Matematik (Book 1) Tingkatan 3

MODULPENTAKSIRAN UASA

Proaktif KSSM

Matematik
Dwibahasa
1Buku
(Bab 1, 3, 5, 7, 9) Ujian Akhir
Sesi Akademik

Bonus untuk
Guru

Praktis DSKP

Tahap Penguasaan (TP)

Standard Pembelajaran (SP) e-RPH

Soalan STEM 3Tingkatan

Ujian Akhir Sesi Akademik

(UASA)

0 Soalan KBAT
90
Sisipan PAK-21

Yong Sze Wei・Ong Jia Sheng

Kandungan

Pentaksiran Mata Pelajaran Matematik Tingkatan 3 KSSM iii - vi

CONTOHBab 1 - 13
1
1 Indeks 12
Indices
Praktis DSKP 14 - 29
Praktis Formatif 1 14
26
Bab
30 - 44
3 Matematik Pengguna: Simpanan dan Pelaburan, 30
Kredit dan Hutang 41
Consumer Mathematics: Savings and Investments,
Credit and Debt 45 - 56
Praktis DSKP 45
Praktis Formatif 2 54
36
Bab
57 - 70
5 Nisbah Trigonometri 57
Trigonometric Ratios 68
Praktis DSKP
Praktis Formatif 3

Bab

7 Pelan dan Dongakan
Plans and Elevations
Praktis DSKP
Praktis Formatif 4

Bab

9 Garis Lurus
Straight Lines
Praktis DSKP
Praktis Formatif 5

Jawapan J1 - J12

ii

Nama: Tarikh:

Bab Indeks

1 Indices

Praktis D S K P

PBD 1 Tatatanda Indeks / Index Notation Hal. Buku Teks: 2-6

A Tulis setiap tatatanda indeks yang berikut dalam bentuk pendaraban berulang.
Write each of the following index notation in the form of repeated multiplication.

SP 1.1.1 Menguasai TP 1
CONTOH
Contoh / Example: 1. (2.5)4 2. (rs)3
102 (rs)3 = rs × rs × rs
(2.5)4 = 2.5 × 2.5 ×
102 = 10 × 10 2.5 × 2.5

3. y 8 ( )4. 3 3 5. (–7)5
5
y8 = y × y × y × y × y × ( )3 3 = 3 × 3 × 3 (–7)5 = (–7) × (–7) × (–7)
y×y×y 5 555 × (–7) × (–7)

B Tulis setiap yang berikut dalam tatatanda indeks.
Write each of the following in index notation. SP 1.1.1 Menguasai TP 1

Contoh / Example: 1. 6 × 6 × 6 × 6 × 6 × 6 2. 1 × 1 × 1 × 1
8×8×8×8×8 66 2 2 2 2

85 ( )1 4
2

3. pqr × pqr 4. 1.29 × 1.29 × 1.29 × 1.29 5. (–jk) × (–jk) × (–jk)
pqr 2
(1.29)4 (–jk)3

TP1 Mempamerkan pengetahuan asas tentang bentuk indeks. 1

C Cari nilai bagi setiap nombor indeks berikut.
Find the value of each of the following index numbers. SP 1.1.2 Menguasai TP 2

Contoh / Example: 1. 34 2. 63
73 = 3×3×3×3 = 6×6×6
= 81 = 216
= 7×7×7
= 343

3. (0.4)2 4. (–1.4)3 ( )5. 1 4
3
= 0.4 × 0.4 = (–1.4) × (–1.4) ×
= 0.16 (–1.4)

= –2.744
CONTOH = 1 × 1 × 1 × 1
3 3 3 3

=1
81

D Tentukan indeks m dalam setiap yang berikut.
Determine the m index in each of the following. SP 1.1.2 Menguasai TP 2

Contoh / Example: 1. 125 = 5m 2. –32 = –2m
8 = 2m
125 = 5 × 5 × 5 –32 = –2 × –2 × –2 ×
8 = 2×2×2 = 53 –2 × –2
= 23
∴m=3 = –25
∴m=3 ∴m=5

3. 512 = 2m 4. 0.027 = 0.3m 5. 6 561 = 9m

512 = 2 × 2 × 2 × 2 × 2 × 0.027 = 0.3 × 0.3 × 6 561 = 9 × 9 × 9 × 9
2×2×2×2 0.3 = 94

= 29 = (0.3)3 ∴m=4
∴m=9 ∴m=3

2 TP2 Mempamerkan kefahaman tentang bentuk indeks.

E Ungkapkan nombor tunggal yang berikut dalam tatatanda indeks.
Express the following single numbers in index notation. SP 1.1.2 Menguasai TP 2

Contoh / Example: 1. 243 (asas / base 3) 2. 625 (asas / base 5)
64 (asas / base 4)
243 = 3 × 3 × 3 × 3 × 3 625 = 5 × 5 × 5 × 5
64 = 4 × 4 × 4 = 35 = 54
= 43

3. 256 (asas / base 2) 4. 343 (asas / base 7) 5. 729 (asas / base 9)

256 = 2 × 2 × 2 × 2 × 2 343 = 7 × 7 × 7 729 = 9 × 9 × 9
×2×2×2 = 73 = 93

= 28
CONTOH
F Tuliskan nombor berikut dalam bentuk indeks dengan menggunakan asas terkecil.
Write the following number in index form by using the smallest base. SP 1.1.2 Menguasai TP 2

Contoh / Example: 1. 81 2. 64 3. 625
16 81 = 34 64 = 26 625 = 54

16 = 24

PBD 2 Hukum Indeks Hal. Buku Teks: 6-24
Laws of Indices
2. (–4)2 × (–4) × (–4)3
A Permudahkan setiap yang berikut. = (–4)2 + 1 + 3
Simplify each of the following. SP 1.2.1 Menguasai TP 3 = (–4)6

Contoh / Example: 1. 35 × 36 × 34
22 × 23
= 22 + 3 = 35 + 6 + 4
= 315
= 25

TP2 Mempamerkan kefahaman tentang bentuk indeks. 3

3. m 4 × m 5 4. p × p 7 × p 6 5. q 3 × q 7 × q 9

= m4+5 = p1+7+6 = q3+7+9
= m9 = p 14 = q 19

B Permudahkan setiap yang berikut.
Simplify each of the following. SP 1.2.2 Menguasai TP 3

Contoh / Example: 1. 45 ÷ 43 2. 67 ÷ 6
26 ÷ 22
= 45 – 3 = 67 – 1
= 26 – 2 = 42 = 66
= 24
CONTOH 5. x10 ÷ x2 ÷ x6
3. a12 ÷ a9 4. b8 ÷ b7 = x 10 – 2 – 6
= x2
= a12 – 9 = b8–7
= a3 =b

C Permudahkan setiap yang berikut.
Simplify each of the following. SP 1.2.1 Menguasai TP 3

