Matematik Tingkatan 2

Khas untuk guru sahaja

Dengan setiap pembelian buku, guru boleh
dapatkan salinan lembut (softcopy) dalam
bentuk PDF untuk memudahkan proses
pengajaran dan pembelajaran di rumah (home-
based learning).

Sekiranya guru memerlukan PDF, sila
hubungi Bahagian Penjualan dan Pemasaran
Cemerlang Publications Sdn. Bhd. di talian:

03-89592001
03-89593001
016-3324137

CONTOHKANDUNGAN 1
5
Format Instrumen PT3 10
Rumus Matematik 16
22
BAB Pola dan Jujukan 28
1 Patterns and Sequences 34
BAB Pemfaktoran dan Pecahan Algebra 41
2 Factorisation and Algebraic Fractions 48
BAB Rumus Algebra 54
3 Algebraic Formulae 60
BAB Poligon 67
4 Polygons 73
BAB Bulatan 80
5 Circles
BAB Bentuk Geometri Tiga Dimensi
6 Three-Dimensional Geometrical Shapes
BAB Koordinat
7 Coordinates
BAB Graf Fungsi
8 Graphs of Functions
BAB Laju dan Pecutan
9 Speed and Acceleration
BAB Kecerunan Garis Lurus
10 Gradient of a Straight Line
BAB Transformasi Isometri
11 Isometric Transformations
BAB Sukatan Kecenderungan Memusat
12 Measures of Central Tendencies
BAB Kebarangkalian Mudah
13 Simple Probability
Penilaian Akhir Tahun
Jawapan

Perbandingan Format Instrumen PT3
Mata Pelajaran Matematik (50)

Bil. Perkara Keterangan

2014 – 2018 Mulai 2019
CONTOH
1 Jenis Instrumen Ujian bertulis Ujian bertulis

2 Jenis Item • Respons Terbuka • Objektif Aneka Pilihan (OAP)

• Respons Terhad • Objektif Pelbagai Bentuk
(OPB)

• Subjektif Respons Terhad

3 Bilangan Soalan 10 soalan Bahagian A: 20 soalan OAP
Bahagian B: 5 soalan OPB
Bahagian C: 6 soalan subjektif

4 Jumlah Markah 100 markah Bahagian A: 20 markah
Bahagian B: 20 markah
Bahagian C: 60 markah
Jumlah: 100 markah

5 Tempoh Ujian 2 jam 2 jam

• Pengetahuan dan kefahaman • Mengingat
tentang konsep, hukum, • Memahami
prinsip dan teorem yang • Mengaplikasi
berkaitan dengan nombor, • Menyelesai masalah
bentuk dan perkaitan

• Kemahiran asas matematik

6 Konstruk • Kemahiran mengendalikan

operasi

• Kemahiran berkomunikasi
dan menaakul secara
matematik

• Kemahiran menyelesaikan
masalah

• Kemahiran menghitung

7 Kaedah Analitik • Dikotomus
Penskoran • Analitik

Rumus Matematik

Perkaitan/ Relations Bentuk Dan Ruang/ Shapes and Space

1 Jarak/ Distance 1 Lilitan bulatan = πd = 2πj
= (x2 – x1)2 + (y2 – y1)2 Circumference of circle = πd = 2πr

2 Luas bulatan = πj 2
Area of circle = πr2
CONTOH
2 Titik tengah/ Midpoint Panjang lengkok Sudut dicangkum pada pusat
Lilitan bulatan 360°
( ) (x, y) = 3 =
x1 + x2 y1 + y2
2 , 2 CircuLmenfegrtehncoef arc
of circle
= Angle subtended at centre
360°

3 Purata laju = Jarak yang dilalui 4 Luas sektor = Sudut dicangkum pada pusat
Masa yang diambil Luas bulatan 360°

Average speed = Distance travelled AArreeaa of sector = Angle subtended at centre
Time taken of circle 360°

4 Min = Hasil tambah nilai data 5 Luas permukaan melengkung silinder = 2πjt
Bilangan data Curved surface area of cylinder = 2πrh

Sum of data 6 Luas permukaan melengkung kon = πjs
Mean = Number of data Curved surface area of cone = πrl

