Page 26 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat
7. Selesaikan masalah yang melibatkan panjang lengkok. SP 1.2.3 TP5
Solve the problems involving the arc lengths.

A
Rajah menunjukkan sebuah sektor AOB berpusat O. Diberi AC = CO = 3.4 cm dan
/ACO = 1.5 rad. Cari perimeter rantau berlorek itu.
The diagram shows a sector AOB with centre O. Given that AC = CO = 3.4 cm and /ACO = 1.5 rad. Find the
perimeter of the shaded region.
O
B C
/ACO = 1.5 rad = 85.94° Panjang lengkok/Arc length = 4.63 × 0.82
π – 1.5 = 3.8 cm
/AOB = = 0.82 rad
2 Perimeter rantau berlorek = 3.4 + 3.8 + (4.63 –3.4)
Jejari/Radius Perimeter of the shaded region = 8.43 cm
AO = √
3.4 + 3.4 – 2(3.4) kos 85.94
2
2
2
= 4.63 cm
(a) A
B Rajah menunjukkan dua sektor, AOD dan BOC berpusat O. Diberi 2OA = 3OB, OA = 9.6
cm dan /AOD = 2.2 rad. Cari perimeter rantau berlorek itu.
The diagram shows two sectors, AOD and BOC with centre O. Given that 2OA = 3OB , OA = 9.6 cm and /AOD
O = 2.2 rad. Find the perimeter of the shaded region.
C
D
Panjang lengkok BC = 6.4(2.2) = 14.08 cm
2
OB = (9.6) = 6.4 cm Arc length
3
Panjang lengkok AD = 9.6(2.2) = 21.12 cm Perimeter rantau berlorek = 21.12 + 14.08 + 2(9.6 – 6.4)
Arc length Perimeter of the shaded region = 41.6 cm

(b) Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
P Rajah menunjukkan satu sektor dengan pusat O. Sebuah bulatan dicangkum di
B 7
dalamnya dengan pusat P. Diberi jejari bulatan ialah 4 cm dan /APB = π rad. Cari
A 8
The diagram shows a sector with centre O. A circle with centre P is inscribed in it. Given that the radius of the
O 7
circle is 4 cm and /APB = π rad, find
8
(i) jejari sektor. (ii) luas rantau berlorek.
the radius of the sector. the area of the shaded region.

7 1 4
(i) /OPB = π × kos / cos 78°459 =
8 2 OP
= 7 π rad OP = 4 = 20.5 cm P 4 B
16 kos 78°459 7 _ π rad
= 78° 459 A 8
\ Jejari sektor = 20.5 + 4
Radius of the sector = 24.5 cm O
π
7
(ii) /POB = – 7 π /APB = 2π – π
2 16 8
= π = 9π
16 8
= 11.25°
\ Luas rantau berlorek / Area of the shaded region

2 9π
4
1
= 22.5° × π(24.5) – 2 3 1 (4)(20.5) sin 78°459 – (4)  2
2
360° 2 2 8
= 9.16 cm 2
5 © Penerbitan Pelangi Sdn. Bhd.

01 Hybrid PBD Mate Tamb Tg5.indd 5 09/11/2021 9:20 AM