Contoh / Example: 1. 8k 4 × 9k 7
4m 3 × 5m 4
= 8 × 9 × k4+7
= 4 × 5 × m3 + 4 = 72k 11
= 20m7

2. 4y 2 × 6y 3. 2t 3 × t 2 × 5t
= 4 × 6 × y2+1 = 2 × 5 × t3+2+1
= 24y 3 = 10t 6

4 TP3 Mengaplikasikan kefahaman tentang hukum indeks untuk melaksanakan tugasan mudah.

D Permudahkan setiap yang berikut.
Simplify each of the following. SP 1.2.2 Menguasai TP 3

Contoh / Example: 1. 8k 3 ÷ 2k 2
20m6 ÷ 5m4
= 8 × k3–2
= 20 × m 6 – 4 2
5
= 4k

= 4m 2

2. 9y 5 ÷ 3y 2 3. 56t 12 ÷ 7t 3 ÷ 4t

= 9 × y5–2
3

= 3y 3
CONTOH = 56 × t 12 – 3 – 1
7×4

= 2t 8

E Permudahkan setiap yang berikut.
Simplify each of the following. SP 1.2.3 Menguasai TP 3

Contoh / Example: 1. (a 2)2 2. (62)5
(42)3
= a2 × 2 = 62 × 5
= 42 × 3 = a4 = 610
= 46

3. (p3)5 4. (72)3 5. (103)4

= p3 × 5 = 72 × 3 = 103 × 4
= p15 = 76 = 1012

TP3 Mengaplikasikan kefahaman tentang hukum indeks untuk melaksanakan tugasan mudah. 5

F Permudahkan setiap yang berikut.
Simplify each of the following. SP 1.2.3 Menguasai TP 3

Contoh / Example: 1. (2 × 6)2 2. (xy)3
(3 × 4)2 = 22 × 62 = x 3y 3

= 32 × 42

3. (4a)3 4. (13x)2 ( )5. 4 3
9
= 43 × a3 = 13 2 × x 2 43
= 64 × a3 = 169x 2 =
= 64a3CONTOH 93
= 64
729

G Permudahkan setiap yang berikut.
Simplify each of the following. SP 1.2.3 Menguasai TP 3

Contoh / Example: 1. (g2h 3)4 2. (23k 4)3
(32f 2)2
= g2 × 4 × h3 × 4 = 23 × 3 × k 4 × 3
= 32 × 2 × f 2 × 2 = g8h12 = 29k12
= 81f 4 = 512k12

H Cari nilai yang berikut.
Find the value of the following. SP 1.2.4 Menguasai TP 2

Contoh / Example: 1. (–6)0 2. [(ab)2]0
60 =1

=1 = 1

TP2 Mempamerkan kefahaman tentang bentuk indeks.

6 TP3 Mengaplikasikan kefahaman tentang hukum indeks untuk melaksanakan tugasan mudah.

I Nyatakan setiap yang berikut dalam bentuk 1 .
an
1
State each of the following in the form of an . SP 1.2.4 Menguasai TP 2

. Contoh / Example: 1. 6–3 2. k –6

7–2

11 1
72 63 k6

J Nyatakan setiap yang berikut dalam bentuk a –n. m–3
State each of the following in the form of a–n. SP 1.2.4 Menguasai TP 2
CONTOH
Contoh / Example: 1. 1 2. 1
1 68 m3
74
6–8
7–4

K Nyatakan setiap yang berikut dalam bentuk n√a .
State each of the following in the form of n√ a . SP 1.2.5 Menguasai TP 2

Contoh / Example: 1 1

1 1. 12 4 2. x 3

15 3

3√15 4√12 3√ x

1

L Nyatakan setiap yang berikut dalam bentuk a n .
1
State each of the following in the form of a n . SP 1.2.4 Menguasai TP 2

Contoh / Example: 1. 3√ 16 2. 5 c
√4 d

1 1 ( )c 1
5
42 16 3
d

Nota Pantas 7

Hubungan antara indeks pecahan dan punca kuasa boleh ditulis sebagai:
The relationship between the fractional indices and the roots can be written as:

1

a n = n√a

TP2 Mempamerkan kefahaman tentang bentuk indeks.

M Nyatakan setiap yang berikut dalam bentuk (n√a )m.
State each of the following in the form of (n√ a )m. SP 1.2.5 Menguasai TP 2

Contoh / Example: 2 5

3 1. 9 3 2. k 2

11 4

(4√11)3 (3√ 9 )2 (2√ k )5

N Hitung nilai setiap nombor dalam bentuk indeks pecahan berikut.
Calculate the value of each of the following fractional index numbers. SP 1.2.5 Menguasai TP 4
CONTOH
Contoh / Example: 3 3

3 1. (625) 4 2. (16) 2

92 3 3

3 = (54) 4 = (42) 2
= 53 = 43
= (32) 2 = 125 = 64

= 33

= 27

5 2 4

3. (81) 4 4. (125) 3 5. (512) 3

5 2 4

= (34) 4 = (53) 3 = (83) 3

= 35 = 52 = 84

= 243 = 25 = 4 096

O Hitung setiap yang berikut.
Calculate each of the following. SP 1.2.6 Menguasai TP 4

Contoh / Example: 1. (52 × 32)2 ÷ (5–1)2
36 ÷ (32 × 2–2)2

= 36 ÷ (32 × 2 × 2–2 × 2) = 52 × 2 × 32 × 2 ÷ 5–1 × 2
= 36 ÷ 34 ÷ 2–4 = 54 × 34 ÷ 5–2
= 54 – (–2) × 34
= 36 – 4 ÷ 1 = 56 × 34
24 = 15 625 × 81
= 1 265 625
= 32 × 24
= 9 × 16
= 144

TP2 Mempamerkan kefahaman tentang bentuk indeks.

8 TP4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum indeks dalam konteks
penyelesaian masalah rutin yang mudah.

2. (23 × 32)3 ÷ (22 × 3)–2 3. 59 ÷ (32 × 53)2 ÷ (35)–2

= 23 × 3 × 32 × 3 ÷ (22 × (–2) × 3–2) = 59 ÷ (32 × 2 × 53 × 2) ÷ 35 × (–2)
= 59 ÷ (34 × 56) ÷ 3–10
= 29 × 36 ÷ (2–4 × 3–2) = 59 – 6 × 3–4 – (–10)
= 53 × 36
= 2 × 39 – (–4) 6 – (–2) = 125 × 729
= 91 125
= 213 × 38

= 8 192 × 6 561

= 53 747 712

P Hitung setiap yang berikut.
Calculate each of the following. SP 1.2.6 Menguasai TP 5
CONTOH
Contoh / Example: 1. 1 × 3 × 35– 1
2
72 52
111

22 × 32 × 62

11 1 13 1
= 2 2 × 3 2 × (3 × 2) 2 = 7 2 × 5 2 × (7 × 5)– 2
111 1
= 7 1 × 3 × 7– 1 × 5– 1
= 22 × 32 × 32 × 22 2 2 2
52
1 1 1 1 71 1 3 1
= + 2 × + 2 = 2 – 2 × – 2
22 32 52

