The document contains 10 mathematics word problems involving the calculation of areas and perimeters of circles, sectors, and combinations of circles and straight lines. The problems require using formulas such as the circumference of a circle formula (2πr), area of a circle formula (πr^2), and calculating areas and perimeters of sectors. Detailed step-by-step workings are shown for each problem.

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