Page 17 - 1202 Question Bank Additional Mathematics Form 5

CHAPTER 1 5. A (b) OL = 18 cos 41.44°
Height above PQ
Paper 1 B = 18 – 18 cos 41.44°
15 cm
Section A C θ = 4.51 cm 1
1. A 9 cm 9. 60 seconds = 33— revolutions
3
O 100 1 5
3
OC
9
60
r rθ (a) —— = — 1 second = —— × —– = — revolution
3
10
5
θ OB 5 3 (a) — × 2π = —–π radians per second
O B OC = — × 15 9 9
r 5
P = 2r + rθ = 23.4 = 9 cm (b) r = 20 cm
10
(a) 2r = 23.4 – 5.4 AB = 15 + 15 – 2(15) cos θ Speed = —–π × 20
2
2
2
2
15 + 15 – 10
2
2
2
r = 9 cm cos θ = ——————– 9 –1
(b) 9θ = 5.4 2(15) 2 = 69.81 cm s –1
θ = 0.6 rad θ = 38° 57ʹ = 0.698 m s
∴ ∠AOB = 0.6 rad θ = 0.68 rad 10.
2
2
2
(b) AC = 15 + 9 – 2(15)(9) cos 38° 57ʹ
2.
AC = 9.8 cm D C
Perimeter of the shaded region 5 50 cm
K = 6 + 15(0.68) + 9.8 A —π rad B
9
= 26 cm 24 cm
θ
11 cm O
L 6.
O (a) Area traversed by the wiper
1
(a) —(11) θ = 160 1 2 5 1 2 5
2
 2
 2
2 = —(74) —π – —(24) —π
θ = 2.64 rad O 2 9 2 9
∴ ∠KOL = 2.64 rad = 4 276.06 cm 2
(b) 12 cm (b) Perimeter of the area traversed by
C the wiper
5
5
 2
 2
A B = 24 —π + 2(50) + 74 —π
9
9
(a) Area of ΔAOB = 72
1 = 271.04 cm
—(12)AB = 72
2
r AB = 12 cm 11. (a) Perimeter of the shaded region
2
 2
π
5
∠AOB = 45° = — rad = 10 —π + 10 sin 72°
Circumference, KL = 11(2.64) 4 + (10 – 10 cos 72°)
= 29.04 cm (b) Area of minor sector AOC = 28.99 cm
1
O ©PAN ASIA PUBLICATIONS
π
2πr = 29.04
(b) Area of the shaded region
 4 2

= —(12) —
2
r = 4.62 cm 2 = —(10) —π
1
2 2
= 18π cm 2 2  2
5
3.
1
B – —(10 cos 72°)(10 sin 72°)
7. P 2
A 12 = 48.14 cm 2
r cm R 11 1
12.
10 2 T
θ rad
9 3
O O π
AB
θ
6
(a) sin — = —— 8 4 Q U –– rad
2 2r 7 6 5 r
θ
2r sin — = AB O
2 1 2
(b) ∠AOB = 1.2 rad = 68.75° ∠POQ = — × 360° = 120° (a) Area = 20 cm 1
3
1
2 π
2  2
2
Difference ∠ROP = — × 30° + 30° = 50° 20 = —r — – —r sin 30°
2
2
6
= Arc length AB – Length of the 3 2 π sin 30°
2 2

chord AB Total angle = 170° = 2.97 rad 20 = r —– – ———–
12
= 5(1.2) – 2(5) sin 34.38° 8. r = ——————–
20
2
= 0.353 cm π sin 30°
2
4. P —– – ———–
12
O r = 41.17 cm
41.44° (b) Length of the chord UT
18 cm
A = 2(41.17) sin 15°
15 cm L
R = 21.31 cm
2
— π rad B Perimeter of the segment
3 30 cm (a) Arc length = 100 cm, r = 18 cm
π
 2
Q = 21.31 + 41.17 —
(a) Perimeter rθ = 100 = 42.87 cm 6
100
2
2
 3 2
 3 2
= 15 —π + 2(30) + 45 —π θ = ——
18
= 185.66 cm θ = 5.56 rad
θ = 318.56°
(b) Area of the paper
1 2 2 1 2 2
 3 2
 3 2
= —(45) —π – —(15) —π
2 2
= 600π cm 2
99
10_1202 QB AMath F5.indd 99 10/01/2022 4:49 PM