MODUL 3

MATEMATIK SPM ENRICHMENT

TOPIC: CIRCLE, AREA AND PERIMETER

TIME: 2 HOURS

1.Diagram 1 shows two sector of circle ORQ and OPS with centre O.

R

12 cm

150O7 cmQ

P

S

227 , calculateDIAGRAM 1

By using =

the perimeter for the whole diagram in cm,

area of the shaded region in cm2.

[ 6 marks ]

Answer :

(a)

(b)2.In diagram 2,ABCD is a rectangle.

21 cm

B

A

14 cm

F

C

DFIGURE 4

E

CF is an arc of a circle with center E whereE is a point on the line DC withEC

= 7 cm. Using 22 , calculate

7

the length, in cm, of arc CF

the area, in cm2, of the shaded region

[ 6 marks ]

Answer :

(a)

(b)Diagram 3 shows two sectors OPQR and OJKL. OPQR and OJKL are three quarters of a circle.

POL and JOR are straight lines. OP = 21cm and OJ= 7 cm.

JQ

PLO

K

R

DIAGRAM 3

Using 227 , calculate

the perimeter, in cm, of the whole diagram,

the area, in cm2, of the shaded region.

[6 marks]

Answer:

(a)

(b)4.In Diagram 4, JK and PQ are arcs of two circles with centre O.OQRT is a square.K

QR

J

PO

T

210

DIAGRAM 4

OT = 14 cm and P is the midpoint of OJ.

Using 227 , calculate

the perimeter, in cm, of the whole diagram,

the area, in cm2 , of the shaded region.

[6 marks]

Answer:

(a)

(b)5.Diagram 5 shows two sectors OLMN and OPQR with the same centre O.

M

LN120PR

O

Q

DIAGRAM 5

OL = 14 cm. P is the midpoint of OL. [Use = 227 ]

Calculate

the area of the whole diagram,

the perimeter of the whole diagram.

[6 marks]

Answer:(a)

(b)In Diagram 6, ABD is an arc of a sector with the centre O and BCD is a quadrant.

A

OD = OB = 14 cm and AOB 45 .

Using 227 , calculate

the perimeter, in cm, of the whole diagram,

the area, in cm2, of the shaded region.

[6 marks]

Answer :

(a)

O

D

DIAGRAM 6

B

C

(b)In Diagram 7, the shaded region represents the part of the flat windscreen of a van which is being wiped by the windscreen wiper AB. The wiper rotates through an angle of 210o about the centre O.

Given that OA = 7 cm and AB = 28 cm.

B

A210o

OA

DIAGRAM 7

Using = 227 , calculate

the length of arc BB ,

the ratio of arc lengths , AA : BB

the area of the shaded region.

Answer:

(a)

B

[7 marks]

(b)

(c)Diagram 8 shows a quadrant ADO with centre O and a sector BEF with centre B. OBC is a right angled triangle and D is the midpoint of the straight line OC. Given OC = OB = BE = 14 cm.

DIAGRAM 8

Using = 227 , calculate

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

(b)the area, in cm2, of the shaded region.

.[6 marks]

Answer:

(a)

(b)9.In Diagram 9, OPQS is a quadrant with the centre O and OSQR is a semicirclewith the centre S.Q

RS

60

T

OP

DIAGRAM 9

Given that OP = 14 cm. Using = 227 , calculate

(a)the area, in cm2, of the shaded region,

(b)the perimeter, in cm, of the whole diagram.

[6 marks]

Answer:

(a)

(b)10. In diagram 10, OABC is a sector of a circle with centre O and radius 14 cm.

B

A

60

C

O

DIAGRAM 10

By using 22 , calculate

7

perimeter, in cm, the shaded area.

area, in cm2, the shaded area.

[7 markah]Answer :

(a)

(b)MODULE 3 - ANSWERS

TOPIC: CIRCLE, AREA AND PERIMETER

1

90

2212 @ 120 2 22 7

(a)

2

360

7

360

7

90

22

120

22

2 12

2 7 12 5

360

7

360

7

57.53

(b)90 22 122 @120 22 72

360

7

3607

90

22

122

120 22 72

1

7 12

360

7

3607

2

122.48

2

FEC 135

(a)

135

2

22 7

360

7

16.5

(b)

L3

13522 7 7

360

7

1

Shaded area (2114)

14 14 L3

2

138.25

3270

22

90

22

2 21

7 2

a)

atau360

360

7

7

27022

2 21 +

9022 7 2 + 14 + 14

360

7

3607

= 138

b)

270

22

21 21

atau2

90

22

7 7

360

7

360

7

270

22 21 21

-2 90

22 7 7

360

7

360

7

K1

K1

N1

K1

K1

N1

K2

K1

N1

K1

K1

N1

K1

K1

N1

K1

K1

= 962.5 cm2

4

a)

60

2

22 28

360

7

60

2

22 28 14 14 14 14 28

360

7

113 13 atau

11333

b)

60

22 28 28 atau

60 2214 14

360

7

360

7

60

22 28 28

60

2214 14 + 14 14

360

7

360

7

504

5

120

22

240 22

a)

14 14atau

7 7

360

7

360

7

120

2214 14 + 240

22 7 7

360

7

360

7

308

b)

120

2

2214atau

240 2 22 7

360

7

360

7

120

2

2214+

240

2 22 7 + 7 + 7

360

7

360

7

6

72 23

45

22

(a)

2

14

360

7

45

22

2

14 14 14 14 14

360

7

70 2

3

(b)

45

22

or

90

22

14 14

3607

14 14

360

7

45

22

90

22

14 14 2 14 14

14 14

360

7

360

7

N1

K1

K1

N1

K1

K1

N1

K1

K1

N1

K1

K1

N1

K1

K1

N1

K1

K1

161N1

7

(i)

210 2 22 35

K1

360

7

128 13 @ 128.33

N1

(ii)

210 2 22 7 : 210 2 22 35

K1

360

7

360

7

1: 5

N1

(iii)

210

22 352 or

210 22 7 2

K1

360

7

3607

210

22 352210 22 7 2

K1

360

7

360

7

2156

N1

845

22

(a)

2

14

or142 142 14

K1

360

7

11 + 14 + 14 + 14 + 5.799

K1

58.80 (2 d. p)

N1

90

22

45

22

(b)

7 7

or

14 x 14K1

360

7

360

745

22

1

90

22

14 14

7 7+

14 14K1

2

360

7

360

7

136.5

N1

9

(a)A1 =

90

2214 14 and A2 =

6022 7 7

360

7

360

7

K1

A1 A2

K1

128 1

N1

3

(b)P1 =

90

2

22 14 or P2 = 180 2 22 7K1

360

7

360

7

P1 + P2 + 14

K1

1058

N1

(a)AB =142 142=392 = 19.80

K1

150 2 22 14 atau 60 2 2214 atau90 2 2214K1

360

7

3607

360

7

Lengkok AC + 14 + 14 + 19.80 atau

Lengkok AB + lengkok BC + 14 + 14 + 19.80

K1

84.47

N1

(b)

15022142atau 1 14 14

K1

360

7

2

15022142-

114 14 atau

K1

360

7

2

770

98

3

158 2 atau 476 atau 158.67

N1

3

3