Piawai Math 2

CONTOH

1Set Halaman S1.1 – S1.16

Bab 1: Pola dan Jujukan
1.1 Pola
1.2 Jujukan
1.3 Pola dan Jujukan
CONTOH
Bab 2: Pemfaktoran dan Pecahan Algebra
2.1 Kembangan
2.2 Pemfaktoran
2.3 Ungkapan Algebra dan Hukum Operasi Asas Aritmetik

Bab 3: Rumus Algebra
3.1 Rumus Algebra

2Set Halaman S2.1 – S2.12

Bab 4: Poligon
4.1 Poligon Sekata
4.2 Sudut Pedalaman dan Sudut Peluaran Poligon

Bab 5: Bulatan
5.1 Sifat Bulatan
5.2 Sifat Simetri Perentas
5.3 Lilitan dan Luas Bulatan

3Set Halaman S3.1 – S3.16

Bab 6: Bentuk Geometri Tiga Dimensi
6.1 Sifat Geometri Bentuk Tiga Dimensi
6.2 Bentangan Bentuk Tiga Dimensi
6.3 Luas Permukaan Bentuk Tiga Dimensi
6.4 Isi Padu Bentuk Tiga Dimensi

Bab 7: Koordinat
7.1 Jarak dalam Sistem Koordinat Cartes
7.2 Titik Tengah dalam Sistem Koordinat Cartes
7.3 Sistem Koordinat Cartes

ii

4Set Halaman S4.1 – S4.16

Bab 1: Pola dan Jujukan
1.1 Pola
1.2 Jujukan
1.3 Pola dan Jujukan

Bab 2: Pemfaktoran dan Pecahan Algebra
2.1 Kembangan
2.2 Pemfaktoran
2.3 Ungkapan Algebra dan Hukum Operasi Asas Aritmetik
CONTOH
Bab 3: Rumus Algebra
3.1 Rumus Algebra

Bab 4: Poligon
4.1 Poligon Sekata
4.2 Sudut Pedalaman dan Sudut Peluaran Poligon

Bab 5: Bulatan
5.1 Sifat Bulatan
5.2 Sifat Simetri Perentas
5.3 Lilitan dan Luas Bulatan

Bab 6: Bentuk Geometri Tiga Dimensi
6.1 Sifat Geometri Bentuk Tiga Dimensi
6.2 Bentangan Bentuk Tiga Dimensi
6.3 Luas Permukaan Bentuk Tiga Dimensi
6.4 Isi Padu Bentuk Tiga Dimensi

Bab 7: Koordinat
7.1 Jarak dalam Sistem Koordinat Cartes
7.2 Titik Tengah dalam Sistem Koordinat Cartes
7.3 Sistem Koordinat Cartes

Jawapan Halaman J1 – J8

ii

TERHAD 1Set

CONTOH Dua jam

Arahan:

1. Buka kertas peperiksaan ini apabila diberitahu.
2. Tulis nama dan angka giliran anda pada ruang yang disediakan.
3. Jawapan anda hendaklah ditulis pada ruang jawapan yang disediakan dalam kertas soalan

ini.

Untuk Kegunaan Pemeriksa

Nama pemeriksa: Markah penuh Markah diperoleh
20
Bahagian A
20 Soalan 4
4
Bahagian B 4
21 4
22 4
23
24 10
25 10
10
Bahagian C 10
26 10
27 10
28 100
29
30
31

Jumlah

© Hak Cipta PEP Publications Sdn. Bhd. S1.1 [Lihat halaman sebelah
Matematik Set 1

TERHAD Bahagian / Section A

[20 markah / marks]
Jawab semua soalan.

Answer all questions.

1. Rajah 1 menunjukkan suatu jujukan nombor. 7. Hitung nilai u jika v = 6 dalam 3u 2 + v = 33.
Diagram 1 shows a number sequence. Calculate the value of u if v = 6 in 3u2 + v = 33.
A1
21, m, 5.25, 2.625, 1.3125 B3
C4
Rajah / Diagram 1 D6

Hitung nilai m.
Calculate the value of m.
A 10.5
B 11.5
C 11.55
D 12.25
CONTOH 8. Faktorkan 49y 2 – 16.
Factorise 49y2 – 16.
A 7y – 4
B 7y 2 – 4
C (7y − 4)(7y + 4)
D (7y − 4)(7y − 4)

