Page 18 - 1202 Question Bank Additional Mathematics Form 5

13. O r C (b) Area of the segment (b) Area of the segment
1
1
1
1
2
2
2
2
β = —(32) (2.02) – —(32) sin 115.82° = —(3) (1.65) – —(3) sin 94.33°
2
2
2
2
r 2
= 573.35 cm 2 = 2.94 cm
β Area of rectangle AOB 20.
A = 25(32 sin 57.91°) B
B 5.6 cm Q
1 = 677.77 cm 2 4 cm
2
(a) 20 = —r β Area required = 677.77 – 573.35 A 4 cm
2
40
β = —– = 104.42 cm 2 P r θ
r 2 16. O
(b) Perimeter of the shaded region (a) rθ = 4
9 cm (r + 4)θ = 5.6
= 2r + rβ 5 cm 4 + 4θ = 5.6
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40
 r 2
= 2r + r —– O A 4θ = 1.6
2

r 2
40
= 2r + —– cm θ = 0.4 rad
4
0.4
14. r = OP = —— = 10 cm
P (a) Circumference of the small gear (b) ∠POQ = 0.4 rad
= 2π(5) (c) Area of ABQP
r T = 10π cm 1 1
2
2
10π = 9θ = —(14) (0.4) – —(10) (0.4)
2
2
O 10
θ Q θ = —–π rad = 19.2 cm 2
— 9
2
(b) Circumference of the big gear 21. O 6 cm
= 2π(9) P Q
(a) rθ = 5 = 18π cm 6 cm 6 cm
5 Number of rotations 9 cm
r = — cm 18π S R
θ = ——
10π
6
(b) OT = — = 1.8 rotations T
θ
∠SOR
4.5
6
5
θ
cos — = — ÷ — 17. (a) sin ——— = —–
2 θ θ 2 6
5 θ θ r cm ∠SOR
= — × — ——— = 48.59°
θ 6 2
5
= — ∠SOR = 97.18°
6 ∠SOR = 1.7 rad
θ = 67.11° (b) Area of the segment STR
1
1
1
θ = 1.17 rad —r θ = 5 ................. 1 = —(6) (1.7) – —(6) sin 97.18°
2
2
2
2
Area of ∆POT 2r + rθ = 9 ................. 2 2 2 2
1
6
2  θ 2
= —r — sin 33.56° rθ = 9 – 2r .......... 3 = 12.74 cm
1 5 6 From 1, 22. M 6 cm O N
2  θ 2 θ 2
= — — — sin 33.56° 1 —– rad 60°
π
2
1 5 6 —r(9 – 2r) = 5 3 120°
2  1.172 1.172
= — —— —— sin 33.56° 9r – 2r = 10
2
2
= 6.06 cm 2 2r – 9r + 10 = 0 60°
(2r – 5)(r – 2) = 0
Area of the shaded region 5 P Q
1
2
= 2(6.06) – —r θ 1 5 2 r = — or r = 2 (a) Arc length PQ
2
2
1
2  2 2
5
1
2
2
2  1.172
2
= 2(6.06) – — —— — — θ = 5 or —(2) θ = 5 = π(6) – Arc length MP
– Arc length QN
8
5
π
π
= 1.44 cm 2 θ = — rad θ = — rad = 6π – 6 — – 6 —
 2  2
2
5
Section B 18. AC = 2r sin θ = 2π cm 3 3
15. Area (b) Perimeter of the shaded region
π 1 1 π π
 2  2
2
2
2
= —(r sin θ) – —r (2θ) – —r sin 2θ 4 = 12 + 6 — + 6 — + 2(6 cos 60°)
2 3 2 2 3 3
O 1 2 2 1
 2 2
2
32 cm = —r (π sin θ – 2θ + sin 2θ) = 12 + 4π + 12 —
19. = 18 + 4π
A B
D = 30.57 cm
25 cm
10 cm
O Paper 2
(a) OD = 32 + 10 – 25 θ 3 cm Section A
= 17 cm 1. (a) Circumference = 2πr
∠AOB P 4.4 cm Q
32 cos ——— = 17 (2π – 1.2)(25) = 2πr
2
(2π – 1.2)(25)
2.2
∠AOB (a) sin — = —— r = ——————
θ
——— = 57.91° 2π
2 2 3
θ
∠AOB = 115.82° — = 47° 10’ r = 2.02 cm
∠AOB = 2.02 rad 2
θ = 94.33°
θ = 1.65 rad
100
10_1202 QB AMath F5.indd 100 10/01/2022 4:49 PM