Module & More Math Tg 5

  Matematik  Tingkatan 5  Bab 2 Matriks

3. Untuk mendarabkan dua matriks, A dan B, bilangan lajur 5. Jika A ialah matriks n × n dan I ialah matriks identiti

matriks A dan bilangan baris matriks B mestilah sama. n × n, maka AI = IA = A.
If A is n × n matrix and I is n × n matrix identity, then AI = IA = A.
To multiply two matrices, A and B, the number of columns of
matrix A and the number of rows of matrix B must be the same. 6. Jika AA–1 = A–1A = I, maka A–1 ialah matriks songsang
bagi A.
A × B = AB If AA–1 = A–1A = I, then A–1 is the inverse matrix of A.
m×n n×p m×p
=

a × [c d e] = ac ad ae 7. Jika A = a  b , maka A mempunyai songsangan jika
bc bd be c  d
b 1×3
BAB 2 2×1 2×3 dan hanya jika ad – bc ≠ 0.

4. Matriks identiti, I ialah matriks n × n dengan unsur pada a  b
pepenjuru utama mempunyai nilai 1 dan unsur lain If A = c  d , then A has an inverse if and only if ad – bc ≠ 0.

mempunyai nilai 0. A–1 = 1 bc d –b
Identity matrix, I is matrix n × n where the elements on its main ad – –c a
diagonal have the value of 1 and the other elements have the
ad – bc dikenali sebagai penentu matriks A, |A|.
value of 0. ad – bc is known as the determinant of matrix A, |A|.

1 0 0 … 0 8. Penyelesaian persamaan linear serentak.
0 1 0 … 0 Solution of simultaneous linear equations.

In × n = 0 0 1 … 0 ca    db x = p
y q
0 0 0 … 1
x 1 d –b p
y = ad – bc –c a q NOTA

6. Hitung setiap penambahan berikut. TP 3

Calculate each of the following additions.

CONTOH

5 –5  +  –3 8 Sudut Kalkulator
2 4 4 –7

Penyelesaian: Biarkan A = 5 –5 , B = –3 8 dan D = A + B
Let 2 4 4 –7 and

= 5 – 3 –5 + 8 Untuk memastikan saintifik kalkulator dalam MODE MAT, tekan MODE MODE MODE 2 .
2 + 4 4–7
To ensure the scientific calculator in MODE MAT, press

= 2 3 Untuk memilih matriks A:  Tekan SHIFT MAT 1 1 (1 untuk A, 2 untuk B, 3 untuk C)
6 –3
To choose matrix A:  Press (1 for A, 2 for B, 3 for C )

Untuk masukkan peringkat matriks A:  Tekan 2 = 2 = (Peringkat 2 × 2)

To key-in the order of matrix A:  Press (Oder 2 × 2)

Untuk masukkan unsur-unsur matriks A:  Tekan 5 = (–) 5 = 2 = 4 = (mengikut urutan a11, a12, a21, a22)
To key-in the elements of matrix A:  Press (follow the sequence a11, a12, a21, a22)

Ulangi langkah di atas untuk martiks B. / Repeat above steps for matrix B.

Untuk mendapat A + B / To obtain A + B :

Tekan  SHIFT MAT 3 1 + SHIFT MAT 32=
Press

2 dipapar (d11)
is displayed for (d11)

Tekan  , 3 dipaparkan (d12) Tekan  , 6 dipapar (d21) Tekan  , –3 dipaparkan (d22)

Press is displayed for (d12) Press  is displayed for (d21) Press is displayed for (d22)

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Matematik  Tingkatan 5  Bab 2 Matriks 

(a) 1 3  +  2 –4 (b) [6  1] + [2  –9] (c) 5  +  2
0 –8 9 3 2 –3
= [6 + 2  1 + (–9)]
= 1 + 2 3 + (–4) = [8  –8] = 5+2
0 + 9 –8 + 3 2 + (–3)

= 3 –1 = 7
9 –5 –1

(d) 3 –1  +  3 –4 2 1 3 –2 7 –3 –4 8 BAB 2
2 –3 4 –2 (e) 3 0  +  4 5 (f) 2 –1  +  1 0

= 3 + 3 –1 + (–4) –6 –7 1 –5 5 6 –5 –3
2 + 4 –3 + (–2)
2 + 3 1 + (–2) 7 + (–4) –3 + 8
= 6 –5 =  3 + 4 0 + 5 =  2 + 1 –1 + 0
6 –5
–6 + 1 –7 + (–5) 5 + (–5) 6 + (–3)

5 –1 3 5
=  7 5 =  3 –1

–5 –12 0 3

(g) 2 –6 3  +  12 –1 5 (h) 7 –5 1  +  63 –6 1 7 0 1 4 6 –1
10 4 0 3 –4 11 3 2 5 0 (i) –1 –2 5  +  –1 8 –3

=  120++12 –6 + (–1) 3 + 5 =  171++63 –5 + (–6) 1+1 2 3 –1 7 10 –5
4 + 3 0 + (–4) 3 + 5 2+0
7 + 4 0 + 6 1 + (–1)
=  132 –7 8 =  1134 –11 2 =  –1 + (–1) –2 + 8 5 + (–3)
7 –4 8 2
2 + 7 3 + 10 –1 + (–5)

11 6 0
=  –2 6 2

9 13 –6

7. Hitung setiap penolakan berikut. TP 3

Calculate each of the following subtractions.

CONTOH (a) 3  –  1 (b) [5  6] – [3  –2]
–4 6
7 1 2 3 = [5 – 3  6 – (–2)]
–3 6  –  7 –5 3–1 = [2  8]
–4 – 6
Penyelesaian: =

= 7 – 2 1–3 = 2
–3 – 7 6 – (–5) –10

= 5 –2
–10 11

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  Matematik  Tingkatan 5  Bab 2 Matriks

4 5 15 – 1 (d) 1 –4  –  –1 5 (e) 5 0 3  –  ––23 –1 1
8 2 3 5 3 –4 1 –4 –4 4 6
(c) – 152 7  –  3 –8

5 = 1 – (–1) –4–5 = 1 5 –– (–2) 0 – (–1) 3–1
3 – 3 5 – (–4) (–3) –4 – 4 –4 – 6
5 – – 1 
= 4 – 15 8 2 = 2 –9 =  74 1 2
0 9 –8 –10
– 12 – 3 7 – (–8)
5 5

BAB 2 –11 9
= 8

–3 15

9 0 3 5 2 –3 5 –1 10 8 7 0 4 –4
(f) –2 3  –  –1 2 (g) 1 –5  –  –2 3 (h) –1 2 1  –  3 1 –5

7 1 2 4 6 9 0 6 5 –1 9 –2 6 –1

9 – 3 0 – 5 2 – 5 –3 – (–1) 10 – 0 8 – 4 7 – (–4)
=  –2 – (–1) 3 – 2 =  1 – (–2) –5 – 3 =  –1 – 3 2 – 1 1 – (–5)

7 – 2 1 – 4 6 – 0 9 – 6 5 – (–2) –1 – 6 9 – (–1)

6 –5 –3 –2 10 4 11
=  –1 1 =  3 –8 =  –4 1 6

5 –3 6 3 7 –7 10

8. Cari nilai x dan nilai y bagi setiap persamaan matriks yang berikut. TP 3

Find the value of x and of y for each of the following matrix equations.

CONTOH (a) [x 4 y + 2] + [–2  –5  6] = [0  –1  11]

[x  –2] – [5  y] = [–4  1] [x – 2  4 – 5  y + 2 + 6] = [0  –1  11]
[x – 2  –1  y + 8] = [0  –1  11]
Penyelesaian:

[x  –2] – [5  y] = [–4  1] x – 2 = 0   ,   y + 8 = 11

[x – 5  –2 – y] = [–4  1] x = 2 y = 3

x – 5 = –4   ,   –2 – y = 1

x = 5 – 4 y = –3

= 1

(b) 3 x –1  +  5y –1 0 = 8 –3 –1 (c) 4 3  –  1 x  =  3 –2
–4 5 6 –9 –2 –2 –4 4 1 –2 – y –4 –3 5 7

–34++5y x + (–1) –1 +0 = 8 –3 –1 4 – 1 –2 3–x = 3 –2
5 + (–9) 6+ (–2) –2 –4 4 1 – (–4) – y – (–3) 5 7

8 x – 1 –1 = 8 –3 –1 3 3 – x = 3 –2
y – 4 –4 4 –2 –4 4 5 1 – y 5 7

x – 1 = –3  ,   y – 4 = –2 3 – x = –2  ,   1 – y = 7

x = –2 y = 2 x = 5 y = –6

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Matematik  Tingkatan 5  Bab 2 Matriks 

9. Hitung setiap pendaraban skalar berikut. TP 3 Sudut Kalkulator

Calculate each of the following scalar multiplications.

CONTOH Masukkan A = 1 ,B= –1 dalam kalkulator saintifik.
Key-in –2 3 in the scientific calculator.
1 3 –4 2 1 –1
 2 1 5 –2 – 2 3 Untuk mendapat / To obtain A – 2B:

= 2(3) 2(–4) = 1 – 2(–1) Tekan  SHIFT MAT 3 1 – 2 SHIFT MAT 32=
2 (1) 2(5) –2 –2(3)
Press 3 dipaparkan / is displayed

= 6 –8 = 3 Tekan  , –8 dipaparkan / is displayed BAB 2
2 10 –8 Press

(a) 1 16 0 –12 (b) 1 15 –25 (c) – 1 10 16 2
4 4 –8 20 5 –10 5 2 –4 4 –6
10 1 –8 14
1 1 1 –20
4 4 4
(16) (0) (–12) 1 1 1 1 1
5 5 2 2 2
= 1 1 1 (15) (–25) – (10) – (16) – (2)
4 4 4
(4) (–8) (20) = 1 (–10) 1 (5) = – 1 (–4) – 1 (4) – 1 (–6)
5 5 2 2 2
4 0 –3 1 1
= 1 –2 5 5 (10) 5 (–20) – 1 (1) – 1 (–8) – 1 (14)
2 2 2

3 –5 –5 –8 –1

= –2 1 = 2 –2 3
2 –4 1
– 2 4 –7

(d) 11 4   + 5 1 –2 (e) –2 –3 + –4 (f)
–5 3 2 3 2 3
1 0 2  + (–1) –21 –0.3 –1
11 + 5(1) 4 + 5(–2) –2(–3) + (–4) 2 –1 3 –2 –4
–5 + 5(2) 3 + 5(3) –2(2) + 3
= = =  1 + 1 0 + 0.3 2+1
2 – 2 –1 + 2 3+4
16 –6 2
= 5 18 = –1 =  2 0.3 3
0 1 7

10. Hitung nilai x dan y bagi setiap yang berikut. TP 3

Calculate the values of x and y of each of the following.

CONTOH (a) x – 3 –4  =  15
4 y –8
4 x – –3   =  11
–5 y –15 x4 + 12 15
– 3y =  –8
Penyelesaian:

4x + 3 =  11 x + 12 = 15   , 4 – 3y = –8
–20 – y –15 x = 3 3y = 12
y = 4
4x + 3 = 11   , –20 – y = –15

4x = 8 y = –20 + 15

x = 2 y = –5

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  Matematik  Tingkatan 5  Bab 2 Matriks

(b) 10 –7   – 2 y –2  =  6 –3 2 x1
x –3 3 –3 –10 3
(c) (–2) y + 2  +  –7  =  –13
10 – 2y –7 +4 6 –3 –2 4 8
x – 6 –3 +6 –10 3
=  –4 + x 1

10 – 2y –3 6 –3 –2y – 4 – 7 =  –13
x – 6 3 –10 3
=  4+4 8

x – 6 = –10   , 10 – 2y = 6 x–4 1

x = –4 2y = 4 –2y – 11 =  –13

BAB 2 y = 2 88

x – 4 = 1   , –2y – 11 = –13
x = 5 2y = 2
y = 1

11. Hitung pendaraban dua matriks berikut. TP 3

Calculate the following multiplication of two matrices.

CONTOH

2 4 –1 3 Tip
–3 5 4 –2
  Langkah-langkah untuk mendarab dua matriks / Steps to multiply two matrices:

Penyelesaian: Katakan / Let P = 2 4 ,Q= –1 3
–3 5 4 –2

2 4 –1 3  Tentukan peringkat matriks. / Determine the order of the matrices.
–3 5 4 –2
=   P × Q = PQ

2 × 2 = 2 × 2 = 2 × 2

= 2 4   –1 3  Hitung hasil tambah bagi hasil darab unsur dalam baris pertama matriks P dengan unsur dalam
–3 5 4 –2 lajur pertama matriks Q (pendaraban unsur ikut anak panah).

= 2 4   –1 3 Calculate the sum of the product of elements in the first row of matrix P and the elements in the first column of
–3 5 4 –2 matrix Q (the multiplication of elements follows the arrows)

= 2 4   –1 3  Hitung hasil tambah bagi hasil darab unsur dalam baris pertama matriks P dengan unsur dalam
–3 5 4 –2 lajur kedua matriks Q (pendaraban unsur ikut anak panah).

Calculate the sum of the product of elements in the first row of matrix P and the elements in the second column
of matrix Q (the multiplication of elements follows the arrows)

 Ulang langkah 2 dan 3 bagi baris kedua matriks P. / Repeat steps 2 and 3 for the second row of matrix P.

= (2)(–1) + (4)(4) (2)(3) + (4)(–2)  
(–3)(–1) + (5)(4) (–3)(3) + (5)(–2)

= 14 –2
23 –19
Sudut Kalkulator

Masukkan A = 2 4 ,B= –1 3 dalam kalkulator saintifik.
Key-in –3 5 4 –2 in the scientific calculator.

Untuk mendapatkan AB / To obtain AB:

Tekan  SHIFT MAT 3 1 SHIFT MAT 32=
Press

14 dipaparkan / is displayed ;

Tekan  ,  –2   dipaparkan;  Tekan  ,  23   dipaparkan;  Tekan  ,  –19  dipaparkan
Press is displayed ;   Press is displayed;   Press is displayed

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Matematik  Tingkatan 5  Bab 2 Matriks 

(a) –2 [–4  1] (b) [4  –1] 5
3 –6

=  (–2)(–4) (–2)(1) =  [(4)(5) + (–1)(–6)]
(3)(–4) (3)(1) = [26]

=  –182 –2
3

(c) [3  1] 2 5 (d) 6 –3 4 BAB 2
–1 3 2 5 –2

=  [(3)(2) + (1)(–1)   (3)(5) + (1)(3)] = ((62))((44))++((–53))((––22))
= [5  18] = 3–20

(e) 0 8 2 –3 (f) 3 –4 0 2
6 1 7 3 6 2 –4 5

=  ((06))((22)) + (8)(7) (0)(–3) + (8)(3) =  ((36))((00))++((–24))((––44)) (3)(2) + (–4)(5)
+ (1)(7) (6)(–3) + (1)(3) (6)(2) + (2)(5)

=  1569 24 =  1–86 –14
–15 22

(g) 1 8 –4 3 1 –1 4 2
3 –1 2 –5 (h) 6 5 –2 –9
–1
4 –7 3 3
= ((13))((33)) + (8)(–5) + (–4)(–1)
+ (–1)(–5) + (2)(–1) (1)(2) + (–1)(–9) + (4)(3)
= (6)(2) + (5)(–9) + (–2)(3)
= –1323
(4)(2) + (–7)(–9) + (3)(3)

23
= –39

80

12. Hitung nilai x dan y bagi setiap yang berikut. TP 3

Calculate the values of x and y of each of the following.

CONTOH (a) 3 [2  y  –4] = 6 3 –12
x 8 4 –16
–2 4 –6
x [y  3] = –10 15

Penyelesaian: 3(2) 3y 3(–4) =  68 3 –12
2x xy –4x 4 –16
–2x(yy ) –x2((33)) =  4 –6
–10 15 6 3y –12 =  68 3 –12
2x xy –4x 4 –16
–2y –6 4 –6
xy 3x =  –10 15 2x = 8   ,   3y = 3

3x = 15  ,   –2y = 4 x = 4 y = 1

x = 5 y = –2

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  Matematik  Tingkatan 5  Bab 2 Matriks

(b) [x 2] –3 2  = [19  –8] (c) 7 x –1 y  =  11 11
5 y –4 2 –3 4 –2 –12

[–3x + 10  2x + 2y] = [19  –8] Gantikan x = –3 –7 – 3x 7y + 4x  =  1–21 11
Substitue –2 –4y + 8 –12

– 3x + 10 = 19   ,   2x + 2y = –8 –7 – 3x = 11  , –4y + 8 = –12

3x = –9 2(–3) + 2y = –8 3x = –18 4y = 20

x = –3 2y = –2 x = –6 y = 5

y = –1

BAB 2 13. Tentukan sama ada matriks Q ialah matriks identiti atau tidak. Berikan justifikasi anda. TP 2

Determine whether matrix Q is an identity matrix. Give your justification.

CONTOH

P= –3 1 ,Q= 1 0 Tip
2 7 0 1
• Semua matriks identiti ialah matriks segi empat
Penyelesaian: sama.

