circle & solid geometry f3

Nama Sekolah / School Name

Test Name / Nama Ujian

Paper 2

Kertas 2

(100 marks)

(100 markah)

Time: 2 hours 30 minutes

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°.

Use π = 22

7 , and give the answer correct to two decimal places.

Calculate

Guna π = 22

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.

[7 marks]

[7 markah]

Answer:

Jawapan:

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°.

Using π = 22

7 , calculate

Dengan menggunakan π = 22

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.


[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

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°.

Use π = 22


Calculate

Guna π = 22


Hitung





[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°.

Using π = 22

7 , calculate


7 , hitungkan

(a) the perimeter, in cm, of the sector OAB,

perimeter, dalam cm, sektor OAB,



[6 marks]

[6 markah]

Answer:

Jawapan:





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°.

Use π = 22


Calculate

Guna π = 22


Hitung





[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

It is given that ∠POQ = 60° and ∠ROS = 45°.

Diberi ∠POQ = 60° dan ∠ROS = 45°.

Using π = 22

7 , calculate


7 , hitungkan

(a) the perimeter, in cm, of the sector OPQ,

perimeter, dalam cm, sektor OPQ,



[6 marks]

[6 markah]

Answer:

Jawapan:





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°.

Use π = 22


Calculate

Guna π = 22


Hitung





[6 marks]

[6 markah]

Answer:

Jawapan:

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°.

Using π = 22

7 , calculate


7 , hitungkan

(a) the perimeter, in cm, of the sector OAB,

perimeter, dalam cm, sektor OAB,



[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°.

Using π = 22

7 , calculate


7 , hitungkan

(a) the perimeter, in cm, of the whole diagram,

perimeter, dalam cm, seluruh rajah itu,



[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

Given that the volume of a hemisphere is 1342

21 cm

3. Find the value of b.

(Use π = 22

7 )

Diberi isi padu setiap hemisfera ialah 1342

21 cm

3. Cari nilai b.

(Guna π = 22

7 )

[4 marks]

[4 markah]

Answer:

Jawapan:



Diagram 11

Rajah 11


7 cm

3. Find the value of k.

(Use π = 22

7 )


7 cm

3. Cari nilai k.

(Guna π = 22

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


21 cm

3. Find the value of a.

(Use π = 22

7 )


21 cm

3. Cari nilai a.

(Guna π = 22

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


21 cm

3. Find the value of h.

(Use π = 22

7 )


21 cm

3. Cari nilai h.

(Guna π = 22

7 )

[4 marks]

[4 markah]

Answer:

Jawapan:



Diagram 14

Rajah 14


7 cm

3. Find the value of k.

(Use π = 22

7 )


7 cm

3. Cari nilai k.

(Guna π = 22

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.

Diagram 15

Rajah 15

Given that the diameter of the hemisphere is 6 cm, calculate the volume, in cm3, of the remaining solid.

(Use π = 22

7 )

Diberi diameter hemisfera itu ialah 6 cm, Hitung isi padu pepejal yang tinggal, dalam cm3.

(Guna π = 22

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.

(Use π = 22

7 )

Diberi isi padu pepejal itu ialah 924 cm3. Cari nilai q.

(Guna π = 22

7 )

[4 marks]

[4 markah]

Answer:

Jawapan:

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.

(Use π = 22

7 )

Diberi diameter hemisfera itu ialah 4 cm, Hitung isi padu pepejal itu, dalam cm3.

(Guna π = 22

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.

(Use π = 22

7 )

Cari isi padu bagi pepejal itu.

(Guna π = 22

7 )

[4 marks]

[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.

(Use π = 22

7 )

Tinggi silinder itu ialah 7 cm manakala tinggi kon itu ialah 7 cm. Cari isi padu bagi pepejal itu.

(Guna π = 22

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.

