1. Nama Sekolah / School Name
Test Name / Nama Ujian
Paper 2 (100 marks) Time: 2 hours 30 minutes
Kertas 2 (100 markah) Masa: 2 jam 30 minit
This paper consists of 20 questions. Answer all questions. Write your answer clearly in the spaces provided
in the question paper. Show your working. It may help you to get marks. If you wish to change your answer,
erase the answer that you have done. Then write down the new answer. The diagrams in the questions
provided are not drawn to scale unless stated. The marks allocated for each question are shown in brackets.
This question paper must be handed in at the end of examination.
Bahagian ini mengandungi 20 soalan. Jawab semua soalan. Jawapan hendaklah ditulis dengan jelas dalam
ruang yang disediakan dalam kertas soalan. Tunjukkan langkah-langkah penting. Ini boleh membantu anda
untuk mendapatkan markah. Sekiranya anda hendak menukar jawapan, padamkan jawapan yang telah
dibuat. Kemudian tuliskan jawapan yang baru. Rajah yang mengiringi soalan tidak dilukiskan mengikut
skala kecuali dinyatakan. Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan.
Kertas soalan ini hendaklah diserahkan di akhir peperiksaan.
1 In Diagram 1, OPQ is a sector of a circle with centre O and RSTU is a semicircle with centre R.
PSRUO is a straight line.
Dalam rajah 1, OPQ ialah sektor kepada bulatan berpusat O dan RSTU ialah semibulatan berpusat R.
PSRUO ialah garis lurus.
Diagram 1
Rajah 1
It is given that PO = 69 cm, RS = 14 cm and ∠POQ = 66°.
Diberi PO = 69 cm, RS = 14 cm dan ∠POQ = 66°.
22
Use π = 7 , and give the answer correct to two decimal places.
Calculate
22
Guna π = dan beri jawapan betul kepada dua tempat perpuluhan.
7
Hitung
(a) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
(b) the perimeter, in cm, of the shaded region.
perimeter, dalam cm, kawasan yang berlorek.
[7 marks]
[7 markah]
Answer:
Jawapan:

2. 2 Diagram 2 shows two sectors OWX and OYZ with the same centre O. OZX is a straight line.
Rajah 2 menunjukkan dua sektor bulatan OWX dan OYZ yang sama-sama berpusat O. OZX ialah garis
lurus.
Diagram 2
Rajah 2
It is given that ∠WOX = 63° and ∠YOZ = 42°.
Diberi ∠WOX = 63° dan ∠YOZ = 42°.
22
Using π = 7 , calculate
22
Dengan menggunakan π = 7 , hitungkan
(a) the perimeter, in cm, of the sector OWX,
perimeter, dalam cm, sektor OWX,
(b) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
[7 marks]
[7 markah]
Answer:
Jawapan:
3 In Diagram 3, OAB is a sector of a circle with centre O and CDEF is a semicircle with centre C.
ADCFO is a straight line.
Dalam rajah 3, OAB ialah sektor kepada bulatan berpusat O dan CDEF ialah semibulatan berpusat C.
ADCFO ialah garis lurus.
Diagram 3
Rajah 3

3. It is given that AO = 19 cm, CD = 3.5 cm and ∠AOB = 78°.
Diberi AO = 19 cm, CD = 3.5 cm dan ∠AOB = 78°.
22
Use π = 7 , and give the answer correct to two decimal places.
Calculate
22
Guna π = 7 dan beri jawapan betul kepada dua tempat perpuluhan.
Hitung
(a) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
(b) the perimeter, in cm, of the shaded region.
perimeter, dalam cm, kawasan yang berlorek.
[6 marks]
[6 markah]
Answer:
Jawapan:
4 Diagram 4 shows two sectors OAB and OCD with the same centre O. ODB is a straight line.
Rajah 4 menunjukkan dua sektor bulatan OAB dan OCD yang sama-sama berpusat O. ODB ialah
garis lurus.
Diagram 4
Rajah 4
It is given that ∠AOB = 60° and ∠COD = 42°.
Diberi ∠AOB = 60° dan ∠COD = 42°.
22
Using π = 7 , calculate
22
Dengan menggunakan π = 7 , hitungkan
(a) the perimeter, in cm, of the sector OAB,
perimeter, dalam cm, sektor OAB,
(b) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
[6 marks]
[6 markah]
Answer:
Jawapan:

4. 5 In Diagram 5, OPQ is a sector of a circle with centre O and RSTU is a semicircle with centre R.
PSRUO is a straight line.
Dalam rajah 5, OPQ ialah sektor kepada bulatan berpusat O dan RSTU ialah semibulatan berpusat R.
PSRUO ialah garis lurus.
Diagram 5
Rajah 5
It is given that PO = 51 cm, RS = 7 cm and ∠POQ = 69°.
Diberi PO = 51 cm, RS = 7 cm dan ∠POQ = 69°.
22
Use π = , and give the answer correct to two decimal places.
7
Calculate
22
Guna π = dan beri jawapan betul kepada dua tempat perpuluhan.
7
Hitung
(a) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
(b) the perimeter, in cm, of the shaded region.
perimeter, dalam cm, kawasan yang berlorek.
[6 marks]
[6 markah]
Answer:
Jawapan:
6 Diagram 6 shows two sectors OPQ and ORS with the same centre O. OSQ is a straight line.
Rajah 6 menunjukkan dua sektor bulatan OPQ dan ORS yang sama-sama berpusat O. OSQ ialah garis
lurus.
Diagram 6
Rajah 6

5. It is given that ∠POQ = 60° and ∠ROS = 45°.
Diberi ∠POQ = 60° dan ∠ROS = 45°.
22
Using π = 7 , calculate
22
Dengan menggunakan π = , hitungkan
7
(a) the perimeter, in cm, of the sector OPQ,
perimeter, dalam cm, sektor OPQ,
(b) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
[6 marks]
[6 markah]
Answer:
Jawapan:
7 In Diagram 7, OPQ is a sector of a circle with centre O and RSTU is a semicircle with centre R.
PSRUO is a straight line.
Dalam rajah 7, OPQ ialah sektor kepada bulatan berpusat O dan RSTU ialah semibulatan berpusat R.
PSRUO ialah garis lurus.
Diagram 7
Rajah 7
It is given that PO = 54 cm, RS = 10.5 cm and ∠POQ = 57°.
Diberi PO = 54 cm, RS = 10.5 cm dan ∠POQ = 57°.
22
Use π = 7 , and give the answer correct to two decimal places.
Calculate
22
Guna π = 7 dan beri jawapan betul kepada dua tempat perpuluhan.
Hitung
(a) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
(b) the perimeter, in cm, of the shaded region.
perimeter, dalam cm, kawasan yang berlorek.
[6 marks]
[6 markah]
Answer:
Jawapan:

6. 8 Diagram 8 shows two sectors OAB and OCD with the same centre O. ODB is a straight line.
Rajah 8 menunjukkan dua sektor bulatan OAB dan OCD yang sama-sama berpusat O. ODB ialah
garis lurus.
Diagram 8
Rajah 8
It is given that ∠AOB = 60° and ∠COD = 45°.
Diberi ∠AOB = 60° dan ∠COD = 45°.
22
Using π = 7 , calculate
22
Dengan menggunakan π = 7 , hitungkan
(a) the perimeter, in cm, of the sector OAB,
perimeter, dalam cm, sektor OAB,
(b) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
[6 marks]
[6 markah]
Answer:
Jawapan:
9 Diagram 9 shows two sectors OAB and OCD with the same centre O. OABE is a quadrant of a circle
with centre O. OBC and OED are straight lines.
Rajah 9 menunjukkan dua sektor bulatan OAB dan OCD yang sama-sama berpusat O. OABE ialah
sukuan bulatan berpusat O. OBC dan OED ialah garis lurus.
Diagram 9
Rajah 9
OE = ED = 21 cm and ∠COD = 60°.
OE = ED = 21 cm dan ∠COD = 60°.

7. 22
Using π = 7 , calculate
22
Dengan menggunakan π = , hitungkan
7
(a) the perimeter, in cm, of the whole diagram,
perimeter, dalam cm, seluruh rajah itu,
(b) the area, in cm2, of the shaded region.
luas, dalam cm2, kawasan yang berlorek.
[6 marks]
[6 markah]
Answer:
Jawapan:
10 Diagram 10 shows five hemisphres arranged side by side in a straight line.
Rajah 10 menunjukkan lima hemisfera yang disusun tepi ke tepi pada satu garis lurus.
Diagram 10
Rajah 10
2
Given that the volume of a hemisphere is 134 cm3. Find the value of b.
21
22
(Use π = )
7
2
Diberi isi padu setiap hemisfera ialah 13421 cm3. Cari nilai b.
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
11 Diagram 11 shows five hemisphres arranged side by side in a straight line.
Rajah 11 menunjukkan lima hemisfera yang disusun tepi ke tepi pada satu garis lurus.
Diagram 11
Rajah 11

