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(b) (c)
B B

C
7 cm ∠BOA = rad 8.2 cm O A
C
π
SP 1.3.1 & 1.3.2 tan q = 7 O θ 6 cm A Luas semibulatan = (8.2) π 3
3
1
2
2
( )
6
2 π
1
Luas sektor OAB = (8.2)
2
q = 49.4°
1
Luas ∆OAB = (6)(7)
( )
2 Luas rantau berlorek 2 π
1
1
2
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= 21 cm 2 = (8.2) π – (8.2) 3
2
2
(
1
Luas sektor = (6) 2 49.4° × π ) = 70.41 cm 2
2 180°
= 15.52 cm 2
Luas rantau berlorek
= 21 – 15.52
= 5.48 cm 2
4. Cari luas tembereng yang berikut.
Find the area of the following segment. TP 5
Contoh/Example Tip SPM
Luas tembereng Luas segi tiga / Area of triangle = 1 AB × h
Area of a segment 2
AC
=
O j = 4 cm = Luas sektor AOB – Segi tiga AOB sin q Tip SPM
1.4 rad B Area of a segment AOB – Triangle AOB 2 j
θ h º AC = j sin q , AB = 2 j sin q
1 2
1 2
C = j q – j sin q° 2 2
2 2 kos q cos q = h
/
A Jika / If q = 1.4 rad dan / and j = 4 cm, 2 2 j
/
luas tembereng/the area of a segment h = j kos q j cos q
Tip SPM = (4) (1.4) – (4) sin ( 1.4 × 180° ) 2 2
1
1
2
2
2
(
q hendaklah dalam darjah = 3.32 cm 2 2 π Luas segi tiga/Area of triangle 2 q q )
1
q
q
)/ (
1
2
untuk trigonometri sinus. = 2 2j sin 2 kos 2 2 2j sin 2 cos 2
Tip
q must be in degree for
SPM
j sin q
trigonometric sine. = 1 2
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 rantau berlorek
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
Luas segi tiga = (8) sin 63.02° = 49.75 cm 2
2
2
= 28.52 cm 2
Luas rantau berlorek
= 35.2 – 28.52
= 6.68 cm 2
10
01 ModulA+ MateTambahan Tg5.indd 10 10/30/20 10:02 AM