2. Kembangkan 4p(p − q). 9. Diberi s = −4 dan t = 14, hitung nilai r dalam
Expand 4p(p − q). r = 5s + 2t.
A 4p – 4q Given s = −4 and t = 14, calculate the value of r
B 4p 2 – 4q in r = 5s + 2t.
C 4p 2 – 4q 2 A –6
D 4p 2 – 4pq B –5
C4
3. Apakah perkara rumus bagi x = 15y + 4?
What is the subject of the formula for D8
x = 15y + 4?
Ax 10. Senaraikan faktor sepunya bagi 8g dan
By 12gh.
List down the common factors for 8g and 12gh.
C4
A 1, 2
D 15 B g, 2g
C 1, 2, g, 2g
4. Sebutan ke-9, T bagi 7, 14, 21, 28 ialah D 1, 2, 4, g, 2g, 4g
9
The 9th term, T9 for 7, 14, 21, 28 is
A 49
11. Nilai y jika x = 2 dan m = –16 dalam sebutan
B 56 y = 3xm – m ialah
The value of y if x = 2 and m = –16 in y = 3xm – m
C 63
is
D 70
A –112
5. (2r + 5)(r − 7) = B –80
A 2(r + 5)(r – 7) C –32
B 2r 2 – 9r – 35 D 48
C 2r 2 – 19r – 35
D 2r 2 – 35 12. Sebutan ke-6, T bagi –2, 6, –18, 54 ialah
6
The 6th term, T6 for –2, 6, –18, 54 is
T A 486
6. Sebutan ke-17, 17 bagi 13, 9, 5, 1 ialah
The 17th term, T17 for 13, 9, 5, 1 is B 162

A –59 B –51 C –147

C –34 D 39 D –1 458

© Hak Cipta PEP Publications Sdn. Bhd. S1.2 Matematik Set 1

TERHAD A 12m – 15n
B m – 15n
13. Apakah faktor sepunya terbesar (FSTB)
bagi 36x 2 dan 24x? C 1 m + 15n
What is the highest common factor (HCF) for 4
36x2 and 24x? 7
A 6x D 20 m + 15n
B 8x
C 12x 19. Diberi 1 c = 2d – 1. Nilai d jika c = –21 ialah
D 24x 3
1
14. Haziq berumur k tahun dan isterinya Given 3 c = 2d – 1. The value of d if c = –21 is
berumur 7 tahun lebih tua daripadanya.
Jumlah umur mereka ialah W. Tuliskan satu
rumus algebra untuk mewakili umur mereka.
Haziq’s age is k years old and his wife’s age is
7 years older than him. Their total age is W. Write
an algebraic formula to represent their age.
A W = 2k + 7
B W = (2k + 7)2
C W = 2k – 7
D 2W = 2k + 7
CONTOH A –7

B –3

C2

D5

20. Rajah 2 menunjukkan harga bagi sekuntum
bunga ros.

Diagram 2 shows the price of a rose.

15. Tuliskan pola bagi jujukan 5, –10, –25, … RM4.60
Write the pattern for the sequence 5, –10, –25, ...
A +5 Rajah / Diagram 2
B – 15
C × (–2) Harga bagi sekuntum bunga melur murah
D × (–5) RM1.20 daripada harga sekuntum bunga
ros. Nia membeli p kuntum bunga ros dan q
16. 18y 2 + 33y − 40 = kuntum bunga melur dengan diskaun 15%.
A 51y 2 – 40y Nyatakan rumus pengiraan wang yang perlu
B 3y (6y + 11y – 40) dibayar, r, oleh Nia.
C (6y − 4)(3y + 10) The price for a jasmine flower is cheap RM1.20
D (6y − 5)(3y + 8) than the price of a rose. Nia buys p roses and q
jasmines with a 15% discount. State the formula
17. Nyatakan pola bagi jujukan 783, 261, 87, ... for calculating the money, r, that Nia has to pay.
State the pattern for the sequence 783, 261, 87, ... A r = 4.6p + 3.8q
B r = 4.6p + 3.6q
A –(–522) C r = 3.96p + 2.95q
B – 522 D r = 3.91p + 2.89q
C ÷3
D ÷ (–3)

18. Permudahkan 3 m + 15n – 2 m.
4 5
3 2
Simplify 4 m + 15n – 5 m.

© Hak Cipta PEP Publications Sdn. Bhd. S1.3 [Lihat halaman sebelah
Matematik Set 1

TERHAD Bahagian / Section B

[20 markah / marks]
Jawab semua soalan.

Answer all questions.

Untuk 21. (a) Tentukan dan tuliskan perkara rumus bagi setiap yang berikut.
Kegunaan Determine and write the subject of the formula for each of the following.
Pemeriksa [2 markah / marks]

Jawapan / Answer:

CONTOH Rumus algebra Perkara rumus
Algebraic expression Subject of the formula

21(a) (i) x = 13 + y
2
(ii) P = 5Q – 2

(b) Tukarkan perkara rumus bagi rumus algebra yang berikut kepada ‘v’ dan padankan

dengan padanan yang betul.

Change the subject of the formulae of the following algebraic expression to ‘v’ and match
[2 markah / marks]
to the correct matching.

Jawapan / Answer:

11w =v
7x

(i) 7v = w
11x

7 =v
11w

21(b) (ii) 11w = 7vx
2
11wx = v
7

Jumlah S1.4 Matematik Set 1
21
4

© Hak Cipta PEP Publications Sdn. Bhd.

