Page 15 - PBD Plus Matematik Tambahan T5 (EG)

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

(b) Diberi luas sektor ialah 40 cm dan perimeternya (c) Diberi luas sektor ialah 6.25 cm dan perimeternya
2
2
ialah 28 cm. Cari nilai yang mungkin bagi jejari ialah 12.5 cm. Cari jejari dan sudut sepadan
sektor dan sudut sepadan tercangkum. tercangkum.
Given the area of the sector is 40 cm and the perimeter is Given the area of the sector is 6.25 cm and the perimeter is 12.5
2
2
28 cm , Find the possible values of the radius of the sector cm. Find the radius and the corresponding subtended angle.
and the corresponding subtended angle.
j + j + jq =12.5
j + j + jq = 28 2j + jq =12.5 dan jq =12.5 – 2j ................... a
2j + jq = 28 dan jq = 28 – 2j ........... a 1 2
j q = 6.25
2
j q = 40
2
1 2 j q = 12.5 .......................... b
2 j(12.5 – 2j) = 12.5
2
j q = 80 .............................. b 2j – 12.5j + 12.5 = 0
2
j(28 – 2j) = 80 4j – 25j + 25 = 0
2
2
j – 14j + 40 = 0 (4j – 5)(j – 5) = 0
(j – 4)(j – 10) = 0 5
j = 4 atau j = 10 cm j = atau j = 5 cm
4
q = 5 rad atau 0.8 q = 8 rad (diabaikan) atau 0.5 rad
2π Penerbitan Pelangi Sdn. Bhd.
Ukuran sudut pada pusat tidak boleh melebihi
360° = 2p rad = 6.28 rad
The measurement of a central angle cannot be greater than
360° = 2p rad = 6.28 rad
Maka, j = 5 cm dan q = 0.5 rad

SP 1.3.2 Menentukan luas tembereng suatu bulatan

11. Tentukan luas tembereng yang berlorek bagi setiap bulatan yang berikut. Beri jawapan betul kepada dua
tempat perpuluhan. TP 5
Determine the area of the shaded segment for each of the following circles. Give your answer correct to two decimal places.
Contoh

Luas sektor = j 2π – 4π 2 (a)
1 2
P 2 3
Q Area of sector 1
 2
O 8.1 cm = ( 8.1) 2 2π A
4 _ π 2 3 1.67 O
3 = 68.71 cm 2 rad 5.5 cm
rad = 120°
3 B
1 2
1
Luas segi tiga = (8.1) sin 120° Luas sektor = j q
2
2
2
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
= 15.05 cm 2
Tip Penting Maka, luas tembereng berlorek = 25.26 – 15.05
Luas tembereng berlorek/ Area of shaded segment Hence, the area of the shaded segment
= luas sektor POQ – luas segi tiga POQ = 10.21 cm 2
area of sector POQ – area of triangle POQ

© Penerbitan Pelangi Sdn. Bhd. 8 TAHAP PENGUASAAN 1 2 3 4 5 6