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