5 Min/ Mean = 7 Luas permukaan sfera = 4πj2
Surface area of sphere = 4πr2

Hasil tambah (Nilai titik tengah kelas × Kekerapan) 8 Isi padu prisma tegak
Jumlah kekerapan = Luas keratan rentas × Panjang
Volume of right prism
SuTmotoafl (Midpoint × Frequency) = Cross sectional area × Length
number of frequency
9 Isi padu silinder = πj2t
6 Kecerunan/ Gradient, m = y2 – y1 Volume of cylinder = πr2h
x2 – x1
1 0 Isi padu kon = 1 πj2t
3
1
Pintasan-y Volume of cone = 3 πr2h
7 Kecerunan, m = – Pintasan-x
4
Gradient, m =– y-intercept 1 1 Isi padu sfera = 3 πj3
x-intercept
Volume of sphere = 4 πr3
3

8 Kebarangkalian suatu peristiwa: 1 2 Isi padu piramid tegak = 1 × Luas tapak × Tinggi
Probability of an event: 3

P(A) = n(A) Volume of right pyramid = 1 × Base area × Height
n(S) 3

9 P(A′) = 1 – P(A) 1 3 Hasil tambah sudut pedalaman poligon
= (n – 2) × 180°
Sum of interior angles of a polygon = (n – 2) × 180°

Bab 1 Pola dan Jujukan

Patterns and Sequences

Bahagian A

1.1 Pola 1.2 Jujukan
Patterns Sequences

CONTOH 1 Rajah 1 menunjukkan satu susunan corak. 4 Antara berikut, yang manakah bukan suatu
Diagram 1 shows an arrangement of patterns. jujukan?

... Which of the following is not a sequence?

Rajah 1/ Diagram 1 A 1, 3, 6, 10, 15, …
B 1, 2, 6, 24, 120, …
C 100, 50, 25, 12.5, 6.25, …
D 100, 80, 50, 40, 30, …

Apakah corak yang seterusnya? Tip Cemerlang
What is the next pattern?

A C

B D Jujukan ialah satu set susunan corak atau nombor yang
menunjukkan suatu pola.
Sequence is a set of arrangement of designs or numbers
which shows a certain pattern.

Tip Cemerlang 5 Antara jujukan berikut, yang manakah

Pola ialah susunan corak yang tertentu dan menunjukkan menunjukkan penambahan 3 kepada sebutan
hubungan yang sistematik antara objek-objek dalam
senarai tersebut. sebelumnya?
Pattern is a specific arrangment of designs and shows a
systematic relation between the objects in the list. Which of the following sequences shows addition

of 3 to its previous term?

A 2, 5, 8, … C 9, 6, 3, …

B 1, 4, 9, … D 27, 9, 3, …

1.3 Pola dan Jujukan
Patterns and Sequences
2 Antara berikut, yang manakah tiada pola?
What of the following does not have pattern? 6 Rajah 3 menunjukkan satu jujukan nombor.
A 1, 3, 5, 7, 9, … Diagram 3 shows a number sequence.
B 1, 1, 2, 3, 5, …
C 1, 5, 9, 11, 15, … 3, 5, 7, 9, 11, …
D 1, 4, 9, 16, 25, …
Rajah 3/ Diagram 3
3 Rajah 2 menunjukkan satu urutan nombor.
Diagram 2 shows a sequence of numbers. Antara berikut, yang manakah ialah pola bagi
jujukan nombor tersebut?
2, 4, N, 16, 32, …
Which of the following is the pattern for the number
Rajah 2/ Diagram 2 sequence?

Antara berikut, yang manakah mewakili N? A 2n + 1 dengan keadaan n = 1, 2, 3, 4, 5, …
Which of the following represents N? 2n + 1 where n = 1, 2, 3, 4, 5, ...
B 2n dengan keadaan n = 0, 1, 2, 3, 4, …
A 12 C 8 2n where n = 0, 1, 2, 3, 4, ...
C n + 2 dengan keadaan n = 0, 1, 2, 3, 4, …
B 10 D 6 n + 2 where n = 0, 1, 2, 3, 4, ...
D 2n dengan keadaan n = 1, 2, 3, 4, 5, …
2n where n = 1, 2, 3, 4, 5, ...

1

Matematik Tingkatan 2 Bab 1

Bahagian B

MODUL HEBAT 1 (PERAK)

7 (a) Padankan jujukan nombor dengan pola (b) Rajah 6 menunjukkan dua jujukan nombor.
Diagram 6 shows two number sequences.
yang betul.
Match the number sequence with the correct
112 8 21
pattern.

[2 markah/ marks]
Jawapan/ Answer:

Bahagi nombor JK L

sebelumnya dengan 5
Divide the previous
number by 5
CONTOH 1 79 15 17
500, 100, 20, 4, … Rajah 6/ Diagram 6
3, 6, 12, 24, … Kuasa dua nombor
sebelumnya 3 5 11 13

Square the previous Berdasarkan nombor yang diberi, tulis
number nilai J, K dan L.