PQ = –3 1 1 0 All identity matrices are square matrices.
2 7 0 1 • AI = IA = A dengan matriks A dan matriks identiti, I

= –23++ 00 00 ++ 17 ialah matriks segi empat sama.
–3 1 AI = IA = A where matrix A and identity matrix, I are square
= 2 7
matrices.

Q ialah matriks identiti kerana PQ = P.

Q is an identity matrix because PQ = P.

(a) P = 5 3 ,Q= 1 0 (b) P = 5 4 ,Q= –1 0
–4 6 0 1 –2 –3 0 –1

PQ = 5 3 1 0 PQ = 5 4 –1 0
–4 6 0 1 –2 –3 0 –1

= 5 + 0 0+3 = –5 + 0 0–4
–4 + 0 0+6 2 + 0 0+3

= 5 3 = –5 –4
–4 6 2 3

Q ialah matriks identiti kerana PQ = P. Q bukan matriks identiti kerana PQ ≠ P.

Q is an identity matrix because PQ = P. Q is not an identity matrix because PQ ≠ P.

14. Diberi E = 3 –2 dan F = 6 –4 , hitung setiap yang berikut. TP 3
–1 4 5 –2

Given E = 3 –2 and F = 6 –4 , calculate each of the following.
–1 4 5 –2

CONTOH (a) I(E – F) (b) IF – IE

E + FI =  IE – IF =  F – E 3 –2
6 –4 –1 4
=E+F =  E – F –2 6 =  5 –2  – 
3 4 5 –4
= 3 –2  + 6 –4 =  –1  –  –2 3 –2
–1 4 5 –2 6 –6
–3 2 = 
–6 6
9 –6 = 
4 2
=

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Matematik  Tingkatan 5  Bab 2 Matriks 

15. Tentukan sama ada B ialah matriks songsang bagi matriks A. TP 3

Determine whether B is the inverse matrix of matrix A.

CONTOH 3 2 2 –1
5 4
3 –1 2 1 (a) A = , B= – 5 3
–5 2 5 3 2 2
A= ,B=

Penyelesaian: AB BA

AB BA 3 2 2 –1 2 –1 3 2
5 4 5 4
= – 5 3 = – 5 3
2 2 2 2
= 3 –1 2 1 = 2 1 3 –1 BAB 2
–5 2 5 3 5 3 –5 2
6 – 5 –3 + 3 6 – 5 4–4
= 10 – 10 –5 + 6 = –5 + 6
6 – 5 3–3 6 – 5 –2 + 2 15 15
= –10 + 10 –5 + 6 = 15 – 15 –5 + 6 – 2 + 2

= 1 0 = 1 0 = 1 0 = 1 0
0 1 0 1 0 1 0 1

AB = BA = I AB = BA = I
Maka, B ialah matriks songsang bagi matriks A. Maka, B ialah matriks songsang bagi matriks A.

Hence, B is the inverse matrix of matrix A. Hence, B is the inverse matrix of matrix A.

(b) A = –3 1 –1 – 1 (c) A = 5 3 ,B= –5 –2
4 –2 ,B= 2 –2 6 3 –6
3
–2 – 2

AB BA AB

= –3 1 –1 – 1 –1 – 1 –3 1 = 5 3 –5 –2
4 2 = 2 4 –2 –2 6 3 –6
–2 3 3
–2 – 2 –2 – 2
–25 + 9 –10 – 18
3 3 3 – 2 –1 + 1 = 10 + 18 4 – 36
2 2 6 – 6 –2 + 3
= 3 – 2 – = –16 –28
–4 + 4 28 –32
–2 + 3 1 0 =
0 1
= 1 0 =
0 1
AB ≠ I
AB = BA = I Maka, B bukan matriks songsang bagi matriks A.
Maka, B ialah matriks songsang bagi matriks A.
Hence, B is not the inverse matrix of matrix A.
Hence, B is the inverse matrix of matrix A.

16. Tentukan sama ada setiap matriks berikut mempunyai matriks songsang atau tidak. TP 2

Determine whether each of the following matrices has inverse matrix.

CONTOH (a) 2 –3 (b) 8 –4
3 4 4 –2
8 5
7 6 ad − bc = (2)(4) − (−3)(3) ad − bc = (8)(–2) − (−4)(4)
= 0
Penyelesaian: = 17 ≠ 0
ad – bc = (8)(6) − (5)(7)
= 13 ≠ 0 ∴ Matriks songsang wujud. ∴ Matriks songsang tidak
∴  Matriks songsang wujud.
Inverse matrix exists. wujud.
   Inverse matrix exists.
Inverse matrix does not exist.

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  Matematik  Tingkatan 5  Bab 2 Matriks

17. Hitung matriks songsang bagi setiap matriks yang berikut. TP 3 Sudut Kalkulator

Calculate the inverse matrix of each of the following matrices.

CONTOH Tip Masukkan A = –2 1 dalam kalkulator saintifik.
Key-in 2 4 in the scientific calculator.

P= –2 1 P−1 = 1 d –b Untuk mendapatkan matriks songsang bagi A:
2 4 ad – bc –c a To obtain the inverse matrix of A:

Penyelesaian: dengan / where ad – bc ≠ 0. Tekan  SHIFT MAT 3 1 x–1 =
Press
P –1 4 –1 P= –2 1 Darabkan unsur b –0.4 dipaparkan
–2 –2 2 4 dan c dengan (–1). is displayed
BAB 2 = 1 Multiply elements b
(–2)(4) – (1)(2) and c with (–1). Tekan  , 0.1 dipaparkan
1 4 –1 Press is displayed
1 4 –1 P−1 = – 10 –2 –2
10 –2 –2 Tekan  , 0.2 dipaparkan
= – Tukarkan kedudukan unsur Press is displayed
a dan d.
= – 2 1 Interchange the positions of Tekan  , 0.2 dipaparkan
5 10 elements a and d. Press is displayed
1
1 5
5

(a) R = 3 0 (b) L = –3 4 (c) M = 6 1
–1 2 5 –5 13 2

R –1 = 1 2 0 L–1 = 1 –5 –4 M–1 = (6)(2) 1 2 –1
(3)(2) – (0)(–1) 1 3 (–3)(–5) – (4)(5) –5 –3 – (1)(13) –13 6

= 1 2 0 =– 1 –5 –4 = (–1) 2 –1
6 1 3 5 –5 –3 –13 6

= 1 0 1 4 = –2 1
3 = 5 13 –6
1 1 3
6 2 1 5

(d) B = 7 2 (e) S = 1 –2 (f) D= 3 –4
5 2 6 3 5 –6

B–1 = 1 2 –2 S–1 = 1 3 2 D–1 = (3)(–6) 1 (–4)(5) –6 4
(7)(2) – (2)(5) –5 7 (1)(3) – (–2)(6) –6 1 – –5 3

= 1 2 –2 = 1 3 2 = 1 –6 4
4 –5 7 15 –6 1 2 –5 3

1 – 1 1 2 = –3 2
2 2 5 15 3
= 5 7 = 2 1 – 5 2
4 4 5 15 2
– –

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Matematik  Tingkatan 5  Bab 2 Matriks 

18. Hitung nilai m jika setiap matriks yang berikut tidak mempunyai matriks songsang. TP 3

Calculate the value of m if each of the following matrices does not have an inverse matrix.

CONTOH (a) 4 –2 (b) m 5
m –5 –2 4
m –3
2 1 ad − bc = 0 ad − bc = 0
(4)(−5) − (−2)(m) = 0
Penyelesaian: −20 + 2m = 0 (m)(4) − (5)(–2) = 0
ad – bc = 0 2m = 20
(m)(1) − (−3)(2) = 0 m = 10 4m + 10 = 0
m + 6 = 0
m = −6 4m = –10
5
m = – 2 BAB 2

19. Selesaikan. TP 3 (a) Diberi A 3 2 = 1 0 . Cari matriks A.
9 5 0 1
Solve.

CONTOH

Diberi A 4 6 = 1 0 . Cari matriks A. Given A 39 2 = 1 0 . Find the matrix A.
3 4 0 1 5 0 1

Given A 43 6 = 1 0 . Find the matrix A.
4 0 1
3 2
Penyelesaian: A ialah matriks songsang bagi 9 5 .

A ialah matriks songsang bagi 4 6 . A is the inverse matrix of 3 2 .
3 4 9 5

A is the inverse matrix of 4 6 . A = —3(—5)—1– —2(9—) 5 –2
3 4 –9 3

A = —4(—4)—1– —6(3—) 4 –6 = – —31 5 –2
–3 4 –9 3

= – —12 4 –6 = – —53 —32
–3 4 3 –1

–2 3
= —32 –2

20. Tulis setiap persamaan linear serentak berikut dalam bentuk matriks. TP 3

Write each of the following simultaneous linear equations in the matrix form.

CONTOH (a) 6x − y = 16 (b) 2x + 3y = 4
7x + 2y = 25 6x + 5y = 20
3x + 4y = 10
8x − 5y = 11 6 –1 x = 16 2 3 x = 4
Penyelesaian: 7 2 y 25 6 5 y 20
3 x + 4 y = 10
8 x −5 y = 11 (c) 4x − y = –11 (d) –3x + 2y = –8
–x + 13y = –13 x – 4y = 6
83 −45 xy = 10
11 4 –1 x –11 –3 2 x –8
–1 13 y –13 1 –4 y 6
= =

31 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 2 Matriks

21. Selesaikan persamaan linear serentak berikut menggunakan kaedah matriks. TP 4

Solve the following simultaneous linear equations using matrix method.

CONTOH (a) 2x − 3y = 13
x − 4y = 9
4x + y = −5
7x + 2y = −8 2 –3 x 13
1 –4 y 9
Penyelesaian: =

4 1 x = –5 Matriks songsang bagi x = 1 –4 3 13
7 2 y –8 Inverse matrix of y – –1 2 9
BAB 2 4 1 (2)(–4) (–3)(1)
7 2

x = (4)(2) 1 (1)(7) 2 –1 –5 = – 51 ––5123 ++ 27
y – –7 4 –8 18

= –10 + 8 = – 51 –255
35 – 32
1 AX = B
= –2 2 A –1AX = A–1B = –51
3 3 X = A–1B ∴  x = 5, y = −1

∴  x = –2, y = 3 Nyatakan nilai x dan nilai y.
State the value of x and of y.

(b) –2x + y = 16 (c) 3x – 4y = 12
5x + 4y = –14 x–y=5

–2 1 x = 16 3 –4 x = 12
5 4 y –14 1 –1 y 5

x = 1 (1)(5) 4 –1 16 x = (3)(–1) 1 (–4)(1) –1 4 12
y (–2)(4) – –5 –2 –14 y – –1 3 5

= – 113 –6840++1248 = ––1122 ++ 2105

= – 113 –7582 = 83

= –46 ∴  x = 8, y = 3
∴  x = –6, y = 4

(d) 7x – 2y = 18 (e) 2x + 3y = 5
8x – 3y = 17 4x + 7y = 13

7 –2 x = 18 2 3 x = 5
8 –3 y 17 4 7 y 13

x = (7)(–3) 1 (–2)(8) –3 2 18 x = 1 7 –3 5
y – –8 7 17 y (2)(7) – (3)(4) –4 2 13

= – 51 –1–5444 + 34 = 21 –3250–+3296
+ 119

= – 51 ––2205 = 21 –64

= 45 = –32
∴  x = 4, y = 5 ∴  x = –2, y = 3

© Penerbitan Pelangi Sdn. Bhd. 32

Matematik  Tingkatan 5  Bab 2 Matriks 

22. Selesaikan setiap yang berikut. (a) Sebuah kedai buku menjual tiga buah buku

Solve each of the following. dan empat batang pen pada harga RM26. Diberi

CONTOH bahawa harga sebuah buku ialah RM4 lebih

Benjamin membeli 5 pinggan nasi goreng dan mahal daripada harga sebatang pen. TP 5 BAB 2
4 pinggan mi goreng dengan harga RM32. Beza
harga antara dua pinggan nasi goreng dengan A bookshop sells three books and four pens at the price
sepinggan mi goreng ialah RM5. of RM26. Given that the price of a book is RM4 more
expensive than the price of a pen.
Benjamin buys 5 plates of fried rice and 4 plates of fried
noodle with the price of RM32. The difference of price (i) Tulis persamaan linear serentak yang
between two plates of fried rice and a plate of fried noodle
is RM5. mewakili situasi di atas.

(i) Tulis persamaan linear serentak yang mewakili Write the simultaneous linear equations that
situasi di atas. represent the situation above.

Write the simultaneous linear equations that represent (ii) Hitung harga, dalam RM, bagi sebuah buku
the situation above.
dan sebatang pen.
(ii) Hitung harga, dalam RM, bagi sepinggan nasi
goreng dan sepinggan mi goreng. Calculate the prices, in RM, of a book and a pen.

Calculate the price, in RM, of a plate of fried rice and a Katakan / Let
plate of fried noodle. x = harga sebuah buku / price of a book
y = harga sebatang pen / price of a pen
Penyelesaian:
(i) 3x + 4y = 26
Katakan / Let x−y=4
x = harga sepinggan nasi goreng
(ii) 3 4 x = 26
price of a plate of fried rice 1 –1 y 4

y = harga sepinggan mi goreng

price of a plate of fried noodle

(i) 5x + 4y = 32 x = 1 (4)(1) –1 –4 26
2x − y = 5 y (3)(–1) – –1 3 4

(ii) 5 4 x = 32 = – 17 ––2266 +– 1162
2 –1 y 5 = – 17 ––4142

x = 1 –1 –4 32 = 26
y (5)(–1) – (4)(2) –2 5 5

= – 1 –32 – 20 Oleh itu, harga sebuah buku ialah RM6 dan harga
13 –64 + 25 sebatang pen ialah RM2.

= – 1 –52 Thus, the price of a book is RM6 and the price of a pen is
13 –39 RM2.

= 4
3

Oleh itu, harga sepinggan nasi goreng ialah
RM4 dan harga sepinggan mi goreng ialah
RM3.

Thus, the price of a plate of fried rice is RM4 and the
price of a plate of fried noodle is RM3.

33 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 2 Matriks

BAB 2 (b) Dalam suatu lawatan sambil belajar, bayaran (c) Rajah di bawah menunjukkan sebuah segi
empat tepat PQRS. TP 6
penyertaan yang dikenakan kepada seorang
The diagram shows a rectangle PQRS.
guru dan seorang murid masing-masing ialah
PQ
RM70 dan RM50. Jumlah guru dan murid yang
y cm
menyertai ialah 40 orang dan jumlah bayaran
SR
penyertaan yang dibayar ialah RM2 200. TP 5
(x + 3) cm
In a study tour, the participation fees charged for a
teacher and a pupil are RM70 and RM50 respectively. Diberi bahawa lebar bagi segi empat tepat ialah
The total number of teachers and pupils who joined is 4 cm lebih pendek daripada panjangnya dan
40 and the total of participation fee paid is RM2 200. perimeter segi empat tepat itu ialah 32 cm.

(i) Tulis persamaan linear serentak yang Given that the width of the rectangle is 4 cm shorter

mewakili situasi di atas. than the length and the perimeter of the rectangle is
32 cm.
Write the simultaneous linear equations that
represent the situation above. (i) Tulis persamaan linear serentak yang
mewakili situasi di atas.
(ii) Hitung bilangan orang guru dan murid yang
Write the simultaneous linear equations that
menyertai lawatan sambil belajar tersebut. represent the situation above.

Calculate the number of teachers and pupils who (ii) Hitung nilai x dan nilai y.
joined the study tour.
Calculate the value of x and of y.
(i) Katakan / Let
x = bilangan orang guru / number of teachers (i) y = x + 3 − 4
y = bilangan orang murid / number of pupils y = x − 1
x − y = 1 …  
x + y = 40
70x + 50y = 2 200

(ii) 1 1 x = 40
70 50 y 2 200

x = (1)(50) 1 (1)(70) 50 –1 40 2(x + 3) + 2y = 32
y – –70 1 2 200 2x + 6 + 2y = 32
2x + 2y = 26 …  

= – 210 2 000 – 2 200 x
–2 800 + 2 200 y

= – 210 –200 (ii) 1 –1 = 1
–600 2 2 26

= 1300 x = (1)(2) 1 2 1 1
y – (–1)(2) –2 1 26

Oleh itu, terdapat 10 orang guru dan 30 orang = 41 –22++2266

murid yang menyertai lawatan sambil belajar = 14 2284

itu.

Thus, there are 10 teachers and 30 pupils who joined

the study tour. = 76

∴ x = 7, y = 6

© Penerbitan Pelangi Sdn. Bhd. 34

PRAKTIS SPM Matematik  Tingkatan 5  Bab 2 Matriks 

2

Kertas 1 6. Harga jual sebuah kamus dan sebuah kalkulator di

8 –1 2S0P1M7 sebuah kedai buku masing-masing ialah RM45 dan
x 2 RM34.40. Puan Faridah membeli 5 buah kamus
1. Jika tidak wujud matriks songsang bagi ,
dan 4 buah kalkulator pada harga RMx. Puan Siti
hitung nilai x.
membeli 6 buah kamus dan 7 buah kalkulator

If the inverse matrix does not exist for 8 –1 , calculate pada harga RMy. Antara berikut, yang manakah BAB 2
x 2
kaedah yang betul untuk menghitung nilai x dan y?
The selling prices of a dictionary and a calculator in a
the value of x. stationery shop are RM45 and RM34.40 respectively.

