Page 19 - 1202 Question Bank Additional Mathematics Form 5

(b) Perimeter of the shaded region (b) Perimeter of the shaded region
π
 2
 2
π
= 6 — + 5 + 1.73 + 1.73 — = (2h cos θ – h)(2θ) + (2h cos θ)θ
α 6 3 + (h – (2h cosθ – h))
+ (6 – 1.73) = 4θh cos θ – 2θh + 2θh cos θ + 2h
25 cm = 15.95 cm – 2h cos θ
(b) Area of the shaded region = 6θh cos θ – 2θh + 2h – 2h cos θ
1 1
2 π
2  2
r = —(5 + 12)(3) – —(6) — = h(6θ cos θ – 2θ + 2 – 2 cos θ)
2
6
π
1
2 π
 2
2.02
6
α
sin — = ——– – —(1.73) — If θ = —,
2
2 25 3 Perimeter of the shaded region
3  2
 2
α = 14.51 cm 2 π π
— = 4° 38ʹ = h 6 — cos 30° – 2 — + 2
2 5. A 6 6
α = 9° 16ʹ B – 2 cos 30° 4
2. A 6 cm E = h(1.94)
= 1.94h cm
C
O D 8. M
θ rad
O N
P r 8 cm
10 cm 3 cm
(a) Area = 48 cm 2 O π
6(AB) = 48 —– rad P
6
(a) 10θ = 3 AB = 8 cm Q
3
8
θ = —– tan ∠AOB = —
10 6 π
 2
θ = 0.3 rad s –1 ∠AOB = 53.13° (a) r — = 3
6
18
(b) Area of sector AOP ∠AOB = 0.93 rad r = —– cm
1
= —r θ (b) ∠EOD = 90° – 53.13° π
2
π
 2
2 = 36.87° OM — = 8
= 50θ Area of the sector EOD 6 48
3
 10 2
36.87°
= 50 —– = ——— × π(6) 2 OM = —–
π
360°
= 15 cm s –1 = 11.58 cm 2 OM = 15.28 cm
2
18
48
5
π
3. sin θ = —– 6. NM = —– – —– = 9.55 cm
π
15
1
θ = 19° 28ʹ P —r cm Q (b) MQ = 2(OM sin 15°)
2

= 2(15.38) sin 15°
∠AOB = 360° – 2(19° 28ʹ) = 29.52 cm
= 321.07° r cm Perimeter of the shaded region
O
= 5.6 rad
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= 29.52 + 9.55 × 2 + 3

2
—π rad
C 3 = 51.62 cm
1
2
(c) Area of MOQ = —(15.28) sin 30°
2
D (a) Length of the chord PQ = 58.37 cm 2
1
r – —r
2
O = 2ABBBBBB 2 Area of the shaded region
15 cm 4 1 18 2 π
 π 2  2
ABBB
3
θ 2.5 cm = 2 —r 2 = 58.37 – — —– —
6
2
A B 4 = 49.78 cm 2
5 cm 5 cm
2AB 3
(a) Arc length ACB = 15(5.6) = ——r 9. A
2
= 84 cm = AB 3r cm
(b) Area of the segment (b) Area of the circle = πr 2 60° a
38° 56ʹ
= ———– × π(15) Area under water a
2
360° = πr – Area of the segment O B
2
1
– —(15) sin 38° 56’ 1 1 60° a
2
3
2 = πr – —r θ – —r sin 120° 4
2
2
2
= 5.75 cm 2 2 2 P
AB 3
1
1
2 2
3  2
 24
Area of the shaded region = πr – —r —π – —r —– 1
2
2
—a
π
= π(15) – 5.75 – —(5) + π(2.5) 2 2 3 2 2 (a) cos 30° = ——
2
2
2
2
2πr
AB 3
2
= 681.47 cm 2 = —–— + —–r 2 OP
3 4 AB 3 1
 2
4. A 5 cm B If r = 25 cm, OP —– = —a
2
2
a
Area under water = 1 579.63 cm 2 OP = —–
6 cm 3 cm E AB 3
π h 7. P
—– rad 1.73 cm OP = OA
D C Arc length AB with centre O
π
12 cm —– rad A
a
3 = —– —–
2π
3
(a) h = 6 sin 30° = 3 cm  2 2
AB 3
3
2πa
AB 3
sin 60° = —– O θ rad 2 θ rad = –—–– × –––
BC h cm T Q
BC = 3.46 cm (a) (i) OP = 2h cos θ 3AB 3 AB 3
2AB 3πa
BE = 1.73 cm (ii) TQ = 2h cos θ – h = —–—–
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10_1202 QB AMath F5.indd 101 10/01/2022 4:49 PM