Top One (Maths F3) Penerbitan Pelangi Sdn Bhd

Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd

Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd

Contents

TER HOTS Challenge 57
57
1 Indices 1 MRSM Cloned Question, TIMSS/PISA Cloned Question,
Indeks Online Quick Quiz (QR Code)
CHAP CHAP CHAP CHAP CHAP
CHAP CHAP CHAP CHAP1.1 Index Notation 1TER 58
2
1.2 Law of Indices 6 6 Angles and Tangents of Circles
8 Sudut dan Tangent bagi Bulatan
PT3 Standard Practice 1
8 6.1 Angles at the Circumfenence and Central Angle 58
HOTS Challenge Subtended by an Arc 60
61
MRSM Cloned Question, TIMSS/PISA Cloned Question, 6.2 Cyclic Quadrilaterals 63
Online Quick Quiz (QR code) 64
6.3 Tangents to Circles 68
TER 9
6.4 Angles and Tangents of Circles 68
2 Standard Form 9
Bentuk Piawai 10 PT3 Standard Practice 6
13
2.1 Significant Figures 15 HOTS Challenge
2.2 Standard Form 15
PT3 Standard Practice 2 MRSM Cloned Question, TIMSS/PISA Cloned Question,
HOTS Challenge Online Quick Quiz (QR Code)
Online Quick Quiz (QR Code)
TER 69
3TER Consumer Mathematics: Savings and 16
Investments, Credit and Debits 7 Plans and Elevations 69
Matematik Pengguna: Simpanan dan Pelaburan, Pelan dan Dongakan 73
Kredit dan Hutang 87
7.1 Orthogonal Projections 92
3.1 Savings and Investments 16 7.2 Plan and Elevations 92
3.2 Credit and Debt Management 22 PT3 Standard Practice 7
PT3 Standard Practice 3 29 HOTS Challenge
HOTS Challenge 32 Online Quick Quiz (QR Code)
Online Quick Quiz (QR Code) 32
TER 93

8 Loci in Two Dimensions
Lokus dalam Dua dimensi

TER 33 8.1 Loci 93
8.2 Loci in Two Dimensions 94
4 Scale Drawings PT3 Standard Practice 8 101
Lukisan Berskala HOTS Challenge 105
MRSM Cloned Question, Online Quick Quiz (QR Code) 105
4.1 Scale Drawings 33
43 TER 106
PT3 Standard Practice 4 47
9 Straight Lines
HOTS Challenge 47 Garis Lurus

MRSM Cloned Question, TIMSS/PISA Cloned Question,
Online Quick Quiz (QR Code)

TER 48 9.1 Straight Lines 106
119
5 Trigonometric Ratios PT3 Standard Practice 9 123
Nisbah Trigonometri
HOTS Challenge 123
5.1 Sine, Cosine and Tangent for Acute Angles in 48
Right-angled Triangles 54 TIMSS/PISA Cloned Question,
Online Quick Quiz (QR Code)
PT3 Standard Practice 5
PT3 Model Paper 124

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CHAPTER

2 Standard Form
Bentuk Piawai

Textbook
pg. 30 – 49

2.1 Significant Figures
Angka Bererti

1. Match each of the following numbers with the correct number of significant figures. PL 1 NOTES

Padankan setiap nombor yang berikut dengan bilangan angka bererti yang betul.

Example Numbers Significant figures

Nombor Angka bererti

3 341 4

(a) 3.02 1

(b) 90 000 3

(c) 3.0002 4

(d) 4.348 5

(e) 0.0045 2

2. Round off each of the following integers to the specified significant figures. PL 2

Bundarkan setiap integer yang berikut kepada angka bererti yang dikehendaki.

Example (a) 948 (b) 590
98 [2 s.f. / a.b.]
[1 s.f. / a.b.] 950 [1 s.f. / a.b.]
100 600 [3 s.f. / a.b.]

(c) 3 863 (d) 622 (e) 8 817
[3 s.f. / a.b.] [2 s.f. / a.b.]

3 860 620 8 820

3. Round off each of the following decimals to the specified significant figures. PL 2

Bundarkan setiap nombor perpuluhan yang berikut kepada angka bererti yang dikehendaki.

Example (a) 3.501 (b) 5.562
4.91 [3 s.f. / a.b.]
[2 s.f. / a.b.] 3.50 [3 s.f. / a.b.]
4.9 5.56 [2 s.f. / a.b.]

(c) 81.979 (d) 9.74 (e) 6.98
[4 s.f. / a.b.] [2 s.f. / a.b.]

81.98 9.7 7.0

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  Mathematics  Form 3  Chapter 2 Standard Form

4. Round off each of the following decimals to the specified significant figures. PL 2

Bundarkan setiap nombor perpuluhan yang berikut kepada angka bererti yang dikehendaki.

Example (a) 0.491 (b) 0.07028
0.0891 0.0703
[2 s.f. / a.b.] [2 s.f. / a.b.] [3 s.f. / a.b.]
0.089 0.49 [1 s.f. / a.b.]
(c) 0.21976
[4 s.f. / a.b.] (d) 0.458 [2 s.f. / a.b.] (e) 0.0299
0.2198 0.46 0.03

5. Calculate each of the following and round off your answer to the specified number of significant figures. PL 3

Hitung setiap yang berikut dan bundarkan jawapan anda mengikut bilangan angka bererti yang dikehendaki.

Example (a) 5.23 × 1 025 ÷ 50
2.12 × 323 + 60 – 13.2 × 10
[2 s.f. / a.b.] = 5.36075 × 103 ÷ 50 [1 s.f. / a.b.]
= 684.76 + 60 – 132 = 107.215 [2 s.f. / a.b.]
= 744.76 – 132
= 612.76 = 100
= 610
[3 s.f. / a.b.] (c) 980 – 6.13 ÷ 0.02
(b) 42.5 × 8.12 ÷ 400
= 980 – 306.5
= 345.1 ÷ 400 = 673.5
= 0.86275 = 670
= 0.863

(d) 55.6 – 34.5 [1 s.f. / a.b.] (e) (31.2 + 0.9) × (4.5 – 0.87) [3 s.f. / a.b.]
109 + 58
= 32.1 × 3.63
= 21.1 = 116.523
167 = 117

= 0.12635

= 0.1

2.2 Standard Form
Bentuk Piawai

6. State each of the following number in standard form. PL 1 PL 2 NOTES

Nyatakan setiap nombor yang berikut dalam bentuk piawai.

Example (a) 545 = (b) 651 00 =
3 540 = 5.45 × 102 6.51 × 104

3.54 × 103 (d) 4 300 000 = (e) 1 037 =
(c) 26 = 4.3 × 106 1.037 × 103

2.6 × 10

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7. State each of the following numbers in standard form. PL 1 PL 2 Mathematics  Form 3  Chapter 2 Standard Form 

Nyatakan setiap nombor yang berikut dalam bentuk piawai. (b) 0.00025 =
2.5 × 10–4
Example (a) 0.0045 = (e) 0.9093 =
0.033 = 4.5 × 10–3 9.093 × 10–1
3.3 × 10–2
(c) 0.000076 = (d) 0.00000056 =
7.6 × 10–5 5.6 × 10–7

8. Convert each of the number in standard form to single number. PL 1 PL 2

Tukar setiap nombor dalam bentuk piawai kepada nombor tunggal.

Example (a) 5 × 103 = (b) 5.5 × 10 =
2 × 104 = 5 000 55
20 000
(d) 8 × 105 = (e) 3.69 × 10 =
(c) 3.46 × 103 = 800 000 36.9
3 460

9. Convert each of the number in standard form to single number. PL 1 PL 2

Tukar setiap nombor dalam bentuk piawai kepada nombor tunggal.

Example (a) 1 × 10–4 = (b) 6.5 × 10–3 =
2 × 10–2 = 0.0001 0.0065
0.02
(d) 9 × 10–6 = (e) 3.68 × 10–6 =
(c) 8.6 × 10–4 = 0.000009 0.00000368
0.00086

10. Calculate each of the following and state the answer in standard form. PL 3

Hitung setiap yang berikut dan nyatakan jawapan dalam bentuk piawai.

Example Common factor (a) 4.44 × 103 + 0.21 × 103
Faktor sepunya = (4.44 + 0.21) × 103
2.3 × 102 + 4.3 × 103 = 4.65 × 103
= (0.23 × 103) + (4.3 × 103)
= (0.23 + 4.3) × 103
= 4.53 × 103

(b) 4.5 × 105 – 3.2 × 104 (c) 640 × 110 ÷ 1 800
= 4.5 × 105 – 0.32 × 105 = 70 400 ÷ 1 800
= (4.5 – 0.32) × 105
= 4.18 × 105 = 7.04 × 104 ÷ 1.8 × 103
= 3.91 × 104 – 3
= 3.91 × 10

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  Mathematics  Form 3  Chapter 2 Standard Form (e) 1.5 × 10–2 × 5.1 × 103 ÷ (4 × 102)
= (1.5 × 5.1 ÷ 4) × 10–2 + 3 – 2
(d) 2.63 × 104 × 3.2 × 10–2 = 1.91 × 10–1
= 2.63 × 3.2 × 104 + (–2)
= 8.42 × 102

11. Solve each of the following problems. PL 5 PL 6

Selesaikan setiap masalah yang berikut.