A –16 C 4 Puan Faridah bought 5 dictionaries and 4 calculators for

B –4 D 16 RMx. Puan Siti bought 6 dictionaries and 7 calculators

for RMy. Which of the following is the correct method to

2. Diberi PQ = R dengan P ialah matriks 2 × 3 dan calculate the values of x and y?

Q ialah matriks m × 4. Nyatakan nilai m yang A 5 6 45.00 = x
4 7 34.40 y
mungkin.
Given PQ = R where P is a 2 × 3 matrix and Q is a m × 4 5 6 34.40 x
matrix. State the possible value of m. B 4 7 45.40 = y

A 2 C 6 C 5 4 45.00 = x
6 7 34.40 y
B 3 D 8

–2 5 3 6 D 5 4 34.40 = x
3 1 2 0 6 7 45.00 y
3. P+ =

Cari matriks P. 7. Diberi bahawa M ialah matriks 2 × 2 dan

Find matrix P. 2S 0P1M8 Given that M is a matrix 2 × 2 and r –6 .
s 4
A 1 11 C –5 –1 M–1 = 1 (6)(1)
3 1 1 1
(4)(–3) –
1
B –1 11 D 5 –1 Cari nilai r dan nilai s.
5 1 –1
Find the value of r and of s.

4. Diberi 15 0 – 2Q = 5 1 – 2 , cari matriks Q. A r = –3, s = –1 C r = 3, s = –1
8 –7 4 5 B r = –3, s = 1 D r = 3, s = 1

–5 3 –1 –5 –6 –10 1
5 4 2 4 9 –8
15 0 – 2Q = 5 1 – 2 8. 2 – 3 + =
8 –7 4 5
Given , find matrix Q. 2S0P1M8

–5 A 11 –19 C 11 17
13 –12 13 –12
A –5 1 C 5 1
–6 –9 –6 9 11 5 11 18
B 13 –4 D 13 –16
5 –1 10 –1
B –6 9 D 14 –16

3 0 3 –2 9. Diberi matriks songsang bagi 3 –4 ialah
–2 1 1 3 –1 q 5 –1
5. = 2S 0P1M9 1p –5 3 . Cari nilai pq.

A 9 –6 C 59 –16 Given that the inverse matrix of 3 –4 is 1 –1 q .
–5 1 Find the value of pq. 5 –1 p –5 3

B 9 –6 D 59 –76 A –92 C 68
–5 7
B 68 D 92

35 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 2 Matriks 2. Puan Tan membeli 3 rim kertas A4 dan 2 rim
Kertas 2
2S0P1M7 kertas A3 pada harga RM42 manakala Puan
Kamala membeli 2 rim kertas A4 dan 5 rim kertas
4 3
1. (a) Cari matriks songsang bagi 8 11 . A3 pada harga RM61. Dengan menggunakan

4 3 kaedah matriks, hitung harga, dalam RM, bagi
8 11
Find the inverse matrix of . satu rim kertas A4 dan satu rim kertas A3.
Puan Tan buys 3 reams of A4 paper and 2 reams of A3
(b) Terdapat dua jenis bungkusan pen di sebuah paper for RM42 while Puan Kamala buys 2 reams of A4
paper and 5 reams of A3 paper for RM61. By using the
kedai buku. Bungkusan A ada 4 batang pen matrix method, calculate the price, in RM, of a ream of A4
paper and of a ream of A3 paper.
merah dan 3 batang pen biru dan harga jual

BAB 2 ialah RM4. Bungkusan B ada 8 batang pen

merah dan 11 batang pen biru dan harga jual

ialah RM11. Dengan menggunakan kaedah Katakan harga satu rim kertas A4 dan satu rim
kertas A3 masing-masing ialah RMx dan RMy.
matriks, cari harga, dalam RM, sebatang pen
Let the prices of a rim of A4 paper and a ream of A3 paper
merah dan sebatang pen biru.
are RMx and RMy respectively.
There are two types of packaging of pens in a book
shop. Packaging A has 4 red pens and 3 blue pens 3x + 2y = 42
and the selling price is RM4. Packaging B has 8 red 2x + 5y = 61
pens and 11 blue pens and the selling price is RM11.
By using matrix method, find the prices, in RM, of a

red pen and a blue pen. 23 2 x = 42
5 y 61
(a) Matriks songsang / Inverse matrix

= 1 (3)(8) 11 –3 x = —15—1––4– 5 –2 42
(4)(11) – –8 4 y –2 3 61

= —210– 11 –3 = —111 8998
–8 4

= 2110 – 3 = 89
20

– 2 1
5 5
Maka, harga satu rim kertas A4 ialah RM8 dan
(b) Katakan harga sebatang pen merah ialah RMx
dan harga sebatang pen biru ialah RMy. harga satu rim kertas A3 ialah RM9.

Let the price of a red pen is RMx and the price of a Thus, the price of a ream of A4 paper is RM8 and the price
of a ream of A3 paper is RM9.
blue pen is RMy.
3. Diberi 4 5 x = –7 , hitung nilai x dan nilai
4x + 3y = 4 3 –1 y –10
8x + 11y = 11

84 3 x = 4 y.
11 y 11
Given that 4 5 x = –7 , calculate the value of
x 11 –3 4 3 –1 y –10
y = —210 –8 4 11
x and of y.

= —210 11 x = 1 –1 –5 –7
12 y (4)(–1) – (5)(3) –3 4 –10

= 0.55 = – —119 57
0.60 –19

Maka, harga sebatang pen merah ialah RM0.55 = –3
dan harga sebatang pen biru ialah RM0.60. 1

Thus, the price of a red pen is RM0.55 and the price ∴ x = −3, y = 1

of a blue pen is RM0.60.

© Penerbitan Pelangi Sdn. Bhd. 36

Matematik  Tingkatan 5  Bab 2 Matriks 

4. (a) Cari matriks songsang bagi 3 –4 . 5. (a) Diberi S = 5 –3 ,  T = a –1 b dan
5 –8 2 –1 a –2 5 jika
2S0P1M8
Find the inverse matrix of 3 –4 . I= 1 0 . Cari nilai dan nilai b ST = I.
5 –8 0 1

(b) Tulis persamaan linear serentak berikut dalam Given that S = 5 –3 ,  T = a –1 b and
2 –1 –2 5
bentuk matriks.

Write the following simultaneous linear equations in 1 0
0 1
matrix form. I = . Find the value of a dan of b if ST = I.

3x – 4y = 7

5x – 8y = 11 BAB 2

Seterusnya, dengan menggunakan kaedah (b) Tulis persamaan linear serentak berikut dalam
bentuk matriks.
matriks, hitung nilai x dan nilai y.
Hence, using matrix method, calculate the value of x Write the following simultaneous linear equations in
matrix form.
and of y.
5x − 3y = 1
(a) Matriks songsang / Inverse matrix 2x − y = 5

= —–2—4 —–1(––—20—) –8 4 Seterusnya, hitung nilai x dan y.
–5 3
Hence, calculate the values of x and y.

= – —14 –8 4 (a) ST = I
–5 3 T = S–1I
T = S–1
2 –1
= —54 – —34 a ––21 b = 1 –1 3
5 –5 – (–6) –2 5

(b) 3 –4 x   =  7 ––2aa 5aba = –1 3
5 –8 y 11 –2 5

x =– 1 –8 47 –a = –1 ab = 3
y 4 –5 3 11 (1)b = 3
a = 1
b = 3
= – —14 –56 + 44
–35 + 33

= – —41 –12 (b) 25 –3 x = 1
–2 –1 y 5

3 x = –1 3 1
= —12 y –2 5 5

∴ x = 3, y = —21 = –1 + 15
–2 + 25

= 14
23

∴ x = 14, y = 23

37 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 2 Matriks

6. Sebuah kilang mengedarkan tempahan stok gula 5x + 6y = 1 200
7x + 4y = 1 240
2S0P1M9 dan tepung kepada Pasar Raya Tulip dan Pasar
Raya Orkid. Pasar Raya Tulip menerima 5 kotak
5 6 x 1 200
gula dan 6 kotak tepung dengan jumlah peket 7 4 y = 1 240

ialah 1 200. Pasar Raya Orkid menerima 7 kotak

gula dan 4 kotak tepung dengan jumlah peket x 1 4 –6 1 200
y 5(4) – 6(7) –7 5 1 240
ialah 1 240. Hitung bilangan peket gula dan =

bilangan peket tepung dalam setiap kotak. = – 212 –48840000 – 7 440
+ 6 200
A factory distributes stock orders of sugar and flour to

Pasar Raya Tulip and Pasar Raya Orkid. Pasar Raya
Tulip receives 5 boxes of sugar and 6 boxes of flour with
BAB 2 the total of 1 200 packets. Pasar Raya Orkid receives = – 212 ––22 620400
7 boxes of sugar and 4 boxes of flour with the total of
1 240 packets. Calculate the number of packets of sugar = 112000

and the number of packets of flour in each box.

Katakan bilangan peket gula ialah x dan bilangan Bilangan gula ialah 120 peket dan bilangan tepung
peket tepung ialah y. ialah 100 peket dalam setiap kotak.

Let the number of packets of sugar is x and the number The number of packets of sugar is 120 and the number of
packets of flour is 100 in each box.
of packets of flour is y.

Sudut KBAT KBAT

Ekstra

Jadual di bawah menunjukkan bilangan buku rujukan dan buku latihan yang dijual di sebuah kedai buku

dalam masa dua bulan.

The table below shows the number of reference books and exercise books sold in a book shop over two months.

Bulan Bilangan buku rujukan Bilangan buku latihan

Month Number of reference books Number of exercise books

Januari / January 50 72

Februari / February y 3x

Diberi jumlah bilangan buku latihan yang dijual ialah —32 jumlah bilangan buku rujukan yang dijual. Jumlah
bilangan buku yang dijual dalam bulan Februari ialah 108 buah. Dengan menggunakan kaedah matriks, hitung
nilai x dan nilai y.

Given the total number of exercise books sold is —23 of the number of reference books sold. The total number of books sold in
February is 108. By using matrix method, calculate the value of x and of y.

Jawapan / Answer :

72 + 3x = 3 (50 + y) 2 –1 x = 2
2 3 1 y 108
48 + 2x = 50 + y
x 1 1 1 2
2x – y = 2  …  y = 5 –3 2 108

3x + y = 108 …  = 51 –26++120186

Kuiz 2

= 2422

∴  x = 22, y = 42

© Penerbitan Pelangi Sdn. Bhd. 38

BAB Matematik Pengguna: Insurans

3 Consumer Mathematics: Insurance

3.1 Risiko dan Perlindungan Insurans

Risk and Insurance Coverage

NOTA IMBASAN

1. Setiap individu terdedah kepada risiko yang akan  Apabila nilai yang diinsuranskan sama dengan
menyebabkan kerugian kewangan. Risiko ialah jumlah insurans yang harus dibeli
kemungkinan berlakunya musibah yang tidak dapat
dielakkan. Bayaran pampasan = jumlah kerugian – deduktibel
Each individual is exposed to risk which will cause financial dengan keadaan jumlah kerugian < nilai yang
burden. Risk is the possibility of a disaster that cannot be
avoided. diinsuranskan
When insured value equals to amount of required insurance
2. Premium bagi insurans hayat Payment of compensation = amount of loss – deductible
Premium of life insurance where amount of loss < insured value

 Apabila nilai yang diinsuranskan kurang daripada

= Nilai muka polisi × Kadar premium per RMx jumlah insurans yang harus dibeli
RMx
Bayaran pampasan
nilai yang
Face value of policy   = diinsuranskan
RMx jumlah insurans
= × Premium rate per RMx × jumlah – (deduktibel)
kerugian

3. Deduktibel ialah sebahagian kerugian yang yang harus dibeli
perlu ditanggung oleh pemegang polisi sebelum
mendapatkan pampasan daripada syarikat insurans. When insured value is less than amount of required insurance
Deductible is a part of loss borne by policyholder before getting
compensation from insurance company. Payment of compensation

insured value amount
amount of required   ×  of loss
 =  – (deductible)

Bayaran pampasan insurance
=  jumlah kerugian – deduktibel
 Mengalami kerugian menyeluruh
Payment of compensation
=  amount of loss – deductible Bayaran pampasan = nilai yang – (deduktibel)
diinsuranskan
Suffered a total loss

4. Ko-insurans ialah perkongsian kos dengan keadaan Payment of compensation = (insured value) – (deductible)
pemegang polisi membayar peratusan tertentu
daripada jumlah kerugian yang dilindungi. 6. Dalam insurans kesihatan, ko-insurans 90/10
Co-insurance is a cost-sharing where the policyholder pays a bermaksud 90% daripada jumlah kos rawatan
certain percentage of the covered total loss. ditanggung oleh syarikat insurans, manakala
10% ditanggung oleh pemegang polisi mengikut
5. Dalam insurans harta, jumlah insurans yang harus persetujuan.
dibeli In health insurance, co-insurance of 90/10 means 90% of the
medical costs borne by insurance company, whereas 10% is
= Peratusan ko-insurans × Nilai boleh insurans harta borne by policyholder.

In property insurance, amount of required insurance
= Co-insurance percentage × Insurable value of property

NOTA

39 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans

1. Kenal pasti syarikat insurans, pemegang polisi, had perlindungan, premium dan risiko yang dilindungi dalam
situasi berikut. TP 1

Identify the insurance company, policyholder, coverage limit, premium and risk insured in the following situations.

BAB 3 CONTOH (a) Puan Sharma mempunyai insurans daripada
Syarikat Insurans Pelita yang melindungi dirinya
Linda mempunyai insurans bernilai RM400 000 ketika dia bercuti di luar negara selama 15 hari.
yang melindungi dirinya daripada risiko membayar Insurans tersebut melindungi kemalangan diri
bil perubatan yang tinggi jika dia dimasukkan bernilai RM300 000 dan Puan Sharma membayar
ke hospital. Setiap bulan dia membayar RM150 RM90 bagi perlindungan tersebut.
kepada Yes Insurance bagi perlindungan tersebut.
Madam Sharma has an insurance from Syarikat
Linda has an insurance worth RM400 000 that covers Insurans Pelita that covers herself while in abroad for
herself from the risk of high medical bills if she being 15 days. The insurance covers personal accident worth
warded in hospital. Every month she pays RM150 to Yes
Insurance for the coverage. RM300 000 and Madam Sharma pays RM90 for the
coverage.
Penyelesaian:
Syarikat insurans / Insurance company : Syarikat insurans / Insurance company:
Yes Insurance Syarikat Insurance Pelita

Pemegang polisi / Policyholder : Pemegang polisi / Policyholder :
Linda Puan Sharma

Had perlindungan / Coverage limit : Had perlindungan / Coverage limit :
RM400 000 RM300 000

Premium / Premium : Premium / Premium :
RM150 RM90

Risiko yang dilindungi / Risk insured : Risiko yang dilindungi / Risk insured :
Belanja perubatan / Medical expenses Kemalangan diri ketika berada di luar negara

Personal accident while in abroad

2. Kenal pasti jenis insurans am bagi setiap pernyataan ini. TP 2 Insurans
kebakaran
Identify the type of general insurance of each of the following statements.
Fire insurance
CONTOH
Insurans
Insurans ini melindungi rumah daripada risiko petir, kebakaran dan letupan. Semua perjalanan
barang seperti perabot di rumah juga akan dilindungi daripada risiko ini.
Travel insurance
The insurance covers house from the risk of lightning, fire and explosion. All things such as furnitures
in the house will also be protected from the risks.

(a) Insurans ini melindungi pemegang polisi yang melancong ke luar negara daripada
risiko seperti kehilangan bagasi, penangguhan penerbangan dan kemalangan diri.

The insurance covers policyholder when travelling to overseas from the risks such as lost
baggage, flight delayed and personal accidents.

(b) Insurans ini melindungi pemegang polisi daripada tuntutan pihak ketiga yang Insurans motor

dihadapi pemegang polisi jika berlaku kemalangan jalan raya. Motor insurance

The insurance covers policyholder from the third party’s claim face by policyholder in the
occuring of road accident.

(c) Insurans ini melindungi pemegang polisi daripada kerugian yang dialami berpunca Insurans

secara langsung daripada kemalangan. kemalangan diri

The insurans covers policyholder from the loss suffered resulting directly from accident. Personal accident
insurance

© Penerbitan Pelangi Sdn. Bhd. 40

Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans 

3. Hitung premium tahunan bagi setiap yang berikut. TP 3

Calculate the annual premium of each of the following.

CONTOH (a) (b)

Nilai muka polisi Nilai muka polisi RM500 000 Nilai muka polisi RM1.2 juta
Face value of the RM300 000 RM2.72
Face value of the Face value of the RM1.2 mil
policy policy policy
RM1.63
Kadar premium RM2.12 Kadar premium Kadar premium
tahunan per tahunan per
tahunan per RM1 000 nilai RM1 000 nilai
muka muka
RM1 000 nilai
Annual premium Annual premium
muka rate per RM1 000 rate per RM1 000
face value face value
Annual premium
rate per RM1 000
face value

Penyelesaian: Premium Premium BAB 3

Premium = RM500 000 × RM2.72 = RM1 200 000 × RM1.63
RM300 000 RM1 000 RM1 000
= RM1 000 × RM2.12
= RM1 360 = RM1 956

= RM636

4. Hitung premium bulanan insurans hayat yang berikut berdasarkan jadual kadar premium tahunan bagi setiap

RM1 000 nilai muka yang diberikan. TP 3

Calculate the monthly premium of the following life insurance based on the annual premium rate schedule per RM1 000 face
value given.

