Page 49 - ACE YR IGCSE A TOP APPR' TO ADD MATH

(b) PQ = 5 + 5 − 2(5)(5) cos (2.6) 19. (a) tan ∠ROS = 5
2
2
2
PQ = 9.636 cm 1.5
[2] tan ∠ROS =1.2793 rad
(c) PRQ = (12)(1.3) ∠POR = π − 2(1.2793)
= 15.6 = 0.583 rad
PSQ = (5)(2.6) [2]
= 13
(b) OR = 5 + 1.5
2
2
2
Difference in length OR = 5.22 cm
= 15.6 − 13 Perimeter of shaded region
= 2.6 cm = PQR + PR
[2] = (5.22)(0.583) + 3
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π
17. (a) BC = 10 × = 6.04 cm
3 [2]
= 10π cm (c) Area of triangle POR
3 1
Perimeter = 2 × 3 × 5
= 10π × 3 = 7.5 cm
2
3
= 10π cm (or 31.42 cm) Area of sector POR
[2] = 1 (5.22) (0.583)
2
(b) Area of sector ABC 2
1 2
2
= 1 (10) 2 π = 7.943 cm
2 3
50π Area of shaded region
= cm = 7.943 − 7.5
2
3 = 0.44 cm
2
sin π = Height of triangle [4]
3 10
20. (a) OQ = OP – PQ
2
2
2
Height of triangle = 5√3 cm
= 2 – 1
2
2
Area of triangle OQ = √3
= 1 × 10 × 5√3 √3
2 SQ = 2 m
= 25√3 cm SP = SQ + PQ
2
2
2
2
√3
2
Area = Area of segment × 3 + Area of triangle = 1 2 2 + 1
2
2
= 3 1 50π – 25√3 + 25√3 SP = √7 m
3
= 70.48 cm 2 [2]
2
[5] –1 PQ
(b) ∠PSQ = tan 1 2
18. (a) ∠PQR = 2 × ∠PQO SQ
1 2
1
= 2 × tan –1 9 1 2
12
= 1.287 rad = tan –1 √3
[1] 2
(b) Area of sector PQO = 0.857 rad
= 1 (12) (0.6435) [1]
2
2 (c) ∠POQ = tan –1 PQ
1 2
= 46.33 cm OQ
2
1 2
Area of triangle PQO = tan –1 1
2
= 1 (12)(12) sin 0.6435 = 0.464 rad
2
= 43.20 cm Area of shaded region
2
= Area of sector POR − Area of triangle POR
Area of shaded region 1 1
= 2 × (46.33 − 43.20) = 2 (0.464 × 2) − (2)(√3)
2
=6.26 cm = 3.59 m
2
2
[4]
[3]
Answers 189
Answers Add Math.indd 189 14/03/2022 12:29 PM