1. PPR Maths nbk
MODUL 3
SKIM TUISYEN FELDA (STF) 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
7 cm
150° O
P Q
S
DIAGRAM 1
22
By using π = , calculate
7
(a) the perimeter for the whole diagram in cm,
(b) area of the shaded region in cm2.
[ 6 marks ]
Answer :
(a)
(b)

2. PPR Maths nbk
2. In diagram 2, ABCD is a rectangle.
A 21 cm B
14 cm
F
D FIGURE 4 E C
CF is an arc of a circle with center E where E is a point on the line DC with EC
22
= 7 cm. Using π = , calculate
7
(a) the length, in cm, of arc CF
(b) the area, in cm2, of the shaded region
[ 6 marks ]
Answer :
(a)
(b)

3. PPR Maths nbk
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.
J Q
P L
O
K
R
DIAGRAM 3
22
Using π = , calculate
7
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm2, of the shaded region.
[6 marks]
Answer:
(a)
(b)

4. PPR Maths nbk
4. In Diagram 4, JK and PQ are arcs of two circles with centre O.
OQRT is a square.
K
Q R
J
P
O
T
210°
DIAGRAM 4
OT = 14 cm and P is the midpoint of OJ.
22
Using π = , calculate
7
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm2 , of the shaded region.
[6 marks]
Answer:
(a)
(b)

5. PPR Maths nbk
5. Diagram 5 shows two sectors OLMN and OPQR with the same centre O.
M
L N
120°
P R
O
Q
DIAGRAM 5
OL = 14 cm. P is the midpoint of OL.
22
[Use π = ]
7
Calculate
(a) the area of the whole diagram,
(b) the perimeter of the whole diagram.
[6 marks]
Answer:
(a)
(b)

6. PPR Maths nbk
6. 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 = 45o .
22
Using π = , calculate O B
7
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm2, of the shaded region.
[6 marks]
D C
DIAGRAM 6
Answer :
(a)
(b)

7. PPR Maths nbk
7. 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′
A′ 210o
O A B
DIAGRAM 7
22
Using π = , calculate
7
(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. PPR Maths nbk
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
22
Using π = , calculate
7
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm2, of the shaded region.
. [6 marks]
Answer:
(a)
(b)

9. PPR Maths nbk
9. In Diagram 9, OPQS is a quadrant with the centre O and OSQR is a semicircle
with the centre S.
Q
R S
60°
T
O P
DIAGRAM 9
22
Given that OP = 14 cm. Using π = , calculate
7
(a) the area, in cm2, of the shaded region,
(b) the perimeter, in cm, of the whole diagram.
[6 marks]
Answer:
(a)
(b)

10. PPR Maths nbk
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
22
By using π = , calculate
7
(a) perimeter, in cm, the shaded area.
(b) area, in cm2, the shaded area.
[7 markah]
Answer :
(a)
(b)

11. PPR Maths nbk
MODULE 3 - ANSWERS
TOPIC: CIRCLE, AREA AND PERIMETER
1
90 22 120 22
(a) × 2 × × 12 @ × 2× × 7 K1
360 7 360 7
90 22 120 22
× 2 × × 12 + × 2 × × 7 + 12 + 5
360 7 360 7 K1
57.53
N1
90 22 120 22 2
(b) × × 12 2 @ × ×7 K1
360 7 360 7
90 22 120 22 1
× × 12 2 + × × 7 2 − × 7 × 12
360 7 360 7 2 K1
122.48
N1
2
(a) ∠FEC = 135o K2
135 22
× 2× ×7
360 7 K1
16.5
N1
135 22
(b) L3 = × ×7×7 K1
360 7
⎛1 ⎞
Shaded area = (21 × 14) − ⎜ × 14 × 14 ⎟ − L3
⎝2 ⎠ K1
= 138.25
N1
3
270 22 90 22
a) × × 2 × 21 atau × ×7×2 K1
360 7 360 7
270 22 90 22
× × 2 × 21 + × × 7 × 2 + 14 + 14 K1
360 7 360 7
= 138 N1
270 22 90 22
b) × × 21 × 21 atau 2× × ×7×7 K1
360 7 360 7
270 22 90 22
× × 21 × 21 - 2 × × ×7×7 K1
360 7 360 7

12. PPR Maths nbk
2
= 962.5 cm N1
4
60 22
a) × 2 × × 28 K1
360 7
60 22
× 2 × × 28 + 14 + 14 + 14 + 14 + 28 K1
360 7
1
113 atau 113⋅33 N1
3
60 22 60 22
b) × × 28 × 28 atau × ×14 ×14 K1
360 7 360 7
60 22 60 22
× × 28 × 28 − × ×14 ×14 + 14 × 14 K1
360 7 360 7
504 N1
5
120 22 240 22
a) × × 14 × 14 atau × ×7×7 K1
360 7 360 7
120 22 240 22
× × 14 × 14 + × ×7×7 K1
360 7 360 7
308 N1
120 22 240 22
b) × 2 × × 14 atau × 2× ×7 K1
360 7 360 7
120 22 240 22
× 2 × × 14 + × 2× ×7 + 7 + 7 K1
360 7 360 7
2
72 N1
3
6
(a) 45 22 K1
×2× × 14
360 7
⎛ 45 22 ⎞ K1
⎜ ×2× × 14 ⎟ + 14 + 14 + 14 + 14
⎝ 360 7 ⎠
2 N1
70
3
(b) 45 22 or 90 × 22 × 14 × 14 K1
× × 14 × 14
360 7 360 7
⎛ 45 22 ⎞ ⎛ 90 22 ⎞ K1
⎜ × × 14 × 14 ⎟ + 2 ⎜14 × 14 − × × 14 × 14 ⎟
⎝ 360 7 ⎠ ⎝ 360 7 ⎠
161 N1

13. PPR Maths nbk
7
210 22
(i) × 2 × × 35 K1
360 7
1
128 @ 128.33 N1
3
210 22 210 22
(ii) × 2× ×7 : × 2 × × 35 K1
360 7 360 7
1: 5 N1
210 22 210 22 2
(iii) × × 35 2 or × ×7 K1
360 7 360 7
210 22 210 22 2
× × 35 2 − × ×7 K1
360 7 360 7
2156 N1
8
45 22
(a) ×2× × 14 or 14 2 + 14 2 − 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 14 K1
360 7 360 7
1 90 22 45 22
× 14 × 14 − × ×7 × 7 + × × 14 × 14 K1
2 360 7 360 7
136.5 N1
9
90 22 60 22
(a) A1 = × × 14 × 14 and A2 = × ×7×7
360 7 360 7 K1
A1 – A2 K1
1
128 N1
3
90 22 180 22
(b) P1 = × 2 × × 14 or P2 = ×2× ×7 K1
360 7 360 7
P1 + P2 + 14 K1

14. PPR Maths nbk
58 N1
10
(a) AB = 14 2 + 14 2 = 392 = 19.80 K1
150 22 60 22 90 22
× 2 × × 14 atau × 2 × × 14 atau × 2 × × 14 K1
360 7 360 7 360 7
Lengkok AC + 14 + 14 + 19.80 atau
Lengkok AB + lengkok BC + 14 + 14 + 19.80 K1
84.47 N1
150 22 1
(b) × × 14 2 atau × 14 × 14 K1
360 7 2
150 22 1
× × 14 2 - × 14 × 14 atau K1
360 7 2
770
– 98
3
2 476
158 atau atau 158.67 N1
3 3