Page 24 - Module & More Matematik Tambahan Tg5

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

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Lebar kad / Breadth of card (c) Luas sektor BAC = × (12) × 1.23
2
= j + 5 10 Area of sector BAC 2
15 10 = 88.56 cm 2
= + 5 10
2 Luas tembereng BAD
= 39.53 cm Area of segment BAD 1
≈ 40 cm 1 2 1 2
(b) Luas kad tidak digunakan = 2 (18) (0.68) − (18) sin38.94°
2
Area of unused card = 8.34 cm 2

= 56 × 40 −  39.53 2 × 1.99 + Jumlah luas berlorek = 8.34 + 88.56 BAB
2
2
Total shaded area
= 96.9 cm
1 15 10 2 (1.99)
2  2 
= 1244.88 cm 2
4. Rajah menunjukkan sebuah bulatan dan sektor
berpusat O. Diberi panjang lengkok PS dan QR
2017
masing-masing ialah 3 cm dan 8 cm, dan SR =
3. Rajah menunjukkan dua sektor, OAB berpusat O 9.5 cm.
dan BAC berpusat A. Diberi OA = 18 cm dan AB = The diagram shows a circle and a sector with centre
2016
12 cm. O. Given the arc length PS and QR are 3 cm and 8 cm
The diagram shows two sectors, OAB with centre O and respectively, and SR = 9.5 cm.
BAC with centre A. Given that OA = 18 cm and AB =
12 cm. Q
P
B A
O
S
R
C
O Cari/Find
(a) jejari bulatan dan sudut q dalam radian,
Cari/Find the radius and the angle q in radian,
(a) ∠BAC dalam radian, (b) luas rantau berlorek itu.
∠BAC in radian, the area of the shaded region.
(b) perimeter rantau berlorek.
the perimeter of the shaded region. (a) 3 = jq …. ➀
(c) luas rantau berlorek itu. 8 = (j + 9.5)q …. ➁
the area of the shaded region. ➁ ÷ ➀ Q
8 j + 9.5 P
6 = j
(a) kos ∠A = B 3 j 3 cm 8 cm
18 12 cm A O
∠A = 70.53° 5j = 9.5 × 3 S
= 1.23 rad j = 5.7 cm 9.5 cm R
(b) Panjang lengkok BC C q = 3 = 0.53 rad
Arc length BC 5.7
= 12 × 1.23 O = 30.37°
= 14.77 cm (b) Luas rantau berlorek = Luas ∆OQR − Luas
∠BOA = 180° − 2(70.53°) sektor POS
= 38.94° = 0.68 rad Area of shaded region = Area OQR – Area of sector
Panjang lengkok BA = 18 × 0.68 1 POS 1
2
Arc length BA = 12.23 cm = 2 (9.5+5.7) sin30.37° − (5.7) (0.53)
2
2
Perimeter = 12.23 + 14.77 + 12 = 49.79 cm 2
= 39 cm

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