Example Malaysia Singapore
32 million
Population Singapura
32 juta
Jumlah Penduduk 4.48 million

Land area (km2) 4.48 juta

Luas tanah (km2) 330 803 700

Based on the table above,

Berdasarkan jadual di atas,

(i) rewrite the population and land area of Malaysia and Singapore in standard forms,

tulis semula jumlah penduduk dan luas tanah Malaysia dan Singapura dalam bentuk piawai,

(ii) express the ratio of the number of people per km2 of Malaysia and Singapore in the form 1: n.

ungkapkan nisbah bilangan penduduk per km2 Malaysia dan Singapura, dalam bentuk 1 : n.

(i) Malaysia Singapore (ii) Number of people per km2 in Malaysia
= 3.2 × 107 ÷ 3.31 × 105
Population 3.2 × 107 Singapura = 0.97 × 102
3.31 × 105 = 9.7 × 10
Jumlah Penduduk 4.48 × 106
Number of people per km2 in Singapore
Land area (km2) 7 × 102 = 4.48 × 106 ÷ 7 × 102
= 0.64 × 104
Luas tanah (km2) = 6.4 × 103

Ratio of number of people per km2
(9.7 × 10 : 6.4 × 103) ÷ 9.7 × 10

Therefore, 1: n = 1: 66

(a) China Africa
Cina Afrika

Population 1.32 billion 832 million
Jumlah Penduduk 1.32 bilion 832 juta

Land area (km2) 9.60 million 26.6 million
Luas tanah (km2) 9.60 juta 26.6 juta

Based on the table above,

Berdasarkan jadual di atas,

(i) rewrite the population and land area of China and Africa in standard forms,

tulis semula jumlah penduduk dan luas tanah Cina dan Afrika dalam bentuk piawai,

(ii) find the number of people per km2 of China and Africa.

cari bilangan penduduk per km2 Cina dan Afrika.

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Mathematics  Form 3  Chapter 2 Standard Form 

(i) China Africa
Cina Afrika

Population 1.32 × 109 8.32 × 108

Jumlah Penduduk

Land area (km2) 9.60 × 106 2.66 × 107

Luas tanah (km2)

(ii) Number of people per km2 in China
= 1.32 × 109 ÷ 9.60 × 106
= 0.1375 × 103
= 1.375 × 102

Number of people per km2 in Africa
= 8.32 × 108 ÷ 2.66 × 107
= 3.13 × 10

PT3 Standard Practice 2

Section A / Bahagian A 5. Which of the following numbers is rounded off to

3 significant figures to obtain 3.08? 2.1.2
Antara nombor-nombor yang berikut, nombor yang manakah
1. Express 6.47 × 10–4 as single number. 2.2.1 dibundarkan kepada 3 angka bererti untuk memperoleh

Ungkapkan 6.47 × 10–4 sebagai nombor tunggal. 3.08?

A 0.000647 C 6 470 A 3.0862 C 3.0713

B 0.0000647 D 647 000 B 3.0785 D 3.8045

2. 3.43 × 10–3 = 2.2.2 Section B / Bahagian B
700 000
C 4.9 × 10–9 1. (a) Round off each number correct to the number
A 4.9 × 10–3 D 4.9 × 10–12
B 4.9 × 10–6 of significant figures stated. 2.1.2
Bundarkan setiap nombor betul kepada bilangan angka
bererti yang dinyatakan.
3. Find the value of 2k – 3s in standard form. Given
[3 marks / 3 markah]
k = 6.14 × 103 and s = 1.09 × 104. 2.2.2
Cari nilai bagi 2k – 3s dalam bentuk piawai. Diberi Answer / Jawapan:

k = 6.14 × 103 dan s = 1.09 × 104. Number of
significant
A 2.042 × 10–4 C –2.042 × 104 Number Answer
figures
B 2.042 × 104 D –2.042 × 10–4 Nombor Jawapan
Bilangan angka
4. Which of the following numbers are written in (i) 0.0183 bererti 0.018

standard form? 2.2.1 2
Antara yang berikut, nombor yang manakah diungkapkan
dalam bentuk piawai?
(ii) 299 530 3 300 000
A 365 000 C 3.65

B 36.5 × 104 D 3.65 × 105 (iii) 43 4 43.00

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  Mathematics  Form 3  Chapter 2 Standard Form Answer / Jawapan:

(b) Express 0.000045 in standard form. 2.2.1 (i) 30.15.–430.54 = 5.43
29.56
Ungkapkan 0.000045 dalam bentuk piawai.
= 1.84 × 10–1
[1 marks / 1 markah]

Answer / Jawapan:
4.5 × 10–5

2. (a) Round off 299 530 correct to three significant (ii) (6.09 × 102) ÷ (2.31 × 10–3)
= 6.09 ÷ 2.31 × 102 – (–3)
figures. 2.1.2 = 2.64 × 105

Bundarkan 299 530 betul kepada tiga angka bererti.

[1 marks / 1 markah] (c) (i) A water fall has a width of 2 356 m, write

Answer / Jawapan: the width in standard form. 2.2.1
Suatu air terjun mempunyai lebar 2 356 m, tulis
+1 lebar itu dalam bentuk piawai.
299 530  300 000
[1 mark / 1 markah]

(b) Fill in the blanks. 2.2.2 Answer / Jawapan:
2 356 = 2.36 × 103 m
Isi tempat kosong. [3 marks / 3 markah]
(ii) It is estimated that by the year 2060, the
Answer / Jawapan: world population will exceed k billion. If
the world population in the year 2060 is
0.000000085 ÷ 1.7 × 10–9 1.05 × 1011, find the value of k. 2.2.3

= 8.5 × 10–8 ÷ 1.7 × 10–9 Jumlah bilangan penduduk dunia dijangka akan
mencapai k billion pada tahun 2060. Jika jumlah
= 8.5 ÷ 1.7 × 10–8 + 9 penduduk dunia pada tahun 2060 ialah 1.05 ×
1011, cari nilai k.
= 50
[2 marks / 2 markah]

Answer / Jawapan:

Section C/ Bahagian C 1.05 × 1011 = k × 109

k = 1.05 × 102 billion

1. (a) Given 69 350 000 = 6.935 × 10k and 0.00000123 2. (a) The speed of a car is 75 kmh–1. At this speed,

= 1.3 × 10l, find the value of k × l. 2.2.2 find the distance, in m, travelled by the car in 9
Diberi 69 350 000 = 6.935 × 10k dan 0.00000123 = 1.3 × 10l,
cari nilai bagi k × l. 000 seconds. Express your answer in standard

[3 marks / 3 markah] form. 2.2.3
Kelajuan sebuah kereta ialah 75 kmh–1. Pada kelajuan
Answer / Jawapan: ini, cari jarak, dalam m, yang dilalui oleh kereta itu
k = 7, l = –6 dalam masa 9 000 saat. Nyatakan jawapan anda dalam
bentuk piawai.
∴ k × l = 7 × (–6) = –42
[3 marks / 3 markah]

Answer / Jawapan:

(b) Evaluate and give your answers in standard 9 000 seconds = 9 000 = 2.5 h
3 600
form. 2.2.2
Nilaikan dan beri jawapan anda dalam bentuk piawai. Distance = Speed × Time

(i) 30.15.–430.54 = 75 kmh–1 × 2.5

[2 marks / 2 markah] = 187.5 km

(ii) (6.09 × 102) ÷ (2.31 × 10–3) = 187 500 m
[2 marks / 2 markah]
= 1.875 × 105 m

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Mathematics  Form 3  Chapter 2 Standard Form 

(b) The diagram below shows a wooden block in Answer / Jawapan:

the shape of cylinder. 2.2.3 Density = Mass
Rajah di bawah menunjukkan sebuah blok kayu Volume
berbentuk silinder.
0.9 kgm–3 = Mass
7.7 × 10–3
14 cm
Mass = (0.9 × 7.7 × 10–3) kg

= 6.93 × 10–3 kg

50 cm = 0.00693 kg

(c) The table below shows the price of some fruits.
Jadual di bawah menunjukkan harga bagi beberapa
buah-buahan.
2.2.3

(i) Calculate the volume, in m3, of the wooden Fruits Price
block. State your answer in standard form.
Buah-buahan Harga
Hitung isipadu, dalam m3, blok kayu itu. Nyatakan
jawapan anda dalam bentuk piawai. Mango / Mangga RM5.99 / kg

  Use / 22 [2 marks / 2 markah]
Guna: π = 7 Mangosteen / Manggis RM9.90 / kg

Answer / Jawapan: Puan Rosmah bough 3.5 kg of mango and 1 kg

Volume of cylinder = πr2h of mangosteen. Calculate the total price that
22
= 7 × (0.07)2 × 0.5 she needs to pay. State your answer in three

= 7.7 × 10–3 m3 significant figures.
Puan Rosmah membeli 3.5 kg mangga dan 1 kg
manggis. Hitung jumlah harga yang perlu dibayarnya.
Nyatakan jawapan anda dalam 3 angka bererti.

