Page 30 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat
11. Tentukan luas tembereng yang berlorek bagi setiap bulatan yang berikut. Beri jawapan betul kepada dua
tempat perpuluhan. SP 1.3.2 TP5
Determine the area of the shaded segment for each of the following circles. Give your answer correct to two decimal places.


Luas sektor = j 2π – 4π 2 (a)
1 2
P
Q Area of sector 2 3
1
 2
O 8.1 cm = ( 8.1) 2 2π A
4 _ π 2 3 1.67 O
3 rad 5.5 cm
= 68.71 cm 2
2π rad = 120°

3 B
1 2
1
Luas segi tiga = (8.1) sin 120° Luas sektor = j q
2
2
2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Area of the triangle Area of sector 1 2
= 28.41 cm 2 = ( 5.5) (1.67)
2
Maka, luas tembereng berlorek = 68.71 – 28.41 = 25.26 cm 2
Hence, the area of shaded segment 1
= 40.30 cm 2 Luas segi tiga = (5.5) sin 95.68°
2
Area of triangle 2
Tip Penting = 15.05 cm 2
Luas tembereng berlorek/ Area of shaded segment Maka, luas tembereng berlorek = 25.26 – 15.05
= luas sektor POQ – luas segi tiga POQ Hence, the area of the shaded segment
area of sector POQ – area of triangle POQ
= 10.21 cm 2
(b) Diberi POQ dan SOR P (c)
adalah dua sektor dengan T P
pusat O dan /POQ = 2SOR 3 cm
= 1.3 rad. Cari jumlah luas O Q O A
rantau berlorek. S
Given that POQ and SOR are two U Q
sectors with centre O and /POQ = R Diberi /PAQ = 45° dan PA = 6 cm dan OP = 5 cm.
2SOR = 1.3 rad. Find the total area
of shaded region. Cari luas rantau berlorek.
Given /PAQ = 45° and PA = 6 cm and OP = 5 cm. Find the area
1
Luas sektor POQ = j q = (3) (1.3) of shaded region.
1 2
2
Area of sector POQ 2 2
= 5.85 cm 2 Perentas PQ = √
6 + 6 – 2(6) kos 45°
2
2
2
1.3 rad = 1.3 × 360° = 74.48° Chord PQ = 4.59 cm
2π 2 2 2
1
Luas segi tiga POQ = (3) sin 74.48° kos/cos /POQ = 5 + 5 – 4.59
2
2(5)
2
Area of triangle POQ 2
= 4.34 cm 2 /POQ = 54.65°
= 0.954 rad
Maka, luas tembereng PTQ = 5.85 – 4.34
Hence, area of segment PTQ = 1.51cm 2 Luas rantau berlorek
1 2 1.3 Area of shaded region
1 2
Luas sektor SOR = j q = ( 3)  2 1 2 1 2 1 2
Area of sector SOR 2 2 2 = 2 (6) (0.785) – (6 )sin 45° + (5 )(0.954)
2
2
= 2.93 cm 2 1 2
1.3 rad = 0.65 × 360° = 37.24° – ( 5) sin 54.65°
2
2 2p = 3.13 cm 2
1
Luas segi tiga SOR = (3) sin 37.24°
2
Area of triangle SOR 2
= 2.72 cm 2
Maka, luas tembereng SUR = 2.93 – 2.72
Hence, area of segment SUR = 0.21cm 2
Jumlah luas rantau berlorek = 1.51 + 0.21
Total area of the shaded region = 1.72 cm
2
9 © Penerbitan Pelangi Sdn. Bhd.

01 Hybrid PBD Mate Tamb Tg5.indd 9 09/11/2021 9:20 AM