= 2×3 = 70 × 5

=6 =5

31 1

2. 81 4 × 16 2 3. (633 ÷ 27) 3

31 11

= (34) 4 × (42) 2 = (633) 3 ÷ 27 3
1
= 33 × 4
= 63 ÷ (33) 3
= 27 × 4
= 63 ÷ 3
= 108
= 21

4. 3 × 3 × 80– 1 5. (124 × 3)3
2 (35 × 45)2
52 42

33 1
= 5 2 × 4 2 × (5 × 42)– 2
33 1 = 124 × 3 × 33
= 5 2 × 4 2 × 5– 2 × 4–1 35 × 2 × 45 × 2

31 3
= 52 – 2 × 42 –1
3 = (4 × 3)12 × 33
310 × 410
= 5 × 42 –1
1

= 5 × (22) 2 = 412 × 312 × 33
310 × 410
=5×2

= 10 = 412 – 10 × 312 + 3 – 10
= 42 × 35
= 16 × 243 = 3 888

TP4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum indeks dalam konteks 9
penyelesaian masalah rutin yang mudah.

TP5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum indeks dalam konteks
penyelesaian masaah rutin yang kompleks.

Q Permudahkan setiap yang berikut.
Simplify each of the following. SP 1.2.6 Menguasai TP 5

Contoh / Example: 55
3x 4y × 4xy 5
(2xy 2)2 × x 3 1. (2x 3 )3 × (3y 4 )2
1
(66x 4y) 2

= 3 × 4 × x 4 + 1y 1 + 5 5
4x 2 + 3y 4
= 23x 5 × 32y 2
1
63x 2y 2

12x 5y 6 5
4x 5y 4
= = 23x 5 × 32y 2

1

= 3x 5 – 5y 6 – 4 (3 × 2)3x 2y 2
= 3y 2
5

23x 5 × 32y 2

1
CONTOH =

33 × 23x 2y 2

= 23 –3 × 32 – 3x 5 – 5 – 1
2
2y 2

= 20 × 3–1x 3y 2

x 3y 2
=

3

3 3. s 3t 4 × (3s –2t)2
(82p–2)3 × (64 2 ÷ 83p)2 1
2. (2p 2)3 (32s 4t 2) 2 ÷ (3–1t)3

86p–6 × 643 ÷ 86p2 = s 3t 4 × 32s –4t 2
= 23p 6 3s 2t ÷ 3–3t 3

86p–6 × (82)3 ÷ 86p2 = 3 s t2 – 1 – 3 3 – 4 – 2 4 + 2 –1 – (–3)
= 8p 6 = 3–2s –3t 8
= t8
86p–6 × 86 ÷ 86p2
= 8p 6 9s 3

= 8 p6 + 6 – 6 – 1 –6 – 2 – 6
= 85p –14

= 32 768
p14

10 TP5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum indeks dalam konteks
penyelesaian masalah rutin yang kompleks.

R Cari nilai m bagi setiap yang berikut.
Find the value of m for each of the following. SP 1.2.7 Menguasai TP 6

Contoh / Example: 1. 33m ÷ 32 = 311
7m = 49 33m – 2 = 311

7m = 72 ∴ 3m – 2 = 11
3m = 13
∴m = 2 m = 13
3

CONTOH2.a2m= a18 3. 2m + 32 = 73
a12
2m = 73 – 9
a2m = a18 – 12 2m = 64
= a6 2m = 26

∴ 2m = 6 ∴ m  = 6
m= 3

3 5. 42m = 166 – 2m

4. 22m × 23 = 16 2 16m = 166 – 2m

3 ∴ m = 6 – 2m
3m = 6
22m + 3 = (24) 2 m=2

22m + 3 = 26

∴ 2m + 3 = 6
2m = 6 – 3
2m = 3

m= 3
2

TP6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum indeks dalam konteks 11
penyelesaian masalah bukan rutin secara kreatif.

Praktis F o r m a t i f 1

Jawab semua soalan.
Answer all questions.

Bahagian A 4. Permudahkan (p4)2 ÷ p .
p 2q –3
1. Antara berikut, pernyataan yang
manakah betul? Simplify (p4)2 ÷ p .
Which of the following statement is true? p2q –3
A (–2) × (–2) × (–2) = 8
B 8 × 8 × 8 × 8 × 8 = 85 A p4q–3
C 6 + 6 + 6 + 6 = 64
D (–5) × (–5) × (–5) = 125 B p3q–3
CONTOH
C p3q3

D p5q3

2. Pilih jawapan yang sama nilai dengan 5. Permudahkan 225x0 ÷ 3x 4.
225x 3 ÷ 9x.

Choose the answer which is same with Simplify 225x0 ÷ 3x4.

225x3 ÷ 9x. B 35x 2 A0 B 75
A 25x 4 D (5x)4
C (5x)2 C 75x D 75x –4

3. Apakah nilai n, jika 36n = 216n – 2? 6. Permudahkan √xy × 3√x × √xy 2 .

What is the value of n, if 36n = 216n – 2? Simplify √xy × 3√x × √xy 2 .

A2 B4 43 1

A x 3y 2 B x 2y

C6 D8 32 11

C x 4y 3 D x 2y 2

Bahagian B

7. (a) Nyatakan bahawa pernyataan yang berikut adalah Benar atau Palsu.

State whether the following statements are True or False. [2 markah / marks]

Jawapan / Answer:

Pernyataan Benar / Palsu
Statement True / False

(i) (23)4 = [(–2)4]3 Benar / True

(ii) 5√16 = 2 Palsu / False

(b) Cari nilai / Find the value of m3 × m –3. [2 markah / marks]
Jawapan / Answer:
m3 × m –3 = m3 + (–3)
= m0
=1

12

Bahagian C [2 markah / marks]
[3 markah / marks]
8. (a) Ringkaskan x 8 ÷ x 6.
Simplify x8 ÷ x6. [2 markah / marks]
Jawapan / Answer: [3 markah / marks]

x8 ÷ x6 = x8–6 PErkaskttrias
= x2
13
(b) Diberi 4m = 82m – 1, cari nilai m.CONTOH
Given that 4m = 82m – 1, find the value of m. 
Jawapan / Answer:

4m = 82m – 1
22m = 26m – 3
∴ 2m = 6m – 3

4m = 3
m= 3
4

(c) Permudahkan a–5 × a 3.
Simplify a–5 × a3.
Jawapan / Answer:

a –5 × a3 = a–5 + 3
= a–2

(d) Nilaikan 2–2 × 34 .
8–3 × 27
2–2 × 34
Evaluate 8–3 × 27 .