(ii) Given the density of the wooden block [3 marks / 3 markah]

is 0.9 kgm–3. Find the mass, in kg, of the Answer / Jawapan:

block in single number. Total price
= RM5.99(3.5) + RM9.90
Diberi ketumpatan blok kayu tersebut ialah
0.9 kgm–3. Cari jisim, dalam kg, blok kayu tersebut = RM30.865
dalam nombor tunggal. = RM30.90

[2 marks / 2 markah]

HOTS Challenge HOTS Extra

Write 3.45 × 10–7 seconds as nanoseconds.
Tulis 3.45 × 10–7 saat sebagai nanosaat.

3.45 × 10–7
= 3.45 × 10–7 × 109
= 3.45 × 102
= 345 nanoseconds

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TER Mathematics Form 3 Chapter 1 Indices

CHAP1 Indices
Indeks

1. (a) False 5. (a) 35 ÷ 32 310 37 33
(b) True (b) 75 ÷ 73
(c) False (c) 28 ÷ 23 72 715 78
(d) True (d) m9 ÷ m8
(e) False (e) h12 ÷ h4 211 25 824

2. (a) 24 27 m17 m2 m
729
 (b)12 h16 h8 h3
8 1
64
(c) 33 16 6. (a) ✓
625 (b) ✗
(d) 93 (c) ✗
(e) 54 (d) ✓
(e) ✗
3. (a) 125 = 5 × 5 × 5 = 53
7. (a) (5 × 2)2
(b) 36 = 6 × 6 = 62 = 52 × 22

(c) 1 000 000 = 10 × 10 × 10 × 10 × 10 × 10 (b) (6 × 2)3
= 106 = 63 × 23

(d) 64 = 8 × 8 = 82 (c) (83 × 92)5
= 83 × 5 × 92 × 5
   (e)12 = 1 × 1 = 12 = 815 × 910
36 6 6 6
(d) (64 × 55)4
4. (a) 22 × 23 = 64 × 4 × 55 × 4
= 22 + 3 = 616 × 520
= 25
(e) (33 × 42 × 5)3
(b) 6 × 62 = 33 × 3 × 42 × 3 × 51 × 3
= 61 + 2 = 39 × 46 × 53
= 63
8. (a) 35 × (34)2 ÷ 39
(c) 47 × 43 = 35 + 4 × 2 – 9
= 47 + 3 = 34
= 410
(b) 87 × 83 ÷ (83)2
(d) 58 × 54 = 87 + 3 – 3 × 2
= 58 + 4 = 84
= 512
(c) (m3)4 × (m5)2 ÷ (m2)8
(e) w7 × w6 = m3 × 4 + 5 × 2 – 2 × 8
= w7 + 6 = m12 + 10 – 16
= w13 = m6

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A

Mathematics Form 3 Chapter 1 Indices (b) 24n × 82n × 32–2n
= 24n × (23)2n × (25)–2n
1 = 24n × 26n × 2–10n
= 24n + 6n – 10n
9. (a) 83 = 38 = 2
= 20
1
=1
(b) 164 = 416 = 2

3

164 =
( )(c)
416 3 = 8

11 1

( )2 (c) 22 × 52 × 102

(d) 2163 = 3216 2 = 36 11 1

= 22 × 52 × (2 × 5)2

1 1 = 1 + 1 × 1 + 1
33 27 2 2
(e) 3–3 = = 22 52

=2×5

= 10

(f) 2–4 = 1 = 1
24 16
1 3
64– 3 ×
92
(d)
(g) (–9)0 = 1 4–3

(43)– 1 × 3
3
(32) 2
=
 (h)1 0 4–3
10
=1 = 43 × – 1 – (–3) × 32 × 3
3 2

10. (a) 3–4 × 38 = 42 × 33
3–3 = 16 × 27

= 3–4 + 8 – (–3) = 432

= 37

27– 2 × 2
3
64 3
12p3q5 (e)
(b) 22p–4q–6 3–4

12 (33)– 2 × 2
4 3
(43) 3
= × p3 – (–4)q5 – (–6) =
3–4

= 3p7q11 = 33 × – 2 – (–4) × 43 × 2
3 3

(c) m9n–3 × m–3 = 32 × 42
m5n–2 = 9 × 16

= m n9 – 3 – 5 –3 – (–2) = 144

= mn–1

= m (f) 1 × 1 ÷ 8
n 8
(83) 2

3 5– 1 15– 3 = 8– 1 + 3 × 1 – 1
2 2 2 2
11. (a) 32
× × = 8– 1 + 3 – 1
2 2
3 1 3
= × 5– 2 × (3 × 5)– 2 = 80
32

= 33 – 3 × 5– 1 – 3 =1
2 2 2 2

= 30 × 5–2

= 1 × 1
52
= 1
25

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(g) (12 × 1 ÷ 8– 2 Mathematics Form 3 Chapter 1 Indices
3
27)2 (b) 5n × 5 = 55 × 53
1 2 5n + 1 = 55 + 3
= (22 × 3 × ÷ (23)– 3 n+1=8
33) 2 n=7

= 2 × 32× 1 – 3 × – 2 (1 + 3) × 1
2 3 2

= 23 × 32

=8×9 (c) 32 = 9n
32 = 32n
= 72 2 = 2n
n=1
12. (a) 2x3y2 × 5xy2
= 2 × 5x3 + 1y2 + 2 (d) (bn)3 = b9
= 10x4y4 b3n = b9
3n = 9
(b) 81m3n2 ÷ 27m4n2 n=3

= 34m3n2 ÷ 33m4n2

= 34 − 3m3 − 4n2 − 2
3
= m (e) 27(5n) = 125(3n)

 5n 125
27
a4 –1 3 =
b
 (c)
(3a3b–1)2 ×   5n 53
3
= 9a3 × 2 b– 4 –1 × 2+1 3 =

= 9a2b–1 n=3

= 9 a2
b

(d) (hk2)2 × h3k–3
h4k–4

= h k1 × 2 + 3 – 4 2 × 2 – 3 – (–4)

= h2 +3 – 4k4 – 3 + 4

= hk5

5

(e) (2r2s–3)3 × r2s4

1

8r 2s2

= 23 × r s2 × 3 + 5 – 1 –3 × 3 + 4 – 2
23 2 2

= r s6+ 5 – 1 –9 + 4 – 2
2 2

= r8s –7

13. (a) 212 ÷ 8 = 2n
212 ÷ 23 = 2n

212 – 3 = 2n
29 = 2n
n=9

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A

Mathematics Form 3 Chapter 1 Indices

PT3 Standard Practice 1 6. A (3a–1b2)2 × 7a3b–3
= 9a–2b4 × 7a3b–3
Section A = (9 × 7)a–2 + 3b4 – 3
= 63ab
 1 4
1. 5m 4 = 3
m5
B 7c × 70 ÷ 49c2 × 3343c 6
1
= ×4 13
m5
= 7c2 × 1 ÷ 72c2 × 7c2
1 3
4 = 7 c1 – 2 + 1 2 – 2 + 2

= m5 =c

Answer: C 1 5 1 1
5 3 5 3
C d2 – = d

13 3

7k2 = × 73k2
 2. 7–1k2
7–1k2 × = 1 3d
33 5
3
(72) 2 k 2
(49k) 2 3 3 1
2 2 53d
= 7 k–1 + 3 – 3 2 + – ≠

= 7–1k2 D e × e × e × e × e × e = e6

= 1 k2 Answer: C
7

Answer: A

3. 3q4 × 5q5 = (3 × 5)q4 + 5 Section B a6
= 15q9
1.
Answer: A a5 × a3

(a3)2 a8

4. 2k + 1 × 2–3k = 32 a–3 2
2k + 1 –3k = 25
1 – 2k = 5 a3
2k = –4
k = –2 3a2 —a13

Answer: D

5. A y6 = y6 – 5 3r 3 s 4 3
y5 =y 4

2. 5r –3s7 × (16r2s–6)2

B y7 × y4 = y7 + 4 3 ×4 —1 ×4
= y11 2
34r4 s 42 3 r2 3 s–6 3
= × × 2 × × 2 × × 2

C y3 + y8 5r –3s7

D (y11)0 = 1 = 81r3s2 × 43 r3s–9
5r –3s7
Answer: B
5 184
= 5 r 3 + 3 – (–3) × S2 – 9 – 7

= 5 184 r9s –14
5

© Penerbitan Pelangi Sdn. Bhd. 4

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Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd

Mathematics Form 3 Chapter 1 Indices

Section C (ii) a = 2, r = 2, n = 10

1. (a) (i) 368 1 × 3 +2 020 = 1 105 +2 020 sn = a(rn – 1)
3 = 3 125 r–1

=55 s10 = 2(210 – 1)
2–1

(ii) (6 – 3.808) ÷ 274 = 2.194 ÷ 274 = 2 046

= 0.008
1
= 125 HOTS Challenge

= 5–3

(b) [(–3k)4m + 6]–3 = 1 Rounds

(–3k)–12m – 18 = (–3k)0 Pusingan 0123456

–12m – 18 = 0

–12m = 18 Players 64 32 16 8 4 2 1

m = – 18 Pemain
12
Match played 32 16 8 4 2 1 –
m = –1.5
Perlawanan dimain

(c) (i) k–1 × k4 Total match played:
= k –1 + 4 32 + 16 + 8 + 4 + 2 + 1 = 63

= k3

5 © Penerbitan Pelangi Sdn. Bhd.