8. 4
Given that the volume of a hemisphere is 567 cm3. Find the value of k.
22
(Use π = )
7
4
Diberi isi padu setiap hemisfera ialah 56 cm3. Cari nilai k.
7
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
12 Diagram 12 shows four hemisphres arranged side by side in a straight line.
Rajah 12 menunjukkan empat hemisfera yang disusun tepi ke tepi pada satu garis lurus.
Diagram 12
Rajah 12
2
Given that the volume of a hemisphere is 13421 cm3. Find the value of a.
22
(Use π = 7 )
2
Diberi isi padu setiap hemisfera ialah 13421 cm3. Cari nilai a.
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
13 Diagram 13 shows four hemisphres arranged side by side in a straight line.
Rajah 13 menunjukkan empat hemisfera yang disusun tepi ke tepi pada satu garis lurus.
Diagram 13
Rajah 13

9. 2
Given that the volume of a hemisphere is 13421 cm3. Find the value of h.
22
(Use π = )
7
2
Diberi isi padu setiap hemisfera ialah 134 cm3. Cari nilai h.
21
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
14 Diagram 14 shows five hemisphres arranged side by side in a straight line.
Rajah 14 menunjukkan lima hemisfera yang disusun tepi ke tepi pada satu garis lurus.
Diagram 14
Rajah 14
4
Given that the volume of a hemisphere is 567 cm3. Find the value of k.
22
(Use π = 7 )
4
Diberi isi padu setiap hemisfera ialah 567 cm3. Cari nilai k.
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
15 Diagram 15 shows a cylindrical solid. A hemisphere shown by the shaded region, is removed from the
solid.
Rajah 15 menunjukkan sebuah pepejal berbentuk silinder. Kawasan berlorek yang berbentuk
hemisfera telah dikeluarkan dari pepejal itu.

10. Diagram 15
Rajah 15
Given that the diameter of the hemisphere is 6 cm, calculate the volume, in cm3, of the remaining solid.
22
(Use π = )
7
Diberi diameter hemisfera itu ialah 6 cm, Hitung isi padu pepejal yang tinggal, dalam cm3.
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
16 Diagram 16 shows a composite solid comprises of a hemisphere and a cone.
Rajah 16 menunjukkan sebuah pepejal gubahan yang terdiri daripada sebuah hemisfera dan sebuah
kon.
Diagram 16
Rajah 16
Given that the volume of the solid is 924 cm3. Find the value of q.
22
(Use π = 7 )
Diberi isi padu pepejal itu ialah 924 cm3. Cari nilai q.
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:

11. 17 Diagram 17 shows a composite solid comprises of a cylinder and a hemisphere.
Rajah 17 menunjukkan sebuah pepejal gabuhan yang terdiri daripada sebuah silinder dan sebuah
hemisfera.
Diagram 17
Rajah 17
Given that the diameter of the hemisphere is 4 cm, calculate the volume, in cm3, of the solid.
22
(Use π = )
7
Diberi diameter hemisfera itu ialah 4 cm, Hitung isi padu pepejal itu, dalam cm3.
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
18 Diagram 18 shows a composite solid comprises of a cuboid and a half cylinder.
Rajah 18 menunjukkan sebuah pepejal gubahan yang terdiri daripada sebuah kuboid dan sebuah
separuh silinder.
Diagram 18
Rajah 18
Find the volume of the solid.
22
(Use π = 7 )
Cari isi padu bagi pepejal itu.
22
(Guna π = 7 )
[4 marks]

12. [4 markah]
Answer:
Jawapan:
19 Diagram 19 shows a composite solid comprises of a cylinder and a right cone.
Rajah 19 menunjukkan sebuah pepejal gubahan yang terdiri daripada sebuah silinder dan sebuah kon
tegak.
Diagram 19
Rajah 19
The height of the cylinder is 7 cm while the height of the cone is 7 cm. Find the volume of the solid.
22
(Use π = 7 )
Tinggi silinder itu ialah 7 cm manakala tinggi kon itu ialah 7 cm. Cari isi padu bagi pepejal itu.
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
20 Diagram 20 shows a cylindrical solid. The shaded region in the shape of a right cone is removed.
Rajah 20 menunjukkan sebuah pepejal berbentuk silinder. Kawasan berlorek yang berbentuk kon tegak
telah dikeluarkan.
Diagram 20
Rajah 20
The height of the cylinder is 14 cm while the height of the cone is 9 cm. Find the volume, in cm3, of the
remaining solid.
22
(Use π = 7 )
Tinggi silinder itu ialah 14 cm manakala tinggi kon itu ialah 9 cm. Cari isi padu, dalam cm3, bagi