Umur Bukan perokok / Non-smoker (RM) Perokok / Smoker (RM)

Age Lelaki / Male Perempuan / Female Lelaki / Male Perempuan / Female

41 2.86 1.92 3.74 2.43
42
43 3.08 2.04 4.06 2.60
44
45 3.35 2.19 4.44 2.81

3.66 2.36 4.87 3.04

4.03 2.56 5.39 3.30

CONTOH (a) Encik Kamal ingin membeli polisi insurans hayat

Encik Leong ingin membeli polisi insurans hayat yang bernilai RM200 000. Dia berumur 41 tahun

yang bernilai RM250 000. Dia berumur 45 tahun dan tidak merokok.

dan merupakan seorang perokok. Encik Kamal wants to buy a life insurance policy worth
RM200 000. He is 41 years old and a non-smoker.
Mr Leong wants to buy a life insurance policy worth
RM250 000. He is 45 years old and a smoker. Premium tahunan / Annual premium

Penyelesaian: = RM200 000 × RM2.86
RM1 000
Premium tahunan / Annual premium
RM250 000 = RM572
= RM1 000 × RM5.39

= RM1 347.50 Premium bulanan / Monthly premium
RM572
= 12

Premium bulanan / Monthly premium = RM47.67
RM1 347.50
= 12

= RM112.29

41 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans

(b) Encik Sanggam ingin membeli polisi insurans (c) Anna ingin membeli polisi insurans hayat yang

hayat yang bernilai RM550 000. Dia berumur bernilai RM600 000. Dia berumur 44 tahun dan

42 tahun dan seorang perokok. tidak merokok.

Mr Sanggam wants to buy a life insurance policy worth Anna wants to buy a life insurance policy worth
RM550 000. He is 42 years old and a smoker. RM600 000. She is 44 years old and a non-smoker.

Premium tahunan / Annual premium Premium tahunan / Annual premium
RM550 000 RM600 000
= RM1 000 × RM4.06 = RM1 000 × RM2.36

= RM2 233 = RM1 416

Premium bulanan / Monthly premium Premium bulanan / Monthly premium
RM2 233 RM1 416
= 12 = 12

= RM186.08 = RM118

BAB 3 5. Selesaikan yang berikut. TP 4 (a) Zulmi ingin membeli polisi insurans hayat

Solve the following. bernilai RM150 000 dan menambah polisi

CONTOH penyakit kritikal. Syarikat insurans tersebut

Puan Gan ingin membeli polisi insurans hayat menawarkan polisi penyakit kritikal dengan

bernilai RM250 000 dan menambah polisi penyakit perlindungan sebanyak 30% nilai muka asas.

kritikal. Syarikat insurans tersebut menawarkan Diberi kadar premium tahunan per RM1 000

polisi penyakit kritikal dengan perlindungan nilai muka bagi polisi insurans hayat itu ialah

sebanyak 40% nilai muka asas. Diberi kadar RM2.12 dan polisi penyakit kritikal ialah RM1.74

premium tahunan per RM1 000 nilai muka bagi bagi umur dan status kesihatan Zulmi, hitung

polisi insurans hayat itu ialah RM1.33 dan polisi premium tahunannya.

penyakit kritikal ialah RM1.84 bagi umur dan status Zulmi wants to buy a life insurance policy worth
RM150 000 and adds on a critical illness policy. The
kesihatan Puan Gan, hitung premium tahunannya. insurance company offers a critical illness policy with
a coverage of 30% of basic face value. Given that the
Madam Gan wants to buy a life insurance policy worth annual premium rate per RM1 000 face value for the
RM250 000 and adds on a critical illness policy. The life insurance policy is RM2.12 and the critical illness
insurance company offers a critical illness policy with a policy is RM1.74 according to the age and health status
coverage of 40% of basic face value. Given that the annual of Zulmi, calculate the annual premium.
premium rate per RM1 000 face value for the life insurance
policy is RM1.33 and the critical illness policy is RM1.84
according to the age and health status of Madam Gan,
calculate the annual premium.

Penyelesaian: Jumlah perlindungan untuk penyakit kritikal
Jumlah perlindungan untuk penyakit kritikal
Amount of coverage for critical illness
Amount of coverage for critical illness
= 30% × RM150 000
= 40% × RM250 000 = RM45 000
= RM100 000

Premium tahunan Puan Gan Premium tahunan Zulmi

Annual premium of Madam Gan Annual premium of Zulmi

= RM250 000 × RM1.33 + RM100 000 × RM1.84 = RM150 000 × RM2.12 + RM45 000 × RM1.74
RM1 000 RM1 000 RM1 000 RM1 000

= RM332.50 + RM184.00 = RM318.00 + RM78.30
= RM396.30
= RM516.50 Premium asas tahunan + Premium
tahunan penyakit kritikal
Annual basic premium + Annual
premium for critical illness

© Penerbitan Pelangi Sdn. Bhd. 42

Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans 

6. Hitung premium kasar bagi polisi komprehensif, polisi pihak ketiga, kebakaran dan kecurian, dan polisi pihak

ketiga berdasarkan Jadual Tarif Motor 2015 di bawah. TP 4

Calculate the gross premium of comprehensive policy, third party, fire and theft policy, and third party policy based on the
Schedule of Motor Tariff 2015 below.

Kapasiti enjin Semenanjung Malaysia Sabah dan Sarawak

tidak melebihi Peninsular Malaysia Sabah and Sarawak

Engine capacity Polisi komprehensif Polisi pihak Polisi komprehensif Polisi pihak
not exceeding ketiga ketiga
Comprehensive policy Comprehensive policy
(cc) Third party policy Third party policy
(RM) (RM)
(RM) (RM)

1400 273.80 +RM26.00 bagi setiap 120.60 196.20 +RM20.30 bagi setiap 67.50
1650 305.50 RM1 000 baki. 135.00 220.00 RM1 000 baki. 75.60
2200 339.10 +RM26.00 for every 151.20 243.90 +RM20.30 for every 85.20
RM1 000 balance. RM1 000 balance.

CONTOH BAB 3

Penggunaan: Semenanjung Malaysia Tip

Usage : Peninsular Malaysia • Diskaun Tanpa Tuntutan (NCD) ialah ganjaran

Jumlah diinsuranskan / Sum insured : RM40 000 untuk pemilik kereta jika tiada sebarang
Kapasiti enjin / Engine capacity : 1630 cc
NCD : 30% tuntutan dibuat dalam tempoh setahun.
No Claim Discount (NCD) is a reward to car owner if

no claim was made on annual basis.

Penyelesaian: Nilai NCD = x% × premium asas
Polisi komprehensif: NCD value = x% × basic premium
• Premium asas bagi polisi pihak ketiga,
Comprehensive policy:
kebakaran dan kecurian
(a) RM1 000 yang pertama
Basic premium for third party, fire and theft policy
The first RM1 000 premium asas polisi komprehensif
RM305.50 Rujuk jadual. =  75% × basic premium of comprehensive policy
(b) RM26 × 39 RM1 014 Refer schedule.

RM40 000 – RM1 000
RM1 000

(c) Premium asas = (a) + (b) RM1 319.50 0.3 × RM1 319.50

Basic premium RM395.85
RM923.65
(d) NCD 30%

(e) Premium kasar = (c) – (d)

Gross premium

Polisi pihak ketiga, kebakaran dan kecurian: Polisi pihak ketiga: Rujuk jadual.
Refer schedule.
Third party, fire and theft policy: Third party policy:

(a) Premium asas 0.75 × RM1 319.50 (a) Premium asas RM135.00
Basic premium = RM989.63 Basic premium

(b) NCD 30% RM296.89 (b) NCD 30% RM40.50

(c) Premium kasar = (a) – (b) RM692.74 (c) Premium kasar = (a) – (b) RM94.50

Gross premium Gross premium

0.3 × RM989.63 0.3 × RM135.00

43 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans

(a) (b)

Penggunaan : Semenanjung Malaysia Penggunaan : Sabah dan Sarawak

Usage : Peninsular Malaysia Usage : Sabah and Sarawak

Jumlah diinsuranskan / Sum insured : RM55 000 Jumlah diinsuranskan / Sum insured : RM35 000
Kapasiti enjin / Engine capacity : 1370 cc Kapasiti enjin / Engine capacity : 2100 cc
NCD : 38.33% NCD : 45%

Polisi komprehensif / Comprehensive policy: Polisi komprehensif / Comprehensive policy:

(a) RM1 000 yang pertama RM273.80 (a) RM1 000 yang pertama RM243.90

The first RM1 000 The first RM1 000

(b) RM26 × 54 RM1 404 (b) RM20.30 × 34 RM690.20
(baki / balance) (baki / balance)

(c) Premium asas = (a) + (b) RM1 677.80 (c) Premium asas = (a) + (b) RM934.10

Basic premium Basic premium

BAB 3 (d) NCD 38.33% RM643.10 (d) NCD 45% RM420.35

(e) Premium kasar = (c) – (d) RM1 034.70 (e) Premium kasar = (c) – (d) RM513.75

Gross premium Gross premium

Polisi pihak ketiga, kebakaran dan kecurian: Polisi pihak ketiga, kebakaran dan kecurian:

Third party, fire and theft policy: Third party, fire and theft policy:

(a) Premium asas 0.75 × RM1 677.80 (a) Premium asas 0.75 × RM934.10
Basic premium = RM1 258.35 Basic premium = RM700.58

(b) NCD 38.33% RM482.33 (b) NCD 45% RM315.26

(c) Premium kasar = (a) – (b) RM776.02 (c) Premium kasar = (a) – (b) RM385.32

Gross premium Gross premium

Polisi pihak ketiga / Third party policy: Polisi pihak ketiga / Third party policy:

(a) Premium asas RM120.60 (a) Premium asas RM85.20
Basic premium Basic premium

(b) NCD 38.33% RM46.23 (b) NCD 45% RM38.34

(c) Premium kasar = (a) – (b) RM74.37 (c) Premium kasar = (a) – (b) RM46.86

Gross premium Gross premium

7. Selesaikan setiap yang berikut. TP 4

Solve each of the following.

CONTOH

Kamil telah membuat tuntuan pampasan kepada syarikat insurans keretanya Tuntutan Kerugian

sepanjang tahun lepas seperti yang ditunjukkan dalam jadual di sebelah. Dalam Claim Loss

polisi tersebut, terdapat peruntukan deduktibel sebanyak RM400. Hitung 1 RM300
2 RM1 800
bayaran pampasan yang diterima bagi setiap kerugian yang dialami Kamil.

Kamil had made claims from his car’s insurance company throughout last year as
shown in the table. In the policy, there was deductible provision of RM400. Calculate the
compensation amount received for each of loss suffered by Kamil.

Penyelesaian:

Tuntuan 1 / Claim 1: Kerugian , Deduktibel Tuntutan 2 / Claim 2: Kerugian . Deduktibel
RM300 < RM400 Loss , Deductible RM1 800 > RM400 Loss . Deductible

Oleh itu, tiada bayaran pampasan. Bayaran pampasan / Compensation payment

Thus, no compensation payment. = RM1 800 – RM400 = RM1 400

© Penerbitan Pelangi Sdn. Bhd. 44

Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans 

(a) Polisi insurans kereta Encik Tony mempunyai deduktibel sebanyak RM500. Tuntutan Kerugian

Jadual di sebelah menunjukkan tuntuan pampasan kepada syarikat Claim Loss

insuransnya dalam setahun. Hitung bayaran pampasan yang diterima bagi 1 RM2 300
2 RM800
setiap kerugian yang dialami Encik Tony. 3 RM150

Mr Tony’s car insurance policy has a deductible of RM500. The table shows the claims
to his insurance company in a year. Calculate the compensation amount received for
each of loss suffered by Mr Tony.

Tuntuan 1 / Claim 1: Tuntutan 2 / Claim 2: Tuntutan 3 / Claim 3:
RM2 300 . RM500 RM800 . RM500
Bayaran pampasan Bayaran pampasan RM150 , RM500

Compensation payment Compensation payment Tiada bayaran pampasan.

= RM2 300 – RM500 = RM800 – RM500 No compensation payment.
= RM1 800 = RM300

(b) Rania telah membeli polisi insurans perubatan dengan deduktibel sebanyak RM5 000 setahun dan had BAB 3

tahunan sebanyak RM200 000. Pada tahun kedua dalam tempoh insuransnya, dia telah membuat

tuntutan dua kali masing-masing sebanyak RM82 000 dan RM3 500 bagi membayar bil hospitalnya.

Hitung bayaran pampasan yang diterima Rania bagi setiap tuntutannya.

Rania has purchased a medical insurance policy with a deductible of RM5 000 per year with an annual limit of
RM200 000. In the second year of her insurance period, she made two claims of RM82 000 and RM3 500 respectively
to pay her hospital bills. Calculate the compensation amount received by Rania for each claim.

Bagi tuntutan RM82 000, Bagi tuntutan RM3 500,

For the claim of RM82 000, For the claim of RM3 500,

RM82 000 . RM5 000 RM3 500 , RM5 000

Bayaran pampasan Tiada bayaran pampasan.

Compensation payment No compensation payment.

= RM82 000 – RM5 000
= RM77 000

(c) Shawati mempunyai polisi insurans perubatan dengan peruntukan deduktibel sebanyak RM30 000
setahun dengan had tahunan bernilai RM500 000. Pada tahun pertama dalam tempoh insuransnya,
Shawati telah menjalani pembedahan dengan kos rawatan sebanyak RM21 400. Pada tahun kedua
dalam tempoh insuransnya, Shawati telah menjalani rawatan kanser dengan kos sebanyak RM280 000.

Hitung bayaran pampasan yang diterima Shawati bagi tahun pertama dan tahun kedua polisi
insuransnya.

Shawati has a medical insurance policy with an allocation of deductible of RM30 000 in a year with an annual limit of
RM500 000. In the first year of her insurance period, she underwent a surgery and the treatment cost was RM21 400.
In the second year of her insurance period, she underwent a cancer treatment with the cost of RM280 000. Calculate
the compensation amount received by Shawati for the first and second year of her insurance policy.

Bagi tahun pertama, Bagi tahun kedua,

For the first year, For the second year,

RM21 400 , RM30 000 RM280 000 . RM30 000

Tiada bayaran pampasan. Bayaran pampasan

No compensation payment. Compensation payment

= RM280 000 – RM30 000
= RM250 000

45 © Penerbitan Pelangi Sdn. Bhd.

BAB 3   Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans (a) Encik Ashvin mempunyai insurans kebakaran

8. Selesaikan. TP 5 TP 6 untuk rumahnya dengan deduktibel sebanyak

Solve. RM8 000. Polisi kebakaran itu mempunyai ko-

CONTOH insurans 80% dan nilai boleh insurans rumah

Rumah Encik Wong mempunyai insurans itu ialah RM1.5 juta. Rumah Encik Ashvin
kebakaran dengan deduktibel sebanyak RM4 000.
Syarikat insurans telah menetapkan ko-insurans mengalami kebakaran dan penilaian kerugian
75% bagi polisi kebakaran tersebut dan nilai boleh
insurans rumah itu ialah RM400 000. Encik Wong adalah sebanyak RM460 000. Hitung bayaran
mengalami kerugian sebanyak RM48 000 akibat
kebakaran di rumahnya. Hitung bayaran pampasan pampasan jika dia telah menginsuranskan
jika dia telah menginsuranskan rumahnya
rumahnya
Mr Wong’s house has a fire insurance with a deductible
of RM4 000. The insurance company has specified the Mr Ashvin has a fire insurance for his house with
co-insurance of 75% for the fire policy and the insurable a deductible of RM8 000. The fire insurance has a
value of the house is RM400 000. Mr Wong suffers a loss of
RM48 000 resulting from a fire in his house. Calculate the co-insurance of 80% and the insurable value of the
amount of compensation if he insures the house house is RM1.5 million. Mr Ashvin’s house caught in a
fire and the loss valuation was RM460 000. Calculate
(i) pada jumlah insurans yang harus dibeli, the amount of compensation if he insures the house

at the amount of required insurance, (i) pada jumlah insurans yang harus dibeli,

(ii) dengan jumlah RM200 000. Kemudian, hitung at the amount of required insurance,
penalti ko-insurans.
(ii) dengan jumlah RM780 000. Kemudian,
at an amount of RM200 000. Hence, calculate the
co-insurance penalty. hitung penalti ko-insurans.

at an amount of RM780 000. Hence, calculate the
co-insurance penalty.

