Page 13 - Modul A+1 Matematik Tambahan Tingkatan 5

(b) (c)
B B

C
7 cm C 8.2 cm O A
BAB 1 ∠BOA = rad
π
3
1
O θ 6 cm A Luas semibulatan = (8.2) π
2
tan q = 7 2
6 = 105.62 cm 2
q = 49.4°
( )
1
1
Luas ∆OAB = (6)(7) Luas sektor AOB = (8.2) 2 π
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2 2 3
= 21 cm 2 = 35.21 cm 2
(
1
Luas sektor = (6) 49.4° × π )
2
2 180° Luas kawasan berlorek
= 15.52 cm 2 = 105.62 – 35.21
Luas kawasan berlorek = 70.41 cm 2
= 21 – 15.52
= 5.48 cm 2
4. Cari luas tembereng bagi setiap yang berikut.
Find the area of segment for each of the following. TP 5
Contoh/Example
Tip SPM
Luas tembereng
Area of the segment Luas segi tiga/Area of triangle = 1 AB × h
= Luas sektor AOB – Luas segi tiga AOB 2
O 4 cm Area of segment AOB – Area of triangle AOB sin q = AC
B 2 j
1 2
1 2
1.4 rad h = j q – j sin q° º AC = j sin q , AB = 2j sin q
2
2
C 2 2
Jika q = 1.4 rad dan j = 4 cm, luas tembereng q q h
/
If q = 1.4 rad and j = 4 cm, the area of the kos cos =
A 2 2 j
segment h = j kos q j cos q
/
(
1
1
Tip SPM = (4) (1.4) – (4) sin 1.4 × 180° ) 2 2
2
2
2 2 π Luas segi tiga/Area of triangle
= 3.32 cm 2 q q q q
1
q hendaklah dalam darjah = 1 ( 2j sin kos )/ ( 2j sin cos )
2

2
untuk trigonometri sinus. 2 2 2 2 2 2
q must be in degrees for = 1 2
j sin q
trigonometric sine. 2
(a) (b)
O Q
1.1 rad
8 cm —
π
3
O 9 cm P
P Q π 2π
∠POQ = π – = rad
3 3
1.1 rad = 1.1 × 180° = 63.02° Luas tembereng
π
( )
1
1
1
2
Luas sektor = (8) (1.1) = 35.2 cm 2 = (9) 2 2π – (9) sin ( 2π × 180° )
2
2 2 3 2 3 π
1
2
Luas segi tiga = (8) sin 63.02° = 49.75 cm 2
2
= 28.52 cm 2
Luas tembereng
= 35.2 – 28.52
= 6.68 cm 2
10
01 Modul Series MateTam Tg5.indd 10 04/10/2021 2:15 PM