TERHAD Untuk

22. (a) Lengkapkan yang berikut. [2 markah / marks] Kegunaan
Complete the following.1 Pemeriksa

Jawapan / Answer: 22(a)

–8, 1, , 19, 28, , 46 2

(b) Perihalkan pola bagi urutan nombor yang berikut. [2 markah / marks]
Describe the pattern of the following sequence. 1

Jawapan / Answer:
(i) 964, 482, 241, 120.5, ...

Pola / Pattern:

(ii) 785, 657, 529, 401, ...

Pola / Pattern:
CONTOH 22(b)

2
Jumlah

22

4

23. (a) Tentukan sama ada setiap yang berikut ialah rumus algebra. Tandakan (✓) dalam

ruang yang disediakan.
Determine whether each of the following is an algebraic formulae. Mark (✓) in the provided
[2 markah / marks]
space. 1

Jawapan / Answer:

Rumus Ya Tidak
Formula Yes No

(i) 5M = 12J + 2K + L

(ii) 1 = 1 – 1
r s t

© Hak Cipta PEP Publications Sdn. Bhd. S1.5 23(a)

2
[Lihat halaman sebelah

Matematik Set 1

TERHAD (b) Isi tempat kosong dengan pekali dan pemboleh ubah yang betul.
[2 markah / marks]
Untuk Fill in the blanks with the correct coefficient and variable.
Kegunaan
Pemeriksa

Jawapan / Answer:

Sebutan Pekali Pemboleh ubah
Term Coefficient Variable

23(b) (i) 27ghi

2 (ii) 8uv
JumlahCONTOH

23

4 [2 markah / marks]
24. (a) Lengkapkan Rajah 3 dengan dua faktor untuk 2p(p – q).
Complete Diagram 3 with two factors of 2p(p – q).

Jawapan / Answer:

(i) 2p(p – q) (ii)
[2 markah / marks]
24(a)
Betul atau Salah
2 Rajah / Diagram 3 True or False

(b) Tentukan betul atau salah. Matematik Set 1
Determine whether true or false.

Jawapan / Answer:

Persamaan
Equation

(i) 5n 2 + 5n = 10n 3

24(b) (ii) 12m – 30 = 6(2m – 5)

2
Jumlah

24

4

© Hak Cipta PEP Publications Sdn. Bhd. S1.6

TERHAD Untuk

25. (a) Padankan yang berikut. [2 markah / marks] Kegunaan
Match the following. Pemeriksa

Jawapan / Answer:

9y(7 – 5x)

(i) 63y – 45xy
(ii) 63xy – 45x
CONTOH 9x(7 – 5y) 25(a)
9x(7y – 5) 2

(b) Bulatkan jawapan yang betul. [2 markah / marks]
Circle the correct answer.

Jawapan / Answer: kl(35 + 1)
(i) 35kl + kl

= 7(5kl + kl)

(ii) 4(a – b) 4ab 25(b)
= 4a – 4b 2

© Hak Cipta PEP Publications Sdn. Bhd. S1.7 Jumlah
25

4
[Lihat halaman sebelah

Matematik Set 1

TERHAD Bahagian / Section C

Untuk [60 markah / marks]
Kegunaan Jawab semua soalan.
Pemeriksa
Answer all questions.

26. (a) Nyatakan faktor-faktor bagi setiap sebutan yang berikut. [2 markah / marks]
State all the factors for each of the following term. 2.2 [2 markah / marks]
[2 markah / marks]
(i) 4b(b + 7) [2 markah / marks]

Jawapan / Answer:
CONTOH
26(a)(i)
2

(ii) (x – 2)(x + 5)
Jawapan / Answer:

26(a)(ii)
2

(iii) 3(y + 8)(y – 3)
Jawapan / Answer:

26(a)(iii)
2

(b) Faktorkan –24w 2 – 11w + 18.
Factorise –24w2 – 11w + 18.

Jawapan / Answer:

26(b) S1.8 Matematik Set 1
2

© Hak Cipta PEP Publications Sdn. Bhd.

TERHAD

(c) Permudahkan 3a – 9a . Untuk
2b 2b Kegunaan
Pemeriksa

Simplify 3a – 9a . [2 markah / marks]
2b 2b

Jawapan / Answer:

26(c)
CONTOH
2
Jumlah

26

10

27. (a) Kenal pasti dan nyatakan pola bagi set nombor genap dan ganjil dalam Rajah 4.
Identify and state the pattern of the set of numbers which is even and odd in Diagram 4.

2, 5, 8, 9, 13, 14, 17, 20, 21, 26 [4 markah / marks]
Rajah / Diagram 4

Jawapan / Answer:

(i) Nombor genap / Even numbers: 27(a)
Pola / Pattern: 4

(ii) Nombor ganjil / Oddn numbers:
Pola / Pattern:

(b) Lengkapkan jujukan berikut berdasarkan pola yang diberi.
Complete the following sequence based on the given patterns.