Darab nombor Based on the given numbers, write the values
sebelumnya dengan 2 of J, K and L.
[3 markah/ marks]
Multiply the previous
number with 2 Jawapan/ Answer:

(b) Rajah 4 menunjukkan kad-kad bernombor. J:
K B AT Diagram 4 shows numbered cards.
K:
500 250 31.25 15.625
L:
Rajah 4/ Diagram 4
9 (a) Tandakan (✓) bagi jujukan nombor dan
Isikan petak kosong pada ruang jawapan (✗) bagi yang bukan.
dengan nombor yang sesuai dari Rajah 4
untuk melengkapkan jujukan nombor. Mark (✓) for number sequence and (✗) for
which is not.
Fill in the boxes in the answer space with a [2 markah/ marks]
suitable number from Diagram 4 to complete
the number sequence. Jawapan/ Answer:
[2 markah/ marks]
(i) 5, 25, 125, 625, …
Jawapan/ Answer:

1 000, , 62.5, , 3.906, ...

(ii) 1, 5, 20, 180, …

8 (a) Rajah 5 menunjukan satu jujukan nombor.
Diagram 5 shows a number sequence. (b) Nyatakan sebutan yang seterusnya bagi
jujukan nombor yang berikut.
3, 9, 27, 81, 243, …
State the next term for the following number
Rajah 5/ Diagram 5 sequences.
[2 markah/ marks]
Nyatakan pola bagi jujukan nombor di atas
dengan menggunakan ungkapan algebra. Jawapan/ Answer:

State the pattern of the number sequence (i) 12, 24, 36, 48, ?
above using algebraic expression.
[1 markah/ mark] (ii) 111, 222, 333, 444, ?

Jawapan/ Answer:


2

Matematik Tingkatan 2 Bab 1

Bahagian C

1 0 (a) Rajah 7 menunjukkan satu jujukan nombor.
Diagram 7 shows a number sequence.

3, 15, 75, 375, 1 875, …

Rajah 7/ Diagram 7 [1 markah/ mark]
(i) Tuliskan pola bagi jujukan nombor dengan menggunakan perkataan. [2 markah/ marks]
Write the pattern of the number sequence using words.

(ii) Nyatakan sebutan ke-8 dan sebutan ke-12.
State the 8th term and the 12th term.
CONTOH
Jawapan/ Answer: Contoh Cemerlang
(i)
Cari sebutan ke-12 bagi jujukan
nombor 3, 6, 9, 12, …
(ii) Sebutan ke-8/ 8th term : Find the 12th term of the number
Sebutan ke-12/ 12th term : sequence 3, 6, 9, 12, …

(b) (i) Rajah 8 menunjukkan Segi Tiga Pascal. Penyelesaian:
Diagram 8 shows the Pascal’s Triangle. Pola: Tambah 3 kepada sebutan
sebelumnya
Pattern: Add 3 to the previous
number
3 + 11(3) = 3 + 33
= 36

1
11
1G1
13H1

Rajah 8/ Diagram 8 [2 markah/ marks]
[1 markah/ mark]
Nyatakan nilai G dan nilai H.
State the value of G and of H.

Jawapan/ Answer:

G: H:

(b) (ii) Tuliskan pola bagi Nombor Fibonacci.
Write the pattern of Fibonnaci Numbers.

Jawapan/ Answer:

Pola/ Pattern:

3

Matematik Tingkatan 2 Bab 1 MODUL HEBAT 1 (EMAS)

(c) Rajah 9 menunjukkan pembahagian sel dalam suatu jangka masa.
Diagram 9 shows the cell division over a period of time.

CONTOH Pada mulanya Selepas 2 jam Selepas 4 jam
Initially After 2 hours After 4 hours

Rajah 9/ Diagram 9 [1 markah/ mark]
(i) Nyatakan pola bagi pembahagian sel. [3 markah/ marks]
State the pattern of cell division.

(ii) Tentukan bilangan sel yang dapat diperhatikan selepas 24 jam.
K BA T Determine the number of cells which can be observed after 24 hours.

