Page 18 - 1202 Bank Soalan Matematik Tambahan Tingkatan 5

13. O j C (b) Luas tembereng (b) Luas tembereng 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
j 2
= 573.35 cm 2 = 2.94 cm
β Luas segi empat tepat AOB 20.
A = 25(32 sin 57.91°) B
B 5.6 cm Q
1 = 677.77 cm 2 4 cm
2
(a) 20 = —j β Luas = 677.77 – 573.35 A 4 cm
2
40
β = —– = 104.42 cm 2 P j θ
j 2 16. O
(b) Perimeter rantau berlorek (a) jθ = 4
9 cm (j + 4)θ = 5.6
= 2j + jβ 5 cm 4 + 4θ = 5.6
40
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 j 2
= 2j + j —– O A 4θ = 1.6
2

j 2
40
= 2j + —– cm θ = 0.4 rad
4
0.4
14. j = OP = —— = 10 cm
P (a) Lilitan gear kecil (b) ∠POQ = 0.4 rad
= 2π(5) (c) Luas ABQP
j 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) Lilitan gear besar 21. O 6 cm
= 2π(9) P Q
(a) jθ = 5 = 18π cm 6 cm 6 cm
5 Bilangan putaran 9 cm
j = — cm 18π S R
θ = ——
10π
6
(b) OT = — = 1.8 putaran T
θ
4.5
∠SOR
5
6
θ
kos — = — ÷ — 17. (a) sin ——— = —–
2 θ θ 2 6
5 θ θ j cm ∠SOR
= — × — ——— = 48.59°
θ 6 2
5
= — ∠SOR = 97.18°
6 ∠SOR = 1.7 rad
θ = 67.11° (b) Luas tembereng STR
1
1
1
θ = 1.17 rad —j θ = 5 ................. 1 = —(6) (1.7) – —(6) sin 97.18°
2
2
2
2
Luas ∆POT 2j + jθ = 9 ................. 2 2 2 2
1
6
2  θ 2
= —j — sin 33.56° jθ = 9 – 2j .......... 3 = 12.74 cm
1 5 6 Daripada 1, 22. M 6 cm O N
2  θ 2 θ 2
= — — — sin 33.56° 1 —– rad 60°
π
2
1 5 6 —j(9 – 2j) = 5 3 120°
2  1.172 1.172
= — —— —— sin 33.56° 9j – 2j = 10
2
2
= 6.06 cm 2 2j – 9j + 10 = 0 60°
(2j – 5)(j – 2) = 0
Luas rantau berlorek 5 P Q
1
2
= 2(6.06) – —j θ 1 5 2 j = — atau j = 2 (a) Panjang lengkok PQ
2
2
1
2  2 2
1
5
2
2
2  1.172
= 2(6.06) – — —— — — θ = 5 atau —(2) θ = 5 = π(6) – Panjang lengkok MP
2
– Panjang lengkok QN
8
5
= 1.44 cm 2 θ = — rad θ = — rad = 6π – 6 — – 6 —
π
 2  2
π
5
2
Bahagian B 18. AC = 2j sin θ = 2π cm 3 3
15. Luas (b) Perimeter rantau berlorek
π 1 1 π π
 2  2
2
2
2
= —(j sin θ) – —j (2θ) – —j sin 2θ 4 = 12 + 6 — + 6 — + 2(6 kos 60°)
2 3 2 2 3 3
O 1 2 2 1
 2 2
2
32 cm = —j (π sin θ – 2θ + sin 2θ) = 12 + 4π + 12 —
19. = 18 + 4π
A B
D = 30.57 cm
25 cm
10 cm
O Kertas 2
(a) OD = 32 + 10 – 25 θ 3 cm Bahagian A
= 17 cm 1. (a) Lilitan = 2πj
∠AOB P 4.4 cm Q
32 kos ——— = 17 (2π – 1.2)(25) = 2πj
2
(2π – 1.2)(25)
2.2
θ
∠AOB (a) sin — = —— j = ——————
——— = 57.91° 2π
2 2 3
θ
∠AOB = 115.82° — = 47° 10’ j = 2.02 cm
∠AOB = 2.02 rad 2
θ = 94.33°
θ = 1.65 rad
100
10A_1202 BS MT Tg5.indd 100 10/01/2022 4:42 PM