Page 22 - PBD Plus Matematik Tambahan T5 (EG)

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

2. Rajah menunjukkan sebuah bulatan dan sektor
KERTAS 2
berpusat O. SP 1.3.3
1. Rajah menunjukkan dua sektor, The diagram shows a circle and a sector with centre O.
OAB berpusat O dan BAC B A Q
berpusat A. P
The diagram shows two sectors, OAB O θ
with centre O and BAC with centre A. C S
Diberi OA = 18 cm dan AB = 12 cm,
cari O R
Given OA = 18 cm and AB = 12 cm, find
(a) /BAC dalam radian. SP 1.1.1 Diberi panjang lengkok PS dan QR masing-
/BAC in radians. masing ialah 3 cm dan 8 cm dan SR = 9.5 cm. Cari
[2 markah / 2 marks] Given the arc length PS and QR are 3 cm and 8 cm respectively
(b) perimeter rantau berlorek. SP 1.2.3 and SR = 9.5 cm. Find
the perimeter of the shaded region. (a) jejari bulatan dan sudut q dalam radian.
[4 markah / 4 marks] the radius of circle and the angle q in radians.
(c) luas rantau berlorek itu. SP 1.3.2 [3 markah / 3 marks]
the area of the shaded region. (b) luas rantau berlorek itu.
[4 markah / 4 marks] the area of the shaded region.
Jawapan / Answer : Jawapan / Answer : [4 markah / 4 marks]
(a) OA = 18 cm
6
kos/cos ∠BAC = (a) 3 = jq .................a
18
∠BAC = 70.53 8 = (j + 9.5) q .................b
= 1.23 rad
b ÷ a:
(b) panjang lengkok BC = 12(1.23) 8 j + 9.5
arc length = 14.77 cm 3 = j
/BOA = 180° – 2(70.53°) 5j = 9.5 × 3
= 38.94° Penerbitan Pelangi Sdn. Bhd.
j = 5.7 cm
= 0.68 rad q = 3 = 0.53 rad.
panjang lengkok BA = 18(0.68) 5.7
arc length = 12.23 cm = 30.37°
Perimeter rantau berlorek = 12.23 + 14.77 + 12
Perimeter of the shaded region = 39 cm (b) Luas rantau berlorek = Luas ∆OQR – Luas
1
(c) Luas sektor BAC = (12) 1.23 sektor POS
2
Area of sector of BAC 2 Area of the shaded region = Area ∆OQR – Area of sector
= 88.56 cm 2 POS
Luas tembereng BAD
1
1
Area of segment of BAD = (9.5 + 5.7) sin 30.37° – (5.7) (0.53)
2
2
1
1
= (18 )(0.68) – (18) sin 38.94° 2 2
2
2
2 2 = 49.79 cm 2
= 8.34 cm 2
Jumlah luas rantau berlorek
Total area of the shaded region
= 8.34 + 88.56
= 96.9 cm 2

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