Page 21 - PBD Plus Matematik Tambahan T5 (EG)

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

(a) jejari, j Jawapan / Answer :
the radius, j
[2 markah / 2 marks] (a) 0.75 rad B
(b) luas yang berlorek. = 0.75 × 360°
the shaded area. 2π 0.75 rad
[3 markah / 3 marks] = 42.97° O C A
Jawapan / Answer : /OCB = 94.06° 4 cm
Q (a) s = jq
P 8 = j OB = 4
θ sin 94.06° sin 42.97°
O j R j = 8 OB = 5.85 cm
q
(b) panjang lengkok = 5.85 (0.75)
arc length = 4.39 cm
(b) OQ = 10 Perimeter rantau berlorek = 4 + 4.39 + (5.85 – 4)
q
Perimeter of the shaded region = 10.24 cm

8
10
2
2
QR = q 2 – q 2
Penerbitan Pelangi Sdn. Bhd.
36
=  = 6 6. Rajah menunjukkan sebuah bulatan berpusat O
Mengaplikasi
dan berjejari 9 cm. SP 1.2.3 KBAT
2
q
q
The diagram shows a circle with centre O and a radius of 9 cm.
 2 2
Luas ∆ORQ = 1 8 6 = 24 C
Area 2 q q q 2 A B
Luas OPQR = 48 2 cm 2
Area q O
 2
Luas sektor OPR = 1 8 2 q
Area of sector 2 q AOB ialah suatu lengkok bagi satu bulatan
= 32 cm 2 berpusat C yang terletak pada lilitan bulatan. Cari.
AOB is an arc of a circle with centre C which is on the
q circumference of the circle. Find
\ Luas berlorek = 48 – 32 (a) /AOB, dalam radian
2
Shaded area q q /AOB, in radians.
3
= 16 3 – 2 4 [2 markah / 2 marks]
q q (b) perimeter bahagian berlorek.
= 16(3 – 2q) cm 2 the perimeter of the shaded region.
q [3 markah / 3 marks]
5. Rajah menunjukkan sebuah sektor BOA berpusat TIP Menjawab
O. SP 1.2.3 KBAT Mengaplikasi Perhatikan CO berserenjang dengan AB dan AOBC ialah
sebuah rombus. Gunakan petua kosinus untuk mencari
The diagram shows a sector BOA with centre O.
panjang perentas.
B It is observed that CO is perpendicular with AB and AOBC is a
rhombus. Use cosine rule to find the length of chord.
0.75 rad
O C A Jawapan / Answer :
Diberi OC = CB = 4 cm, cari (a) AO = OB = CA = CB = CO A B
Given that OC = CB = 4 cm, find Maka ∠AOC = 60° =∠COB θ 9
(a) jejari sektor itu 2π O
the radius of the sector. q = 120° = 3 rad
[4 markah / 4 marks]
(b) perimeter rantau berlorek. (b) panjang perentas AB
the perimeter of the shaded region. length of chord of AB
9 + 9 – 2(9) kos 120°
[2 markah / 2 marks] = √
2
2
2
= 15.59 cm
TIP Menjawab 2p
Gunakan petua sinus. panjang lengkok = 9  2 = 6π
3
arc length
Using sine rule.
\ Perimeter bahagian berlorek = 15.59 + 6π
Perimeter of the shaded region = 34.44 cm 2
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