13. pepejal yang tinggal.
22
(Guna π = 7 )
[4 marks]
[4 markah]
Answer:
Jawapan:
Answer:
1 (a) Area of sector OPQ
Luas sektor OPQ
66° 22
= × × 692
360° 7
= 2743.24 cm2
Area of semicircle RSTU
Luas sektor RSTU
1 22
= 2 × 7 × 142
= 308 cm2
Area of the shaded region
Luas kawasan berlorek
= 2743.24 − 308
= 2435.24 cm2
(b) Length of arc PQ
Panjang lengkok PQ
66° 22
= × 2 × × 69
360° 7
= 79.51 cm
Length of arc 79.51
Panjang lengkok 79.51
1 22
= × 2 × × 142
2 7
= 44 cm
Perimeter
= 79.51 + 44 + 69 + (69 − (14 × 2))
= 233.51 cm
2 (a) Length of arc WX
Panjang lengkok WX
63° 22
= × 2 × × 28
360° 7
154
= 5 cm
Perimeter
154
= 5 + 28 × 2
4
= 865 cm

14. (b) Area of sector OWX
Luas sektor OWX
63° 22
= 360° × 7 × 282
2156
= 5 cm2
Area of sector OYZ
Luas sektor OYZ
42° 22
= 360° × 7 × 212
1617
= 10 cm2
Area of the shaded region
Luas kawasan berlorek
2156 1617
= 5 − 10
1
= 2692 cm2
3 (a) Area of sector OAB
Luas sektor OAB
78° 22
= 360° × 7 × 192
= 245.82 cm2
Area of semicircle CDEF
Luas sektor CDEF
1 22
= × × 3.52
2 7
= 19.25 cm2
Area of the shaded region
Luas kawasan berlorek
= 245.82 − 19.25
= 226.57 cm2
(b) Length of arc AB
Panjang lengkok AB
78° 22
= × 2 × × 19
360° 7
= 25.88 cm
Length of arc 25.88
Panjang lengkok 25.88
1 22
= 2 × 2 × 7 × 3.52
= 11 cm
Perimeter
= 25.88 + 11 + 19 + (19 − (3.5 × 2))
= 67.88 cm
4 (a) Length of arc AB
Panjang lengkok AB
60° 22
= 360° × 2 × 7 × 14
44
= 3 cm

15. Perimeter
44
= 3 + 14 × 2
2
= 423 cm
(b) Area of sector OAB
Luas sektor OAB
60° 22
= × × 142
360° 7
308
= 3 cm2
Area of sector OCD
Luas sektor OCD
42° 22
= × × 72
360° 7
539
= 30 cm2
Area of the shaded region
Luas kawasan berlorek
308 539
= 3 − 30
7
= 8410 cm2
5 (a) Area of sector OPQ
Luas sektor OPQ
69° 22
= × × 512
360° 7
= 1566.79 cm2
Area of semicircle RSTU
Luas sektor RSTU
1 22
= 2 × 7 × 72
= 77 cm2
Area of the shaded region
Luas kawasan berlorek
= 1566.79 − 77
= 1489.79 cm2
(b) Length of arc PQ
Panjang lengkok PQ
69° 22
= 360° × 2 × 7 × 51
= 61.44 cm
Length of arc 61.44
Panjang lengkok 61.44
1 22
= 2 × 2 × 7 × 72
= 22 cm
Perimeter
= 61.44 + 22 + 51 + (51 − (7 × 2))
= 171.44 cm
6 (a) Length of arc PQ

16. Panjang lengkok PQ
60° 22
= 360° × 2 × 7 × 21
= 22 cm
Perimeter
= 22 + 21 × 2
= 64 cm
(b) Area of sector OPQ
Luas sektor OPQ
60° 22
= 360° × 7 × 212
= 231 cm2
Area of sector ORS
Luas sektor ORS
45° 22
= 360° × 7 × 142
= 77 cm2
Area of the shaded region
Luas kawasan berlorek
= 231 − 77
= 154 cm2
7 (a) Area of sector OPQ
Luas sektor OPQ
57° 22
= × × 542
360° 7
= 1451.06 cm2
Area of semicircle RSTU
Luas sektor RSTU
1 22
= 2 × 7 × 10.52
= 173.25 cm2
Area of the shaded region
Luas kawasan berlorek
= 1451.06 − 173.25
= 1277.81 cm2
(b) Length of arc PQ
Panjang lengkok PQ
57° 22
= 360° × 2 × 7 × 54
= 53.74 cm
Length of arc 53.74
Panjang lengkok 53.74
1 22
= 2 × 2 × 7 × 10.52
= 33 cm
Perimeter
= 53.74 + 33 + 54 + (54 − (10.5 × 2))
= 173.74 cm
8 (a) Length of arc AB
Panjang lengkok AB