Penyelesaian: (i) Bayaran pampasan
(i) Bayaran pampasan
Amount of compensation Amount of compensation
= RM48 000 – RM4 000
= RM44 000 = RM460 000 – RM8 000

= RM452 000

(ii) Jumlah insurans yang harus dibeli (ii) Jumlah insurans yang harus dibeli

Amount of required insurance Amount of required insurance

= 0.75 × RM400 000 = 0.80 × RM1 500 000
= RM300 000 = RM1 200 000

RM780 000 , RM1 200 000

Bayaran pampasan

RM200 000 , RM300 000 Amount of compensation

Bayaran pampasan / Amount of compensation = RM780 000 × RM460 000 – RM8 000
RM1 200 000
RM200 000
= RM300 000 × RM48 000 – RM4 000 = RM299 000 – RM8 000

= RM32 000 – RM4 000 = RM291 000

= RM28 000 Penalti ko-insurans
Co-insurance penalty
Penalti ko-insurans / Co-insurance penalty = RM460 000 – RM299 000
= RM48 000 – RM32 000 = RM161 000
= RM16 000

Atau / Or

Penalti ko-insurans / Co-insurance penalty

 =
1– RM200 000 × RM48 000
RM300 000

= RM16 000

© Penerbitan Pelangi Sdn. Bhd. 46

Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans 

9. Hitung jumlah kos yang ditanggung syarikat insurans dan jumlah kos yang ditanggung pemegang polisi bagi

situasi berikut. TP 5

Calculate the amount of cost incurred by insurance company and the amount of cost borne by policyholder for the following
situations.

CONTOH (a) Puan Tan mempunyai polisi insurans perubatan

Puan Roslina mempunyai polisi insurans perubatan dengan peruntukan deduktibel sebanyak RM750

dengan peruntukan deduktibel sebanyak RM600 dan fasal penyertaan peratusan ko-insurans

dan fasal penyertaan peratusan ko-insurans 80/20. 70/30. Kos perubatan yang dilindungi dalam

Kos perubatan yang dilindungi dalam polisi Puan polisi Puan Tan ialah RM74 000.

Roslina ialah RM25 000. Madam Tan has a medical insurance policy with
a deductible provision of RM750 and co-insurance
Puan Roslina has a medical insurance policy with percentage participation of 70/30. The medical costs
a deductible provision of RM600 and co-insurance covered in Madam Tan’s policy is RM74 000.
percentage participation of 80/20. The medical costs
covered in Puan Roslina’s policy is RM25 000. Kos perubatan selepas deduktibel

Penyelesaian: Medical costs after deductible BAB 3
Kos perubatan selepas deduktibel
= RM74 000 – RM750
Medical costs after deductible
= RM73 250
= RM25 000 – RM600
Jumlah kos yang ditanggung oleh syarikat
= RM24 400
insurans
Jumlah kos yang ditanggung oleh syarikat insurans
Amount of cost incurred by insurance company
Amount of cost incurred by insurance company 70
80 = 100 × RM73 250
= 100 × RM24 400
= RM51 275
= RM19 520

Jumlah kos yang ditanggung Puan Roslina Jumlah kos yang ditanggung Puan Tan

Amount of cost borne by Puan Roslina Amount of cost borne by Madam Tan
30
= 20 × RM24 400 + RM600 = 100 × RM73 250 + RM750
100
= RM22 725
= RM5 480

PRAKTIS SPM 3

Kertas 1 2. Berikut ialah kepentingan insurans kecuali

1. Antara berikut, yang manakah bukan maksud The following are the importances of insurance except
risiko?
A Membayar bil perubatan.
Which of the following is not the meaning of risk?
Pays medical bills.
A Kemungkinan berlakunya peristiwa
berbahaya. B Sebagai pampasan kepada kematian

The possibility of a dangerous event happens. pemegang polisi.

B Ketidakpastian berlakunya kerugian. As a compensation of the death of policyholder.

The uncertainty in the occurrence of loss. C Sebagai pampasan jika pemegang polisi

C Kemungkinan berlakunya kejadian yang mengalami kecacatan.
menguntungkan.
As a compensation if policyholder suffers disability.
The possibility of a profitable event happens.
D Mengganti kemusnahan kereta dengan kereta
D Kemungkinan berlakunya kerugian.
baharu.
The possibility of loss happens.
Replace the destroyed car with a new car.

47 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans

3. Encik Jason ingin membeli insurans hayat dengan Chin Ze’s medical insurance has a deductible of RM800

nilai muka sebanyak RM150 000. Kadar premium and 80/20 co-insurance. If the amount of the medical
cost borne by Chin Ze is RM5 200 in his claim, calculate
tahunan bagi setiap RM1 000 nilai muka yang the actual amount of the claim.

ditawarkan kepada Encik Jason ialah RM2.12. A RM6 080 C RM26 000
B RM22 000 D RM30 000
Berapakah premium bulanan yang perlu dibayar

oleh Encik Jason?

Mr Jason wants to buy a life insurance with a face value
of RM150 000. The annual premium rate per RM1 000
of face value offered to Mr Jason is RM2.12. What is the

monthly premium needs to be paid by Mr Jason? Kertas 2

A RM26.50 C RM150.00 1. Berikut merupakan maklumat kereta Puan Teoh
yang menetap di Kulim, Kedah.
B RM73.58 D RM318.00
The following is the information of Madam Teoh’s car
BAB 3 4. Puan Alia telah menerima pampasan daripada who lives in Kulim, Kedah.

syarikat insurans keretanya sebanyak RM1 490. Jumlah diinsuranskan / Sum insured : RM80 000
Kapasiti enjin / Engine capacity : 1625 cc
Jika polisi insurannya itu mempunyai deduktibel NCD : 30%

sebanyak RM500, hitung jumlah sebenar kerugian Dengan merujuk Jadual Tarif Motor 2015 pada
muka surat 43, hitung premium kasar bagi
yang dialami oleh Puan Alia.
By referring to the Schedule of Motor Tariff 2015 on page
Puan Alia had received a compensation of RM1 490 from 43, calculate the gross premium of
her car’s insurance company. If her insurance policy has
a deductible of RM500, calculate the actual loss suffered (a) polisi komprehensif,
by Puan Alia. the comprehensive policy,
(b) polisi pihak ketiga.
A RM990 C RM1 440 the third party policy.
B RM1 090 D RM1 990

5. Jadual di bawah menunjukkan tuntutan Firdaus Jawapan / Answer :

kepada syarikat insurans motosikalnya dalam (a) Polisi komprehensif:
Comprehensive policy:
tempoh setahun.

The table below shows Firdaus’s claims to his
motorcycle’s insurance company in a year.

Bulan / Month Tuntutan / Claim (RM) (a) RM1 000 yang pertama RM305.50
April / April 690 RM2 054
Jun / June 260 The first RM1 000
450
Oktober / October (b) RM26 × 79

Jika polisi insuransnya mempunyai deduktibel (c) Premium asas = (a) + (b) RM2 359.50

sebanyak RM300, nyatakan nisbah jumlah Basic premium

bayaran pampasan yang diterima Firdaus kepada (d) NCD 30% RM707.85

jumlah kerugian sebenar yang dialami Firdaus (e) Premium kasar = (c) – (d)

bagi tahun itu. Gross premium RM1 651.65
If his insurance policy has a deductible of RM300, state

the ratio of the total compensation payment received by

Firdaus to the total actual loss suffered by Firdaus in that (b) Polisi pihak ketiga:

year. Third party policy:

A 4 : 7 C 27 : 70

B 13 : 28 D 41 : 140 (a) Premium asas RM135.00
Basic premium
6. Insurans perubatan Chin Ze mempunyai
deduktibel sebanyak RM800 dan ko-insurans (b) NCD 30% RM40.50
80/20. Jika jumlah kos perubatan yang perlu
ditanggung Chin Ze ialah RM5 200 dalam suatu (c) Premium kasar = (a) – (b) RM94.50
tuntutannya, hitung jumlah sebenar tuntutan itu.
Gross premium

© Penerbitan Pelangi Sdn. Bhd. 48

2. Jian Ming ingin membeli insurans hayat bernilai Matematik  Tingkatan 5  Bab 3 Matematik Pengguna: Insurans  BAB 3
RM200 000 daripada AB Insurance.
3. Hazli mempunyai polisi insurans perubatan
Jian Ming wants to buy a life insurance worth RM200 000 dengan peruntukan deduktibel sebanyak RM500
from AB Insurance. dan fasal penyertaan peratusan ko-insurans x/y
dengan x ialah peratusan kos yang ditanggung
(a) Nyatakan pemegang polisi, syarikat insurans oleh syarikat insurans dan y ialah peratusan kos
dan nilai muka polisi. yang ditanggung pemegang polisi. Kos perubatan
rawatan yang dilindungi dalam polisi Hazli ialah
State the policyholder, insurance company and the RM17 900. Hitung nilai x dan nilai y jika jumlah
face value of the policy. kos yang ditanggung Hazli ialah RM4 975.

(b) Hitung premium bulanan yang perlu dibayar Hazli has a medical insurance policy with a deductible
oleh Jian Ming jika kadar premium tahunan provision of RM500 and co-insurance percentage
yang dikenakan oleh syarikat insurans ialah participation of x/y where x is the percentage of the cost
RM2.34 bagi setiap RM1 000 nilai muka covered by insurance company and y is the percentage
mengikut umur dan status kesihatan Jian of the cost borne by the policyholder. The medical
Ming. treatment costs covered in Hazli’s policy is RM17 900.
Calculate the value of x and of y if the amount of cost
Calculate the monthly premium needs to pay by borne by Hazli is RM4 975.
Jian Ming if the annual premium rate charged
by the insurance company is RM2.34 for every Jawapan / Answer :

RM1 000 face value according Jian Ming’s age and Jumlah kos yang ditanggung Hazli = RM4 975
health status.
Amount of cost borne by Hazli
Jawapan / Answer :
y × RM17 900 + RM500 = RM4 975
(a) Pemegang polisi / Policyholder : Jian Ming 100 10y0 =
RM4 975−RM500
Syarikat insurans / Insurance company : RM17 900
AB Insurance

Nilai muka / Face value : RM200 000

(b) Premium tahunan / Annual premium = 0.25

= RM200 000 × RM2.34 = RM468 y = 25
RM1 000
x = 100 − 25

= 75

Premium bulanan / Monthly premium
RM468
= 12 = RM39 ∴ x = 75, y = 25

Sudut KBAT KBAT

Ekstra

Rumah Puan Jess telah mengalami kebakaran dan Jawapan / Answer:
kerugian yang dicatatkan sebanyak RM92 000. Katakan x = jumlah insurans yang telah dibeli
Rumah itu mempunyai insurans kebakaran dengan
ko-insurans 80% dan deduktibel RM3 000. Jika nilai Let x = amount of insurance purchased
boleh insurans rumah itu ialah RM1.75 juta dan
jumlah pampasan yang diterima oleh Puan Jess ialah Jumlah insurans yang harus dibeli
RM63 240, hitung jumlah insurans yang telah dibeli
oleh Puan Jess untuk rumahnya itu. Amount of required insurance

Puan Jess’s house caught in fire and the loss was RM92 000. = 0.8 × RM1 750 000 = RM1 400 000
The house has a fire insurance with co-insurance of 80% and
deductible of RM3 000. If the insurable value of the house is RM63 240 + RM3 000 = RM66 240 , RM92 000
RM1.75 million and the amount of compensation received by
Puan Jess was RM63 240, calculate the amount of insurance Jumlah insurans yang telah dibeli , RM1 400 000
purchased by Puan Jess for her house.
The amount of insurance purchased , RM1 400 000
x
RM1 400 000 × RM92 000 − RM3 000 = RM63 240

x = RM1 008 000 Kuiz 3

49 © Penerbitan Pelangi Sdn. Bhd.

BAB Matematik Pengguna: Percukaian

4 Consumer Mathematics: Taxation

4.1 Percukaian

Taxation

NOTA IMBASAN

1. Langkah-langkah menghitung cukai pendapatan:  Apabila cukai yang perlu dibayar , PCB tahunan
Steps to calculate income tax: Lebihan potongan annual
=  PCB – cukai yang perlu dibayar basis
 Hitung pendapatan bercukai
Calculate chargeable income When tax payable , PCB
Excess deduction
Pendapatan bercukai / Chargeable income =  PCB – tax payable

jumlah

    = pendapatan –
pengecualian – pelepasan 3. Cukai jalan dihitung dengan merujuk jadual kadar
cukai cukai cukai jalan.
tahunan Road tax is calculated by referring to road tax rates table.

         
total annual – tax – tax
income exemption relief
4. Cukai pintu / Property assessment tax
 Hitung cukai pendapatan dengan merujuk jadual
kadar cukai. = kadar cukai pintu × nilai tahunan
property assessment tax × annual value
Calculate income tax by referring to tax rates table.
dengan / where
Cukai pendapatan / Income tax
= cukai dasar  +  cukai atas baki  –  rebat cukai nilai tahunan = anggaran sewa bulanan × 12 bulan
base tax  +  tax on the next balance  –  tax rebate annual value = estimated monthly rental × 12 months

5. Cukai tanah / Quit tax

kadar cukai tanah setiap jumlah keluasan

2. Potongan cukai bulanan (PCB) = unit keluasan   ×  tanah
Monthly tax deduction (PCB)
  Apabila cukai yang perlu dibayar . PCB quit rent rate per unit area total land area
Bayaran cukai yang tidak mencukupi
= cukai yang perlu dibayar – PCB tahunan 6. Kadar cukai jualan untuk barangan ialah 5% atau 10%
annual dan kadar cukai perkhidmatan ialah 6%.
When tax payable . PCB basis The sales tax rates on goods are 5% or 10% and the service tax
Insufficient tax payment rate is 6%.
= tax payable – PCB


1. Nyatakan tujuan percukaian di Malaysia. TP 1

State the purposes of taxation in Malaysia.

Sebagai sumber pendapatan negara Tujuan Untuk mengawal penjualan barangan
atau perkhidmatan tertentu
As a source of government revenue percukaian
To control sales of certain goods or services
Sebagai alat pelaksanaan polisi kerajaan Purposes of
taxation Sebagai alat kewangan untuk
As government policy implementation tool menstabilkan ekonomi

As a financial tool to stabilise the economy

© Penerbitan Pelangi Sdn. Bhd. 50

Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian 

2. Nyatakan jenis cukai dan pihak berkuasa yang mengutip cukai berikut. TP 2

State the type of taxes and the authority that collects the taxes.

Huraian Jenis cukai Pihak berkuasa

Description Type of taxes The authority

CONTOH Pejabat Tanah Negeri

Cukai yang dikenakan kepada pemilik tanah. Cukai tanah State Land Office

The tax levied on the owner of land. Quit rent Lembaga Hasil Dalam Negeri

(a) Cukai yang dikenakan atas pendapatan Cukai pendapatan Inland Revenue Board

terperoleh individu. Income tax

The tax imposed on the income earned by individual.

(b) Cukai yang dikenakan kepada pemilik Cukai pintu Majlis bandaran atau majlis
rumah. daerah
Property assessment tax
The tax levied on the owner of house. Municipal council or district council
Cukai jualan dan
(c) Cukai yang dikenakan terhadap barangan perkhidmatan Jabatan Kastam Diraja
dan perkhidmatan.
Sales and service tax Royal Malaysian Customs
The tax levied on goods and services. Department

(d) Cukai yang dikenakan terhadap pengguna Cukai jalan Jabatan Pengangkutan Jalan BAB 4
jalan raya yang memiliki kenderaan.
Road tax Road Transport Department
The tax levied on road user who owns vehicle.

3. Hitung pendapatan bercukai bagi setiap yang berikut. TP 3

Calculate the chargeable income of each of the following.

CONTOH (a)

Pendapatan tahunan RM48 400 Pendapatan tahunan RM65 400
RM4 320
Annual income Annual income RM3 000
RM31 500
Pendapatan dikecualikan cukai RM2 160 Pendapatan dikecualikan cukai

Tax exempted income Tax exempted income

Pengecualian cukai RM1 320 Pengecualian cukai

Tax exemption Tax exemption

Pelepasan cukai / Tax relief RM5 500 Pelepasan cukai / Tax relief

Penyelesaian: Pendapatan bercukai / Chargeable income
= RM65 400 – RM4 320 – RM3 000 – RM31 500
Pendapatan bercukai / Chargeable income = RM26 580
= RM48 400 – RM2 160 – RM1 320 – RM5 500
= RM39 420

(b) RM189 600 (c) RM83 700
RM32 000 RM7 200
Pendapatan tahunan RM8 500 Pendapatan tahunan RM2 500
RM41 025 RM24 325
Annual income Annual income

Pendapatan dikecualikan cukai Pendapatan dikecualikan cukai

Tax exempted income Tax exempted income

Pengecualian cukai Pengecualian cukai

Tax exemption Tax exemption

Pelepasan cukai / Tax relief Pelepasan cukai / Tax relief

Pendapatan bercukai / Chargeable income Pendapatan bercukai / Chargeable income
= RM189 600 – RM32 000 – RM8 500 – RM41 025 = RM83 700 – RM7 200 – RM2 500 – RM24 325
= RM108 075 = RM49 675

51 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian

4. Hitung cukai pendapatan individu bagi setiap yang berikut mengikut Kadar Cukai Taksiran 2020 yang
diberikan. TP 3

Calculate the individual income tax of each of the following according to the given Tax Rates for Assessment 2020.

