Page 29 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

(b) Diberi luas sektor berlorek ialah 7.5 cm . (c) P
2
Given the area of the shaded sector is 7.5 cm . Diberi luas sektor berlorek ialah
2
P 30 cm dan panjang lengkok
2
O 18 cm
___
5π major PQ ialah
6 18 cm.
O Q Q Given the area of the shaded sector is 30
cm and the major arc length is 18 cm.
2
5π
 2
2
7.5 = 2 6 18 = πj – 30 Tip Penting
πj 2 2π 2πj πj 2 Guna/ Use
2
5π j = 2(πj – 30) panjang lengkok = luas sektor
7.5 = 3 2 18 lilitan bulatan luas bulatan
arc length
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πj 2 2π 2 πj – 30 = 9j circum ference = area of sector
area of circle
7.5 × 2 × 3 πj – 9j – 30 = 0
j = 2
2
(–9) – 4π(–30)
5π j = 9 √
j = 1.69 cm 2π
j = 4.84 cm
10. Tentukan sudut tercangkum, q dalam radian, di pusat bulatan bagi setiap yang berikut. Beri jawapan betul
kepada dua tempat perpuluhan jika perlu. SP 1.3.1 TP4
Determine the subtended angle, q in radians, at the centre of the circle for each of the following. Give your answer correct to two decimal
places where necessary.

Diberi luas sektor berlorek (a) Diberi luas sektor berlorek
4.8 cm ialah 23 cm 2 ialah 15.7 cm dan berjejari
2
Given the area of the shaded sector 4.2 cm.
O θ O θ Given the area of the shaded sector is
is 23 cm . 2
15.7 cm and with radius 4.2 cm.
2
23 = q 15.7 = 2π – q
π(4.8) 2 2π π(4.2) 2 2π
q = 23 × 2 q = 2π – 15.7 × 2
(4.8) 2 (4.2) 2
= 1.99 rad = 4.50 rad

(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.
2
2
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
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
2
j q = 40
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
j – 14j + 40 = 0 (4j – 5)(j – 5) = 0
2
(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
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
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