Jawapan/ Answer:

(i) Pola/ Pattern :

(ii)

Contoh Cemerlang

Sebuah baldi diisikan dengan 2  air. Kadar kebocoran air bagi baldi itu ialah 200 m bagi setiap 5 minit. Hitung isi
padu air, dalam m, dalam baldi itu selepas 0.5 jam.
A pail is filled with 2  of water. The rate of water leakage of the pail is 200 m per 5 minutes. Calculate the volume of
water, in ml, in the pail after 0.5 hour.
Penyelesaian:
Jujukan nombor: 200 m, 400 m, 600 m, …
Number sequence: 200 m, 400 m, 600 m, …
0.5 jam ialah sebutan ke-(30 ÷ 5), maka (200 � 6) m air telah dikeluarkan daripada baldi.
0.5 hour is (30 ÷ 5)th term, therefore (200 � 6) m of water has leaked out of the pail.
Isi padu air dalam baldi/ Volume of water in the pail: 2 000 – (200 � 6) = 2 000 – 1 200
= 800 m

4

Bab 2 Pemfaktoran dan
Pecahan Algebra

Factorisation and Algebraic Fractions

Bahagian A

2.1 Kembangan 2.2 Pemfaktoran
Expansion Factorisation

CONTOH 1 Kembangkan/ Expand: 5 Tentukan faktor sepunya terbesar bagi sebutan

–3(s – 4) 4mn, 12m2n2 dan 32mn3.
Determine the highest common factor of the terms
A –3s + 12 C –3s – 4
B –3s – 12 D 3s – 4 4mn, 12m2n2 dan 32mn3.

A 4 C 4mn

B 4m D mn

2 Kembangkan/ Expand: 6 Faktorkan/ Factorise:

(3x – 2)(x + 4) 12p2 – 4q

A 3x2 + 12x – 8 C 3x2 + 10x – 8 A 4(p2 – q) C 4(3p2 – q)
B 3x2 – 2x – 8 D 3x2 + 10x + 8 B 4(p2 + q) D 4(3p2 + q)

Tip Cemerlang 7 Faktorkan/ Factorise:

Semua ungkapan mesti didarab antara satu sama lain. 49r 2 – 4
All expressions must be multiplied with each other.
A (7r – 2)2 C (49r – 2)(r + 2)
(x + 2)(x – 2) B (7r + 2)2 D (7r – 2)(7r + 2)

8 Faktorkan/ Factorise:

3 Kembangkan/ Expand: 6p2 + 13p – 63

(2p – 3q)(p + q) – 12pq MODUL HEBAT 17 (PERAK)

A 2p2 + 13pq + 3q2 A (2p + 7)(3p – 9)
B 2p2 – 13pq – 3q2 B (2p + 9)(3p + 7)
C 2p2 + 13pq – 3q2 C (2p – 9)(3p – 7)
D 2p2 – 13pq + 3q2 D (2p + 9)(3p – 7)

4 Rajah 1 menunjukkan sebuah segi empat sama. Tip Cemerlang
Diagram 1 shows a square.

3w – 2 Kaedah pendaraban silang boleh digunakan untuk
pemfaktoran dan juga untuk menyemak kembangan
dua ungkapan.
Cross multiplication method can be used for factorisation
and also to check the expansion of two expressions.

Rajah 1/ Diagram 1 9 Faktorkan/ Factorise:

Cari luas segi empat sama dalam sebutan w. 3xz – 9wz + 21wy – 7xy
Find the area of the square in terms of w.
A (3w + x)(7y + 3z) C (3w + x)(7y – 3z)
MODUL HEBAT 17 (GANGSA) B (3w – x)(7y – 3z) D (3w – x)(7y + 3z)

A 9w2 – 12w + 4 C 3w2 – 4w + 4
B 9w2 + 12w – 4 D 3w2 + 4w + 4

5

Matematik Tingkatan 2 Bab 2 12 Permudahkan/ Simplify:

10 Rajah 2 menunjukkan sebuah segi tiga. 5xz 3xy
KB AT Diagram 2 shows a triangle. y z

–



A x(5z2 – 3y2) C 5z – 3y2
yz yz

(2x + 1) cm x(5z – 3y) (5xz – 3y2)
B yz D yz


Rajah 2/ Diagram 2 13 Permudahkan/ Simplify:

CONTOH Diberi luas segi tiga ialah x2 + 5x +1 cm2. x2 – 4x + 4 ÷ 4x2 – 16
2 x2 – 16 8x – 32
Cari tinggi segi tiga itu dalam sebutan x.
5x
It is given the area of the triangle is x2 + 2 +1 cm2.
MODUL HEBAT 17 (PERAK)

Find the height of the triangle in terms of x. x – 2 2(x – 2)
x + 4 x2 + 6x + 8
MODUL HEBAT 17 (GANGSA) A C