Pendapatan bercukai (RM) Pengiraan (RM) Kadar (%) Cukai (RM)

Chargeable income (RM) Calculations (RM) Rate (%) Tax (RM)

0 – 5 000 5 000 pertama / First 5 000 0 0
5 001 – 20 000 1
5 000 pertama / First 5 000 3 0
20 001 – 35 000 15 000 berikutnya / Next 15 000 8 150
14
35 001 – 50 000 20 000 pertama / First 20 000 21 150
15 000 berikutnya / Next 15 000 450
50 001 – 70 000
35 000 pertama / First 35 000 600
70 001 – 100 000 15 000 berikutnya / Next 15 000 1 200

50 000 pertama / First 50 000 1 800
20 000 berikutnya / Next 20 000 2 800

70 000 pertama / First 70 000 4 600
30 000 berikutnya / Next 30 000 6 300

CONTOH (a)

BAB 4 Pendapatan bercukai RM34 600 2 jenis rebat cukai: Pendapatan bercukai RM57 100
2 types of tax rebate:
Chargeable income (i) Rebat RM400 jika pendapatan Chargeable income

Zakat atau fitrah RM150 bercukai < RM35 000 Zakat atau fitrah RM360
Rebate of RM400 if the chargeable
Zakat or fitrah Zakat or fitrah
income < RM35 000
Penyelesaian: (ii) Jumlah bayaran zakat / fitrah Cukai dasar / Base tax = RM1 800
Cukai dasar / Base tax = RM150 Amount of zakat / fitrah paid
Cukai atas baki berikutnya
Cukai atas baki berikutnya Rujuk kadar cukai. RM34 600 terletak
pada banjaran RM20 001 – RM35 000. Tax on the next balance
Tax on the next balance Refer to the tax rates. RM34 600 lies in
the range of RM20 001 – RM35 000. = (RM57 100 – RM50 000) × 14%
= (RM34 600 – RM20 000) × 3% = RM994
= RM438

Cukai pendapatan / Income tax Cukai pendapatan / Income tax
= RM1 800 + RM994 – RM360
= RM150 + RM438 – RM150 – RM400 Pendapatan bercukai < RM35 000 = RM2 434
Chargeable income < RM35 000
= RM38
Zakat / fitrah

(b) (c) (d)

Pendapatan bercukai RM33 900 Pendapatan bercukai RM83 900 Pendapatan bercukai RM67 380

Chargeable income Chargeable income Chargeable income

Cukai dasar / Base tax = RM150 Cukai dasar / Base tax = RM4 600 Zakat atau fitrah RM410

Cukai atas baki berikutnya Cukai atas baki berikutnya Zakat or fitrah

Tax on the next balance Tax on the next balance Cukai dasar / Base tax = RM1 800

= (RM33 900 – RM20 000) × 3% = (RM83 900 – RM70 000) × 21% Cukai atas baki berikutnya
= RM417 = RM2 919
Tax on the next balance
Cukai pendapatan / Income tax Cukai pendapatan / Income tax
= RM150 + RM417 – RM400 = RM4 600 + RM2 919 = (RM67 380 – RM50 000) × 14%
= RM167 = RM7 519 = RM2 433.20

Cukai pendapatan / Income tax
= RM1 800 + RM2 433.20 – RM410
= RM3 823.20

© Penerbitan Pelangi Sdn. Bhd. 52

Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian 

5. Tentukan sama ada situasi potongan cukai bulanan (PCB) yang berikut mempunyai bayaran cukai yang tidak
mencukupi atau lebihan potongan. Kemudian, hitung jumlah tersebut. (Rujuk Kadar Cukai Taksiran pada
Soalan 4). TP 4

Determine whether the following monthly tax deduction (PCB) situation has an insufficient tax payment or excess deduction.
Hence, calculate the amount. (Refer Tax Rates for Assessment in Questions 4).

CONTOH (a)

Gaji tahunan / Annual salary RM72 000 Gaji tahunan / Annual salary RM60 000

PCB RM170 PCB RM10

Derma kepada badan kebajikan Pelepasan cukai / Tax relief

Donate to welfare centre RM1 500 Individu / Individual RM9 000

Zakat/Fitrah RM380 Rawatan ibu / Mother’s treatment RM3 500

Pelepasan cukai / Tax relief Gaya hidup / Lifestyle RM2 500

Individu / Individual RM9 000 Yuran tadika / Kindergarten fees RM1 000

Ibu dan bapa / Parents RM2 200 Isteri / Wife RM4 000

Gaya hidup / Lifestyle RM1 800 Anak / Child RM2 000

KWSP / EPF RM4 000 Insurans hayat dan KWSP RM5 300

Perkeso RM250 Life insurance and EPF

Penyelesaian: Perkeso RM250 BAB 4

Pendapatan bercukai / Chargeable income Pendapatan bercukai / Chargeable income
= RM72 000 – RM1 500 – (RM9 000 + RM2 200 + = RM60 600 – (RM9 000 + RM3 500 + RM2 500

RM1 800 + RM4 000 + RM250) + RM1 000 + RM4 000 + RM2 000 + RM5 300
= RM53 250 + RM250)
= RM33 050
Cukai pendapatan / Income tax
Cukai pendapatan / Income tax
= RM1 800 + (RM53 250 – RM50 000) × 14% – RM380 = RM150 + (RM33 050 – RM20 000) × 3% – RM400
= RM1 800 + RM455 – RM380 = RM150 + RM391.50 – RM400
= RM141.50
= RM1 875
Pendapatan bercukai < RM35 000
Jumlah PCB tahunan / Total annual PCB Chargeable income < RM35 000

= RM170 × 12 Jumlah PCB tahunan / Total annual PCB
= RM2 040 = RM10 × 12
= RM120
Cukai pendapatan , Jumlah PCB tahunan
Cukai pendapatan . Jumlah PCB tahunan
Income tax , Total annual PCB
Income tax . Total annual PCB
Lebihan potongan / Excess deduction

= RM2 040 – RM1 875

= RM165

Oleh itu, RM165 akan dipulangkan kepada Bayaran cukai yang tidak mencukupi
pembayar cukai oleh LHDN.
Insufficient tax payment
Thus, RM165 will be refunded to the taxpayer by IRB.
= RM141.50 – RM120
Tip = RM21.50

Cukai pendapatan / Income tax Oleh itu, RM21.50 akan dibayar oleh pembayar
= cukai dasar + cukai atas baki – rebat cukai cukai kepada LHDN.

base tax + tax on the next balance – tax rebate Thus, RM21.50 will be paid by the taxpayer to IRB.

53 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian

6. Hitung cukai pendapatan menggunakan taksiran cukai bersama dan taksiran cukai berasingan. Kemudian,
tentukan taksiran yang lebih sesuai. (Rujuk Kadar Cukai Taksiran pada Soalan 4). TP 6

Calculate the income tax using joint tax assessment and separate tax assessment. Hence, determine which assessment is
more suitable. (Refer Tax Rates for Assessment in Questions 4).

CONTOH

Jadual di sebelah Perkara / Item Encik Kuan Puan Elly
menunjukkan RM72 000 RM50 500
maklumat Gaji tahunan / Annual salary RM1 500 RM1 000
pendapatan Encik Puan Elly
Kuan dan isterinya, Derma yang diluluskan / Approved donation Encik Kuan RM9 000
Puan Elly. RM9 000 RM3 000
Pelepasan cukai / Tax relief Had / Limit RM2 300 RM1 000
The table shows the RM1 800 RM3 500
information on salaries of Individu / Individual RM9 000 RM5 400
Encik Kuan and his wife,
Puan Elly Gaya hidup / Lifestyle RM2 500

Penyelesaian: Anak / Child RM2 000

Insurans dan KWSP / Insurance and EPF RM7 000

Perkara / Item Taksiran Cukai Bersama Taksiran Cukai Berasingan

Jumlah pendapatan Joint Tax Assessment Separate Tax Assessment

Total income Encik Kuan & Puan Elly Encik Kuan Puan Elly
72 000 + 50 500
BAB 4 (–) Jumlah pengecualian (Derma) = RM122 500 RM72 000 RM50 500
1 500 + 1 000
(–) Total exemption (Donation) = RM2 500 RM1 500 RM1 000

Pelepasan cukai / Tax relief RM9 000 RM9 000 RM9 000
(–) Individu / Individual RM2 500 RM2 300 RM2 500
(–) Gaya hidup / Lifestyle RM2 000 RM1 800 RM1 000
(–) Anak / Child
(–) Insurans dan KWSP RM7 000 RM5 400 RM3 500
  Insurance and EPF
Pendapatan bercukai RM99 500 RM52 000 RM33 500

Chargeable income RM4 600 RM1 800 RM150
Baki / Balance
Cukai dasar / Base tax = 99 500 – 70 000 Baki / Balance Baki / Balance
= RM29 500 = 52 000 – 50 000 = 33 500 – 20 000
Cukai atas baki = RM2 000 = RM13 500

Tax on the next balance

29 500 × 21% 2 000 × 14% 13 500 × 3%
= RM6 195 = RM280 = RM405

Rebat cukai / Tax rebate – – RM400
Cukai pendapatan
4 600 + 6 195 1 800 + 280 150 + 405 – 400
Income tax = RM10 795 = RM2 080 = RM155

2 080 + 155 = RM2 235

Taksiran cukai bersama . taksiran cukai berasingan. Tip Pendapatan bercukai
Chargeable income
Maka, taksiran cukai berasingan lebih sesuai. Had tuntutan pelepasan cukai hanya dikira < RM35 000
sekali sahaja dalam taksiran bersama.
Joint tax assessment . separate tax assessment. The claimable limit is calculated only once in
Thus, separate tax assessment is more suitable. joint assessment.

© Penerbitan Pelangi Sdn. Bhd. 54

Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian 

(a) Perkara / Item Encik Izani Puan Alia
RM58 600 RM48 000
Jadual di sebelah Gaji tahunan / Annual salary RM1 200
RM800
menunjukkan Derma yang diluluskan / Approved donation RM230 RM65
Encik Izani Puan Alia
maklumat Zakat/Fitrah RM9 000 RM9 000
RM2 700 RM2 500
pendapatan Encik Pelepasan cukai / Tax relief Had / Limit RM4 000 RM1 800
RM3 000 RM3 000
Izani dan isterinya, Individu / Individual RM9 000

Puan Alia. Gaya hidup / Lifestyle RM2 500

The table shows the Ibu dan bapa / Parents RM3 000
information on salaries of
Encik Izani and his wife,
Puan Alia

Insurans dan KWSP / Insurance and EPF RM7 000

Perkara / Item Taksiran Cukai Bersama Taksiran Cukai Berasingan

Jumlah pendapatan Joint Tax Assessment Separate Tax Assessment

Total income Encik Izani & Puan Alia Encik Izani Puan Alia
58 600 + 48 000
(–) Jumlah pengecualian (Derma) = RM106 600 RM58 600 RM48 000
800 + 1 200
(–) Total exemption (Donation) = RM2 000 RM800 RM1 200

Pelepasan cukai / Tax relief RM9 000 RM9 000 RM9 000 BAB 4
(–) Individu / Individual RM2 500 RM2 500 RM2 500
(–) Gaya hidup / Lifestyle RM3 000 RM3 000 RM1 800
(–) Ibu dan bapa / Parents
(–) Insurans dan KWSP RM7 000 RM3 000 RM3 000
  Insurance and EPF
Pendapatan bercukai RM83 100 RM40 300 RM30 500

Chargeable income RM4 600 RM600 RM150
Baki / Balance
Cukai dasar / Base tax = 83 100 – 70 000 Baki / Balance Baki / Balance
Cukai atas baki = RM13 100 = 40 300 – 35 000 = 30 500 – 20 000
= RM5 300 = RM10 500
Tax on the next balance

13 100 × 21% 5 300 × 8% 10 500 × 3%
= RM2 751 = RM424 = RM315

Rebat cukai / Tax rebate – – RM400
Zakat/Fitrah 230 + 65 = RM295
RM230 RM65
Cukai pendapatan 4 600 + 2 751 – 295
= RM7 056 600 + 424 – 230 150 + 315 – 400
Income tax = RM794 – 65
= RM0

794 + 0 = RM794

Taksiran cukai bersama . taksiran cukai berasingan. Maka, taksiran cukai berasingan lebih sesuai.

Joint tax assessment . separate tax assessment. Thus, separate tax assessment is more suitable.

55 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian

7. Hitung cukai jalan bagi setiap motosikal berikut berdasarkan kadar cukai jalan yang diberikan. TP 3

Calculate the road tax of each of the following motorcycles based on the given road tax rates.

Kapasiti enjin (cc) Kadar di Semenanjung Malaysia Kadar di Sabah dan Sarawak

Engine capacity (cc) Rates in Peninsular Malaysia Rates in Sabah and Sarawak

< 150 RM2.00 RM2.00
151 – 200 RM30.00 RM9.00
201 – 250 RM50.00 RM12.00
251 – 500 RM180.00 RM30.00
501 – 800 RM250.00 RM40.00
Melebihi / Exceeding 800 RM350.00 RM42.00

CONTOH (a) Kapasiti enjin / Engine capacity : 1200 cc
Kawasan / Area : Selangor
Kapasiti enjin / Engine capacity : 700 cc
Kawasan / Area : Johor Cukai jalan / Road tax = RM350

Penyelesaian:
Cukai jalan / Road tax = RM250

(b) Kapasiti enjin / Engine capacity : 420 cc (c) Kapasiti enjin / Engine capacity : 1080 cc
Kawasan / Area : Sarawak Kawasan / Area : Sabah

BAB 4 Cukai jalan / Road tax = RM30 Cukai jalan / Road tax = RM42

8. Hitung cukai jalan bagi setiap kereta berikut berdasarkan kadar cukai jalan yang diberikan. TP 3

Calculate the road tax of each of the following cars based on the given road tax rates.

Kapasiti Semenanjung Malaysia Sabah & Sarawak
enjin (cc)
Peninsular Malaysia
Engine
capacity (cc) Kadar asas Kadar progresif Kadar asas Kadar progresif

< 1000 Base rate Progressive rate Base rate Progressive rate
1001 – 1200
1201 – 1400 RM20.00 – RM20.00 –
1401 – 1600
RM55.00 – RM44.00 –
1601 – 1800
RM70.00 – RM56.00 –
1801 – 2000
RM90.00 – RM72.00 –

RM0.40 setiap cc melebihi RM0.32 setiap cc melebihi
1600 cc
RM200.00 1600 cc RM160.00
RM0.32 each cc exceeding 1600 cc
RM0.40 each cc exceeding 1600 cc

RM0.50 setiap cc melebihi RM0.25 setiap cc melebihi
1800 cc
RM280.00 1800 cc RM224.00
RM0.25 each cc exceeding 1800 cc
RM0.50 each cc exceeding 1800 cc

CONTOH (a) Kapasiti enjin / Engine capacity : 1825 cc
Kawasan / Area : Sarawak
Kapasiti enjin / Engine capacity : 1750 cc
Kawasan / Area : Melaka Cukai jalan / Road tax
= RM224 + (1825 cc – 1800 cc) × RM0.25
Penyelesaian: = RM224 + RM6.25
Cukai jalan / Road tax = RM230.25
= RM200 + (1750 cc – 1600 cc) × RM0.40
= RM200 + RM60
= RM260

© Penerbitan Pelangi Sdn. Bhd. 56

Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian 

(b) Kapasiti enjin / Engine capacity : 1570 cc (c) Kapasiti enjin / Engine capacity : 1950 cc
Kawasan / Area : Negeri Sembilan Kawasan / Area : Perak

Cukai jalan / Road tax Cukai jalan / Road tax
= RM90 = RM280 + (1950 cc – 1800 cc) × RM0.50
= RM280 + RM75
= RM355

9. Diberi anggaran sewa bulanan sesebuah rumah, lengkapkan jadual berikut jika kadar cukai pintu di kawasan

itu ialah 8.8%. TP 3

Given the estimated monthly rental of a house, complete the following table if the property assessment tax rate of the area is
8.8%.

Anggaran sewa Nilai tahunan Cukai pintu setahun Cukai pintu setiap
bulanan
Annual value Property assessment tax setengah tahun
Estimated monthly rental per year
Property assessment tax for
each half-year

CONTOH RM400 × 12 8.8% × RM4 800 RM422.40 ÷ 2
RM400 = RM4 800 = RM422.40 = RM211.20

RM800 RM800 × 12 8.8% × RM9 600 RM844.80 ÷ 2
= RM9 600 = RM844.80 = RM422.40
BAB 4
RM1 000 RM1 000 × 12 8.8% × RM12 000 RM1 056 ÷ 2
= RM12 000 = RM1 056 = RM528

RM1 250 RM1 250 × 12 8.8% × RM15 000 RM1 320 ÷ 2
= RM15 000 = RM1 320 = RM660

10. Hitung cukai tanah bagi setiap keluasan tanah berikut. TP 3

Calculate the quit rent of each of the following land areas.