Jawapan / Answer:

2–2 × 34 2–2 × 34
8–3 × 27 = 2–9 × 33

= 2–2 – (–9) × 34 – 3
= 27 × 3
= 128 × 3

= 384

Jawapan Jawapan
untuk QR

Bab 1 D 1. 8 × k 3 – 2 = 4k
2
Praktis D S K P
2. 9 × y 5 – 2 = 3y 3
PBD 1 3

A 1. (2.5)4 = 2.5 × 2.5 × 2.5 × 2.5 3. 56
2. (rs)3 = rs × rs × rs × t 12 – 3 – 1 = 2t 8
3. y 8 = y × y × y × y × y × y × y × y 7×4

( )4. 3 3 = 3 × 3 × 3 E 1. a 2 × 2 = a 4 2. 62 × 5 = 610
5 555 3. p  3 × 5 = p 15 4. 7 2 × 3 = 76
5. 10 3 × 4 = 1012
5. (–7)5 = (–7) × (–7) × (–7) × (–7) × (–7)

B 1. 66 ( )2. 1 4
2
3. pqr 2 4. (1.29)4
5. (–jk)3
CONTOH F 1. 22 × 62 2. x 3y 3
4. 132 × x 2 = 169x 2
3. 43 × a 3 = 64a 3
64
5. 43 = 729
93
C 1. 3 × 3 × 3 × 3 = 81
G 1. g 2 × 4 × h 3 × 4 = g 8h 12
2. 6 × 6 × 6 = 216 2. 2 3 × 3 × k 4 × 3 = 512k 12

3. 0.4 × 0.4 = 0.16

4. (–1.4) × (–1.4) × (–1.4) = –2.744 H 1. 1 2. 1
1×1×1×1= 1
5. 3 3 3 3 81 1 1
I 1. 63 2. k 6
D 1. 53, m = 3 2. –25, m = 5
3. 29, m = 9 4. (0.3)3, m = 3 J 1. 6–8 2. m–3
5. 94, m = 4
K 1. 4√12 2. 3√ x
E 1. 243 = 3 × 3 × 3 × 3 × 3
= 35 1 ( )2. c 1
5
2. 625 = 5 × 5 × 5 × 5 L 1. 16 3
= 54 d

3. 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 M 1. (3√ 9 )2 2. (2√ k )5
= 28
3 3
4. 343 = 7 × 7 × 7
= 73 N 1. (54) 4 = 125 2. (42) 2 = 64
5
5. 729 = 9 × 9 × 9 2
= 93 3. (34) 4 = 243
4. (53) 3 = 25
4

5. (83) 3 = 4 096

F 1. 34 2. 26 3. 54 O 1. 52 × 2 × 32 × 2 ÷ 5–1 × 2
= 1 265 625
PBD 2 2. (–4)2 + 1 + 3 = (–4)6
4. p 1 + 7 + 6 = p 14 2. 23 × 3 × 32 × 3 ÷ (22 × (–2) × 3–2)
A 1. 35 + 6 + 4 = 315 = 53 747 712
3. m 4 + 5 = m 9
5. q 3 + 7 + 9 = q 19 3. 59 ÷ (32 × 2 × 53 × 2) ÷ 35 × (–2)
= 91 125

B 1. 45 – 3 = 42 2. 67 – 1 = 66 P 1. 1 × 3 × (7 × – 1 = 5
3. a12 – 9 = a 3 4. b 8 – 7 = b 2
5. x 10 – 2 – 6 = x 2 72 52 5)

31

2. (34) 4 × (42) 2 = 108

C 1. 8 × 9 × k 4 + 7 = 72k 11 11
2. 4 × 6 × y 2 + 1 = 24y 3
3. 2 × 5 × t 3 + 2 + 1 = 10t 6 3. (633) 3 ÷ 27 3 = 21

4. 3 × 3 × (5 × – 1 = 10
2
52 42 42)

5. 12 4 × 3 × 33 = 3 888
35 × 2 × 45 × 2

J1

5 2–2 × 34 2–2 × 34
8–3 × 27 2–9 × 33
Q 1. 23x 5 × 32y 2 = x 3y 2 (d) =
3
1
= 2–2 – (–9) × 34 – 3
63x 2y 2 = 27 × 3
= 128 × 3
2. 86p–6 × 643 ÷ 86p 2 = 32 768 = 384
23p 6 p14

3. s 3t 4 × 32s –4t 2 = t8 Bab 3
3s 2t ÷ 3–3t 3 9s 3
Praktis D S K P
R 1. 33m – 2 = 311 2. a 2m = a18 – 12
= a6 PBD 1
∴ 3m – 2 = 11
3m = 13 ∴ 2m = 6 A 1. Akaun Simpanan Semasa, Akaun
m=3 Simpanan Tetap
m = 13 Current Account, Fixed Deposit Account
3
2. Hartanah, Amanah Saham, Saham
3. 2m = 73 – 9 Real Estate, Unit Trust, Shares
CONTOH
2m = 64 B 1. I = 2 000 × 2.0 × 6 = 20
100 12
2m = 26
2 000 + 20 = 2 020
∴m=6
3 2. I = 5 000 × 1.5 × 2 = 150
100
4. 22m + 3 = (24) 2
5 000 + 150 = 5 150
22m + 3 = 26 3. I = 7 000 × 2.5 × 3 = 43.75

∴ 2m + 3 = 6 100 12
2m = 6 – 3 7 000 + 43.75 = 7 043.75
2m = 3
3 ( )C 1. Nilai matang = RM10 000 1 + 0.042 12(5)
m = 2 Matured value= RM12 332.26 12
( )2. Nilai matang = RM20 000 1 + 0.032 6(6)
5. 16m = 166 – 2m Matured value= RM24 221.05 6
∴ m = 6 – 2m
3m = 6 ( )3. Nilai matang = RM5 600 1 + 0.034 12(1.25)
m =2 Matured value= RM5 842.78 12

Praktis Formatif 1 D 1. ROI
= (RM50 000 + RM250) – RM30 000 × 100%
Bahagian A RM30 000
= 67.5%
1. B 2. C 3. C 4. D
2. ROI
5. D 6. A = (RM25 000 + RM2 000) – RM20 000 × 100%
RM20 000
Bahagian B = 35%

7. (a) (i) Benar / True 3. ROI
(ii) Palsu / False = (RM1 500 + RM300) – RM2 000 × 100%
RM2 000
(b) 1 = –10%

Bahagian C
8. (a) x 8 ÷ x 6 = x 8 – 6

= x2
(b) 4m = 82m – 1

22m = 26m – 3

∴ 2m = 6m – 3

4m = 3 E

m= 3 Saham . Hartanah = Unit amanah
4 Shares Real estates Unit trust

(c) a –5 × a 3 = a–5 + 3 F Unit amanah Saham Hartanah
= a–2 Unit trust Shares Real estates
. .

