MODULE 3-Circle Area and Perimeter
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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. By using = 7 22 , calculate (a) the perimeter for the whole diagram in cm, (b) area of the shaded region in cm 2 . [6 marks ] Answer : (a) (b) O P Q R S 150° 7 cm 12 cm DIAGRAM 1
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spm math, MODULE 3-Circle Area and Perimeter
Transcript of MODULE 3-Circle Area and Perimeter
MODUL 3MATEMATIK SPM “ENRICHMENT”
TOPIC: CIRCLE, AREA AND PERIMETERTIME: 2 HOURS
1. Diagram 1 shows two sector of circle ORQ and OPS with centre O.
By using =7
22, calculate
(a) the perimeter for the whole diagram in cm,
(b) area of the shaded region in cm2.
[ 6 marks ]
Answer :
(a)
(b)
OP Q
R
S
150°7 cm
12 cm
DIAGRAM 1
2. In diagram 2, ABCD is a rectangle.
FIGURE 4
CF is an arc of a circle with center E where E is a point on the line DC with EC
= 7 cm. Using7
22 , calculate
(a) the length, in cm, of arc CF
(b) the area, in cm2, of the shaded region[ 6 marks ]
Answer :
(a)
(b)
C
BA
D E
F
14 cm
21 cm
F
3. 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.
DIAGRAM 3
Using22
7 , calculate
(a) the perimeter, in cm, of the whole diagram,(b) the area, in cm2, of the shaded region.
[6 marks]Answer:
(a)
(b)
OP
Q
R
J
L
K
4. In Diagram 4, JK and PQ are arcs of two circles with centre O.OQRT is a square.
DIAGRAM 4
OT = 14 cm and P is the midpoint of OJ.
Using7
22 , calculate
(a) the perimeter, in cm, of the whole diagram,(b) the area, in cm2 , of the shaded region.
[6 marks]Answer:
(a)
(b)
OP
J
T
RQ
K
210
5. Diagram 5 shows two sectors OLMN and OPQR with the same centre O.
OL = 14 cm. P is the midpoint of OL.
[Use =7
22]
Calculate
(a) the area of the whole diagram,(b) the perimeter of the whole diagram.
[6 marks]Answer:(a)
(b)
P120
O
Q
R
L
M
N
DIAGRAM 5
6. In Diagram 6, ABD is an arc of a sector with the centre O and BCD is aquadrant.
OD = OB = 14 cm and 45AOB .
Using7
22 , calculate
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm2, of the shaded region.
[6 marks]
Answer :
(a)
(b)
O
A
B
CD
DIAGRAM 6
A BO
210o
B
A
7. In Diagram 7, the shaded region represents the part of the flat windscreen of a vanwhich is being wiped by the windscreen wiper AB. The wiper rotates through anangle of 210o about the centre O.Given that OA = 7 cm and AB = 28 cm.
DIAGRAM 7
Using π =7
22, calculate
(a) the length of arc BB ,
(b) the ratio of arc lengths , AA : BB
(c) the area of the shaded region. [7 marks]
Answer:
(a)
(b)
(c)
8. 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 =7
22, 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.
DIAGRAM 9
Given that OP = 14 cm. Using π =7
22, calculate
(a) the area, in cm2, of the shaded region,
(b) the perimeter, in cm, of the whole diagram.
[6 marks]
Answer:
(a)
(b)
O P
Q
R S
60°
T
C
B
A
O
60
10. In diagram 10, OABC is a sector of a circle with centre O and radius 14 cm.
DIAGRAM 10
By using7
22 , calculate
(a) perimeter, in cm, the shaded area.
(b) area, in cm2, the shaded area.[7 markah]
Answer :
(a)
(b)
MODULE 3 - ANSWERSTOPIC: CIRCLE, AREA AND PERIMETER
1
(a) 77
222
360
120@12
7
222
360
90 K1
53.57
51277
222
360
12012
7
222
360
90
K1
N1
(b) 22 77
22
360
120@12
7
22
360
90 K1
48.122
1272
17
7
22
360
12012
7
22
360
90 22 K1
N1
2
(a) 135FEC K2
5.16
77
222
360
135
K1
N1
(b) 777
22
360
1353 L K1
25.138
14142
1)1421( 3
LareaShaded
K1
N1
3
a) 2127
22
360
270 atau 27
7
22
360
90 K1
2127
22
360
270 + 27
7
22
360
90 + 14 + 14 K1
= 138 N1
b) 21217
22
360
270 atau 2 77
7
22
360
90 K1
21217
22
360
270 - 2 77
7
22
360
90 K1
= 962.5 cm2 N1
4
a) 287
222
360
60 K1
2814141414287
222
360
60 K1
3
1113 atau 11333 N1
b) 28287
22
360
60 atau 1414
7
22
360
60 K1
28287
22
360
60 1414
7
22
360
60 + 14 × 14 K1
504 N1
5
a) 14147
22
360
120 atau 77
7
22
360
240 K1
14147
22
360
120 + 77
7
22
360
240 K1
308 N1
b) 147
222
360
120 atau 7
7
222
360
240 K1
147
222
360
120 + 7
7
222
360
240 + 7 + 7 K1
3
272 N1
6
(a) 147
222
360
45 K1
14141414147
222
360
45
K1
3
270 N1
(b) 14147
22
360
45 or 1414
7
22
360
90 K1
1414
7
22
360
90141421414
7
22
360
45 K1
161 N1
7
(i) 357
222
360
210 K1
1283
1@ 128.33 N1
(ii) 77
222
360
210 : 35
7
222
360
210 K1
1: 5 N1
(iii) 2357
22
360
210 or 27
7
22
360
210 K1
22 77
22
360
21035
7
22
360
210 K1
2156 N1
8
(a)360
45 2
7
22 14 or 141414 22 K1
11 + 14 + 14 + 14 + 5.799 K158.80 (2 d. p) N1
(b) 777
22
360
90 or
360
45
7
22 14 x 14 K1
777
22
360
901414
2
1 +
360
45
7
22 14 14 K1
136.5 N1
9
(a) A1 = 14147
22
360
90 and A2 = 77
7
22
360
60
K1A1 – A2 K1
1283
1N1
(b) P1 = 147
222
360
90 or P2 = 7
7
222
360
180 K1
P1 + P2 + 14 K1
58 N110
(a) AB = 22 1414 = 392 = 19.80 K1
147
222
360
150 atau 14
7
222
360
60 atau 14
7
222
360
90 K1
Lengkok AC + 14 + 14 + 19.80 atauLengkok AB + lengkok BC + 14 + 14 + 19.80 K184.47 N1
(b) 2147
22
360
150 atau 1414
2
1 K1
2147
22
360
150 - 1414
2
1 atau K1
3
770– 98
3
2158 atau
3
476atau 158.67 N1