A

CHAPTER

2 Standard Form
Bentuk Piawai

1. Numbers Significant (d) 55.6 – 34.5
figures 109 + 58
Nombor
Angka bererti = 21.1
(a) 3.02 167
(b) 90 000 1
(c) 3.0002 = 0.12635
(d) 4.348 3
(e) 0.0045 = 0.1
4
(e) (31.2 + 0.9) × (4.5 – 0.87)
5 = 32.1 × 3.63
= 116.523
2
= 117

2. (a) 950 6. (a) 545 = 5.45 × 102
(b) 600 (b) 65 100 = 6.51 × 104
(c) 3860 (c) 26 = 2.6 × 10
(d) 620 (d) 4 300 000 = 4.3 × 106
(e) 8820 (e) 1 037 = 1.037 × 103

3. (a) 3.50 7. (a) 0.0045 = 4.5 × 10–3
(b) 5.56 (b) 0.00025 = 2.5 × 10–4
(c) 81.98 (c) 0.000076 = 7.6 × 10–5
(d) 9.7 (d) 0.00000056 = 5.6 × 10–7
(e) 7.0 (e) 0.9093 = 9.093 × 10–1

4. (a) 0.49 8. (a) 5 × 103 = 5 000
(b) 0.0703 (b) 5.5 × 10 = 55
(c) 0.2198 (c) 3.46 × 103 = 3 460
(d) 0.46 (d) 8 × 105 = 800 000
(e) 0.03 (e) 3.69 × 10 = 36.9

5. (a) 5.23 × 1 025 ÷ 50 9. (a) 1 × 10–4 = 0.0001
= 5.36075 × 103 ÷ 50 (b) 6.5 × 10–3 = 0.0065
= 107.215 (c) 8.6 × 10–4 = 0.00086
= 100 (d) 9 × 10–6 = 0.000009
(e) 3.68 × 10–6 = 0.00000368
(b) 42.5 × 8.12 ÷ 400
10. (a) 4.44 × 103 + 0.21 × 103
= 345.1 ÷ 400 = (4.44 + 0.21) × 103
= 0.86275 = 4.65 × 103
= 0.863
(b) 4.5 × 105 – 3.2 × 104
(c) 980 – 6.13 ÷ 0.02 = 4.5 × 105 – 0.32 × 105
= 980 – 306.5 = (4.5 – 0.32) × 105
= 673.5 = 4.18 × 105
= 670

© Penerbitan Pelangi Sdn. Bhd. 6

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Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd

(c) 640 × 110 ÷ 1 800 Mathematics Form 3 Chapter 2 Standard Form
= 70 400 ÷ 1 800
= 7.04 × 104 ÷ (1.8 × 103) PT3 Standard Practice 2
= 3.91 × 104 – 3
= 3.91 × 10 Section A
1. 6.47 × 10–4 = 0.000647
(d) 2.63 × 104 × 3.2 × 10–2 Answer: A
= 2.63 × 3.2 × 104 + (–2)
= 8.42 × 102 2. 3.43 × 10–3 = 3.43 × 10–3
700 000 7 × 105
(e) 1.5 × 10–2 × 5.1 × 103 ÷ (4 × 102)
= (1.5 × 5.1 ÷ 4) × 10–2 + 3 – 2 = 0.49 × 10–3 – 5
= 1.91 × 10–1 = 4.9 × 10–9

Answer: C

11. (a) (i) 3. 2k – 3s
= 2(6.14 × 103) – 3(1.09 × 104)
China Africa = 1.228 × 104 – 3.27 × 104
= –2.042 × 104
Cina Afrika
Answer: C
Population 1.32 × 109 8.32 × 108

Jumlah Penduduk

Land area (km2) 4. Answer: D

Luas tanah (km2) 9.60 × 106 2.66 × 107

(ii) Number of people per km2 in China 5. A 3.0862 = 3.09 (3 s.f.)
= 1.32 × 109 ÷ 9.60 × 106 B 3.0785 = 3.08 (3 s.f.)
= 0.1375 × 103 C 3.0713 = 3.07 (3 s.f.)
= 1.375 × 102 D 3.8045 = 3.80 (3 s.f.)

Number of people per km2 in Africa Answer: B
= 8.32 × 108 ÷ 2.66 × 107
= 3.13 × 10 Section B Number of
1. (a) significant
Number Answer
figures
Nombor Jawapan
Bilangan
(i) 0.0183 angka bererti 0.018
(ii) 299 530 300 000
(iii) 43 2
43.00
3

4

(b) 0.000045 = 4.5 × 10–5

+1

2. (a) 299 530  300 000
(b) 0.000000085 ÷ 1.7 × 10–9
= 8.5 × 10–8 ÷ 1.7 × 10–9

= 8.5 ÷ 1.7 × 10–8 + 9
= 50

7 © Penerbitan Pelangi Sdn. Bhd.

A

Mathematics Form 3 Chapter 2 Standard Form

Section C (b) (i) Volume of cylinder = πr2h
22
1. (a) k = 7, l = – 6 = 7 × (0.07)2 × 0.5

∴ k × l = 7 × (–6) = – 42 = 7.7 × 10–3 m3

(b) (i) 5.43 = 5.43 (ii) Density = Mass
30.1 – 0.54 29.56 Volume

= 1.84 × 10–1 0.9 kgm–3 = Mass
7.7 × 10–3
(ii) (6.09 × 102) ÷ (2.31 × 10–3)
= 6.09 ÷ 2.31 × 102 – (–3) mass = (0.9 × 7.7 × 10–3) kg
= 2.64 × 105
= 6.93 × 10–3 kg

= 0.00693 kg

(c) (i) 2 356 = 2.36 × 103 m (c) Total price

(ii) 1.05 × 1011 = k × 109 = RM5.99(3.5) + RM9.90
k = 1.05 × 102 billion
= RM30.865

= RM30.90

2. (a) 9 000 seconds = 9 000 = 2.5 h HOTS Challenge
3 600
3.45 × 10–7
Distance= Speed × Time = 3.45 × 10–7 × 109
= 3.45 × 102
= 75 kmh–1 × 2.5 = 345 nanoseconds

= 187.5 km

= 187 500 m

= 1.875 × 105 m

© Penerbitan Pelangi Sdn. Bhd. 8

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CHAPTER Consumer Mathematics: Saving and Investments, Credits and
Debts
3
Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang

1. Savings Investment

Simpanan Pelaburan

Fixed deposits Shares Current account Real estates Unit trust

Simpanan tetap Saham Simpanan semasa Hartanah Unit amanah saham

2. (a) Simple interest, I  3.
(a) Matured value = 5 000 1 + 0.032 4(3)
4
= RM2 000 × 1.6 × 3
100 12 = RM5 501.69
= RM8
(b) Matured value
Total saving = RM2 000 + RM8
= RM2008  =
15 000 1 + 0.043 1(10)
1
1.8
(b) Simple interest, I = 3 000 × 100 × 2 = RM22 852.53

= RM108 (c) Matured value

Total saving = RM3 000 + RM108  =
= RM3 108 8 000 1 + 0.035 12(7)
12

(c) Simple interest, I = 5 000 × 2.2 × 9 = RM10 217.33
100 12
= RM82.50 4. (a) ✓
(b) ✓
Total saving = RM5 000 + RM82.50 (c) ✗
= RM5 082.50

5. (a) Total return = RM7 000 + RM100 – RM6 500

= RM600
600
ROI = 6 500 × 100%

= 9.23%

(b) 12.5% = (13 200 + 2x) – 12 000 × 100%
12 000
0.125(12 000) + 12 000 = 13 200 + 2x

2x = 300

x = RM150

(c) (i) ROI = RM1 500 – RM2 000 × 100%
RM2 000

= –25%

(ii) Change of government policy could cause loss in investment.

9 © Penerbitan Pelangi Sdn. Bhd.