A (x + 1) cm C (2x + 1) cm B 2(xx+–42) D 2(x – 2)
B (x + 2) cm D (2x + 5) cm x+2

Ungkapan Algebra dan Hukum Operasi 14 Permudahkan/ Simplify:

2.3 Asas Aritmetik x2 – 16 × x2 – 6x + 9 – 12 – 7x
Algebraic Expressions and Laws of Basic x2 – 9 3x + 12 3x + 9
Arithmetic Operations


11 Permudahkan/ Simplify: MODUL HEBAT 17 (PERAK)

(x + 2)2 – 5(2x – 1) A x2 9 x2 – 7x
3x + C 3x + 9
A (x + 3)2 C (x + 3)(x – 3)
B (x – 3)2 D (3x + 1)(x – 1) x2 – 5x D x2 – 7x + 12
B 3x + 9 3x + 9

Bahagian B

1 5 (a) Rajah 3 menunjukkan satu kaedah (b) Padankan ungkapan algebra kepada

penyelesaian bagi pemfaktoran. jawapan yang betul.
Diagram 3 shows a solving method of Match the algebraic expressions to the correct

factorisation. answer.

2m –4 –4m [3 markah/ marks]
Jawapan/ Answer:

m 5 10m x2 + 25
2m2 –20 6m
(x – 5)2 x2 + 10x + 25

Rajah 3/ Diagram 3 (x – 5)(x + 5) x2 – 20x + 25

Nyatakan kaedah ini. [1 markah/ mark] (x + 5)2 x2 – 25
State this method. x2 – 10x + 25
Jawapan/ Answer:


6

Matematik Tingkatan 2 Bab 2

1 6 (a) Tuliskan faktor sepunya terbesar bagi (b) Lengkapkan operasi berikut.
ungkapan algebra berikut. Complete the following operation.

Write the highest common factor for the [2 markah/ marks]
following algebraic expressions. Jawapan/ Answer:
[2 markah/ marks]
x2 – 16 ÷ 3x – 12
Jawapan/ Answer: y2 – 81 y2 – 9y

(i) 4pq, 12p2, –30q2 = (x – 4)(x + 4) � 3(x – 4)
(y – 9)(y + 9)
(ii) 24x2y2, 3xyz, 12xy2z2
+ 4y
= 3y + 27
(b) Lengkapkan operasi berikut.CONTOH
Complete the following operation.

[2 markah/ marks] 18 (a) Tandakan (✓) bagi ungkapan yang betul
dan (✗) bagi ungkapan yang salah.
Jawapan/ Answer:
Mark (✓) for the correct expressions and (✗) for
6p(3q – 5p) – 9pq = – 30p2 – 9pq the incorrect expressions.
[2 markah/ marks]
= –30p2 +
Jawapan/ Answer:
= –3p( )
3x(y + 5z2) = 3xy + 5z2
Tip Cemerlang
x(x + 8) + 15 = (x + 5)(x + 3)
Kemaskan ungkapan dengan menyelesaikan sebutan
serupa.
Tidy the expression by solving the like terms. (b) Rajah 5 menunjukkan sebuah prisma
dengan keratan rentas berbentuk
17 (a) Rajah 4 menunjukkan panjang dan lebar trapezium.

permaidani. Diagram 5 shows a prism with cross section of
Diagram 4 shows the length and width of a a trapezium.

carpet. x

(2n – m) m y – 2x
A
(2m – n) m 2y B

Rajah 4/ Diagram 4 3z
xy
Isi petak kosong dengan jawapan yang Rajah 5/ Diagram 5
betul. Nyatakan A dan B bagi ungkapan untuk
luas permukaan dengan betul.
Fill in the boxes with the correct answer. State A and B for the expressions of the surface
[2 markah/ marks] area correctly.

Jawapan/ Answer: [2 markah/ marks]
Jawapan/ Answer:
= (2m – n)( )
xy + xy2
= – 2m2 – 2n2 + mn
2xy
= ( ) m2
3yz – 6xz

3yz + 6xz


7

Matematik Tingkatan 2 Bab 2 [1 markah/ mark]
[1 markah/ mark]
Bahagian C [2 markah/ marks]

1 9 (a) Kembangkan/ Expand:
(i) 3g(5 + h)

(ii) (p + 3)(5p – 2q)

(iii) 3m2(m – n)(2m + n)


Jawapan/ Answer:
(i) (ii)

(iii)
CONTOH
(b) Rajah 6 menunjukkan ladang Pak Wan yang berbentuk segi empat tepat.
Diagram 6 shows Pak Wan's rectangular farm.

