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Area of triangle POQ (b) Area of segment
3
1
= 1 (8) sin π 2 = Area of triangle POQ + area of major sector POQ
2
1
2 7 = 12(8)(8) sin 1.35 + (8) (2π − 1.35)
2
= 31.198 cm 2
2
= 189.09 m
2
Area of shaded region [4]
= (43.08 − 31.198) cm
2
= 11.88 cm 9. Perimeter
2
[5] = QO + OP + PQ
1
r
3
7. tan ∠BOA = AB = r + r + r 1 2
OA = 2r + r 2
AB = 5 tan 0.8 3
= 5.148 cm Area
1
1 2
OB = OA + AB = r 2 1 r
2
2
2
OB = 25 + 26.504 2 3
2
1
OB = 7.177 cm = r
3
6
CB = OB − OC
1
2
r
= 7.177 − 5 2r + = 2 1 2
r
3
= 2.177 cm 3 2 3 6
r
CA = (5)(0.8) 2r + = r 3
3
= 4 cm 6r + r = r
3
2
2
Perimeter of ABC r – r – 6 = 0
= AB + BC + CA (r – 3)(r + 2) = 0
= 5.148 + 2.177 + 4 r = 3 or r = –2 (rejected)
= 11.325 cm [4]
OD 10. Area of AOD
cos ∠BOA =
OA = (6) 2 2 π
1
1 2
OD = 5 cos 0.8 2 3
= 12π cm
2
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= 3.484 cm
Area of triangle DOA Area of ABCD
1
= (OD)(OA) sin ∠BOA = 12π × 4
2 3 2
1
= (3.484)(5) sin 0.8 = 16π cm
2 Area of BOC
= 6.247 cm = Area of AOD + Area of ABCD
2
2
Area of sector COA = 28π cm
1
1
= (5) (0.8) Area of BOC = (OB) 2 2 π
2
1 2
2 2 3
2
= 10 cm 2 π
28π = (OB) 1 2
3
Area of ACD OB = 2√21 cm or OB = –2√21 (invalid)
= 10 − 6.247
= 3.753 cm 2 Perimeter of the shaded region
[9] = AB + DC + AD + BC
2
2
π
π + 2√21
2
2
2
8. (a) 10 = 8 + 8 − 2(8)(8) cos ∠POQ = (2√21 – 6) + (2√21 – 6) + 6 1 2 1 2
3
3
cos ∠POQ = 7 = 38.1 cm
32 [5]
∠POQ = 1.35 rad
11. (a) Area of secor QOR
Perimeter of segment = 1 (6) (1.3)
2
= Arc PQ + Line PQ 2
= 8(2π − 1.35) + 10 = 23.4 cm
2
= 49.47 m [2]
[4]
Answers 187
Answers Add Math.indd 187 14/03/2022 12:29 PM