J2

G Unit amanah Simpanan Simpanan tetap Faedah yang dikenakan / Interest charged
Unit trust Saving Fixed deposit
. = ( )= RM5 000 × 17 × 15
100 365

H 1. Kos purata sesyer / Average cost per share = RM34.93

= (2 000 × RM3.15) + (3 000 × RM3.55) ∴ Bayaran bulan hadapan
2 000 + 3 000
Payment for next month
= RM3.39 = RM5 000 + RM50 + RM34.93

2. = RM5 084.93
Kos purata sesyer / Average cost per share
3. Bayaran minimum / Minimum payment
= (3000×RM1.65)+(5000×RM1.60)+(4000×RM1.70) = RM3 000 × 5%
3 000 + 5 000 + 4 000
= RM150
= RM1.65
Faedah yang dikenakan / Interest charged

( )= (RM3 000 – RM150) × 15 × 15
100 365
PBD 2
= RM17.57
A 1. a, b
CONTOH 2. c, d ∴ Bayaran bulan hadapan

B 1. Mudah untuk pembelian tanpa tunai Payment for next month

Easy for cashless purchases = RM3 000 – RM150 + RM17.57
2. Mudah terbelanja melebihi kemampuan
4. Jumlah faedah pada kad kredit
Easily overpriced Total interest of the credit card
= SGD165 × 0.035 × RM3.01
C 1. RM500 × 5% = RM500 × 0.05 = RM17.38
= RM25
∴ Jumlah pembayaran / Total payment
Bayaran minimum = RM50 = SGD165 × RM3.01 + RM17.38
Minimum payment = RM514.03
2. RM1 500 × 5% = RM1 500 × 0.05
F 1. 100 000 + (100 000 × 0.05 × 8) = 140 000
= RM75
140 000 =1 458.33
Bayaran minimum = RM75 8 × 12
Minimum payment

( )D 18 × 15 = RM2.59; 0
1. (400 – 50) × 100 365 2. 5 000 + (5 000 × 0.04 × 4) = 5 800

( )2. 3 000 × 18 × 15 = RM22.19 5 800 = 120.83
100 365 4 × 12

3 000 × 1% = RM30 3. 80 000 + (80 000 × 0.06 × 10) = 128 000

( )3. (1 200 – 60) × 18 × 15 = RM8.43; 0 128 000 = 1 066.67
100 365 10 × 12

G 1. I = Prt
400 = y × 0.05 × 4
E 1. Caj bayaran lewat / Late payment charge 400
= RM700 × 1% y = 0.05 × 4

= 7 (, 10)

= RM10 y = 2 000

Faedah yang dikenakan / Interest charged Bayaran ansuran / Monthly instalment

( )= RM700 × 18 20 = 2 000 + 400
100 365 4 × 12
×

∴ Bayaran bulan hadapan = RM50

Payment for next month 2. A = 1 600
12 × 15
= RM700 + RM10 + RM6.90

= RM716.90 A = 288 000

2. Caj bayaran lewat / Late payment charge A = P + Prt x
= RM5 000 × 1% 100
288 000 = 180 000 + (180 000 × × 15)

= RM50

x= 288 000 – 180 000 × 100
180 000 × 15

x=4

J3

3. (a) Jumlah bayaran balik (ii) ROI
(b) Total repayment
(c) = 95 000 + (95 000 × 0.08 × 6) = 3 600 + 225 – (3 000 × 1.25) × 100%
= RM140 600 3 000 × 1.25
Bayaran ansuran bulanan
Monthly instalment = 2%
= 140 600
(iii) Dia telah mendapat keuntungan
6 × 12
= RM1 952.78 kerana pulangan pelaburannya
Jumlah bayaran balik
Total repayment adalah lebih daripada 0%.
= 95 000 + (95 000 × 0.08 × 5)
= RM133 000 She get a profit as the return of investment
is more than 0%.
Tempoh pinjaman yang lebih berbaloi
ialah 5 tahun kerana mempunyai (b) Jumlah bayaran balik / Total repayment
jumlah bayaran balik yang lebih = 80 000 + (80 000 × 0.04 × 11)
rendah.
= RM115 200
The period which is more cost-effective
is 5 years because it has a lower total Jadikan y sebagai bayaran yang perlu
repayment amount.
dibuat selepas 2 bulan.
Bayaran ansuran bulanan bagi
tempoh 5 tahun Let y be the total payment needed after 2
Monthly instalment for period of 5 years months.
CONTOH = 133 000
(11 × y – 2 = 700
5 × 12 12)
= RM2 216.67
y = 700 × [(11 × 12) – 2]
Wang yang tinggal selepas tolak kos
hidup = 700 × 130
Money left after substract the living cost
RM5 000 – RM3 000 = RM2 000 = RM91 000

Encik Manu perlu memilih tempoh Jumlah pembayaran yang dibuat oleh
6 tahun kerana dia tidak dapat
menampung bayaran ansuran bagi Puan Tay dalam 2 bulan tersebut
tempoh 5 tahun. Total payment Puan Tay did in that 2 months
= RM115 200 – RM91 000
Encik Manu needs to choose the period of
6 years as he cannot afford the monthly = RM24 200
instalment for period of 5 years.
7. (a) Jumlah bayaran balik / Total repayment
= RM12 000 + (RM12 000 × 0.07 × 2)

= RM12 000 + RM1 680

= RM13 680

∴ Ansuran bulanan / Monthly instalment
= RM13 680 ÷ (2 × 12)

= RM13 680 ÷ 24

= RM570

(b) (i) Jumlah pulangan / Total returns
= 12 × RM2 000

= RM24 000

∴ ROI = RM24 000 × 100%
RM400 000
Praktis Formatif 2

Bahagian A = 6%

1. A 2. C 3. B 4. B (ii) Jumlah pulangan / Total returns
= RM400 000(1 + 0.04 )(2)(1) –
Bahagian B 2

5. (a) (i) Palsu / False RM400 000
(ii) Betul / True
= RM416 160 – RM400 000
(b) (i) Akaun simpanan tetap
= RM16 160
Fixed deposit account
(ii) Akaun semasa / Current account ∴ ROI = RM16 160 × 100%
RM400 000

= 4.04%

Bahagian C (iii) Puan Divya adalah pelabur yang

6. (a) (i) k × 1.25 × 0.06 = 225 bijak. Nilai pulangan pelaburannya

k = 225 lebih tinggi daripada Puan Arinah.
1.25 × 0.06
Puan Divya is a wise investor. Her value
= 3 000 of investment returns is higher than Puan
Arinah.
J4

Bab 5 √ 3 + 6 √ 3
2 2
( ) ( )2.
Praktis D S K P 5 kos / cos 30° + 6 sin 60° = 5

PBD 1    =  11√ 3
2
A 1. (a) Sisi bertentangan / Opposite side
(b) Hipotenus / Hypotenuse 3. (tan 60°)2 – sin 30° = (√ 3 )2 – 1
(c) Sisi sebelah / Adjacent side 2