(x – 4) m

Rajah 6/ Diagram 6 [1 markah/ mark]
[2 markah/ marks]
(i) Diberi bahawa panjang ladang Pak Wan adalah dua kali lebarnya.
Nyatakan panjang ladang Pak Wan dalam sebutan x.
It is given that the length of Pak Wan's farm is twice its width.
State the length of Pak Wan's farm in terms of x.


(ii) Cari luas, dalam m2, ladang Pak Wan dalam sebutan x.
Find the area, in m2, of Pak Wan's farm in terms of x.

Jawapan/ Answer:

(i) Panjang/ Length :

(ii)

8

Matematik Tingkatan 2 Bab 2

(c) Rajah 7 menunjukkan luas kawasan sawah padi Zukri.
Diagram 7 9sh-0o1wNs the area of Zukri's paddy fields.

Luas/ Area : (18p2 – 12pq + 2q2) m2

CONTOH Rajah 7/ Diagram 7

(i) Diberi bahawa lebar sawah padi Zukri adalah setengah panjangnya. [2 markah/ marks]

Cari lebar, dalam m, sawah padi Zukri dalam sebutan p dan q.
It is given that the width of Zukri's paddy field is half its length.
Find the width, in m, of Zukri's paddy field in terms of p and q.



(ii) Zukri menyediakan (9p3 – 6p2q + pq2) m3 air untuk menanam pokok padi. [1 markah/ mark]

KBA T Cari ketinggian, dalam, m, paras air di sawah padi Zukri dalam sebutan p.
Zukri prepared (9p3 – 6p2q + pq2) m3 of water to plant the rice plants.
Find the height, in m, of the water level of Zukri's paddy fields in terms of p.


Jawapan/ Answer:

(i)

(ii)

Contoh Cemerlang

Panjang sisi keratan rentas dan isi padu bagi sebuah kuboid masing-masing ialah (x + 5) cm dan (5x 3 + 50x2 + 125x) cm3
Cari panjang kuboid itu.
The side length of the cross section and volume of a cuboid is (x + 5) cm and (5x3 + 50x2 + 125x) cm3 respectively.
Find the length of the cuboid.

Penyelesaian: Panjang kuboid/ Length of the cuboid = 5x
5x3 + 50x2 + 125x = 5x(x2 + 10x + 25)
= 5x(x + 5)2

9

Jawapan Langkah Penyelesaian
Lengkap (Bahagian A)

Bab 1 1 6 (a) (i) 2

(ii) 3xy

Bahagian A (b) 18pq, 9pq, 10p – 3q

1 B 2 C 3 C 17 (a) 2n – m, 4mn, –2m2 + 5mn – 2n2
4 D 5 A 6 A
(b) y(y – 9), xy

18 (a) 3x(y + 5z2) = 3xy + 5z2 ✗
✓
CONTOHBahagian B
x(x + 8) + 15 = (x + 5)(x + 3)
7 (a)
(b) A = 3yz – 6xz;
B = xy + xy2

(b) 250, 15.625 Bahagian C

8 (a) 3n dengan keadaan n = 1, 2, 3, 4, 5, … 1 9 (a) (i) 3g(5 + h) = 15g + 3gh

3n where n = 1, 2, 3, 4, 5, ... (ii) (p + 3)(5p – 2q) = 5p2 – 2pq + 15p – 6q

(b) J = 3, K = 5, L = 13 (iii) 3m2(m – n)(2m + n)

9 (a) (i) ✓ (ii) ✗ = 3m2(2m2 – mn – n2)

(b) (i) 60 (ii) 555 = 6m4 – 3m3n – 3m2n2

Bahagian C (b) (i) (2x – 8) m

(ii) (2x – 8)(x – 4) = 2x2 – 8x – 8x + 32

1 0 (a) (i) Darab nombor sebelumnya dengan 5 = 2x2 – 16x + 32
Multiply the previous number with 5
(ii) Sebutan ke-8/ 8th term: 234 375 (c) (i) 18p2 – 12pq + 2q2 = 2(9p2 – 6pq + q2)
Sebutan ke-12/ 12th term: 146 484 375
(b) (i) G = 2; H = 3 = 2(3p – q)2
(ii) Tambah dua sebutan sebelumnya
Adding the previous two terms ∴ Lebar/ Width = (3p – q) m
(c) (i) 2n dengan keadaan n = 0, 1, 2, …
2n where n = 0, 1, 2, ... (ii) 9p3 – 6p2q + pq2 = p(9p2 – 6pq + q2)
18p2 – 12pq + 2q2 2(9p2 – 6pq + q2)
(ii) Jujukan nombor pada setiap dua jam:
Number sequence in every two hours: = p m
2
1, 2, 4, … p
24 jam ialah sebutan ke-12. ∴ Ketinggian/ Height = 2 m
24 hours is the 12th term.
∴ 211 = 2 048 sel/ cells Bab 3

