Page 68 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 1 Circular Measure
SPM Highlights C Solving problems involving areas of

sectors
The diagram below shows a circle and a sector of
a circle with a common centre O. The radius of the 15
circle is r cm.
The diagram below shows sector AOB with centre O
M and sector PQR with centre P. Sector AOB has radius
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
of 8 cm, OQ = OP = 3 cm and ∠AOB is a right angle.
P
θ A
O Q B
N
Q R
P
O
Given that PM = 6 and arc length of MN and PQ are
2
21 cm and 7 cm respectively. Calculate Calculate the area, in cm , of the shaded region.
(a) the values of r and q, [Use π = 3.142]
(b) the area, in cm , of the shaded region.
2
[Use π = 3.142] Solution
1
A sector AOB = r q
2
Solution 2
1
(a) Arcs MN and PQ subtend the same angle θ at = × 8 × 1 3.142 2
2
the centre, therefore 2 2
1
s s = × 8 × 1.571
2
MN = PQ 2
r + 6 r = 50.27 cm
2
21 = 7
r + 6 r A = × 3 × 3
1
21r = 7(r + 6) triangle OPQ 2
21r = 7r + 42 = 4.5 cm
2
14r = 42 For ∆OPQ
r = 3 cm ∠QPR = ∠POQ + ∠OQP
q = s 1.571
r = 1.571 + 1 2 2
= 7 rad = 2.3565 rad
3
2
1 PQ = OP + OQ 2
2
(b) A = r sin q 2 2
triangle MON 2 = 3 + 3
7
= 1 × (6 + 3) × sin 1 2 = 4.243 cm
2
2 3
1
= 29.28 cm A sector PQR = r q Form 5
2
2
2
A = 1 r q 1
2
2
sector POQ 2 = × (4.243) × 2.3565
1 7 2
2
= × 3 × = 21.21 cm
2
2 3
= 10.5 cm Area of the shaded region
2
Area of the shaded region = A – A – A
= A – A sector AOB triangle OPQ sector PQR
triangle MON sector POQ = 50.27 – 4.5 – 21.21
= 29.28 – 10.5 2
= 18.78 cm = 24.56 cm
2
Try Question 9 in ‘Try This! 1.3’
229