(i) Mendarab nombor sebelumnya dengan –2. [1 markah / mark]
Multiply the previous number with –2.
27(b)(i)
Jawapan / Answer: 1

45, –90, , –360 [Lihat halaman sebelah
Matematik Set 1
© Hak Cipta PEP Publications Sdn. Bhd. S1.9

TERHAD (ii) Menolak 24 daripada nombor sebelumnya. [1 markah / mark]
Subtract 24 from previous number. [2 markah / marks]
Untuk
Kegunaan Jawapan / Answer:
Pemeriksa
40, , –8, –32

27(b)(ii)
1
CONTOH
(c) Tentukan sebutan ke-5, T bagi jujukan berikut.
5
Determine the 5th term, T5 of the following sequence.

(i) 2 592d, 432d, 72d, ...

Jawapan / Answer:

27(c)(i)
2

(ii) –13, 1, 15, ... [2 markah / marks]
Jawapan / Answer:

27(c)(ii) S1.10 Matematik Set 1
2

Jumlah
27

10

© Hak Cipta PEP Publications Sdn. Bhd.

TERHAD

28. (a) Ungkapkan setiap pemboleh ubah dalam tanda berkurung yang berikut sebagai Untuk
perkara rumus. Kegunaan
Pemeriksa
Express each of the variables in the bracket as a subject of the formula.

(i) √ 8r = c [r ] [2 markah / marks]

Jawapan / Answer:

CONTOH 28(a)(i)

2

(ii) 5t 2 = k 2 + 12 [k] [2 markah / marks]
Jawapan / Answer:

28(a)(ii)

2

(iii) 2 + 1 = 1 [y] [2 markah / marks]
p 9 y

Jawapan / Answer:

© Hak Cipta PEP Publications Sdn. Bhd. S1.11 28(a)(iii)

2
[Lihat halaman sebelah

Matematik Set 1

TERHAD (b) Diberi 5y 2 = x 2 + 3nf, cari nilai n apabila y = – 4, x = 5 dan f = –2.
Given 5y2 = x2 + 3nf, find the value of n when y = – 4, x = 5 and f = –2.
Untuk [2 markah / marks]
Kegunaan Jawapan / Answer:
Pemeriksa

CONTOH28(b)
2

(c) Diberi 5a = 1 (1 – c)2, cari nilai b apabila a = 10 dan c = 11.
b

Given 5a = 1 (1 – c)2, find the value of b if a = 10 and c = 11. [2 markah / marks]
b

Jawapan / Answer:

28(c)

2
Jumlah

28

10

29. (a) Permudahkan / Simplify

(i) 5x + 5y × 6r – 6s [3 markah / marks]
r–s 8

Jawapan / Answer:

29(a)(i) S1.12 Matematik Set 1
3

© Hak Cipta PEP Publications Sdn. Bhd.

TERHAD [3 markah / marks] Untuk
Kegunaan
(ii) (x 2 + 3x – 10) ÷ (6x – 12)
Jawapan / Answer: Pemeriksa

(b) Kembangkan / Expand
(i) 3j (4k + 2)(k – 7)
Jawapan / Answer:
CONTOH 29(a)(ii)
3

[2 markah / marks]

29(b)(i)

2

(ii) (2p – q)2(p + q) [2 markah / marks]
Jawapan / Answer:

© Hak Cipta PEP Publications Sdn. Bhd. S1.13 29(b)(ii)

2
Jumlah

29

10
[Lihat halaman sebelah

Matematik Set 1

TERHAD

Untuk 30. (a) Tentukan sebutan kesepuluh, T bagi
Kegunaan 10
Pemeriksa Determine the tenth term, T10 for 1.3

(i) 3, 15, 27, 39, ... [2 markah / marks]
[2 markah / marks]
Jawapan / Answer:

CONTOH30(a)(i)
2

(ii) 215, 187, 159, 131, ...
Jawapan / Answer:

30(a)(ii)
2

(b) Tentukan sebutan kedua belas, T bagi 1, 4, 9, 16, … [2 markah / marks]
12
Determine the twelfth term, T12 for 1, 4, 9, 16, …

Jawapan / Answer:

30(b) S1.14 Matematik Set 1
2

© Hak Cipta PEP Publications Sdn. Bhd.

TERHAD

(c) (i) Lukiskan sebutan ketiga. Untuk
Draw the third term. Kegunaan
Pemeriksa

Sebutan pertama Sebutan kedua Sebutan keempat
Fourth term
First term Second term

CONTOH [2 markah / marks]

Jawapan / Answer:

30(c)(i)

(ii) Terangkan pola di 30(c)(i). 2
Describe the pattern in 30(c)(i).
[2 markah / marks]
Jawapan / Answer: 30(c)(ii)

2
Jumlah

30

10

31. (a) Hasil bahagi (3x – 15y) dengan 1 ialah m.
4
1
3 The quotient of (3x – 15y) and 4 is m.