A

Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts

6. (a) Return / Pulangan (b) (i) Average cost per share
(b) Uncertainty / Ketidakpastian
(c) Cash / Wang tunai 5 000(1.68) + 4 000(1.56) +

= 3 000(1.52)
5 000 + 4 000 + 3 000

= RM1.60

7. Saving Real Shares (ii) Cost averaging strategy is used.
account estate This strategy can reduce the
Saham investment risk by bringing down
Akaun Hartanah average purchasing cost.
simpanan

Liquidity High Low Moderate (c) (i) Average cost per share

Kecairan Tinggi Rendah Sederhana 2 000(0.55) + 2 000(0.53) +

8. (a) 0.75 = RM2 100 = 2 000(0.56)
total unit 2 000 + 2 000 + 2 000

acquired = RM0.55

Total unit acquired = RM2 100 (ii) Jeremy’s strategy is more profitable
0.75 because the average cost per share
is lower compared to Firdaus’s
= 2 800 units purchase.

 10 000 = x 1 +
9. (a) 0.0325 4(5)
4

10 000 = x(1.008125)20

x = 10 000
1.00812520

x = RM8 505.74

(b) (i) 1.35 = amount paid
2 000

Amount paid = RM2 700

(ii) 8% = 2 000x – 2 700 × 100%
2 700

0.08(2 700) + 2 700 = 2 000x

x = RM1.46 per share

 (c)
(i) MV = 12 000 1 + 0.032 1(2)
1

= RM12 780.29

ROI = 12 780.29 – 12 000 × 100%
12 000

= 6.50%

(ii) 2(6.50%) = [8 000(1.27) + 180] – x × 100%
x

0.13x = 10 340 – x

1.13x = 10 340

x = RM9 150.44

© Penerbitan Pelangi Sdn. Bhd. 10

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Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd

10. (a) FALSE Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts
(c) TRUE
(b) TRUE
(d) TRUE

11. Make personal budget. Always late in making repayment. Start saving.

Membuat belanjawan peribadi. Selalu membuat pembayaran balik dengan lewat. Memulakan simpanan.

Smart management of credits and debts.

Pengurusan kredit dan hutang yang bijak.

Use auto billing to pay bill. Manage credit card wisely. Excessive loan.

Membayar bil secara automatik. Menguruskan kad kredit dengan bijak. Meminjam dengan banyak.

12. Late payment charge = RM0

A Current amount in April statement
= RM1 710 + RM30.60 + 0
Weakness = RM1 740.60
of credit
card usage (b) (i) Finance charge
1.5 46
Kekurangan = RM2 500 × 100 × 30
penggunaan
= RM57.50
kad kredit
D Late payment charge
C
= (RM2 500 + RM57.50) × 1%
= RM25.58

13. (a) Finance charge Current amount in May statement
= RM2 500 + RM57.50 + RM25.58
Caj kewangan = RM2 583.08

(b) Late payment charge (ii) Additional repayment amount
= RM2 583.08 – RM2 500
Caj bayaran lewat = RM83.08

(c) Total amount of outstanding balance (c) (i) USD1.00 = RM3.92

Jumlah baki tertunggak

14. (a) Outstanding balance Toy price in Ringgit Malaysia
= RM1 800 – RM90
= RM1 710 = USD78 × RM3.92
USD1.00

Finance charge = RM305.76

 = 1.5 15 Current amount on July statement
RM1 800 × 100 × 30
= RM305.76 + (RM305.76 × 1%)
= RM308.82

 + RM1 710 × 1.5 × 20
100 30

= RM13.50 + RM17.10
= RM30.60

11 © Penerbitan Pelangi Sdn. Bhd.

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Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts

(ii) Outstanding balance (b) (i) Total repayment
= RM308.82 – RM16
= RM292.82 = RM40 000 + (RM40 000 × 0.05 × 4)

= RM48 000

Finance charge Instalment = RM48 000
4 × 12
 = 18 18
RM308.82 × 100 × 365 = RM1 000

+ (RM292.82 × 18 20 (ii) Total repayment
100 365
× = RM40 000 + (RM40 000 × 0.05 × 3)

= RM2.74 + RM2.89 = RM46 000

= RM5.63 Instalment = RM46 000
3 × 12
Late payment charge = RM0 = RM1 277.78

Current amount Amount James needs to top up
= RM292.82 + RM5.63+ RM0
= RM298.45 = RM1 277.78 – RM1 000

= RM277.78

15. (a) Total repayment Total
= RM110 000 + (RM110 000 × 0.05 × 9)
= RM159 500 (c) (i) 4 200 × 1 = repayment
4 7 × 12

Instalment = RM159 500 Total
9 × 12
= RM1 476.85 1 050 = repayment
84

(b) Total repayment Total repayment = RM88 200

= RM80 000 + (RM80 000 × 0.07 × 5) Let r = annual interest rate
RM88 200 = RM64 000 +
= RM108 000
(RM64 000 × r × 7)
Instalment = RM108 000 r = 0.054
5 × 12 r = 5.4%
= RM1 800
(ii) Total repayment
(c) Total repayment
= RM64 000 + (RM64 000 × 0.035
= RM50 000 + (RM50 000 × 0.06 × 4) × 5)

= RM62 000 = RM75 200

Instalment = RM62 000 Instalment = RM75 200
4 × 12 5 × 12
= RM1 291.67
= RM1 253.33
16. (a) (i) Total repayment
Dahniah cannot shorten the loan
= RM81 000 + (RM81 000 × 0.08 × 5)
period because the instalment for 5
= RM113 400 1
years is more than 4 of her salary.
Instalment = RM113 400
5 × 12
= RM1 890

(ii) Let t = new loan period

= RM81 000 + (RM81 000

× 0.08 × t)
= RM81 000 + 6 480t

1 290 = RM81 000 + 6 480t
t × 12
15 480t = 81 000 + 6 480t

t = 9 years

© Penerbitan Pelangi Sdn. Bhd. 12

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Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd

Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts

PT3 Standard Practice 3 3. Answer: D

Section A 4. Answer: D

1. Answer: C 5. P = RM230, r = 5.5% = 0.055

2. Initial capital Interest = RM177.10
= (RM2.05 × 6 000) + (RM1.00 × 2 700)
= RM15 000 Interest = Prt
177.10 = 230(0.055)t
Dividend 177.10 = 12.65t

= RM135 × 3 t = 14 years
= RM405
Answer: B

Capital gains 6. Finance charge
= RM16 500 – RM15 000 1.5 50
= RM1 500 = RM1 200 × 100 × 30

= RM30

Total returns Late payment charge

= RM1 500 + RM405 = (RM1 200 + RM30) × 1%
= RM12.30
= RM1 905

ROI = 1 905 × 100 Total outstanding in August
15 000 = RM1 200 + RM30 + RM12.30
= RM1 242.30
= 12.7%

Answer: A Answer: C

Section B Simple interest Duration Simple interest Total savings
1. Prinsipal rate per annum rate
Tempoh Jumlah simpanan
Prinsipal Kadar faedah mudah Faedah mudah
setahun 1 year diperoleh (RM)
(RM) 4months
6.8% I = 24 820 389 820
(a) 365 000 4 Bulan 2 619.17
(b) 2 500 14.3% I = 119.17

2. Savings How easily a savings or an asset can be converted into cash.

Simpanan Keupayaan simpanan atau aset diubah menjadi wang tunai.

Investment An action of gaining profit in the future from income and capital
gain.
Pelaburan
Tindakan mendapat pulangan pada masa hadapan melalui pendapatan dan
Liquidity keuntungan modal.

Kecairan Money that is kept in the bank or other than bank like piggy
bank.
Interest
Wang yang disimpan di bank mahupun bukan bank seperti dalam tabung.
Faedah
Charge on the money borrowed or invested.

Caj ke atas wang yang dipinjam atau dilaburkan.

13 © Penerbitan Pelangi Sdn. Bhd.

A

Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts

Section C HOTS Challenge

1. (a) Finance charge

= RM900 × 1.5 × 48 (a) Total repayment
100 30
= RM90 000 + (RM90 000 × 0.034 × 8)
= RM21.60
= RM114 480

Late payment charge Instalment = RM114 480
8 × 12
= (RM900 + RM21.60) × 1% = RM1 192.50
= RM9.22

Late payment charge = RM10.00  (b) 1(8)
Matured value = RM10 000 1 + 0.034
Current amount in November statement 1
= RM900 + RM21.60 + RM10 = RM13 066.65
= RM931.60
Car loan interest for RM10 000
(b) P = RM20 000, r = 5 per annum,
t = 3 years 100 = RM10 000 + (10 000 × 0.034 × 8)
= RM12 720

(i) Half yearly, n = 2 Shamsiah should not use the money in fixed
deposit account because the interest gained
 = 1 + 0.05 2(3) in fixed deposit account is higher than the
RM20 000 2 interest in car loan.