2. (a) Hipotenus / Hypotenuse =3– 1
(b) Sisi sebelah / Adjacent side 2
(c) Sisi bertentangan / Opposite side
= 21
PR 2
QR
B 1. sin θ = AB
AE
kos / cos θ PQ H 1. (a) kos / cos 60° =
QR
= 1 AB
2 20
tan θ PR =
PQ
CONTOH = 1
2
2. sin θ = YZ AB = 20 ×
XZ
= 10 cm

kos / cos θ = XY (b) kos / cos 30° = BC
XZ CD
YZ
tan θ = XY √ 3 =5
sin θ 2 CD
7
C 1. = 12 CD = 5×2
√ 3
2. sin θ = 3
5 = 10 cm
√ 3
3. sin θ = √ 24
7 AD
2. tan 60° = DE

D 1. kos / cos θ = 12 √ 3 = AD
13 4

2. kos / cos θ =2 AD = 4√ 3 cm
√ 13
4√ 3
3. kos / cos θ = 9 AC = 2
19

E 1. tan θ = 23 AC = 2√ 3 cm
16
kos / cos x° = 2√ 3
7 4
2. tan θ = 18
√ 3
√ 645 = 2
14
3. tan θ = x = 30°

F 1. sin 30° = 1 ; sin 45° = 1 ; sin 60° = √ 3 I 1. 48.3° = 48° + 0.3°
2 √ 2 2 = 48° + (0.3 × 60)ʹ
= 48° + 18ʹ
2. kos 30° = √ 3 ; kos 45° = 1 ; kos 60° = 1 = 48° 18ʹ
cos 30° 2 cos 45° √ 2 cos 60° 2
2. 9.7° = 9° + 0.7°
3. tan 30° = 1 ; tan 45° = 1; tan 60° = √ 3 = 9° + (0.7 × 60)ʹ
√ 3 = 9° + 42ʹ
1 = 9° 42ʹ
( )G 1 √ 2
1. sin 30° – 5√ 2 sin 45° = 2 – 5√ 2 3. 21.2° = 21° + 0.2°

= 1 –5 = 21° + (0.2 × 60)ʹ
2 = 21° + 12ʹ
= 21° 12ʹ
= –4 1
2

J5

J 1. 12° 45ʹ = 12° + 45ʹ 2. kos / cos 84° = 37
= 12° + (45 ÷ 60)° H
= 12° + 0.75° 37
= 12.75° H = kos / cos 84°

2. 17° 6ʹ = 17° + 6ʹ H = 353.97 cm
= 17° + (6 ÷ 60)°
= 17° + 0.1° tan 84° = O
= 17.1° 37

3. 19° 30ʹ = 19° + 30ʹ O = 37 tan 84°
= 19° + (30 ÷ 60)° O = 352.03 cm
= 19° + 0.5°
= 19.5° 353.97 cm 84°
37 cm

K 1. kos / cos θ = 4
7

= 0.5714 352.03 cm
CONTOH
θ = kos–1 / cos–1(0.5714) 3. sin 52°= O
52
= 55.15°

2. sin θ = 5 O = 52 sin 52°
12
O = 40.98 cm
= 0.4167 A
kos / cos 52° = 52
θ = sin–1 (0.4167)

= 24.63° A = 52 kos / cos 52°
A = 32.01 cm
3. tan θ = 6
3

=2 52°
32.01 cm
θ = tan–1 2 52 cm

= 63.43°

L 1. sin 62° = x 40.98 cm
20

x = 20 sin 62° 1. tan 48° = 1.6 – 0.5
x = 17.66 cm x
N

2. tan 35° = 7 x = 1.1
x tan 48°
7
x = tan 35° = 0.99 m

x = 10.00 cm ∴ Jarak / Distance = 0.99 m

3. tan 26° = x O 1. Sudut di antara garis BE dan satah ABCD
12 ialah ∠DBE.
The angle between line BE and plane ABCD is
x = 12 tan 26° ∠DBE.
x = 5.85 cm E AD

M 1. sin 35° = 12 4 cm
H
20.92 cm
H= 12 D B B 9 cm C
sin 35° 12 cm
BD = √92 + 42
H = 20.92 cm 35° = √ 97

tan 35° = 12 17.14 cm tan ∠DBE = 6
A √ 97

A = 12 tan ∠DBE = tan–1 6
tan 35° √ 97

A = 17.14 cm tan ∠DBE = 31.35°

J6

2. (a) sin ∠DFE = 3 (b) Sudut di antara satah BPC dan satah
5 ABCD ialah ∠PQR.

∠DFE = sin–1 3 The angle between plane BPC and plane
5 ABCD is ∠PQR.

∠DFE = 36.87° tan ∠PQR = 9
12
(b) AB = √ 52 – 32
= √ 16 tan ∠PQR = tan–1 3
= 4 cm 4

kos / cos 25° = BC tan ∠PQR = 36.87°
12
7. (a) tan x = 1, SR = QR = 16 cm
BC = 12 kos / cos 25° PR = PQ + QR
= 10.88 cm = 5 + 16

∴ AC = 4 + 10.88 = 21 cm
         =  14.88 cm
tan y = 16
CONTOH3. AB = √ 36 21
= 6 cm
y = tan–1 16
21
Katakan titik tengah AB = O
Let the midpoint of AB = O y = 37.3°
(b) Jika θ mewakili sudut antara tapak tangga
Sudut di antara garis EF dan satah ABF
ialah ∠EFO. dengan permukaan tanah, maka
The angle between line EF and plane ABF is
∠EFO. If θ represents the angle between the base of

FF the ladder and the ground, then
1.7
kos / cos θ = 12

9 cm θ = kos–1 / cos–1 1.7
12

= 81.9°

Tangga itu tidak selamat untuk diguna-

A 3 cm O 3 B kan kerana 81.9° berada di luar julat 70°
cm
E 3 cm O hingga 80°. Pekerja itu perlu menggerak-

OF = √ 92 – 32 sin ∠EFO = EO kan tapak tangga jauh sedikit daripada
= √ 72 OF
dinding.

3 The ladder is not safe to use as 81.9° lies
sin ∠EFO = sin–1 √ 72 outside the range 70° to 80°. The worker
needs to move the base of the lader slightly
sin ∠EFO = 20.70°
away from the wall.

Praktis Formatif 3 Bab 7

Bahagian A Praktis D S K P

1. C 2. A 3. B 4. A PBD 1

Bahagian B (b) 65.38° A 1. Bukan / No 2. Ya / Yes
5. (a) 30° 3. Bukan / No 2.

Bahagian C B 1.

6. (a) tan 44° = x
5
C 1. A/E H/I D/L
x = 5 tan 44°
x = 4.83
∴ Tinggi bangunan / Height of building

= 4.83 + 1.75

= 6.58 m B/F G/J C/K

Panjang semua sisi tidak berubah.
The length of all sides do not change.

