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. TP 4
Determine the area of the shaded segment for each of the following circles. Give your answer to two decimal places.
1
(a)
CONTOH
BAB
P
Q
5.3 cm Q 8.1 cm O
O
2.3 rad 4 rad
3
P
Penyelesaian: Luas berlorek/Shaded area
Luas berlorek = luas sektor POQ – luas segi tiga POQ
= luas sektor POQ – luas segi tiga POQ
Shaded area = area of sector POQ − area of triangle POQ area of sector POQ – area of triangle POQ
Luas sektor/Area of sector

1 1 Luas sektor = j 2π − 4π 
1 2
= j q = (5.3) (2.3) = 32.3 cm 2 2 3
2
2
2 2 Area of sector
 
2.3 = 1 (8.1) 2 2π
2.3 rad = × 360° = 131.78° 2 3
2π
= 68.71 cm 2
Luas segi tiga/Area of triangle 2π
1 rad = 120°
= ×(5.3) × sin131.78° = 10.47 cm 2 3
2
2 Luas segi tiga = × (8.1) × sin 120°
1
2
Maka, luas tembereng Area of triangle 2
Hence, the area of the segment = 28.41 cm 2
Maka, luas tembereng = 68.71 − 28.41
= 32.3 − 10.47 Penerbitan Pelangi Sdn. Bhd.
= 21.83 cm 2 Hence, the area of segment = 40.3 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
2
4.3 + 4.3 − 4.65 2
2
Luas sektor POS = luas sektor QOR Kos/ cos ∠AOB = 2(4.3) 2
area of sector POS = area of sector QOR ∠AOB = 65.46°
1 j q = 1 (3.5) (π − 1.4) = 1.14 rad
2
2
2 2 50° = 0.87 rad
= 10.67 cm 2
1.4 Luas tembereng = 1 (5.5) (0.87) − 1 (5.5) 2
2
1.4 rad = × 360° 2 2
2π Area of segment 1
2
= 80.21° sin50° + 2 (4.3) (1.14)−
2
Luas segi tiga = 1 × (3.5) × sin80.21° 1 2
Area of triangle 2 2 (4.3) sin65.46°
= 6.04 cm 2 2
Maka, luas tembereng = 2(10.67 − 6.04) = 3.7 cm
Hence, the area of segment = 9.26 cm 2
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