Keluasan tanah Kadar cukai tanah setiap unit keluasan Cukai tanah setahun

Land area Quit rent rate per unit area Quit rent per year

CONTOH RM0.50 per m2 150 × RM0.50 = RM75
230 × RM0.22 = RM50.60
150 m2

(a) 230 m2 RM0.22 per m2

(b) 185 m2 RM0.92 per m2 185 × RM0.92 = RM170.20

(c) 305 m2 RM1.44 per m2 305 × RM1.44 = RM439.20

11. Hitung cukai jualan yang berikut. TP 3 (a) Harga barang / Goods price : RM349.99
Kadar cukai jualan / Sales tax rate : 5%
Calculate the sales tax of the following.
Cukai jualan / Sales tax
CONTOH = RM349.99 × 5%
= RM17.50
Harga barang / Goods price : RM145.60
Kadar cukai jualan / Sales tax rate : 5%

Penyelesaian:
Cukai jualan / Sales tax
= RM145.60 × 5%
= RM7.28

57 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian

12. Hitung cukai perkhidmatan yang berikut. TP 3 (a) Perkhidmatan / Service : Restoren / Restaurant
Jumlah yang dikenakan : RM36.45
Calculate the service tax of the following. Amount charged
Kadar cukai perkhidmatan / Service tax rate : 6%
CONTOH
Cukai perkhidmatan / Service tax
Perkhidmatan / Service : Hotel = RM36.45 × 6%
Jumlah yang dikenakan / Amount charged : RM480 = RM2.19
Kadar cukai perkhidmatan / Service tax rate : 6%

Penyelesaian:
Cukai perkhidmatan / Service tax = RM480 × 6%
= RM28.80

13. Selesaikan setiap yang berikut.

Solve each of the following.

(a) Encik Fahmi merupakan seorang pengawai bank dan menerima gaji sebanyak RM5 200 sebulan. Isteri
Encik Fahmi merupakan seorang suri rumah dan mempunyai seorang anak berumur 23 tahun yang
sedang menuntut ijazah di UKM. Jadual di bawah menunjukkan pengecualian dan pelepasan cukainya.

Encik Fahmi is a bank officer and received a salary of RM5 200 a month. Encik Fahmi’s wife is a housewife and has a
23-years-old daughter who studying a degree in UKM. The table below shows the his tax exemption and relief. TP 5

BAB 4 Individu / Individual RM9 000
Ibu dan bapa / Parents (RM3 000 terhad/limited) RM4 800
Gaya hidup / Lifestyle (RM2 500 terhad/limited) RM2 350
Isteri / Wife (RM4 000 terhad/limited) RM4 200
Anak / Child (RM8 000 terhad/limited) RM3 600
Insurans hayat dan KWSP / Life insurance and EPF (RM7 000 terhad/limited) RM5 020

Encik Fahmi telah membayar zakat sebanyak RM80 dan menderma sebanyak RM500 kepada badan

kebajikan yang diluluskan.

Encik Fahmi has paid zakat amounting to RM80 and donated RM500 to an approved charity welfare.

(i) Hitung pendapatan bercukai Encik Fahmi.

Calculate the chargeable income of Encik Fahmi.

(ii) Hitung jumlah rebat cukai yang layak diterima oleh Encik Fahmi.

Calculate the total tax rebate eligible for Encik Fahmi.

(iii) Hitung cukai pendapatan Encik Fahmi bagi taksiran tahun 2020.

Calculate Encik Fahmi’s income tax for year assessment of 2020.

(i) Pendapatan bercukai / Chargeable income
= (RM5 200 × 12) – RM500 – (RM9 000 + RM3 000 + RM2 350 + RM4 000 + RM3 600 + RM5 020)
= RM62 400 – RM500 – RM26 970
= RM34 930

(ii) Pendapatan bercukai , RM35 000, Encik Fahmi layak mendapat rebat cukai RM400.

Chargeable income , RM35 000, Encik Fahmi eligibles to receive tax rebate of RM400.

Jumlah rebat cukai / Total tax rebate = RM400 + RM80 = RM480

(iii) Cukai dasar / Base rate = RM150
Cukai atas baki / Tax on the next balance = (RM34 930 – RM20 000) × 3%
= RM447.90

Cukai pendapatan / Income tax = RM150 + RM447.90 – RM480
= RM117.90

© Penerbitan Pelangi Sdn. Bhd. 58

Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian 

(b) Rajah di sebelah menunjukkan resit pembelian Restoran Rainbow
Deeja di sebuah restoran. TP 4 Kuala Lumpur

The diagram shows Deeja’s purchasing receipt in a INVOICE
restaurant.
02/10/2020  12:42 PM R785224
(i) Hitung cukai perkhidmatan yang dikenakan.
Qty Description Price Amount
Calculate the service tax levied.
1 Pasta 18.80 18.80
(ii) Berapakah jumlah yang perlu dibayar oleh 3 Mushroom soup 5.50 16.50
Deeja? 1 Salmon grill 24.50 24.50
1 Chicken chop 21.60 21.60
What is the total amount needs to be paid by Deeja? 2 Cuppucino 9.40 18.80
3 Coffee 4.80 14.40
(i) Jumlah sebelum cukai / Subtotal before tax
= 18.80 + 16.50 + 24.50 + 21.60 + 18.80 + 14.40 Subtotal Before Tax RM
= RM114.60 Service Tax 6% RM
Round Adjustment RM
Cukai perkhidmatan / Service tax
= RM114.60 × 6% TOTAL RM
= RM6.88
BAB 4
(ii) Jumlah yang perlu dibayar
Total amount needs to pay
= RM114.60 + RM6.88 + RM0.02
= RM121.50

PRAKTIS SPM 4

Kertas 1 C Jabatan Pengangkutan Jalan

1. Antara berikut, yang manakah bukan tujuan Road Transport Department
percukaian?
D Lembaga Hasil Dalam Negeri
Which of the following is not the purpose of taxation?
Inlang Revenue Board
A Mengagih semula ketidakseimbangan
pendapatan rakyat. 3. Antara berikut, yang manakah benar tentang
cukai jualan?
Redistribute income inequality of people.
Which of the following is true about sales tax?
B Menggalakkan pembelian barangan buatan
Malaysia. A Cukai yang dikenakan kepada pemilik
kenderaan.
Encourage the purchase of goods made in Malaysia.
The tax that levied to vehicle owner.
C Menambah beban rakyat.
B Cukai yang dikutip daripada pelanggan
Add people’s burden. selepas menerima perkhidmatan tersebut.

D Mengawal aktiviti tidak sihat seperti The tax that is collected from the customer after
perjudian. receiving the services.

Control unhealthy activities such as gambling. C Cukai yang dikenakan sekali sahaja pada
peringkat pengeluaran atau pengimportan.
2. Cukai jualan dan perkhidmatan akan dibayar
kepada The tax that levied only once at the stage of
manufacturing or importation.
Sales and service tax will be paid to
D Barang-barang disita jika gagal membayar
A Pejabat Tanah Negeri tunggakan cukai.

State Land Office Properties will be seized if fail to pay the tax.

B Jabatan Kastam Diraja Malaysia

Royal Malaysian Customs Department

59 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 4 Matematik Pengguna: Percukaian

4. Pendapatan bercukai Wan berjumlah RM18 150 Kadar cukai jalan / Road tax rate

termasuk pengecualian cukai dan pelepasan Kereta Kadar asas Kadar progresif

cukai. Jika pendapatan penggajian tahunan dan Car Base rate Progressive rate

pengecualian cukainya masing-masing ialah 1783 cc RM200.00 RM0.40 setiap cc
melebihi 1600 cc
RM37 200 dan RM300, hitung jumlah pelepasan 2982 cc
RM0.40 of each cc
cukai Wan. exceeding 1600 cc
Wan’s chargeable income amounting RM18 150

including the tax exemption and tax relief. If his annual RM880.00 RM2.50 setiap cc
employment income and tax exemption are RM37 200 melebihi 2500 cc
and RM300 respectively, calculate the amount of Wan’s
RM2.50 of each cc
tax relief. exceeding 2500 cc

A RM18 300 C RM19 050

B RM18 750 D RM19 380

5. Puan Nies telah membuat potongan cukai bulanan Hitung jumlah cukai jalan yang perlu dibayar oleh

(PCB) sebanyak RM145 daripada pendapatannya Encik Devdan.

setiap bulan. Selepas mengira cukai yang perlu Calculate the total road tax needs to be paid by Mr
Devdan.
dibayarnya, terdapat lebihan potongan yang akan

dipulangkan oleh LHDN. Antara berikut, yang Jawapan / Answer :

manakah cukai pendapatan yang mungkin bagi

Puan Nies? Cukai jalan bagi kereta 1783 cc

Mrs Nies has made monthly tax deduction (PCB) for Road tax for 1783 cc car
RM145 of her monthly income. After calculating the tax
BAB 4 = RM200.00 + (1783 – 1600) × RM0.40
that has to be paid, there is an excess deduction that = RM273.20

will be refunded by IRB. Which of the following is the Cukai jalan bagi kereta 2982 cc

possible income tax of Mrs Nies? Road tax for 2982 cc car

A RM1 630 C RM1 850 = RM880.00 + (2982 – 2500) × RM2.50
= RM2 085
B RM1 740 D RM2 300
Jumlah cukai jalan / Total road tax
Kertas 2 = RM273.20 + RM2 085
= RM2 358.20
1. Jadual berikut menunjukkan maklumat dua buah
kereta milik Encik Devdan dan kadar cukai kereta
tersebut.

The table below shows the information on the two cars
owned by Mr Devdan and the tax rates of the cars.

Sudut KBAT KBAT

Ekstra

Jadual di bawah menunjukkan kadar cukai yang If the income tax is RM108 975 and no tax rebate, what is
the value of x?
akan dikenakan kepada seorang pembayar cukai.
Jawapan / Answer:
The table below shows the tax rate levied to a tax payer.

Pendapatan Pengiraan Kadar Cukai Cukai pendapatan = cukai dasar + cukai atas baki

bercukai Calculation Rate Rate Income tax = base rate + tax on the next balance

Chargeable (RM) (%) (RM) 108 975 = 83 650 + (x – RM400 000) × 0.25
income 108 975 – 83 650
0.25
400 000 pertama 83 650 x – 400 000 =

RMx The first 400 000 = 101 300

200 000 berikutnya 25 x = 101 300 + 400 000
Next 200 000
= 501 300 Kuiz 4

Jika cukai pendapatannya ialah RM108 975 dan tiada
rebat cukai, apakah nilai x?

© Penerbitan Pelangi Sdn. Bhd. 60

BAB Kekongruenan, Pembesaran dan Gabungan
Transformasi
5
Congruency, Enlargement and Combined Transformations

5.1 Kekongruenan

Congruency

NOTA IMBASAN

1. Dua poligon adalah kongruen jika panjang sisi sepadan (iv) Sudut-Sudut-Sisi / Angle-Angle-Side (AAS)

dan sudut sepadan mempunyai ukuran yang sama. K LR S • ∠MKL = ∠TRS (A)
Two polygons are congruent if the lengths of corresponding sides MT • ∠KLM = ∠RST (A)
and the corresponding angles have the equal measurement. • LM = ST (S)

2. Sifat kekongruenan segi tiga: (v) Sudut-Sudut-Sudut / Angle-Angle-Angle (AAA)
Properties of triangle congruency:
K L R S • ∠MKL = ∠TRS (A)
(i) Sisi-Sisi-Sisi / Side-Side-Side (SSS)

K L R S • KL = RS (S) • ∠KLM = ∠RST (A)
• LM = ST (S)
• MK = TR (S) M T • ∠KML = ∠RTS (A)
• Luas ∆KLM = Luas ∆RST
MT
(ii) Sisi-Sudut-Sisi / Side-Angle-Side (SAS) Area of ∆KLM = Area of ∆RST

K L R S • KL = RS (S) (vi) Sisi-Sisi-Sudut / Side-Side-Angle (SSA)
• ∠KLM = ∠RST (A)
• LM = ST (S) K L R S • KL = RS (S)

MT • LM = ST (S)
(iii) Sudut-Sisi-Sudut / Angle-Side-Angle (ASA)
M T • ∠KML = ∠RTS (A)
K LR S • Luas ∆KLM = Luas ∆RST
M
• ∠MKL = ∠TRS (A) Area of ∆KLM = Area of ∆RST
• KL = RS (S)

T • ∠KLM = ∠RST (A)

1. Senaraikan semua sisi sepadan dan sudut sepadan bagi poligon kongruen berikut. TP 1

List all the corresponding sides and the corresponding angles of the following congruent polygons.

CONTOH (a) Q U

Q R T
P W
V
S

P R ST Sisi sepadan Sudut sepadan

Penyelesaian: Sudut sepadan Corresponding sides Corresponding angles

Sisi sepadan Corresponding angles PQ = UV ∠SPQ = ∠UVW
QR = TU ∠PQR = ∠TUV
Corresponding sides ∠QPR = ∠QTS RS = TW ∠QRS = ∠WTU
∠PQR = ∠TQS SP = VW ∠RSP = ∠VWT
PQ = QT ∠QRP = ∠QST
PR = ST
QR = QS

61 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi

2. Tentukan sama ada setiap pasangan poligon berikut adalah kongruen atau bukan. Berikan sebab anda. TP 2

Determine whether each of the following pairs of polygons are congruent. Give your reason.

CONTOH (a) B Q (b)

F K 7.8 cm L A BP Q

38° A CP R
7.8 cm 60° 4.2 cm 6 cm 82° 4.2 cm
ED 65° 115°
38° M TS
E 6 cm G DC SR

Penyelesaian: ∴ Bukan kongruen. Panjang sisi ∠DAB = ∠BCD = 360° – 2(65°)
pentagon ABCDE dan PQRST tidak 2
∠EGF = 180° – 38° – 60° = 115°
= 82° sama.
= ∠PSR = ∠PQR
Not congruent. The side lengths of
∠KLM = 180° – 38° – 82° pentagon ABCDE and PQRST are not ∠SPQ = ∠QRS = 360° – 2(115°)
= 60° equal. 2

= 65°

∴ Kongruen. Semua sisi sepadan = ∠ADC = ∠ABC

dan sudut sepadan adalah sama. ∴  Kongruen. Semua sisi sepadan

Congruent. All corresponding sides dan sudut sepadan adalah sama.
and angles are equal.
Congruent. All corresponding sides
and angles are equal.

3. Diberi bahawa segi tiga ABC dan PQR adalah kongruen. Senaraikan semua sisi dan sudut sepadan yang mungkin

berdasarkan sifat kekongruenan segi tiga yang diberikan. TP 3

It is given that triangles ABC and PQR are congruent. List all the possible corresponding sides and angles based on the

BAB 5 property of triangle congruence given. Tip

CONTOH Bilangan sisi sepadan dan sudut sepadan yang mungkin bergantung kepada sifat

Sisi-Sudut-Sisi / Side-Angle-Side kekongruenan segi tiga dan sisi atau sudut yang diberikan.
Penyelesaian: The number of possible corresponding sides and angles depends on the given property of triangle
congruency and side or angle.

A R Q A R Q
B B
C P C P

Sisi / Side AC = PQ Sisi / Side AC = PQ
Sudut / Angle ∠ACB = ∠PQR Sudut / Angle ∠BAC = ∠QPR

Sisi / Side BC = QR Sisi / Side AB = PR

(a) Sudut-Sisi-Sudut / Angle-Side-Angle

A R Q A RQ
C BP C BP

Sudut / Angle ∠BAC = ∠QPR Sudut / Angle ∠BAC = ∠QPR
Sisi / Side AB = PQ
Sisi / Side AC = PR
Sudut / Angle ∠ABC = ∠PQR
Sudut / Angle ∠ACB = ∠PRQ

© Penerbitan Pelangi Sdn. Bhd. 62

Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi 

(b) Sudut-Sudut-Sisi / Angle-Angle-Side

AR Q AR Q

C B P C B P

Sudut / Angle ∠ABC = ∠PQR Sudut / Angle ∠ABC = ∠PQR
Sudut / Angle ∠BAC = ∠QPR Sudut / Angle ∠BAC = ∠QPR

Sisi / Side AC = PR Sisi / Side BC = QR

AR Q AR Q

C B P C B P

Sudut / Angle ∠ABC = ∠PQR Sudut / Angle ∠ABC = ∠PQR
Sudut / Angle ∠ACB = ∠PRQ Sudut / Angle ∠ACB = ∠PRQ

Sisi / Side AC = PR Sisi / Side AB = PQ

4. Diberi bahawa luas segi tiga PQR dan STU adalah sama. Nyatakan sifat kekongruenan segi tiga yang digunakan

untuk menentukan segi tiga PQR dan STU adalah kongruen berdasarkan sisi dan sudut sepadan yang diberikan.

It is given that the areas of triangles PQR and STU are equal. State the property of triangle congruence used to determine that
triangles PQR and STU are congruent based on the corresponding sides and angles given. TP 3

Sisi dan sudut sepadan Segi tiga PQR Segi tiga STU Sifat kekongruenan segi tiga BAB 5

Corresponding sides and angles Triangle PQR Triangle STU Property of triangle congruence

CONTOH Q ST

PQ = SU PR Sisi-Sisi-Sisi
QR = TU Q
PR = ST Side-Side-Side
PR
(a) QR = TU Q U
∠RPQ = ∠TSU
∠PQR = ∠TUS PR ST
Q
(b) PQ = SU Sudut-Sudut-Sisi
PR = ST PR
∠QPR = ∠TSU Q Angle-Angle-Side

(c) ∠QPR = ∠TSU PR U
∠PQR = ∠TUS ST
∠PRQ = ∠STU
Sisi-Sudut-Sisi
(d) PQ = SU
∠QPR = ∠TSU Side-Angle-Side
∠PQR = ∠TUS
U
ST

Sudut-Sudut-Sudut

Angle-Angle-Angle

U

ST

Sudut-Sisi-Sudut

Angle-Side-Angle

U

63 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi

5. Jawab setiap yang berikut.

Solve each of the following.