(i) Berdasarkan pernyataan tersebut, tulis suatu rumus algebra.
[2 markah / marks]
Based on the statement, write an algebraic formula.

Jawapan / Answer:

© Hak Cipta PEP Publications Sdn. Bhd. S1.15 31(a)(i)

2
[Lihat halaman sebelah

Matematik Set 1

TERHAD (ii) Ungkapkan x dalam 31(a)(i) sebagai perkara rumus. [2 markah / marks]
Express x in 31(a)(i) as subject of the formula.
Untuk
Kegunaan Jawapan / Answer:
Pemeriksa

31(a)(ii)CONTOH
2

(iii) Hitung nilai x jika diberi m = 24 dan y = 3. [2 markah / marks]
Calculate the value of x if given m = 24 and y = 3.

Jawapan / Answer:

31(a)(iii)
2

(b) Diberi rumus pertukaran suhu Celsius, C daripada Farenheit, F ialah C = 5 (F – 32).
9
5
Given the formula of Celsius, C to Farenheit, F is C = 9 (F – 32).

(i) Hitung nilai suhu Celsius, dalam °C, apabila suhu Farenheit air ialah 50°F.

Calculate the value of degree Celsius, in °C, when the value of degree Farenheit of
[2 markah / marks]
water is 50°F.

Jawapan / Answer:

31(b)(i)

2 (ii) Hitung nilai suhu Farenheit, dalam °F, apabila air membeku.
Calculate the value of degree Farenheit, in °F, when water is freezing.
31(b)(ii) [2 markah / marks]
Jawapan / Answer:
2
Jumlah

31

10

© Hak Cipta PEP Publications Sdn. Bhd. S1.16 Matematik Set 1

Set 1 (b) 8w 9 –27w

Bahagian A –3w 2 16w
–24w 2 +18 –11w
1. A 2. D 3. A 4. C 5. B \ –24w 2 – 11w + 18
6. B 7. B 8. C 9. D 10. D
11. B 12. A 13. C 14. A 15. B = (8w + 9)(–3w + 2)
16. D 17. C 18. D 19. B 20. D
CONTOH 3a 9a 3a – 9a
Bahagian B (c) 2b – 2b = 2b
21. (a) (i) x
= –6a
(ii) P 2b
–3a
(b) (i) 11wx = v = b
7
11w
(ii) 7x =v 27. (a) (i) 2, 8, 14, 20, 26
2, 8, 14, 20, 26

22. (a) 10; 37 +6 +6 +6 +6
(b) (i) ÷ 2
(ii) – 128 Tambah 6 / Add 6

23. (a) (i) Ya / Yes (ii) 5, 9, 13, 17, 21
(ii) Tidak / No 5, 9, 13, 17, 21

(b) (i) 27; ghi +4 +4 +4 +4
(ii) 8; uv
Tambah 4 / Add 4
24. (a) 1; 2; p; 2p; (p – q)
(Pilih mana-mana dua sahaja) (b) (i) –90 × –2 = 180
(Select any two only) (ii) 40 – 24 = 16

(b) (i) Salah / False (c) (i) 2 592d, 432d, 72d, 12d, 2d
(ii) Betul / True
÷6 ÷6 ÷6 ÷6
25. (a) (i) 9y(7 – 5x)
(ii) 9x(7y – 5) \ T = 2d
5
(b) (i) kl (35 + 1)
(ii) 4a – 4b (ii) –13, 1, 15, 29, 43

Bahagian C +14 +14 +14 +14
26. (a) (i) 1, 4, b, 4b, 4b(b + 7), b(b + 7),
\ T = 43
4(b + 7), (b + 7) 5
(ii) 1, (x – 2), (x + 5), (x – 2)(x + 5)
(iii) 1, 3, 3(y + 8), 3(y – 3), (y + 8), (y – 3), 28. (a) (i) √ 8r = c
8r = c 2
(y + 8)(y – 3), 3(y + 8)(y – 3) c2
r = 8

(ii) 5t 2 = k 2 + 12
k 2 = 5t 2 – 12
k = √ 5t 2 – 12

J1

(iii) 2 + 1 = 1 30. (a) (i) TTTTTT\175896 0T======1036587957931=9+++++1+11111111222222======5689713917511
p 9 y

18 + p = 1
9p y

y = 9p p
18 +

(b) 5y 2 = x 2 + 3nf
5(–4)2 = (5)2 + 3n(–2) (ii) TTTTTT\715698 0T======1071114–97350=931–––––––32222782882888======4–1–9717935073
80 = 25 – 6n
6n = 25 – 80

CONTOH n = – 55
6

(c) 5a = 1 (1 – c)2
b

5(10) = 1 (1 – 11)2 (b) T = 122 = 144
b 12

(–10)2 (c) (i)
b
50 =

b = 100
50
b = 2


29. (a) (i) 5x + 5y × 6r – 6s (ii) Petak hitam dan petak putih
r–s 8 disambung hujung ke hujung secara
5(x + y) 6(r – s) berselang seli. Pola petak hitam ialah
= r–s × 8 1, 3, ... manakala pola petak putih
ialah 2, 4, ...
= 30(x + y) The black squares and the white squares
8 are joined end to end alternately. The
15(x + y) pattern of black squares is 1, 3, .... while
= 4 the pattern of white square is 2, 4, ...