Bab 2 Bahagian A

Bahagian A 1 C 2 D 3 D 4 B 5 A
6 C 7 C 8 B 9 D 10 D
1 A 2 C 3 B 4 A 5 C 1 1 A 12 B
6 C 7 D 8 D 9 B 10 B
11 B 12 A 13 C 14 A Bahagian B

Bahagian B 13 (a) q2 = 3x + 2y;
y3 + 3 = x2 + 2
1 5 (a) Kaedah pendaraban silang (b) p + q = s;
Cross multiplication method p + q + r = 180°
(b) 14 (a) V
(b)

J1

Matematik Tingkatan 2 Jawapan

15 (a) M1 = 3, M2 = 5, V1 = 10, V2 = 6 (c) Panjang sisi/ Side length = 25
dan/ and
= 5 cm
M1 = 15, M2 = 2, V1 = 6, V2 = 45
(b) x + y = 15; x – 2 = 5 4 – y = 5
x = 7 y = –1
62 – 2x = 6y
x + y = 7 + (–1)
Bahagian C = 6

16 (a) (i) 3x2 – 2 = ab Bab 4

3x2 = ab + 2 Bahagian A

x2 = ab + 2 1 B 2 B 3 B 4 A 5 B
3 6 A 7 D 8 C 9 A 10 C
11 B

Bahagian B

1 2 (a) Heksagon
Hexagon
(b)
CONTOH x = ab + 2
3

(ii) y = x+3
x–3

xy – 3y = x + 3

xy – x = 3 + 3y

x( y – 1) = 3 + 3y

x = 3 + 3y
y–1

(b) (i) 4x + y = z3, y = 10, z = 2
5

4x + 10 = 23 1 3 (a) Sisi empat/ Quadrilateral: JKLM
5
Pentagon/ Pentagon: JKLMN
4x + 2 = 8

4x = 6 Heksagon/ Hexagon: JKLMNO

x = 6 (b) x = y + z
4


= 3 14 ✓ ✗
2

(ii) x = 5, y – x = 22 → y = 27 ✓✗

x = 3 + y+9 –z Bahagian C
5 = 3 + 27 + 9 – z
z = 3 + 36 – 5 1 5 (a) (i) Poligon G: Dekagon
= 4 Polygon G: Decagon
Poligon H: Heksagon
(c) (i) E = 1 mv2 Polygon H: Hexagon
2 (ii) 1. Semua sisi adalah sama panjang
All sides are of equal length
(ii) 725 = 1 m(52) 2. Semua sudut pedalaman adalah sama
2 All interior angles are equal

1 450 = 25m (b) (i)

m = 58 kg Paksi simetri
Axes of symmetry
17 (a) (i) LP = 1 tp
2

LQ = tp

(ii) LP = 1 tp
2

LP = 1 LQ
2

2LP = LQ (ditunjukkan/ shown)

(b) (i) x = 3p + 7q + 14
(ii) p = 2, q = 5
(ii) 6 bucu/ vertices
x = 3(2) + 7(5) + 14 (iii) (6 – 2) � 180° = 4(180°)
= 6 + 35 + 14
= RM55 (ditunjukkan/ shown) = 720°

J2

Matematik Tingkatan 2 Jawapan

(c) ∠RQS = 180° ÷ 3 = 60° Bahagian C

Bilangan sisi poligon: 1 2 (a) πd = 70 cm
Number of sides of polygon: 3.142 � d = 70 cm

360° = 6 d = 22.28 cm
60°

1 6 (a) (i) (14 – 2) � 180° = 12(180°) ∴ Diameter/ Diameter = 22.28 cm

= 2 160° (b) (i), (ii)

(ii) Sudut pedalaman/ Interior angle:

2 160° = 154.3°
14

Sudut peluaran/ Exterior angle:
CONTOH
180° – 154.3° = 25.7° 60°

(b) (i) x + 3x = 180°
4x = 180°
x = 45°

(ii) Bilangan sisi/ Number of sides:

360° =8
45°

(iii) Oktagon/ Octagon

(c) Sudut pedalaman pentagon sekata: (iii) Panjang lengkok/ Length of arc:
Interior angle of regular pentagon:
2 � 22 � 2.5 � 60° = 2 13 cm
(5 – 2) � 180° 540° 7 360° 21
5 = 5
(c) Jejari permukaan air/ Radius of water surface:
= 108°
Sudut pedalaman heksagon sekata: 272 � j 2 = 452 4 cm
Interior angle of regular hexagon: 7

(6 – 2) � 180° j 2 = 144
6
= 720° j = 12 cm
6
Jejari gelas sfera/ Radius of the spherical glass:
= 120°
52 + 122 = 169
x + 108° + 120° + 108° = 360°
= 13 cm
x + 336° = 360°
Lilitan gelas/ Circumference of the glass:
x = 24°

22 � 2 � 13 = 81 5 cm
7 7
Bab 5
1 3 (a) (i) Luas/ Area = 154
Bahagian A
272 � j 2 = 154
1 D 2 A 3 D 4 D 5 C j 2 = 49
6 B 7 A 8 C 9 B
j = 7 cm
Bahagian B
(ii) Lilitan/ Circumference = 2 � 22 � 7
10 (a) Jejari/ Radius 7
(b) = 44 cm

(b) Luas segi empat tepat/ Area of rectangle:

42 � 42 = 42 � 21
2
= 882 cm2

Luas bulatan/ Area of circle:

22 � 42 2 = 22 � 10.52
7 4 7

11 (a) 7; 360°; 6.6 = 346.5 cm2

(b) Lengkok ABC = Lengkok DEF Luas kawasan berlorek:
Arc ABC = Arc DEF Area of shaded region:

Garis FJ = Garis JI 882 – 2(346.5) = 882 – 693
Line FJ = Line JI
= 189 cm2


J3

Matematik Tingkatan 2 Jawapan

(c) Panjang AC/ Length of AC: Luar dasar/ Area of base:

252 – 72 = 576 22 � (14)2 = 616 cm2
= 24 cm 7

Luas segi tiga/ Area of triangle: (c) (i) Segi tiga bersudut tegak
Right-angled triangle
1 �7 � 24 = 84 cm2
2 (ii)

Luas sukuan bulatan/ Area of quadrant:

90° � 22 � 42 = 12.57 cm2
360° 7

Luas kawasan berlorek/ Area of shaded region:
84 – 12.57 = 71.43 cm2
CONTOH
Bab 6

Bahagian A

1 D 2 A 3 C 4 A 5 B (iii) Luas permukaan/ Surface area:
6 D 7 C 8 C 9 D 10 A 1
1 1 B 2 2 �4�3 + 6 � (4 + 5 + 3)

Bahagian B = 12 + 6(12)

1 2 (a) Piramid/ Pyramid = 84 unit2
(b) Bucu/ Vertices: 16
Permukaan rata/ Flat surfaces: 10 Bab 7
Sisi/ Edges: 24
Bahagian A

13 (a) 1 A 2 B 3 D 4 C 5 C
6 B 7 D 8 B 9 A 10 B
11 D

(b) 1.4, 2.5, 17.16 Bahagian B

14 ✓ ✗ 1 2 (a) Jarak dari paksi-x: 2 unit
✗ ✓ Distance from x-axis: 2 units
Jarak dari paksi-y: 3 unit
Distance from y-axis: 3 units
(b) (2, 0), (–1, 3)

Bahagian C 1 3 (a) Jarak GH = Jarak HI
Distance of GH = Distance of HI
15 (a) (i) 1. Hemisfera/ Hemisphere
2. Kon/ Cone (b)
(ii) Bilangan permukaan rata: 0
Number of flat surface
Bilangan permukaan lengkung: 2
Number of curve surface

(b) Panjang lengkok/ Length of arc: 1 4 ✓
✓
2 � 22 � 21 � 360° – 120° ✗
7 360° ✗

= 2 � 22 � 21 � 240°
7 360°
Bahagian C
= 88 cm
1 5 (a) (i) Jarak mencancang: 12 unit
Panjang lengkok adalah sama dengan lilitan Vertical distance: 12 units
Jarak mengufuk: 5 unit
dasar kon Horizontal distance: 5 units
Length of arc is equivalent to the circumference of the
base of cone (ii) 122 + 52 = 169
= 13 unit/ units
2 � 22 � j = 88 cm
7 j = 14

J4

CONTOH