Page 18 - Module & More Matematik Tambahan Tg5

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

11. Tentukan luas tembereng yang berlorek bagi suatu bulatan yang berikut. Berikan jawapan anda kepada dua
tempat perpuluhan.
Determine the area of the shaded segment for each of the following circles. Give your answer to two decimal places. TP 4

CONTOH (a) 1

P
Q
5.3 cm Q 8.1 cm O BAB
O
2.3 rad 4 rad
3
P
Penyelesaian: Luas berlorek
Luas berlorek = luas sektor POQ – luas segi tiga POQ
Shaded area = area of sector POQ − area of triangle POQ = luas sektor POQ – luas segi tiga POQ

Luas sektor/Area of sector Luas sektor = 1 j 2π − 4π 
2
1 1 2 3
= j q = (5.3) (2.3) = 32.3 cm 2 Area of sector 1 2 2π
2
2
2 2 = (8.1) ( )
2.3 2 3
2.3 rad = × 360° = 131.78° 2
2π = 68.71 cm
2π
Luas segi tiga/Area of triangle 3 rad = 120°
1
= ×(5.3) × sin131.78° = 10.47 cm 2 Luas segi tiga = 1 × (8.1) × sin 120°
2
2
2 2
Maka, luas tembereng Area of triangle = 28.41 cm 2
Hence, the area of the segment Maka, luas tembereng = 68.71 − 28.41
= 32.3 − 10.47 Hence, the area of segment = 40.3 cm 2
= 21.83 cm 2
(b) Q (c)
P
1.4 rad O 4.3 cm
O B
A
3.5 cm 50
R 5.5 cm
S P
Diberi POR dan SOQ adalah garis lurus yang
melalui pusat O. 2 2 2
Given that POR and SOQ are straight lines passing Perentas AB = 5.5 + 5.5 − 2(5.5) kos50°
through the centre O. Chord AB = 4.65 cm
4.3 + 4.3 − 4.65 2
2
2
Luas sektor POS = luas sektor QOR Kos/ cos ∠AOB = 2(4.3) 2
1 j q = 1 (3.5) (π − 1.4) ∠AOB = 65.46°
2
2
2 2 = 1.14 rad
= 10.67 cm 2 50° = 0.87 rad
1.4
1.4 rad = × 360° 1 2 1 2
2π Luas tembereng = 2 (5.5) (0.87) − 2 (5.5)
= 80.21° Area of segment sin50° + 1 (4.3) (1.14)−
2
1 2
Luas segi tiga = × (3.5) × sin80.21°
2
2 1 (4.3) sin65.46°
2
= 6.04 cm 2 2
Maka, luas tembereng = 2(10.67 − 6.04) = 3.7 cm 2
= 9.26 cm 2
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