Page 11 - Module & More Matematik Tambahan TG5

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

(d) (e) R
Q =
P
=
1 5.3 cm
Q O
O rad P
BAB 12 1.4 rad
7.3 cm

∠POQ = 150° Panjang lengkok PQR = 2π(5.3) − 5.3(1.4π)
Panjang jejari/Arc length, j Arc length PQR = 10 cm
(7.3 × 360°) ∠POQ = (2π − 1.4π) ÷ 2
= = 1.99 cm
(210 × 2π) = 0.3π
Maka/Hence, = 54°
Penerbitan Pelangi Sdn. Bhd.
2
PQ = 1.99 + 1.99 − 2(1.99) kos150° Panjang perentas PQ
2
2
= 3.84 cm Length of chord PQ
2
2
2
150° = 5.3 + 5.3 − 2(5.3) kos 54°
Panjang lengkok = × 2π(1.99) = 4.81 cm
Arc length 360° Perimeter = (4.81 × 2 + 10) cm
= 5.2 cm = 19.62 cm
Perimeter = 5.2 cm + 3.84 cm
= 9.04 cm
7. Selesaikan masalah yang melibatkan panjang lengkok. TP 4
Solve the problems involving the arc lengths.
(a)
CONTOH A

4.7 cm 9.2 cm
P Q
A B O 0.7 rad D B
O

Rajah di atas menunjukkan sebuah sektor
AOB berpusat O dan mempunyai jejari 9.2 cm.
Rajah di atas menunjukkan dua bulatan berpusat Cari perimeter rantau berlorek itu.
O dan masing-masing mempunyai jejari 8 cm dan The diagram above shows a sector AOB with centre
14 cm. Cari perimeter rantau berlorek itu. O and radius 9.2 cm. Find the perimeter of the
The diagram above shows two circles with centre O shaded region.
and with radii 8 cm and 14 cm respectively. Find the 0.7 × 360°
perimeter of the shaded region. 0.7 rad = 2π = 40.1°
Penyelesaian: sin 40.1° = AD
s 4.7 9.2
∠POQ = = AD = 5.93 cm
j 14
2
= 0.336 rad OD = 9.2 − 5.93 2
Panjang lengkok AB = 7.03cm
Arc length AB Panjang lengkok = 9.2 × 0.7
= 8(0.336) = 2.69 cm Arc length = 6.44 cm
Perimeter = 2.69 + 6(2) + 4.7 Perimeter = 5.93 + 6.44 + (9.2 – 7.03)
= 19.39 cm = 14.54 cm

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01 M&M Mate T Tg5.indd 8 09/12/2022 12:20 PM