= RM23 193.87

(ii) Quarterly, n = 4

 = 1 + 0.05 4(3)
RM20 000 4

= RM23 215.09

(c) Down payment
= RM3 699 × 10%
= RM369.90

Amount financed
= RM3 699 – RM369.90
= RM3 329.10

Finance charges
= 8% × RM3 329.10
= RM266.33

© Penerbitan Pelangi Sdn. Bhd. 14

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Top One (MATHS F3) Penerbitan Pelangi Sdn BhdCHAP
TER

4 Scale Drawings
Lukisan Berskala

1. (i) 1 (b) Scale = length of drawing
actual length
(ii) drawing / lukisan
(iii) 10 = 40
4
(iv) actual object / objek sebenar
(v) cm = 10
1

2. Diagrams Scale drawings Scale = 1
1
Rajah Lukisan berskala Skala
10
1
I Yes 1:2 = 1 : 10

II No – height of drawing
actual height
III No – (c) Scale =

IV No – = 1.4 4
× 100

V Yes 1 : 1 = 1
2 35

= 1 : 35

3. (a) 10 cm : 1 km 5. (a) 3 = 1
= 10 cm : 1 × 100 000 cm actual length 10 000
= 1: 10 000
actual length = 3 × 10 000
(b) 2 cm : 1m
= 2 cm : 1 × 100 cm = 30 000 cm
= 1: 50
= 0.3 km
(c) 5 cm : 2 km
= 5 cm : 2 × 100 000 (b) 5 = 1
= 1 : 40 000 actual length 2 000

actual length = 5 × 2 000

= 10 000 cm

= 100 m

4. (a) Scale = height of drawing (c) 9 = 1
actual height actual length 1

= 2 2 1
600 2
actual length = 9 ×
= 1
300 = 4.5 cm

= 1 : 300

15 © Penerbitan Pelangi Sdn. Bhd.

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Mathematics Form 3 Chapter 4 Scale Drawings

6. (a) length of drawing = 1 8. (a) Scale = 1 : 1
actual length 200 000 2
length of drawing 1
length of drawing = 1 length of object = 1
12 000 000 200 000
2
= 60 cm 8 1
of 1
length object =

(b) Length of square field = 900 2 1
= 30 m 2
Length of object = 8 ×

length of drawing = 1 = 4 cm
actual length 300

length of drawing = 1 1.5 cm
3 000 300 3 cm

Length of drawing = 3 000
300
1.0 cm 1.0 cm
0.5 cm
= 10 cm
1.0 cm

(c) Diameter of coin = 5 cm 1.5 cm

length of drawing = 1 4 cm
actual length 1

2

length of drawing = 1 9. (a) Length of square A = 49
5 1 = 7 cm

2

Diameter on the drawing = 5 × 2 Length of square B = 196
= 10 cm = 14 cm

length of drawing 1 length of drawing = 14
actual length 100 actual length 7
7. (a) =
= 1
1 7
base of drawing = 12 × 100 × 100
14
= 12 cm
1
height of drawing = 6 × 100 × 1 = 1
100
= 6 cm 2

Therefore, the scale used is 1 : 1
2

(b)
4 1
actual distance = 3 000 000

actual distance = 4 × 3 000 000 cm

= 12 000 000 ÷ 100 000 km

= 120 km

© Penerbitan Pelangi Sdn. Bhd. 16

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Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd

(c) 30 = 1 Mathematics Form 3 Chapter 4 Scale Drawings
height of building 600
PT3 Standard Practice 4

height of building = 30 × 600 Section A
= 18 000 cm
1. 2 cm : 1 m = (2 cm : 100 cm) ÷ 2
= 18 000 ÷ 100 m = 1 : 50

= 180 m Answer: A

10. (a) (i) Dimension:
Actual length = 2.5 + 1.5 + 4 = 8 m
Actual width = 6 m 2. 8 = 1
Actual height 7 500

(ii) Area of the living room = 4 × 3 Actual height = (8 × 7 500) cm
= 12 m2 = 60 000 cm

= 600 m

(iii) Area of bedroom 1 Answer: B
= (3 × 4) – (1 × 0.5)
= 11.5 m2 3. Answer: B

Total cost of carpet 4. Radius of the swimming pool
= 11.5 × RM60 = (5 – 0.75 – 0.75) ÷ 2
= RM690 = 1.75 cm

(iv) Height of bathroom 1 1.75 = 1
Actual radius 400
= 2 ÷ (1 × 0.5)
=4m

Actual radius = 1.75 × 400

= 700 cm

=7m

Area of the swimming pool = pr2
22
= 7 × (7)2

= 154 m2

Answer: D

5. Length of KL in drawing

135 = 1 × 15 × KL
2

KL = 135 × 2
15

= 18 cm

Actual length of KL

= (18 × 10 000) cm
= 180 000 cm

= 1.8 km

Answer: C

17 © Penerbitan Pelangi Sdn. Bhd.

A1

Mathematics Form 3 Chapter 4 Scale Drawings

6. height of painting = 1 Section C
(1.1 × 100) 20 1. (a) (i)
Length of LM = 152 – 122
height of painting = (1.1 100) ÷ 20
= 9 cm
= 5.5 cm
Length of drawing 1
Answer: B 9 = 1

3

Section B Length of drawing = 9 × 3

= 27 cm

1. Object Scale drawing Scale
Objek Lukisan berskala Skala
(ii) Length of KL in drawing = 1
(a) 12 1

1:3 3

18 cm Length of KL in drawing = 12 × 3

6 cm = 36 cm

(b) Area of KLM in drawing
1
5 cm = 2 × 27 × 36

10 cm 1 : 1 = 486 cm2
2

(b) (i) 1 cm
1 cm
(c)
7 m2 1 : 1
7 m2

(d)

3 m 15 cm 1 : 20 (ii)

The scale drawing is bigger 2 cm
than the object.
2 cm
Lukisan berskala adalah lebih
2. (a) besar daripada objek.

1:3 The scale drawing is the
same as the object.
1 : 1 (c) 0.6825 = 1
2 Lukisan berskala adalah sama Actual height 600 000
saiz dengan objek.
Actual height = 0.6825 × 600 000
The scale drawing is = 409 500 cm
smaller than the object.
= 4.095 km
Lukisan berskala adalah lebih
kecil daripada objek.

(b) 5 cm : 1 m = (5 cm : 100 cm) ÷ 5
= 1 : 20

© Penerbitan Pelangi Sdn. Bhd. 18

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4. (a) (i) Scale Mathematics Form 3 Chapter 4 Scale Drawings
= area of drawing : Actual area
HOTS Challenge

= 400 : 25 Length of bedroom 1 in the plan
= (16 – 5 – 2(2.5))
= 20 : 5 = 6 cm

= 1 : 1
4

(ii) New scale Width of bedroom 1 in the plan
= Length of drawing : actual length =3+3
=2:4 = 6 cm
=1:2

(b) length of square= 324 6 = 1
= 18 cm Actual length 50

Actual length = 6 × 50 cm

length of drawing = 1 = 300 cm
actual length 4
=3m

18 = 1 Actual width = 3 m
actual length 4
Area of bedroom 1 = 3 × 3
actual length = 4 × 18 = 9 m2

= 72 cm

(c) (i) Height of drawing = 40
actual length 2.8 × 1 000

= 1
70

Scale = 1 : 70

(ii) length of 42 board = 1
white 70

length of white board = 42 × 70
= 2 940 mm

= 294 cm

Length of white board : width of
white board
=3:2
= 294 : 196

Width of white board = 196 cm

Total length of red ribbon needed

= (294 × 2) + (196 × 2)
= 980 cm

19 © Penerbitan Pelangi Sdn. Bhd.

A1

CHAPTERMathematics Form 3 Chapter 5 Trigonometric Ratio

5 Trigonometric Ratios
Nisbah Trigonometri

1. (a) (i) adjacent side / sisi bersebelahan 6. (a) True
(ii) opposite side/ sisi bertentangan (b) False
(iii) hypotenuse / hipotenus (c) False
(d) False
(b) (i) opposite = BC (e) True
adjacent AB

(ii) opposite = BC 7. (a) sin 30° = PQ
hypotenuse AC PR

(iii) adjacent = AB PQ = 0.50
hypotenuse AC PR

PQ = 0.50
10
2. (a) tan θ
sin θ 5 PQ = 0.50 × 10
cos θ 13
= 5 cm
5
12 8. (a) cos x = AB
BC
12
13 = 12
13

3. (a) sin θ = O = 6 = 3 AB : BC = 12 : 13
H 10 5 = 24 : BC

A 8 4 BC = 13 × 2
H 10 5 = 26 cm

(b) cos θ = = = AC = 262 – 242
= 10 cm
O 6 3
(a) tan θ = A = 8 = 4

(b) sin x = 3
4
3
4. (a) cos 29° = 0.8746 9 = 4
(b) tan 80° = 5.671 AC
(c) sin 50° = 0.7660 9× 4
= 3
5. (a) θ = cos–10.55
= 56.63° = 12 cm

(b) θ = tan–10.15 (c) Let MB be a line perpendicular to BC
= 8.53°
AM = 102 – 82 = 6 cm
(c) θ = sin–10.87
= 60.46° tan z = MB
AM
8
= 6

= 4
3

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Mathematics Form 3 Chapter 5 Trigonometric Ratio

9. (a) cos ∠BAC = 4 PT3 Standard Practice 5
5

AB = 4 cm, AC = 5 cm Section A / Bahagian A

AD = AC2 + CD2 1. cos x = 3
5
= 52 + 122
= 13 cm NP = 3
10 5
cos ∠ADC = 12
13 NP = 3 × 10
5

(b) (i) tan x = tan ∠DAE NP = 6 cm
CE 4
BC = 3 NK = 2 × 6 cm
= 12 cm
4 = DE + 7
3 12
NL = 102 – 62
DE + 7 = 16 = 8 cm

DE = 9 cm

(ii) x= tan–1 4 tan y = NL
3 NK

= 53.13° = 8
12

(c) cos x = 3 = 2
5 3

AB = 3 Answer: C
30 5

AB = 3 × 30 = 18 cm 2. Assume the length of sides AB and BC is
5 2 cm and AD is the height of the equilateral
triangle.
BD = 302 – 182
= 24 cm AD = 22 – 12
= 3
tan y = DC
BD

= 10 sin 60° = AD
24 AB

y = tan–1 10  3
24 2
=
= 22.62°

Answer: B

3. SW = UW = 8 cm

sin x = UW = 8
VW 17

 x 8
= sin–1 17

x = 28.07°

Answer: D

21 © Penerbitan Pelangi Sdn. Bhd.