(ii) (x 2 + 3x – 10) ÷ (6x – 12)
x + 3x
= 2 6x – – 10
12
(x + 5)(x – 31. (a) (i) (3x – 15y) ÷ 1 =m
= 6(x – 2) 2) 4

(x + 5) 4(3x – 15y) = m
6
= (ii) 4(3x – 15y) = m

(b) (i) 3j(4k + 2)(k – 7) 12x – 60y = m
= (12jk + 6j ) × (k – 7) 12x = m + 60y
= 12jk 2 – 84jk + 6jk – 42j m + 60y
= 12jk 2 – 78jk – 42j x = 12

(ii) (2p – q)2(p + q) (iii) x= m + 60y
= (2p – q) × (2p – q) × (p + q) 12
= (4p2 – 4pq + q2) × (p + q)
= 4p3 + 4p 2q – 4p2q – 4pq2 + q2p + q3 x= 24 + 60(3)
= 4p3 – 3pq 2 + q 3 12

x = 204
12

J2

x = 17 25. (a) (i) E dan / and H
(ii) F dan / and G
(b) (i) C= 5 (F – 32)
9 (b) (i) 7
(ii) 9
5
= 9 (50 – 32) Bahagian C

= 90 26. (a) 160o πj12 – 160o πj22
9 360° 360°

= 10oC    =4 22 4 22
9 7 23)2 9 7 × 212
(ii) Apabila air membeku, C = 0°C × × (21 + – ×

When water is freezing, C = 0°C = 2 704.25 – 616
CONTOH
5 (F – 32) = 0 = 2 088.25 cm2
9
F – 32 = 0 (b) OT = OQ – TQ
F = 32oF
= 17 – 2
Set 2
= 15 cm
Bahagian A TR = √OR 2 – OT 2

1. D 2. C 3. D 4. A 5. D = √172 – 152
6. A 7. D 8. A 9. C 10. B =8
11. D 12. B 13. A 14. B 15. C PR = 8 + 8
16. B 17. D 18. D 19. B 20. C
= 16 cm

Bahagian B (c) πj 2 = 2 464 cm2

21. (a) (i) Jejari / Radius 22 × j2 = 2 464
(ii) Tembereng minor / Minor segment 7
j = √ 784
j = 28
(b) SOR; TOP; SOT; POQ; QOR
(Pilih mana-mana dua sahaja) \ Diameter = 28 + 28
(Select any two only)
= 56 cm

22. (a) (i) 360o 27. (a) (i) 2c = 180o – 64o
(ii) 1 080o
c = 116°
(b) (i) Mempunyai sisi yang sama panjang 2
Has equal sides length c = 58o
(ii) 4d = 360o – 65o – 67o – 88o
(ii) Mempunyai sudut pedalaman yang
sama saiz d = 140°
Has equal size of interior angles 4
d = 35o

23. (a) (i) 108o (b) 2x + x = 720o – 90o – 145o – 120o – 155o
(ii) 150o
x = 210°
(b) (i) Betul / True 3
(ii) Salah / False x = 70o

28. (a) Sudut sektor major AOB

24. (a) 11 cm Angle of major sector AOB
(b) 77 cm2
(c) 49 cm = 360o – 82o
(d) 3 773 cm2
= 278o
\ Panjang lengkok major AB

Length of major arc AB

= 278° × 2 × 22 × 56
360° 7

J3

= 271.82 cm (b) 2πj = 2 × 22 × 63
7
(b) 2πj = 132 cm
= 396 cm
2 × 22 × j = 132
7 2
(c) πj 2 = 1 018 7 cm2
j 7
= 132 × 44 22 2
7 j2 7
= 21 × = 1 018

\ Diameter = 21 + 21 j = √ 324
j = 18 cm
= 42 cm

(c) Jejari / Radius = 98 ÷ 2 31. (a) Bilangan sisi poligon
CONTOH
= 49 cm Number of polygon sides

\ Luas bulatan / Area of circle 360°
40°
= 22 × 492 =
7
=9
= 7 546 cm2
\ Jumlah sudut pedalaman

29. (a) (i) 2j = (180o – 136o) ÷ 2 The sum of interior angles

2j = 44° = (9 – 2) × 180o
2
= 1 260o

j = 22° (b) (i) Sudut peluaran pentagon sekata
2
j = 11o Exterior angle of regular pentagon