J7

2. D C B 1. Pelan / Plan

B GH D/K

2 cm 1 cm 2 cm

6 cm

A/E B 45°

Panjang sisi AB, BC, CD dan AD tidak AA/II EF CC/JJ
berubah manakala panjang sisi AE, BE, CE AA/BB
dan DE berubah. CC/DD CC/AA DD/BB
Length of sides AB, BC, CD and AD do not 2 cm
change but length of sides AE, BE, CE and DE E/G F/H F/E H/G
change.
3 cm
5 cm

PBD 2 7cm OH Pelan / Plan B/F
A/E/N
A 1.

I J/K J/I K
Dongakan depan / Front elevation Dongakan sisi / Side elevation

I/J L/K 6 cm 2. Pelan / Plan

B G/J D/L K

2 cm 2 cm 2 cm

CC//G/M/ M 4 cm 45°
D/H
5 cm

I 3 cm CC/AA DD/BB DD/CC 6 cm BB/AA

3cm NT 3.5 cm AA EE/HH CC/FF 45°
I

HH//G F/E AA/BB CC/DD CC/AA DD/BB
G/E L H/F
2 cm 2 cm
3.5 cm

F/L EF//EF L/G
I/H 1 cm
E/G 5 cm K/J
1 cm

JK MN H/J 4 cm I/K Dongakan sisi / Side elevation
Dongakan depan / Front elevation Dongakan sisi / Side elevation Dongakan depan / Front elevation

2. Pelan / Plan DH C 1. K H
6 cm JG I
B/F/M L
4cmO
A/E O N 1 cm 5 cm E F 3 cm
G M
CD
C 4 cm

3cmC AB

45° 2. B G H D
L
II/KK 3 cm J 2 cm L

4 cm EF
C
AA/BB 6 cm CC/DD CC/AA 3 cm DD/BB A

3 cm R V U
T
J//I I 4 cm
2.5 cm I/E/F JJ//O/P/ P G/O/E H/P/F 6 cm Q
2.5 cm
3 cm 4 cm G/H S

J

K/M 5 cm L/N L/K 7 cm N/M P
Dongakan depan / Front elevation N
Dongakan sisi / Side elevation I M OK

Praktis Formatif 4 2. D

Bahagian A
1. D

J8

Bahagian B 3. y = 2x + 8 4. y= 2 x – 4
3 3
3. (a) Panjang sisi tidak berubah
m=2 m= 2
Length of the side unchanged 3
(b) Panjang sisi berubah c= 8
3 c = –4
Length of the side changes
(c) Panjang sisi tidak berubah 5. y = – 4 x – 1
5
Length of the side unchanged
(d) Panjang sisi berubah m = – 4
5
Length of the side changes

Bahagian C c = –1

4. (a) 3 cm 3 cm B 1. m = – 3 , c = 3 2. m= 1 , c = 2
4 2

3 cm 3 cm ∴ y = – 3 x + 3 ∴y= 1 x + 2
4 2
Pelan / Plan 3 cm

Dongakan depan
Front elevation
CONTOH 3. m= 5 – (–1) = 6 4. m= –1 – (–3)
1 – (–6) 7 3–1 =1

y= 6 x + c y=x+c
7 –1 = 3 + c

3 cm 5= 6 (1) + c c = –4
7 ∴y=x–4
Dongakan sisi
3 cm Side elevation c= 29
7

3 cm ∴y= 6 x+ 29
7 7
(b) Daripada isi padu pepejal:

From the volume of solid 5. m = 0 6. m = Tidak tertakrif
∴y=2 Undefined
1 (x + 1 + x + 3)(5)(4) = 80 ∴x=4
2
    P e la n  / P la n                        x = 2
C 1. 4y = x – 3 2. 7y = 2x – 14
1 3
C/G D H 4 4 y= 2 x – 2
7
y= x –

3. 4y = –5x – 8 4. 2x – 5y = 10
         5y = 2x – 10
5 cm y = – 5 x – 2
4
y= 2 x – 2
5

45° 5. 5x + 3y = 15
         3y = 15 – 5x
AA/EE 3 cm B 2 cm F

y = – 5 x + 5
3
AA/CC BB/DD BB/AA DD/CC

4 cm D 1. 3x + y – 7 = 0 2. 4x – 2y – 3 = 0
3. x + 2y = 4 4. 2x – 3y = 6
x + 2y – 4 = 0
5. –4x + 5y = 20 2x – 3y – 6 = 0
4x – 5y + 20 = 0
E/G F/H F/E H/G
Dongakan depan / Front elevation Dongakan sisi / Side elevation E 1. (a) 7(6) – 3 = 39

Bab 9 2. m = –7 A tidak terletak pada garis lurus
A does not lie on the straight line
Praktis D S K P c= 4 (b) 7(4) – 3 = 25
PBD 1 7
B terletak pada garis lurus
A 1. m = 5 B lies on the straight line
c = –9 2. (a) 4(2) + 5 = 13

A terletak pada garis lurus J9
A lies on the straight line

(b) 4(5) + 5 = 25 H 1. y = –6x + 13
B terletak pada garis lurus ∴ m = –6
B lie on the straight line
Persamaan baharu / New equation
3. (a) 2 (12) + 6 = 14 y = –6x + c
3
Pada titik A / At point A
A tidak terletak pada garis lurus –7 = –6(4) + c
A does not lie on the straight line
c = 17
(b) 2 (9) + 6 = 12 ∴ y = –6x + 17
3
2. 3y = 5x – 7
B terletak pada garis lurus
B lies on the straight line y= 5 x – 7
3 3

F 1. 16 = 6(k) – 8 2. k = 5(3) + 3 ∴m= 5
6k = 24 k = 18 3
k=4
Persamaan baharu / New equation
3. k = 4(6) – 9
k = 15 5
CONTOH y= 3 x + c

G 1. L1 : y = 6x, m1 = 6 Pada titik A / At point A
L2 : 2y = 8x – 9
9= 5 (6) + c
3
y = 4x – 9 , m2 = 4
2 c = –1

m1 ≠ m2, maka L1 tidak selari dengan L2. ∴y= 5 x – 1
m1 ≠ m2, thus L1 is not parallel to L2. 3

2. L1 : 3x + y = 4 3. –7y = 5x – 7

      y = –3x + 4, m1 = –3 y = – 5 x + 1
7
L2 : 3y = 5 – 6x
5 ∴ 5
y= 3 – 2x, m2 = –2 m = – 7

m1 ≠ m2, maka L1 tidak selari dengan L2. Persamaan baharu / New equation
m1 ≠ m2, thus L1 is not parallel to L2.
y = – 5 x + c
7

3. L1 : y = –4x + 7, m1 = –4 Pada titik A / At point A
L2 : –y = 4x – 6 5
y = –4x + 6, m2 = –4 26 = – 7 (3) + c
m1 = m2, maka L1 selari dengan L2.
m1 = m2, thus L1 is parallel to L2. c = 197
7

