Page 19 - 1202 Bank Soalan Matematik Tambahan Tingkatan 5

(b) Perimeter rantau berlorek (b) Perimeter rantau berlorek
 2
π
π
 2
= 6 — + 5 + 1.73 + 1.73 — = (2h kos θ – h)(2θ) + (2h kos θ)θ
α 6 3 + (h – (2h kosθ – h))
+ (6 – 1.73) = 4θh kos θ – 2θh + 2θh kos θ + 2h
25 cm = 15.95 cm – 2h kos θ
(b) Luas rantau berlorek = 6θh kos θ – 2θh + 2h – 2h kos θ
1
1
2 π
2  2
j = —(5 + 12)(3) – —(6) — = h(6θ kos θ – 2θ + 2 – 2 kos θ)
2
6
π
1
2 π
 2
2.02
6
α
sin — = ——– – —(1.73) — Jika θ = —,
2
2 25 3 Perimeter rantau berlorek
3  2
α = 14.51 cm 2 π π
 2
— = 4° 38ʹ = h 6 — kos 30° – 2 — + 2
2 5. A 6 6
α = 9° 16ʹ B – 2 kos 30° 4
2. A 6 cm E = h(1.94)
= 1.94h cm
C
O D 8. M
θ rad
O N
P j 8 cm
10 cm 3 cm
(a) Luas = 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) j — = 3
6
18
(b) Luas sektor AOP ∠AOB = 0.93 rad j = —– cm
1
= —j θ (b) ∠EOD = 90° – 53.13° π
2
π
 2
2 = 36.87° OM — = 8
= 50θ Luas sektor 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
48
18
5
3. sin θ = —– 6. NM = —– – —– = 9.55 cm
π
π
15
1
θ = 19° 28ʹ P —j cm Q (b) MQ = 2(OM sin 15°)
2
= 2(15.38) sin 15°

∠AOB = 360° – 2(19° 28ʹ) = 29.52 cm
= 321.07° j cm Perimeter rantau berlorek
O
6 ©PAN ASIA PUBLICATIONS
= 5.6 rad

= 29.52 + 9.55 × 2 + 3
2
—π rad
C 3 = 51.62 cm
1
2
(c) Luas MOQ = —(15.28) sin 30°
2
D (a) Panjang perentas PQ = 58.37 cm 2
1
j – —j
2
2
O = 2ABBBBBB Luas rantau berlorek
 π 2  2
15 cm 4 1 18 2 π
ABBB
3
θ 2.5 cm = 2 —j 2 = 58.37 – — —– —
6
2
A B 4 = 49.78 cm 2
5 cm 5 cm
2AB 3
(a) Panjang lengkok ACB = 15(5.6) = ——j 9. A
2
= 84 cm = AB 3j cm
(b) Luas tembereng (b) Luas bulatan = πj 2 60° a
38° 56ʹ
= ———– × π(15) Luas di bawah air a
2
360° O B
2
1 = πj – Luas tembereng
2
– —(15) sin 38° 56ʹ 1 1 60° a
3
2 = πj – —j θ – —j sin 120° 4
2
2
2
= 5.75 cm 2 2 2 P
2 2
1
1
AB 3
Luas rantau berlorek = πj – —j —π – —j —– 1
3  2
 24
2
2
—a
π
= π(15) – 5.75 – —(5) + π(2.5) 2 2 3 2 2 (a) kos 30° = ——
2
2
2
2
2πj
2
= 681.47 cm 2 = —–— + —–j OP
AB 3 2
3 4 AB 3 1
 2
4. A 5 cm B Jika j = 25 cm, OP —– = —a
2
2
a
Luas di bawah air = 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 Panjang lengkok AB berpusat O
π
12 cm —– rad A
a
3 = —– —–
2π
3
(a) h = 6 sin 30° = 3 cm  2 2
AB 3
3
AB 3
2πa
sin 60° = —– O θ rad 2 θ rad = –—–– × –––
BC h cm T Q
BC = 3.46 cm (a) (i) OP = 2h kos θ 3AB 3 AB 3
2AB 3πa
BE = 1.73 cm (ii) TQ = 2h kos θ – h = —–—–
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10A_1202 BS MT Tg5.indd 101 10/01/2022 4:42 PM