= 360°
5
(ii) Sudut pedalaman heksagon
= 72o
Interior angle of hexagon

= (6 – 2) × 180o (ii) Sudut peluaran heksagon sekata
6
Exterior angle of regular hexagon

= 120o = 360°
\ 3k = 120o 6

k = 120° = 60o
3
k = 40o (iii) Sudut peluaran dekagon sekata

Exterior angle of regular decagon

(b) (i) Sudut pedalaman oktagon sekata = 360°
10
Interior angle of regular octagon

= (8 – 2) × 180o = 36o
8

= 135o Set 3

(ii) Sudut pedalaman dekagon sekata Bahagian A

Interior angle of regular decagon 1. B 2. C 3. A 4. D 5. B
6. C 7. D 8. B 9. C 10. A
= (10 – 2) × 180o 11. B 12. B 13. C 14. B 15. A
10 16. C 17. B 18. D 19. A 20. B

= 144o

30. (a) 140° × 2πj = 90 Bahagian B
360°
21. (a) ✓
7 × 2 × 3.142 × j = 90 (b) ✗
18 (c) ✗
j = 36.83 (d) ✓

\ OP = 36.83 cm

J4

22. (a) (i) Salah / False (ii) √(3 – 2)2 + [5 – (–9)]2
(ii) Betul / True
= √12 + 142
(b) (i) 294 cm2 = √197
(ii) 5 544 cm2 = 14.04 unit / units

23. (a) Silinder / Cylinder (iii) √(3 – 8)2 + (6 – 1)2
(b) Kon / Cone
(c) Piramid / Pyramid = √ –52 + 52
(d) Sfera / Sphere
= √ 50
= 7.07 unit / units

24. (a) 5 unit / units (b) 3+p = 7
(b) 6.4 unit / units 2
(c) 11.4 unit / units
(d) 4 unit / units
CONTOH 3 + p = 14
p = 11
q
+ 10 = 6
2
q + 10 = 12
25. (a) (i) (ii)

q=2

28. (a) (i) RP = √RQ 2 – PQ 2

= √292 – 202

(b) (i) 4πk 2 = 21 cm
\ Isi padu prisma
4 πk 3
3 Volume of prism

(ii) = 1 × 20 × 21 × 18
2

Bahagian C = 3 780 cm3

26. (a) (i) Luas kertas / Area of paper (ii) Luas permukaan prisma

= 14 × 10 Surface area of prism

= 140 cm2 = (2 × 1 × 20 × 21) + (18 × 20) +
= 2πjt 2

\ 2 × 22 × j × 10 = 140 (18 × 29) + (18 × 21)
7
j = 2.23 cm = 1 680 cm2

(b) 1 × 4 πj 3 = 155 232 mm3
2 3
(ii) Isi padu silinder / Volume of cylinder

= 22 × (2.23)2 × 10 1 × 4 × 22 × j 3 = 155 232
7 2 3 7
j = 3√ 74 088
= 156.29 cm3 j = 42 mm

(b) 4πj 2 = 49 896 cm2 \ Diameter = 42 + 42

4 × 22 × j2 = 49 896 = 84 mm
7
j = √ 3 969
j = 63 cm 29. (a) (i) –6 + 8 = 1 , –4 + 6 = 1
2 2
\ Koordinat pejabat pos = (1, 1)

27. (a) (i) √ [–5 – (–8)]2 + [–6 – (–11)]2 Coordinates of the post office

= √ 32 + 52

= √ 34
= 5.83 unit / units

J5

(ii) √(–6 – 8)2 + (–4 – 6)2 (b) (i) 2+x =0
= √–142 + (–10)2 2
2+x=0
= √296 x = –2
= 17.2 unit / units 5+y
= 17.2 km 2 =0

(b) √(4 – k)2 + (5 – 17)2 = 13 5+y=0
(4 – k)2 + 144 = 169 y = –5
(4 – k)2 = 25 \ Koordinat / Coordinates W
4 – k = √ 25 = (x, y)
k = –1
= (–2, –5)
CONTOH
30. (a) (i) Luas permukaan kon (ii) √ [2 – (–2)2] + [5 – (–5)2]

Surface area of cone = √42 + 102

   =22× 202 + 22 × 20 × = √ 116
7 7 101 = 10.77 unit / units

= 7 605.71 mm2 Set 4

(ii) Tinggi kon / Height of cone Bahagian A

= √1012 – 202 1. C 2. C 3. A 4. C 5. B
6. C 7. C 8. A 9. B 10. B
= 99 mm 11. D 12. A 13. D 14. A 15. D
16. D 17. A 18. C 19. C 20. C
\ Isi padu kon / Volume of cone

