Page 15 - Modul A+1 Matematik Tambahan Tingkatan 5

1.4 Aplikasi Sukatan Membulat Buku Teks
Application of Circular Measures m.s. 20-22

1. Selesaikan setiap masalah yang berikut.
Solve each of the following problems. TP 6
BAB 1 Contoh/Example (a) Rajah menunjukkan sebuah sektor dengan pusat O

dan berjejari 9.3 cm. Diberi bahawa Q terletak pada
Rajah menunjukkan sekeping jubin berbentuk segi lilitan bulatan.
empat sama ABCD. Diberi bahawa panjang sisi AB The diagram shows a sector with centre O and a radius of
ialah 14 cm. Keempat-empat kawasan berlorek itu 9.3 cm. Given that Q lies on the circumference.
adalah serupa. OPQ dan BPQ adalah sektor masing-
masing dengan pusat O dan B.
©PAN ASIA PUBLICATIONS
The diagram shows a square tile ABCD. Given that the length Q R
of side AB is 14 cm. The 4 shaded regions are identical. OPQ O
and BPQ are sectors with centres O and B respectively. 140°
48°
A P B
P
Cari
S O Q Find
(i) perimeter kawasan berlorek,
the perimeter of the shaded region,
(ii) luas kawasan berlorek.
D C
R the area of the shaded region.
Jika terdapat 150 keping jubin yang sama pada suatu
lantai, cari luas yang dilitupi oleh kawasan berlorek itu. R
If there are 150 same tiles on the floor, find the area covered Q 22°
by the shaded regions. 48° O
140°
P
48°
H
P
(i) ∠POQ = 180° – 2(48°)
= 84°
O Q
Luas sektor OPQ Panjang perentas PQ
Area of the sector OPQ = ! 9.3 + 9.3 – 2(9.3) kos 84°
2
2
2
1
= × π(7) 2 = 12.45 cm
4
Luas ∆OPQ Panjang lengkok PQ
p
Area of ∆OPQ = 9.3 × 84° × 180°
1
= × 7 × 7 = 13.63 cm
2
1
= (7) 2 ∠QOR = 360° – 84° – 140°
2 = 136°
Luas tembereng PHQ Panjang perentas QR
Area of the segment PHQ = ! 9.3 + 9.3 – 2(9.3) kos 136°
2
2
2
1
1
= × π(7) – (7) 2 = 17.25 cm
2
4 2
Maka, luas kawasan berlorek setiap jubin Panjang lengkok QR
p
Hence, the area of the shaded region of each tile = 9.3 × 136° × 180°
1
= 8 [ 1 × π(7) – (7) 2 ] = 22.07 cm
2
4
2
= 8(13.98) Maka, perimeter kawasan berlorek
= 111.88 cm 2 = 12.45 + 17.25 + 13.63 + 22.07
= 65.4 cm
Jumlah luas kawasan berlorek
Total area of the shaded regions (ii) Luas kawasan berlorek
]
1
= 111.88 × 150 = [ 1 × 9.3 × 84° × p – (9.3) sin 84°
2
2
= 1.6782 × 10 cm 2 2 180° 2
4
p
1
+ [ 1 × 9.3 × 136° × 180° – (9.3) sin 136° ]
2
2
2
2
= 93 cm 2
12
01 Modul Series MateTam Tg5.indd 12 04/10/2021 2:15 PM