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(b) QP = OP – OQ – 2(OP)(OQ) cos ∠QOP 13. (a) 28 = 20 + 20 − 2(20)(20) cos ∠POQ
2
2
2
2
2
2
QP = 6 + 6 − 2(6)(6) cos (π − 1.3) ∠POQ = 1.55 rad
2
2
2
QP = 9.55 cm Reflex angle ∠POQ = 2π − 1.55
∠QOS = 4.733 rad
∠QPS =
2 [3]
= 1.3 rad (b) Area of triangle POQ
2 = 1 (20)(20) sin 1.55
= 0.65 rad 2
= 200 cm 2
Area of sector QPS [2]
= 1 (9.55) (0.65) (c) Area of the unshaded region
2
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2 = Area of circle − Area of major sector POQ
= 29.64 cm − Area of triangle POQ
2
[5] 2 1 2
(c) Area of segment PQ = π(20) − (20) (4.733) – 200
2
= Area of sector POQ − Area of triangle POQ = 110 cm
2
1 1 [3]
= (6) (π – 1.3) – (6)(6) sin(π – 1.3)
2
2 2 14. (a) 13 = 20 + 20 − 2(20)(20) cos ∠BOC
2
2
2
2
= 15.805 cm ∠BOC = 0.662 rad
Area of shaded region ∠OBC = π – ∠BOC
= Area of semicircle PQR – Area of sector QPS 2
− Area of segment PQ = 1.2398 rad
= 1 (6) π – 29.64 – 15.805 Perimeter of sector BAC
2
2 = BA + BC + AC
= 11.1 cm = 13 + 13 + (13)(1.2398)
2
[4] = 42.12 cm
[4]
12.
P Q (b) Area of segment BC
= Area of sector BOC − Area of triangle BOC
1 1
= 2 (20) (0.662) − (20)(20) sin (0.662)
2
2
U = 9.46 cm
2
R Area of sector BAC
2
T = 1 (13) (1.2398)
2
= 104.76 cm
2
S Area of shaded region
= Area of segment BC + Area of sector BAC
PQ = TS = √180 – 20 = 9.46 + 104.76
2
2
= 80√5 cm = 114.22 cm
2
cos ∠QRU = 20 [5]
180 1
2
∠QRU = 1.459 rad 15. 20 = r (1)
2
Arc length QS r = 2√10 cm (length of rectangle)
= 100(2π − 1.459 × 2) Height of rectangle
= 336.519 cm
= (2√10) + (2√10) – 2(2√10)(2√10) cos (1)
2
2
Arc length PT = 6.064 cm
= 80[2π − 2(π − 1.459)]
= 233.44 cm Area of rectangle
= 38.35 cm
2
Length of rope [4]
= 80√5 × 2 + 336.519 + 233.44 16. (a) q = 1.3 rad × 2
= 927.73 cm = 2.6 rad
[6] [1]
Cambridge IGCSE
TM
188 Ace Your Additional Mathematics

Answers Add Math.indd 188 14/03/2022 12:29 PM