A1

Mathematics Form 3 Chapter 5 Trigonometric Ratio

4. sin x = 3 2. (a) sin x = AB ✓
5 AC ✓
✓
3 = 3 cos x/ kos x = BC
AC 5 AC

AC = 5 cm

tan y = 1.6 tan x = AB
BC
CD
5 = 1.6

CD = 8 cm (b) tan x= 0.4
x = tan–1 0.4
Answer: B = 21.8°

5. Answer: A Section C / Bahagian C

6. Angle between the line BK and the Plane 1. (a) (i) ∠PMQ
ADK = ∠BKA (ii) ∠PLQ

AK = AD2 + DK2 (b) (i) cos x = 8
= 242 + 72 17
8
= 25 cm NL = 17
76.5
10
tan ∠BKA = 25 NL = 36 cm

 ∠BKA= 10
tan–1 25 (ii) tan y = 1

∠BKA = 21.8° NL = 1
LM

Answer: D 36 = 1
LM

LM = 36 cm

Section B tan z = 36
1. (a) tan 45° 36
1
(b) sin 60° 2 z = tan–1 1
cos 60°
1 z = 45°
(c) kos 60° 2
(d) sin 45°
3
2

1

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Mathematics Form 3 Chapter 5 Trigonometric Ratio

(c) (i) cos 40° = AC HOTS Challenge
22

AC = 22 cos 40°

AC = 16.85 m Let the length of ladder be x

BC cos 65° = 1.6
22 x
(ii) sin 40° =
1.6
BC = 22 sin 40° 0.4226 = x

BC = 14.14 m x = 3.79 m

tan 52° = DC
14.14

DC = 14.14 tan 52°

DC = 18.10 m

23 © Penerbitan Pelangi Sdn. Bhd.

A1

CHAPTER

6 Angles and Tangents of Circles
Sudut dan Tangent bagi Bulatan

1. Values 2. Relationship
Diagrams Circles /Bulatan between x
nilai and y
Rajah x y
x(°) O Hubungan
(a) 40 antara x dan y
28
x 42 x + y = 90°
O
80° x y x=y
O x=y
(b)
y
O
x° O
x
(c)

x
28°

(d) 2 cm xy

x 90 x = 2y
O
126°

6 cm

3. (a) x = 60° ÷ 2 = 30° 4. (a) ∠DCB = 150° ÷ 2 = 75°
(b) Major ∠AOC x = 180° – 75° = 105°
= 360° – 100°
= 260° Major ∠DOB = 360° – 150°
x = 260° ÷ 2 = 130° = 210°

(c) Minor ∠POQ = 360° – 280° = 80° x = 210° ÷ 2 = 105°
∠PRQ = 80° ÷ 2 = 40°

x = 180° – 40° =140°

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(b) ∠ACD = ∠ABD = 50° Mathematics Form 3 Chapter 6 Angles and Tangents of Circles
∠BCD = 50° + 60° = 110°
x = 180° – 110° (c) x = 40°
= 70° y = 75°

(c) ∠ABC = 180° – 85° = 95° 9. (a) ∠CGF = 60°
x = 180° – 40° – 95° ∠FGE = 70°
= 45° x = 180° – 60° – 70°
= 50°
5. (a) ∠DAE = ∠AEB = 30°
∠ADE = 180° – 30° – 70° (b) ∠SOT = 20° × 2 = 40°
= 80° ∠OTS = (180° – 40°) ÷ 2 = 70°
x = ∠ADE = 80° ∠QST = 110°
∠QTS = 180° – 110° – 20° = 50°
(b) 2x + x = 180° x = 70° – 50° = 20°
3x = 180°
x = 60° (c) ∠EDF = 40°
∠EDB = 60° + 40° = 100°
y = 2 × 60° = 120°
∠EFB = 180° – 100° = 80°
(c) ∠PQR = 180° – 70° – 45° = 65° ∠DFB = 80° – 38° = 42°
x = 180° – 65° = 115° ∠DEB = ∠DFB = 42°
x = ∠DEB = 42°
6. (a) Yes
(b) No
(c) No
(d) Yes
(e) No

7. (a) x = 90° – 43° = 47°
y = 43°
x + y = 90°

(b) x = 90°
y = 40°

(c) ∠ROQ = ∠POQ = 65°
x = 360° – 2(65°) = 230°
y = 90° – 65° = 25°

8. (a) x = 60°
y = 180° – 60° – 35°
= 85°

(b) x = 70°
y = 80° – 60°
= 20°

25 © Penerbitan Pelangi Sdn. Bhd.

A1

Mathematics Form 3 Chapter 6 Angles and Tangents of Circles

PT3 Standard Practice 6 5. x = ∠FEC = 50°

Section A ∠FOB = 360° – 78° – 90° – 50°
= 142°
1. ∠POQ = 180° – 35° – 35° = 110°
∠QOR = 180° – 63° – 63° = 54° y = 142° = 71°
x = 360° – 110° – 54° 2
= 196°
Answer: B ∴ x + y = 50° + 71° = 121°

2. OQ = PQ – OP Answer: D
=8–5
= 3 cm Section B ✓
1. ✓
QT = 52 – 32 Statement ✓
= 4 cm
= 40 mm Pernyataan

∴ ST = 2 × QT = 80 mm (a) ∠OFE ≠ 90°
Answer: C
(b) OB || PC

(c) ∠k = ∠m

(d) ∠m + ∠n = 180°

3. ∠ODC = ∠OCD = y 2.
(a)
= 180° – 100°
2
y = 40°
30° x = 30°
∠ABC + ∠ADC = 180° x x = 60°
120° + x + 40° = 180°
(b)
x = 180° – 120° – 40°
30°
x = 20° xO

∴ x + y = 20° + 40° = 60°

Answer: C

4. ∠MNL = 68° – 35° = 33° (c)

x = 90° – 33° = 57° xO x = 100°
100° x = 100°
∠NOK = 68° × 2
= 136° (d)

∠NKO= 180° – 136° = 22° 50°
2
O
∠NKL = ∠MNL
22° + y = 33° x

y = 11°

∴ y + x = 57° + 11° = 68°

Answer: A

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Section C Mathematics Form 3 Chapter 6 Angles and Tangents of Circles

1. (a) (i) ON = (5 – 1) cm 2. (a) ∠MOL= 38° × 2
= 4 cm = 76°

NB = 52 – 42 x = 180° – 90° – 76°
= 3 cm x = 14°
y = 180° – 76°
BD = 32 + 92 y = 104°
= 90
(b) x = ∠LKM × 2
(ii) tan ∠NDB = 3 = 30° × 2
9 = 60°

 ∠NDB 3 ∠KML = 90°
= tan–1 9 y = 180° – 90° – 30°
y = 60°
= 18.43°
(b) (i) x = 180° – 38°
∠ADB = 2 × ∠NDB = 142°
= 36.9°
(ii) ∠MGP = x (minor arc)
(b) (i) x = 90° – 50° = 142°
= 40°
∠MGP = 360° – 142° (major arc)
(ii) ∠LOA = 65° × 2 = 218°
= 130°
y = 218° ÷ 2
∠OAL = 180° – 130° = 109°
2
= 25° (iii) x + y = 142° + 109°
= 251°
∠BAL = ∠BLK
y + 25° = 50° (c) x = 180° – 90° – 70°
= 20°
y = 25°

(iii) x + y = 40° + 25° HOTS Challenge
= 65°
S = 2πr = 60 cm A
(c) ∠CAB = 40°
2 × 22 × OE = 60 cm l 30 cm
∠ABC = (180° – 40°) ÷ 2 7
= 70°
OE = 60 × 7 ÷ 44 = 9.55cm
∠CBE = 180° – 70°
= 110° AE = 302 + 9.552 = 31.48cm

x = 180° – 110° – 40° B/E r O
= 30°

27 © Penerbitan Pelangi Sdn. Bhd.