= 1 × 22 × 202 × 99 Bahagian B
3 7 21. (a) (i) b

= 41 485.71 mm3 (ii) S

(b) Jejari / Radius = 14 ÷ 2 (b) (i) Prisma / Prism
= 7 cm (ii) Piramid / Pyramid

Isi padu silinder / Volume of cylinder 22. (a) P : Lilitan / Circumference
= 3.142 × 72 × 21 Q : Jejari / Radius
= 3 233.12 cm3
(b) (i) ✓
Isi padu hemisfera (ii) ✗

Volume of hemisphere

= 1 × 4 × 3.142 × 73
2 3

= 718.47 cm3 23. (a) Sudut peluaran
Exterior angles
\ Jumlah isi padu / Total volume
= 3 233.12 cm3 + 718.47 cm3 (b) Sudut pedalaman
= 3 951.59 cm3 Interior angles

31. (a) (i) –5 + 5 = 0 , –4 + 2 = –1 (c) Sudut pedalaman
2 2 Interior angles
\ Titik tengah / Midpoint = (0, –1)
(d) Sudut peluaran
(ii) 3 + 9 = 6 , 6 +4 =5 Exterior angles
2 2
\ Titik tengah / Midpoint = (6, 5) 24. (a) (i) Ya / Yes
(ii) Ya / Yes
(iii) –4 + 8 = 2 , 6 + 0 = 3
2 2 (b) (i) (0, 5)
\ Titik tengah / Midpoint = (2, 3) (ii) (5, 0)

J6

25. (a) (i) ✓ \ Luas bulatan / Area of circle
(ii) ✗ = πj 2
(iii) ✓
= 22 × 352
(b) 5 7

= 3 850 cm2

Bahagian C (8k – 6)
2
26. (a) (i) Bilangan sisi 28. (a) Jejari / Radius = m

Number of sides = 4k – 3 m

= 360° \ Luas bulatan / Area of circle
30° = πj 2 m2
= π × (4k – 3)2 m2
= 12 = (4k – 3)(4k – 3)π m2
= (16k 2 – 24k + 9)π m2
CONTOH (ii) Bilangan sisi

Number of sides

= 360° (b) Panjang lengkok PR / Length of arc PR
120°

=3 = 70° × 2 × 22 × 36
360° 7
(iii) Bilangan sisi
= 44 cm
Number of sides \ Perimeter bagi sektor OPR

= 360° Perimeter of sector OPR
40°
= 36 cm + 36 cm + 44 cm
=9
= 116 cm
(b) (i) v + v + 15 = w
2v + 15 = w (c) –5 + x = 2
2
(ii) v = 97 –5 + x = 4
w = 2(97) + 15 x=9
= 209 6+y
\ 209 biji donat / donuts 2 = –2

6 + y = –4
y = –10
27. (a) (i) √ [(–2) – (–4)]2 + [4 – (–1)]2

= √ 4 + 25 \ Koordinat / Coordinates H = (x, y)

= √ 29 29. (a) 5(c + 3) = (9, –10)
= 5.39 unit / units = 5c + 15

(ii) √ [3 – (–4)]2 + [1 – (–2)]2 (b) (i) 8 048f, –4 024f, 2 012f, –1 006f, 503f

= √ 49 + 9 ÷ (–2) ÷ (–2) ÷ (–2) ÷ (–2)

= √ 58 \ T5 = 503f
= 7.62 unit / units (ii) 15, –2, –19, –36, –53

(b) 2x + 2y × 9m – 9n –17 –17 –17 –17
m–n 6
2(x + y) 9(m – n) \ T5 = –53
= m–n × 6
(c) Koordinat / Coordinates F = (4, 0)
18(x + y) 6+x
= 6 2 =4

= 3(x + y) 6+x=8

(c) Jejari / Radius = 70 ÷ 2 x=2
= 35 cm

J7

10 + y = 0 (b) (i) 4πj 2 = 4 × 22 × 152
2 7
10 + y = 0
y = –10 = 2 828.57 cm2

\ Koordinat / Coordinates S = (x, y) (ii) 4 πj 3 = 4 × 22 × 153
3 3 7
= (2, –10)
= 14 142.86 cm3

30. (a) (i) x 4 4x 31. (a) (i) Isi padu kon / Volume of cone

x –5 –5x = 1 × 22 × 82 × 15
x 2 –20 –x 3 7
\ x 2 – x – 20
= (x + 4)(x – 5) = 1 005.71 cm3
CONTOH
(ii) Tinggi sendeng kon

(ii) 2y 7 7y Slant height of cone

= √152 + 82

y –3 –6y = 17 cm
2y 2 –21 y
\ 2y 2 + y – 21 \ Luas permukaan kon

= (2y + 7)(y – 3) Surface area of cone

   =22×82 + 22 × 8 × 17
7 7

= 628.57 cm2

(b) Luas permukaan piramid
Surface area of pyramid

= (9 × 9) + (4 × 36)
= 225 cm2

J8

CONTOH