17. 60° 22
= 360° × 2 × 7 × 28
88
= 3 cm
Perimeter
88
= 3 + 28 × 2
1
= 85 cm
3
(b) Area of sector OAB
Luas sektor OAB
60° 22
= 360° × 7 × 282
1232
= 3 cm2
Area of sector OCD
Luas sektor OCD
45° 22
= 360° × 7 × 142
= 77 cm2
Area of the shaded region
Luas kawasan berlorek
1232
= − 77
3
2
= 3333 cm2
9 (a) Length of arc AB
Panjang lengkok AB
30° 22
= 360° × 2 × 7 × 21
= 11 cm
Length of arc CD
Panjang lengkok CD
60° 22
= 360° × 2 × 7 × 42
= 44 cm
Perimeter
= 11 + 44 + 21 × 4
= 139 cm
(b) Area of sector OAB
Luas sektor OAB
30° 22
= × × 212
360° 7
231
= 2 cm2
Area of sector OCD
Luas sektor OCD
60° 22
= × × 422
360° 7
= 924 cm2
Area of sector OBE
Luas semibulatan OBE

18. 30° 22
= 360° × 7 × 212
= 231 cm2
Area of the shaded region
Luas kawasan berlorek
231
= 2 + 924 − 231
1
= 8082 cm2
10 2 2 22
134 = ( )(r)3
21 3 7
2816 21
r3 = 21 × 44
= 64
3
r = 64
=4
b=5×2×4
= 40 cm
11 4 2 22
567 = 3 ( 7 )(r)3
396 21
r3 = 7 × 44
= 27
3
r = 27
=3
k=5×2×3
= 30 cm
12 2 2 22
13421 = 3 ( 7 )(r)3
2816 21
r3 = 21 × 44
= 64
3
r = 64
=4
a=4×2×4
= 32 cm
13 2 2 22
13421 = 3 ( 7 )(r)3
2816 21
r3 = 21 × 44
= 64
3
r = 64
=4
h=4×2×4
= 32 cm

19. 14 4 2 22
567 = 3 ( 7 )(r)3
396 21
r3 = 7 × 44
= 27
3
r = 27
=3
k=5×2×3
= 30 cm
15 Volume of cylinder
Isipadu silinder
= 770 cm3
Volume of hemisphere
Isipadu hemisfera
4
= 567 cm3
Volume of remaining solid
Isipadu pepejal yang tinggal
4
= 770 − 567
3
= 7137 cm3
16 2 3 1 2
3 πr + 3 πr q = 924
2 1
πr2(3 r + 3 q) = 924
2 1 924
3 r + 3 q = πr2
1 924 2
3 q = πr2 - 3 r
924 2
q = ( πr2 - 3 r) × 3
924 7 2
= ( 2 × - × 7) × 3
7 22 3
14
= (6 - ) × 3
3
4
=3×3
= 4 cm
17 Volume of cylinder
Isipadu silinder
= 11 704 cm3
Volume of hemisphere
Isipadu hemisfera
16
= 16 cm3
21
Volume of the solid
Isipadu pepejal

20. 16
= 11 704 + 1621
16
= 11 720 cm3
21
18 Volume
Isi padu
1 22
= 14 × 12 × 16 + 2 × 7 × 72 × 12
= 2 688 + 924
= 3 612 cm3
19 Volume
Isi padu
22 1 22
= 7 × 192 × 7 + 3 × 7 × 92 × 7
22 1 22
= 7 × 361 × 7 + 3 × 7 × 81 × 7
= 7 942 + 594
= 8 536 cm3
20 Volume of the cylinder
Isipadu silinder
= 2744π cm3
Volume of the cone
Isipadu kon
= 147π cm3
Volume of the remaining solid
Isipadu pepejal yang tinggal
= 2744π − 147π
22
= 2597 × 7
= 8162 cm3