A1

CHAPTERMathematics Form 3 Chapter 7 Plan and Elevations

7 Plans and Elevations
Pelan dan Dongakan

1. View from Orthogonal (b) Plan / Pelan
projections
Pandangan Plan / Pelan
dari Unjuran ortogon
Y 4 cm
Plan/Pelan 2 cm

X 2 cm 4 cm

X

XY

Y 2 cm 2 cm
2 cm
4 cm

2. (a) ✓  ✗  ✓ (b) ✗  ✓  ✓
3. (a) (b)

4. (c) Plan / Pelan
(a)
Plan / Pelan Plan / Pelan Plan / Pelan 1.25 cm 1.25 cm

2 cm 1.5 cm 1.5 cm

4 cm 2 cm X 4 cm
4 cm
Y 2 cm 2 cm
X Y 2.5 cm
X
Y 2.5 cm

X Y

0.8 cm 1.5 cm 1.5 cm
2 cm
4 cm 4 cm

2 cm 2 cm

28 4 cm

2.5 cm

2 cm

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5. (a) Mathematics Form 3 Chapter 7 Plan and Elevations

QSuuakduraannItI II Quadrant I
FDroonngtakealenvdaetpioann
Sukuan I
A 4 cm B
SDidoengealkeavnastiiosin
4 cm
A/B

6 cm 6 cm

C D C/D
45°

A/C B/D

Plan / Pelan Quadrant IV
QSuuakduraannItIIIII
Sukuan IV

29 © Penerbitan Pelangi Sdn. Bhd.

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Mathematics Form 3 Chapter 7 Plan and Elevations

(b) QSuuakduraannItI II E QSuuakduraannIt I
SDidoengealkeavnastiiosin FDroonngtakealenvdaetpioann
A
5 cm A/E

2.17 cm

C/B D/F B/F 2.5 cm C/D
E D
45°
F

5 cm

Quadrant III B 1.25 cm A 1.25 cm C

Sukuan III Plan / Pelan
QSuuakduraannItVIV

© Penerbitan Pelangi Sdn. Bhd. 30

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(c) Quadrant II Mathematics Form 3 Chapter 7 Plan and Elevations

B Sukuan II Quadrant I
C/A
Side elevation Sukuan I

Dongakan sisi Front elevation

5 cm Dongakan depan

Q B/Q

R/P A/P C/R

D/E S/T E / T 2 cm D / S
45° P T Q SR

5 cm

Quadrant III AE B DC

Sukuan III Plan / Pelan
Quadrant IV

Sukuan IV

31 © Penerbitan Pelangi Sdn. Bhd.

A1

Mathematics Form 3 Chapter 7 Plan and Elevations

6. (a)

QSuuakduraannItI II QSuuakduraannIt I
SDidoengealkeavnastiiosin FDroonngtakealenvdaetpioann

A A

E/B 2 cm

BE

4 cm

D/C C 6 cm D
6 cm 45°

CB 6 cm

AE D
3 cm

Quadrant III Plan / Pelan
Quadrant IV
Sukuan III
Sukuan IV

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(b) QSuuakduraannItI II Mathematics Form 3 Chapter 7 Plan and Elevations
SDidoengealkeavnastiiosin
P QSuuakduraannIt I
FDroonngtakealenvdaetpioann
A P/A

4 cm B/E T/E 4 cm
Q/T Q/B
4 cm 4 cm

R/S 6 cm C/D S/D 4 cm R/C
A B/C
45°
E/D

6 cm

T/S 2 cm P 2 cm Q/R

Quadrant III Plan / Pelan
Quadrant IV
Sukuan III
Sukuan IV

(c)

33 © Penerbitan Pelangi Sdn. Bhd.

A1

Mathematics Form 3 Chapter 7 Plan and Elevations QSuuakduraannIt I
SDidoengealkeavnastiiosin
Quadrant II
P
Sukuan II
6 cm
Front elevation

Dongakan depan

P

3 cm B/A D/A 3 cm
C/D C/B
3 cm
3 cm F/G

F/E 6 cm G/H E/H
E D A/H 45°

6 cm

P

FC B/G
2 cm
4 cm
Quadrant IV
Plan / Pelan
Quadrant III Sukuan IV

Sukuan III

© Penerbitan Pelangi Sdn. Bhd. 34

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Mathematics Form 3 Chapter 7 Plan and Elevations

7. (a) QSuuakduraannItI II QSuuakduraannIt I
SDidoengealkeavnastiiosin FDroonngtakealenvdaetpioann
F/E
G/H E/H F/G

7 cm

B/A 6 cm C/D A/D 6 cm B/C
Quadrant III 45° D 2 cm C

Sukuan III H G

E 4 cm F
A B
6 cm
Plan / Pelan
QSuuakduraannItVIV

35 © Penerbitan Pelangi Sdn. Bhd.

A1

Mathematics Form 3 Chapter 7 Plan and Elevations QSuuakduraannIt I 4 cm
SDidoengealkeavnastiiosin
(b)
6 cm
QSuuakduraannItI II
FDroonngtakealenvdaetpioann 4 cm

6 cm

4 cm

4 cm

45°

4 cm QSuuakduraannItVIV
6 cm

Plan / Pelan
QSuuakduraannItIIIII

© Penerbitan Pelangi Sdn. Bhd. 36

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Mathematics Form 3 Chapter 7 Plan and Elevations

(c) QSuuakduraannItI II C/B A/B QSuuakduraannIt I D/C
FDroonngtakealenvdaetpioann SDidoengealkeavnastiiosin
D/A
5 cm 5 cm

P P
6 cm
6 cm

H/E G/F E/F H/G
A/E B/F 45°

5 cm P

D/H 5 cm C/G

Plan / Pelan QSuuakduraannItVIV
QSuuakduraannItIIIII

37 © Penerbitan Pelangi Sdn. Bhd.

A2

Mathematics Form 3 Chapter 7 Plan and Elevations

8. (a) (i) Length/Panjang = 5 × 5 = 25 cm PT3 Standard Practice 7
Width/Lebar = 2 × 5 = 10 cm
Height/Tinggi = 5 × 5 = 25 cm Section A
1. Answer: B
(ii) Volume/Isi padu 2. Answer: B
3. Answer: C
= 1 (10 × 25 × 5) + (10 × 25 × 20) 4. Answer: B
3 5. Answer: D
6. Answer: A
= 5416 2 cm3
3

20 cm 25 cm Section B
10 cm 25 cm 1.
Diagram
Othogonal
Rajah projection

(a) Unjuran
ortogon

✓

Y

(b)

Y ✓

(c)

Y

(d)

Y

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2. (a) The length of the edge KP on Mathematics Form 3 Chapter 7 Plan and Elevations
the pyramid is the same as its
(b) The actual plan
orthogonal projection.
✗ 1 cm
Panjang sisi KP pada piramid 1 cm
itu adalah sama dengan unjuran 4 cm G/C
ortogonnya. L/D 2 cm K H

(b) The size of ∠PNM on the ✓
pyramid is the same as its
orthogonal projection.

Saiz ∠PNM pada piramid itu adalah
sama dengan unjuran ortogonnya.

(c) The size of ∠KLP on the pyramid I/A J E F/B
is the same as its orthogonal
projection. ✗

Saiz ∠KLP pada piramid itu adalah Plan of solid in scale 1 : 2
sama dengan unjuran ortogonnya.
1 cm
(d) The length of the edge PL on 1 cm
the pyramid is the same as its
✗ L/D K H G/C
orthogonal projection.

Panjang sisi PL pada piramid
tersebut adalah sama dengan unjuran
ortogonnya.

Section C I/A J E F/B
3. (a) G/H

E/H F/G (c)

M/E 3 cm Z/W

I/L J/K

A/D B/C N/O Y/X 5 cm
B/A 6 cm C/D

39 © Penerbitan Pelangi Sdn. Bhd.

A2

Mathematics Form 3 Chapter 7 Plan and Elevations Object B:
Plan/Pelan:
HOTS Challenge
Front Elevation/Dongakan depan:
Object A:
Plan/Pelan:

Front Elevation/Dongakan Depan:

Side Elevation/Dongakan sisi:

Side Elevation/Dongakan sisi:

© Penerbitan Pelangi Sdn. Bhd. 40

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TER

8 Loci in Two Dimensions
Lokus dalam Dua Dimensi

1. (a) The path traced by a moving wheel (c)
of a bicycle on a horizontal plane is a
straight line. O

(b) The path traced by a moving swing is
an arc of a circle.

(c) The locus of a spinning ball is a sphere.
(d) The locus of a rolling rod is a cylinder.

2. (a)

3. (a) Q R

O

(b) P S
Q
O (b)

PR

S

41 © Penerbitan Pelangi Sdn. Bhd.

A2