4. L1 : 2x – 6y + 5 = 0 ∴ y = – 5 x + 197
7 7

y= 1 x + 5 , m1 = 1 I 1. 3y = 5x + 6 .....①
3 6 3 y = 1 – 3x .....②

L2 : 3y = x + 9
    y =
1 x + 3, m2 = 1 Gantikan / Substitute ② dalam / into ①
3 3
3(1 – 3x) = 5x + 6
m1 = m2, maka L1 selari dengan L2. 3 – 9x = 5x + 6
m1 = m2, thus L1 is parallel to L2.
x = – 3
14
x 1
5. L1 : y = 2 + 7, m1 = 2 Gantikan / Substitute x = – 3 dalam / into ②
14
L2 : 4y = 2x – 7
( )y 3
y= 1 x – 7 , m2 = 1 = 1 – 3 – 14
2 4 2
23
m1 = m2, maka L1 selari dengan L2. y= 14
m1 = m2, thus L1 is parallel to L2.
Titik persilangan / The point of intersection
( )=
– 3 , 23
14 14

J10

2. x + y – 1 = 0 .....① Gantikan / Substitute x = 1 dalam / into ②
4x + 5y = 6.....② 2

Dari / From ①: x = 1 – y .....③ ( )51 = 2y + 8
2
Gantikan / Substitute ③ dalam / into ②
4(1 – y) + 5y = 6 2y = – 11
2
4 – 4y + 5y = 6
y=2 y = – 11
4
Gantikan / Substitute y = 2 dalam / into ③
x=1–2 Titik persilangan / The point of intersection
x = –1
Titik persilangan / The point of intersection ( )= 1 , – 11
2 4
= (–1, 2)
3. 2x + 2y = 7 .....①
3. 9 – 2x = 4y .....① 5x – 4y = 11 .....②
x = 1 – 6y .....②
① × 2 4x + 4y = 14 .....③
② + ③ 9x = 25
Gantikan / Substitute ② dalam / into ①CONTOH
x = 25
9 – 2(1 – 6y) = 4y 9

y = – 7 Gantikan / Substitute x = 25 dalam / into ①
8 9
( )2 25
Gantikan / Substitute y = – 7 dalam / into ② 9 + 2y = 7
8
13
( )x= 1 – 6 – 7 2y = 9
8
13
x= 25 y = 18
4
Titik persilangan / The point of intersection
Titik persilangan / The point of intersection
( )= 25, 13
( )=25 , – 7 9 18
4 8

J 1. 3x + 9y = 7 .....① K 1. 2y = –x + 8 2y = –3x + 12
5x – 3y = 1 .....②
x08 x04
② × 3 15x – 9y = 3 .....③ y40 y60
① + ③ 18x = 10
y
5
x= 9

5 10
9
Gantikan / Substitute x = dalam / into ① 8
6 2yy ϭ –3xx ϩ 12
( )35 + 9y = 7
9

9y = 16 4
3

y= 16 2
27 2y ϭ –x ϩ 8

Titik persilangan / The point of intersection O 2 4 6 8 10 xx

( )= 5 , 16 Titik persilangan = (2, 3)
9 27 The point of intersection = (2, 3)

2. 6x – 4y = 14 .....① 2. y = –2x + 12 3y – 4x = 6
5x = 2y + 8 .....②
x0 x0
② × 2 10x = 4y + 16 .....③ y 12 6 y2 –1.5
③ – ① 4x = 2 0 0
      x =
1
2

J11

yy Praktis Formatif 5
12
10 Bahagian A

3yy Ϫ 4xx ϭ 6 1. D 2. B 3. D 4. C 5. C

Bahagian B

8 6. (a) 2; –1 (b) 2 ; – 2
5 5

6 (c) 2; 4 (d) 5 ; 3
3 4 4
4 yy ϭ –2xx ϩ 12
2 7. (a) Ya / Yes (b) Tidak / No

(c) Tidak / No (d) Ya / Yes

xx Bahagian C

–4 –2 O 2468 8. (a) y = –3x + 3
2
Titik persilangan = (3, 6)
The point of intersection = (3, 6) ∴ m = –3
CONTOH Persamaan baharu / New equation:
10 – 0 5
L 1. (a) m= 6–0 = 3 y = –3x + c

(b) b–0 = 5 Pada titik (3, 7) / At point (3, 7)
8–5 3 7 = –3(3) + c
c = 16
b=5 ∴ Maka / Thus, y = –3x + 16

(c) y= 5 x + c (b) (i) mBC = 9–7 =– 2
3 2–7 5

Pada titik S / At point S, 9–6 3
5 (ii) mCD = mAB = 2–0 = 2
0= 3 (5) + c

25 Persamaan baharu / New equation:
3 3
c = – y= 2 x + c

Persamaan garis lurus ST ialah Pada titik C / At point C
The equation of straight line ST is
7= 3 (7) + c
5 25 2
y= 3 x – 3
c=– 7
6 3 2
2. (a) mOP = 8 = 4 , cOP = 0
Maka / Thus, y = 3 x – 7
2 2
persamaan garis lurus OP: y = 3 x
4 (iii) Persamaan garis AD:
Equation of line AD:
the equation of straight line OP 2
y=– 5 x + 6
mST = – 9 =– 3 , cST = 9
6 2
Gantikan y = 3 x – 7 ke dalam
persamaan garis lurus ST: y = – 3 x + 9 2 2
2
Substitute y = 3 x– 7 into
the equation of straight line ST 2 2

(b) y= 3 x .....① y=– 2 x + 6
4 5

y=– 3 x + 9 .....② 3 x – 7 =– 2 x + 6
2 2 2 5

Gantikan / Substitute ① ke dalam / into ②,                1190x = 19
2
3 x = – 3 x + 9
4 2                     x = 5
Gantikan / Substitute x = 5 ke dalam / into
x=4 2
Gantikan / Substitute x = 4 dalam / into ①, y=– 5 x + 6

y= 3 (4) = 3 y=– 2 (5) + 6
4 5

∴ Titik persilangan = (4, 3) y=4
∴ D = (5, 4)
The point of intersection = (4, 3)

J12

MODULPENTAKSIRAN UASA 3

Proaktif KSSM ModulPentaksiran PROAKTIF KSSM Matematik (Buku 1)

Dihasilkan dalam dua Buku, iaitu Buku 1 dan Buku 2
Memudahkan murid membuat latih tubi sebelum menghadapi
Pentaksiran Bilik Darjah
Menepati format, standard dan keperluan Pentaksiran Bilik Darjah
(PBD)
Merujuk Buku Teks dan Dokumen Standard Kurikulum dan
Pentaksiran (DSKP)
Persembahannya diolah secara integrasi untuk memudahkan murid
memahami kesemua elemen yang diterapkan
Jawapan lengkap disertakan

Judul-judul dalam siri ini:

Subjek Tingkatan 1 2 3

Matematik (Buku 1&2) Dwibahasa

Sains Dwibahasa

Pendidikan Islam

Reka Bentuk & Teknologi

Pendidikan Jasmani &
Pendidikan Kesihatan

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