(Use π = 22

7 )

Tinggi silinder itu ialah 14 cm manakala tinggi kon itu ialah 9 cm. Cari isi padu, dalam cm3, bagi

pepejal yang tinggal.

(Guna π = 22

7 )

[4 marks]

[4 markah]

Answer:

Jawapan:

Answer:

1 (a) Area of sector OPQ

Luas sektor OPQ

= 66°

360° ×

22

7 × 69

2

= 2743.24 cm2

Area of semicircle RSTU

Luas sektor RSTU

= 1

2 ×

22

7 × 14

2

= 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°

360° × 2 ×

22

7 × 69

= 79.51 cm

Length of arc 79.51

Panjang lengkok 79.51

= 1

2 × 2 ×

22

7 × 14

2

= 44 cm

Perimeter

= 79.51 + 44 + 69 + (69 − (14 × 2))

= 233.51 cm

2 (a) Length of arc WX

Panjang lengkok WX

= 63°

360° × 2 ×

22

7 × 28

= 154

5 cm

Perimeter

= 154

5 + 28 × 2

= 864

5 cm

(b) Area of sector OWX

Luas sektor OWX

= 63°

360° ×

22

7 × 28

2

= 2156

5 cm

2

Area of sector OYZ

Luas sektor OYZ

= 42°

360° ×

22

7 × 21

2

= 1617

10 cm

2



= 2156

5 −

1617

10

= 2691

2 cm

2

3 (a) Area of sector OAB

Luas sektor OAB

= 78°

360° ×

22

7 × 19

2

= 245.82 cm2

Area of semicircle CDEF

Luas sektor CDEF

= 1

2 ×

22

7 × 3.5

2

= 19.25 cm2



= 245.82 − 19.25

= 226.57 cm2

(b) Length of arc AB

Panjang lengkok AB

= 78°

360° × 2 ×

22

7 × 19

= 25.88 cm

Length of arc 25.88


= 1

2 × 2 ×

22

7 × 3.5

2

= 11 cm

Perimeter

= 25.88 + 11 + 19 + (19 − (3.5 × 2))

= 67.88 cm

4 (a) Length of arc AB

Panjang lengkok AB

= 60°

360° × 2 ×

22

7 × 14

= 44

3 cm

Perimeter

= 44

3 + 14 × 2

= 422

3 cm

(b) Area of sector OAB

Luas sektor OAB

= 60°

360° ×

22

7 × 14

2

= 308

3 cm

2

Area of sector OCD

Luas sektor OCD

= 42°

360° ×

22

7 × 7

2

= 539

30 cm

2



= 308

3 −

539

30

= 847

10 cm

2


Luas sektor OPQ

= 69°

360° ×

22

7 × 51

2

= 1566.79 cm2


Luas sektor RSTU

= 1

2 ×

22

7 × 7

2

= 77 cm2



= 1566.79 − 77

= 1489.79 cm2


Panjang lengkok PQ

= 69°

360° × 2 ×

22

7 × 51

= 61.44 cm

Length of arc 61.44


= 1

2 × 2 ×

22

7 × 7

2

= 22 cm

Perimeter

= 61.44 + 22 + 51 + (51 − (7 × 2))

= 171.44 cm

6 (a) Length of arc PQ

Panjang lengkok PQ

= 60°

360° × 2 ×

22

7 × 21

= 22 cm

Perimeter

= 22 + 21 × 2

= 64 cm

(b) Area of sector OPQ

Luas sektor OPQ

= 60°

360° ×

22

7 × 21

2

= 231 cm2

Area of sector ORS

Luas sektor ORS

= 45°

360° ×

22

7 × 14

2

= 77 cm2



= 231 − 77

= 154 cm2


Luas sektor OPQ

= 57°

360° ×

22

7 × 54

2

= 1451.06 cm2


Luas sektor RSTU

= 1

2 ×

22

7 × 10.5

2

= 173.25 cm2



= 1451.06 − 173.25

= 1277.81 cm2


Panjang lengkok PQ

= 57°

360° × 2 ×

22

7 × 54

= 53.74 cm

Length of arc 53.74


= 1

2 × 2 ×

22

7 × 10.5

2

= 33 cm

Perimeter

= 53.74 + 33 + 54 + (54 − (10.5 × 2))