(a) Dalam rajah di bawah, segi tiga BCD ialah imej (b) Rajah di bawah menunjukkan segi tiga EFG dan
bagi segi tiga ABD di bawah pantulan pada garis KLM yang dilukis pada suatu satah Cartes. TP 5
BD. TP 4
The diagram below shows triangles EFG and KLM
In the diagram below, triangle BCD is the image of drawn on a Cartesian plane.
triangle ABD under a reflection in the line BD.
y
D
E4

17 cm G2 L

K

–6 –4 –2 O 24 6 x

A 3x cm B 15 cm C F –2 M

Hitung:
Calculate:
Tunjukkan bahawa segi tiga EFG dan KLM adalah
(i) nilai x,
the value of x, kongruen menggunakan Sisi-Sisi-Sisi.
(ii) ∠ADC.
Show that triangles EFG and KLM are congruent using

Side-Side-Side.

(i) Segi tiga ABD dan BCD adalah kongruen. EF = [–3 – (–2)]2 + [4 – (–2)]2 = 6.1 unit / units
EG = [–3 – (–5)]2 + (4 – 2)2 = 2.8 unit / units
Triangles ABD and BCD are congruent. FG = [–5 – (–2)]2 + [2 – (–2)]2 = 5 unit / units

3x = 15
 x = 5

(ii) ∠ADB = ∠CDB KL = (7 – 1)2 + (2 – 1)]2 = 6.1 unit / units

BAB 5 sin ∠ADB = 15 LM = (7 – 3)2 + [2 – (–1)]2 = 5 unit / units
17
KM = (1 – 3)2 + [1 – (–1)]2 = 2.8 unit / units
  15
∠ADB = sin–1 17 EF = KL, EG = KM, FG = LM
Maka, segi tiga EFG dan KLM adalah kongruen
= 61°55 menggunakan Sisi-Sisi-Sisi.

∠ADC = 2 × 61°55 Thus, triangles EFG and KLM are congruent using
Side-Side-Side.
= 123°51

5.2 Pembesaran

Enlargement

NOTA IMBASAN Enlargement is a transformation where all the points of object
move from the centre of enlargement with a constant ratio, that
1. Dua objek adalah serupa apabila is known as a scale factor, k.
Two objects are similar when
(i) semua sudut sepadan adalah sama AЈ
all corresponding angles are equal A
(ii) semua nisbah sisi sepadan adalah sama
all ratios of corresponding sides are equal Pusat pembesaran, P B BЈ
Centre of enlargement, P
2. Pembesaran ialah satu transformasi dengan semua titik C
objek bergerak dari pusat pembesaran dengan nisbah CЈ
malar, yang dikenali sebagai faktor skala, k.

© Penerbitan Pelangi Sdn. Bhd. 64

Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi 

k = PA 3. Bagi suatu pembesaran, luas imej
PA = k2 × luas objek

= panjang sisi sepadan imej For an enlargement, area of the image
length of corresponding side of image = k2 × area of the object

panjang sisi objek Tip
length of side of object

6. Tentukan sama ada setiap pasangan poligon berikut adalah serupa atau tidak. Jumlah sudut pedalaman poligon
Total interior angles of a polygon
Berikan sebab anda. TP 2 = (n – 2) × 180°

Determine whether each of the following pairs of polygons are similar. Give your reason.

CONTOH (a) B T
6 cm 3 cm
B 13.2 cm A 130° 100° C S
Q4.2 cm R 12 cm 4 cm 100° P
13.2 cm
86° D 5.4 cm 140° 6 cm R 130° 3.6 cm
10 cm 4 cm Q
52° 86°
82° 140° 9 cm P 7.92 cm S E 8 cm D

A 7 cm C ∠C = ∠S,  ∠D = ∠R,  ∠A = ∠P,  ∠B = ∠T
∠E = 540° – 90° – 130° – 100° – 90°
Penyelesaian: AB = 10 = 5 = 130° = ∠Q
∠B = ∠S, ∠C = ∠Q RS 6 3

∠D = 360° – 86° – 82° – 140° BD = 13.2 = 5 AB = 6 = 2
= 52° = ∠P PS 7.92 3 PT 3

∠R = 360° – 140° – 52° – 86° DC = 9 = 5 BC = CD = 12 = 3
= 82° = ∠A PQ 5.4 3 ST RS 4
Tidak serupa kerana nisbah
AC = 7 = 5 DE = 8 = 2 BAB 5
QR 4.2 3 QR 4 sisi sepadan tidak sama.

Serupa kerana semua sudut sepadan adalah sama AE = 13.2 = 11 Not similar because the ratios
PQ 3.6 3 of the corresponding sides are
dan semua nisbah sisi sepadan adalah sama. not equal.

Similar because all the corresponding angles are equal
and all the ratios of the corresponding sides are equal.

(b) B Q (c) P
P 80° B
70° A
30° C
A Q

30°
R

DC

∠A = ∠R Tip S R

∠C = 180° – 70° – 30° Dua segi tiga adalah AD = 3
= 80° = ∠P serupa jika kedua-duanya PS 4
mempunyai sudut sepadan
∠Q = 180° – 80° – 30° yang sama. DC = 2
= 70° = ∠B Two triangles are similar SR 4
if they have the same
corresponding angles. = 1
2

Serupa kerana semua sudut sepadan adalah sama. Tidak serupa kerana nisbah sisi sepadan tidak sama.

Similar because all the corresponding angles are equal. Not similar because the ratios of the corresponding sides are

not equal.

65 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi

7. Perihalkan setiap pembesaran berikut. TP 3 Tip

Describe each of the following enlargements. Apabila imej pembesaran bertentangan dengan pusat
pembesaran, faktor skala mempunyai tanda negatif.
CONTOH y When the image is on opposite of the centre of enlargement, the
F scale factor has a negative sign.
E
(a)
4
y
2 FЈ KЈ
EЈ
4
–6 –4 –2 O 2 46 x
P 2 NЈ

G –6 –4 –2 O 2 46 x
–2 GЈ
K N LЈ
MЈ

–4 –2 GЈ

Penyelesaian: Kesalahan Lazim LM
P –4

Faktor skala Pembesaran tidak diperihalkan dengan
lengkap tanpa menyatakan pusat
Scale factor pembesaran dan faktor skala. Faktor skala
EF Enlargement is not defined completely without
= EF stating the centre of enlargement and the Scale factor
scale factor. KL
= 2 = KL ∴  Pembesaran pada pusat (–5, –4)
4 dengan faktor skala 3.
1 ∴  Pembesaran pada pusat (5, –3) = 6
= 2 dengan faktor skala 1 . 2 Enlargement at centre (–5, –4) with a
scale factor of 3.
BAB 5 2 =3

Enlargement at centre (5, –3) with a
1
scale factor of 2 .

(b) y (c) y
L4
6 K
L P2
M
4
LЈ

2 MЈ M MЈ x
K KЈ P 46
–6 –4 –2 O
x 2 KЈ

–6 –4 –2 O 2 46 –2 LЈ

NЈ

–2

N –4
–4

Faktor skala = PM ∴  Pembesaran pada pusat Faktor skala = KM ∴  Pembesaran pada pusat
PM KM (–1, 2) dengan faktor skala
Scale factor 1 Scale factor
2 (1, 1) dengan faktor skala 2 . 4 –2.
= 4 =– 2
Enlargement at centre (1, 1) with Enlargement at centre (–1, 2)
1 1 = –2 with a scale factor of –2.
= 2 a scale factor of 2 .

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Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi 

8. Kenal pasti imej pembesaran bagi setiap faktor skala berikut. TP 2 Faktor skala Imej

Identify the image of enlargement for each of the following scale factors. Scale factor Image

DB 0.5 A
C A Objek –0.5 C
P Object 1.5 B
–1.5 D

9. Lukis imej bagi setiap objek M berikut di bawah pembesaran pada pusat P dengan faktor skala, k yang diberikan.

Draw the image of each of the following object M under an enlargement at centre P with the given scale factor, k. TP 4

CONTOH (a) k = 1.5 (b) k = –2

k=2

M

MЈ M P
M M M
P
P
BAB 5
10. Lukis objek bagi setiap imej D berikut di bawah pembesaran pada pusat P dengan faktor skala, k yang diberikan.

Draw the object of each of the following image D under an enlargement at centre P with the given scale factor, k. TP 4

CONTOH (a) k = –2 (b) k = 3

k=– 1
2

D DP PD D

D

P
DЈ

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  Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi

11. Hitung nilai x bagi setiap yang berikut. TP 3

Calculate the value of x of each of the following.

CONTOH (a) Luas objek / Area of object = 18 m2
Luas imej / Area of image = 72 m2
Luas objek / Area of object = 27 cm2 Faktor skala / Scale factor = x
Luas imej / Area of image = 243 cm2
Faktor skala / Scale factor = x 72
18
Penyelesaian: Tip x2 =

x2 = 243 Luas imej =4
27 Area of image  x = ±4
=9 Luas objek
 x = ±9 k2 = Area of object = 2 atau / or –2

  = 3 atau / or –3

(b) Luas objek / Area of object = x cm2 (c) Luas objek / Area of object = 20 cm2
Luas imej / Area of image = x cm2
Luas imej / Area of image = 80 cm2 Faktor skala / Scale factor = 5
2
Faktor skala / Scale factor = – 3

 – 2 2 80 52 = x
3 x 20
=
 x = 25 × 20
80 × 9
  x = 4 = 500

= 180

12. Selesaikan setiap yang berikut.

Solve each of the following.
BAB 5

4 cm
(a) Dalam rajah di sebelah, poligon PRSTU ialah imej bagi poligon PQXWV di bawah P 5 cm Q R
suatu pembesaran pada pusat P. Diberi bahawa PQ = QX dan VW = WX, hitung
6 cm X
In the diagram, polygon PRSTU is the image of polygon PQXWV under an enlargement at V W

centre P. Given that PQ = QX and VW = WX, calculate  TP 5 4.8 cm S

(i) faktor skala, k pembesaran itu, UT
3.6 cm
the scale factor, k of the enlargement,

(ii) perimeter kawasan berlorek,

the perimeter of the shaded region,

(iii) luas kawasan berlorek.

the area of the shaded region.

(i) Faktor skala k = PR (iii) Luas PQXWV / Area of PQXWV
PQ 1
Scale factor = 2 ×3×4+6×2 P A 3 cm Q
X
k = 4.8 + 6 1.8 = PR
6 5
= 6 + 12 6 cm

= 1.8 PR = 9 cm = 18 cm2

(ii)   k = TU QR = 9 – 5 Luas PRSTU / Area of PRSTU V 2 cm W
VW = 4 cm
= 1.82 × 18
1.8 = 3.6 = 58.32 cm2
VW
VW = WX, RS = PR Luas kawasan berlorek
VW = 2 cm Perimeter
= 4 + 9 + 2(3.6) + 4.8 + 2(2) + 5 Area of the shaded region
= 34 cm
= 58.32 – 18

= 40.32 cm2

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Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi 

(b) Daniel ingin melukis sebuah lukisan pokok pada dinding biliknya Plastik lutsinar
Transparent sheet
menggunakan konsep pembesaran seperti yang ditunjukkan
Mentol
dalam rajah di sebelah. Jika jarak antara mentol dengan plastik Bulb

lutsinar itu ialah 12 cm dan faktor skala pembesaran tersebut

ialah 20, hitung jarak antara plastik lutsinar itu dengan dinding

tersebut dalam meter. TP 6

Daniel wants to draw a tree on his room’s wall using the concept of
enlargement as shown in the diagram. If the distance between the bulb
and the transparent sheet is 12 cm and the scale factor of the enlargement
is 20, calculate the distance between the transparent sheet and the wall in
metre.

Katakan jarak antara plastik lutsinar itu dengan dinding ialah d cm.

Let the distance between the transparent sheet and the wall is d cm.

20 = d + 12
12
d + 12 = 240

d = 228 cm

Oleh itu, jarak antara plastik lutsinar itu dengan dinding ialah 2.28 m.

Thus, the distance between the transparent sheet and the wall is 2.28 m.

5.3 Gabungan Transformasi BAB 5

Combined Transformation

NOTA IMBASAN

1. Gabungan transformasi AB bermaksud transformasi B diikuti dengan dengan transformasi A.
Combined transformation of AB means that tansformation B followed by transformation A.

2. Apabila diberi objek dan gabungan transformasi AB:
When object and combined transformation AB are given:

Objek Transformasi B Imej Transformasi A Imej
Object Transformation B Image Transformation A Image

3. Apabila diberi imej dan gabungan transformasi AB: Objek
When image and combined transformation AB are given: Object

Imej Transformasi A Objek Transformasi B
Image Transformation A Object Transformation B

4. Apabila imej di bawah gabungan transformasi AB dan BA adalah sama, maka gabungan transformasi ini memenuhi sifat
kalis tukar tertib.
When the images under combined transformations AB and BA are the same,then the combined transformation satisfies the commutative law.

5. Apabila memerihalkan gabungan transformasi AB, perihalkan transfomasi B dan diikuti tansformasi A.
When describing the combined transfomation AB, describe the transformation B followed by transformation A.

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  Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi

13. Poligon A dilukis pada suatu satah Cartes. Lukis imej bagi poligon A di bawah gabungan transformasi MN yang

diberikan dan labelkan dengan A. TP 4

Polygon A is drawn on a Cartesian plane. Draw the image of poligon A under a combined transformation MN given and label
it as A.

CONTOH (a) M = Pantulan pada garis y = x

1 2M = Translasi 5 Reflection in the line y = x
/ Translation 10
N = Putaran 90° ikut arah jam pada (9, 7)
N = Putaran 90° lawan arah jam pada (6, 8)
Rotation of 90° clockwise about (9, 7)
Rotation of 90° anticlockwise about (6, 8)
y y=x

y 14

12 (6, 8) Aࣳࣳ 12 Aࣳ M
N M +10 10 A (9, 7)
10
8 Aࣳࣳ
8 N
A
6
6 4
2
4

2

–4 –2 O x O x
–2 2 4 6 Aࣳ 8 10 12 14 2 4 6 8 10 12 14 16

+5

BAB 5 1 2(b) M = Translasi / Translation 13 (c) M = Putaran 90° ikut arah jam pada (20, 20)
–8 Rotation of 90° clockwise about (20, 20)

N = Pembesaran pada pusat (5, 10) dengan N = Pembesaran pada pusat (30, –10) dengan

faktor skala – —21 faktor skala 2
Enlargement at centre (30, –10) with a scale
Enlargement at centre (5, 10) with a scale factor factor of 2


of – —12 . y
60
y Aࣳࣳ
50
14 +13 M
12 Aࣳ
40

10 (5, 10) M 30
–8 Aࣳ

8 A 20 (20, 20)
N A
N 10
6 x
–30 –20 –10 O 10 20 30 40 50 60
4 Aࣳࣳ –10
(30, –10)
2

O x
2 4 6 8 10 12 14 16 18

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Matematik  Tingkatan 5  Bab 5 Kekongruenan, Pembesaran dan Gabungan Transformasi 

14. Poligon B ialah imej bagi suatu objek di bawah gabungan transformasi MN. Lukis objek bagi poligon B dan
labelkan dengan B. TP 4

Polygon B is an image of an object under a combined transformation MN. Draw the object of polygon B and label it as B.

CONTOH (a) M = Pembesaran pada pusat (4, 10) dengan
M = Pembesaran pada pusat (5, 1) dengan faktor
faktor skala 3
skala 3 Enlargement at centre (4, 10) with a scale factor
of 3
Enlargement at centre (5, 1) with a scale factor of 3.
N  = Pantulan pada garis x = 8
N  = Pantulan pada garis lurus PQ Reflection in the line x = 8

Reflection in the line PQ y
10 x = 8
y
14 Q 8 BЈ B
6
12

10 4
2
8B BЈЈ BЈЈ
6 –4 –2 O x
–2
–4 2 4 6 8 10 12 14

4 (5, 1) BЈ x
24 68 10 12 14 16 18
2
P

O

(b) M = Putaran 90° ikut arah jam pada (2, 1) (c) M = Putaran 90° ikut arah jam pada (10, 3) BAB 5
Rotation of 90° clockwise about (2, 1)
1 2 –11 Clockwise rotation of 90° about (10, 3)
N = Translasi / Translation 5
N  = Pantulan pada garis y = 4

Reflection in the line y = 4

y y B
10 10
BЈЈ (10, 3)
8 8
6 +11 6 68 x
BЈ 4 BЈЈ 10 12 14 16 18
2 –5 y=4
4 BЈ
–6 –4 –2 O x 2
–2 2 4 6 8 10 12
–4 O 24
B –2
–4

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