) Locus Y is the circumference of a
3 cm 2 cm 2 cm circle with the radius of 1.5 cm and
E 1.5 cm F E/F (ii) 123 buah centre K.
I/J H/G 4 cm J/G 3 cm I/H
2.5 cm
4 cm
N A/D 5 cm B/C D/C N/A/B 8. (a) 6 cm K
45° 1.5 cm Locus Y
L/E/D F/C 2.6 cm
J G 1 cm 8 cm
3 cm 2.6 cm 4 cm 2. Locus H is a perpendicular bisector to
3 cm 5 cm the straight line connecting point C and
M/N K/I/A H/B point D.
126
F3 Answers .indd 126 28/02/2023 12:04 PM
Mathematics Form 3 Answers
7. D 2. (a) M
Locus H A
X
⊗
C D
B C
⊗
8. K P
P
L
3. Locus L is a pair of parallel lines and 3 cm ⊗ Y
2 cm from the line MN. (b) 1.5 cm
K N
Q 4 cm R ⊗
Locus L P
2 cm 2 cm
9. P L M
T
T
M N ⊗ ⊗ X Y
U V
2 cm X (c)
T ⊗ ⊗ T
D C
Q
Locus L 10. 1.5 cm 2.5 cm
⊗
4. Locus M is a straight line in the middle 1.9 cm R
of line PQ and line XY, and parallel to 1.5 cm
both lines. Canteen 50 m A B
Library The distance between R and A is
Walkway 22 m.
P X 100 m
3. (a)
Summative Practice 8 E Y D
⊗
Section A F C
1. C 2. A 3. D
A B
Q Locus M Y Section B Lokus X
1. (a) (i) 3 (ii) 7
(b) • •
5. (b) C
D • •
Locus P ⊗
Section C P
1. (a) (i) Point S and point Q
A B (ii) Point R 2 cm
(iii) Point P
(b) (i) Point A is always equidistant
from point P and point Q. A B
C (ii) Point B is always 8 cm from
point X.
(c) (c) (i), (ii)
6.
F E D V U
3 cm
X X W
⊗ ⊗ ⊗ ⊗ A ⊗
Y ⊗ H H
A B C 2 cm
Jalan Mersing
S T
127
F3 Answers .indd 127 28/02/2023 12:04 PM
Mathematics Form 3 Answers
(b) (i) 10x – y = 5
Chapter Summative Practice 9
9 Straight Lines (ii) x + y = 1
1 –5
2 Section A
TIMSS 1. C 2. A 3. C 4. B 5. B
(c) (i) 2x – 5y = –30 6. A
(a) – 5 x y
4 (ii) –15 + 6 = 1 Section B
5
(b) (d) (i) x + 4y = 48 1. x y
2 y = 3x + 4 8 + 12 = 1
(ii) x + y = 1
48 12 4x – 3y = 6 3x – y = –4
Formative Practice 9.1 9. (a) Yes
(b) No 2y = –3x + 24 4x + 3y = 12
1. (a) y = x – 3 (c) Yes
(b) y = –2x + 5 (d) Yes x 3 + y 4 = 1 y = 4 3 x – 2
2. (a) y = 4x + 5 10. m = – 2
(b) y = –2x – 3 5 2. (2, 9) (–3, 0) (–4, –1) (5, 12)
11. k = –7
3. 4, y = 4x + 4 3
4. (a) x = 2 12. (a) Yes (b) No
y (c) Yes (d) Yes 3. (a) y = 5
x = 2 13. – 1 (b) y = x + 5
x 2
14. (a) y = –5x + 8 Section C
(b) y = 4x – 4
(c) y = 4 x + 2 1. (a) (i) S = (–6, 8); Q = (15, 0)
(2, –4) 3 (ii) y = 4x + 32
1
15. (a) y = 2x – 9 (b) (i) y = – x + 1
2
(b) y = –4 (b) y = –2x + 3 (ii) (2, 0)
y 2 4 (iii) y = – 4x – 13; (–5, 7)
(c) y = – x –
5 5 5
x 2. (a) (i) ( , 11)
2
16. (a) y = 3 x + 3
8 (ii) y = –2x + 16
1 (b) (i) (–8, 0)
(b) y = x – 2
y = – 4 2 (ii) y = –3x + 6
1
(2, –4) (c) y = 8 x – 4 (c) (i) y = – x + 33
5 (ii) (17, 4) 4 4
17. (a) y = 4x – 9
5. (a) –3, –7 3 7 1
(b) y = – x + 3. (a) (i)
2 10 5 6
(b) , 5
3 (c) y = 1 x + 6 (ii) (6, 9); x = 6
(c) 7, 4 2 (b) (i) y = – 5 x + 35
18. (a) y = –3x + 15 12 12
1
(d) – , 8 (b) y = 2x + 2 (ii) (7, 13)
2 4 17
1 9 (c) y = – x – 3 (c) (i) 2
3
6. (a) y = – x + (ii) y = x + 2
2 4
3 11 19. (a) (2, –1) 2 8
(b) y = x – 4. (a) (i) h = 4 ; y = – x +
7 7 (b) ( 1 , 5) 3 3
(c) y = – 3 x + 3 2 (ii) 5 1 units 2
32 8 (c) (–3, 1) 3
(d) y = 5 x – 1 (d) (1, –7) (b) –2
4 8 20. (a) (–1, 4) (c) (i) y = 5 x – 7
(b) (4, –2) 2
4
7. (a) y = – x + 4 1 (ii) (–4, –7)
3 (c) (– , 1) 5. (a) (i) 1
2 3
(b) y = x – 2 2 (ii) 4
7 (d) ( 3 , 2) 21
1 (iii)
(c) y = – x + 2 2
4 21. (a) 2 9 54 3 4
5 (b) y = – x + (iii) y = 5 x – 5
(d) y = x + 5 7 7
4 1
22. (a) (10, 0) (b) (i) h = 7; y = 3 x + 6
8. (a) (i) 3x – y = –8 (b) y = 1 x + 7 (ii) (–2, 7)
x y 2
(ii) + = 1
– 8 8 (c) (4, 9)
3
128
F3 Answers .indd 128 28/02/2023 12:04 PM
NOTES
129
F3 Answers .indd 129 28/02/2023 12:04 PM
NOTES
130
F3 Answers .indd 130 28/02/2023 12:04 PM
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