= 173.74 cm


Panjang lengkok AB

= 60°

360° × 2 ×

22

7 × 28

= 88

3 cm

Perimeter

= 88

3 + 28 × 2

= 851

3 cm


Luas sektor OAB

= 60°

360° ×

22

7 × 28

2

= 1232

3 cm

2

Area of sector OCD

Luas sektor OCD

= 45°

360° ×

22

7 × 14

2

= 77 cm2



= 1232

3 − 77

= 3332

3 cm

2


Panjang lengkok AB

= 30°

360° × 2 ×

22

7 × 21

= 11 cm

Length of arc CD

Panjang lengkok CD

= 60°

360° × 2 ×

22

7 × 42

= 44 cm

Perimeter

= 11 + 44 + 21 × 4

= 139 cm


Luas sektor OAB

= 30°

360° ×

22

7 × 21

2

= 231

2 cm

2

Area of sector OCD

Luas sektor OCD

= 60°

360° ×

22

7 × 42

2

= 924 cm2

Area of sector OBE

Luas semibulatan OBE

= 30°

360° ×

22

7 × 21

2

= 231 cm2



= 231

2 + 924 − 231

= 8081

2 cm

2

10 134

2

21 =

2

3 (

22

7 )(r)

3

r3 =

2816

21 ×

21

44

= 64

r = 3

64

= 4

b = 5 × 2 × 4

= 40 cm

11 56

4

7 =

2

3 (

22

7 )(r)

3

r3 =

396

7 ×

21

44

= 27

r = 3

27

= 3

k = 5 × 2 × 3

= 30 cm

12 134

2

21 =

2

3 (

22

7 )(r)

3

r3 =

2816

21 ×

21

44

= 64

r = 3

64

= 4

a = 4 × 2 × 4

= 32 cm

13 134

2

21 =

2

3 (

22

7 )(r)

3

r3 =

2816

21 ×

21

44

= 64

r = 3

64

= 4

h = 4 × 2 × 4

= 32 cm

14 56

4

7 =

2

3 (

22

7 )(r)

3

r3 =

396

7 ×

21

44

= 27

r = 3

27

= 3

k = 5 × 2 × 3

= 30 cm

15 Volume of cylinder

Isipadu silinder

= 770 cm3

Volume of hemisphere

Isipadu hemisfera

= 564

7 cm

3

Volume of remaining solid

Isipadu pepejal yang tinggal

= 770 − 564

7

= 7133

7 cm

3

16 2

3 πr

3 +

1

3 πr

2q = 924

πr2(2

3 r +

1

3 q) = 924

2

3 r +

1

3 q =

924

πr2

1

3 q =

924

πr2 -

2

3 r

q = (924

πr2 -

2

3 r) × 3

= (924

72 ×

7

22 -

2

3 × 7) × 3

= (6 - 14

3 ) × 3

= 4

3 × 3

= 4 cm

17 Volume of cylinder

Isipadu silinder

= 11 704 cm3

Volume of hemisphere

Isipadu hemisfera

= 1616

21 cm

3

Volume of the solid

Isipadu pepejal

= 11 704 + 1616

21

= 11 72016

21 cm

3

18 Volume

Isi padu

= 14 × 12 × 16 + 1

2 ×

22

7 × 7

2 × 12

= 2 688 + 924

= 3 612 cm3

19 Volume

Isi padu

= 22

7 × 19

2 × 7 +

1

3 ×

22

7 × 9

2 × 7

= 22

7 × 361 × 7 +

1

3 ×

22

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

= 2597 × 22

7

= 8162 cm3

circle & solid geometry f3

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