Page 32 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

1.4 Aplikasi Sukatan Membulat Buku Teks
PBD
PBD Application of Circular Measures ms. 20 – 27
PBD
13. Selesaikan masalah yang berikut. SP 1.4.1 TP5
Solve the following problems.

(a)
A B
P

A O B

Q
D C
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Rajah menunjukkan sekeping jubin bersegi empat Rajah menunjukkan satu corak yang dibina
sama dengan sisi 20 cm. A dan C masing-masing daripada dua sektor serupa menyentuh satu
ialah pusat bagi sektor BAD dan BCD. Cari sama lain di O dan dengan pusat masing-
The diagram shows a piece of square tile with side of 20 cm. A and masing A dan B. Diberi panjang perentas PQ
C are the centres of the sectors BAD and BCD respectively. Find ialah 16 cm dan AB = 24 cm. Cari
(i) luas rantau berlorek. The diagram shows a pattern made up of two similar sectors
the area of the shaded region. with centres A and B respectively and touch each other at O.
(ii) jumlah luas, dalam m , diliputi oleh rantau Given that the chord PQ is 16 cm and AB = 24 cm. Find
2
berlorek jika dimensi bilik ialah 6 m × 5 m (i) /PAQ dalam radian.
the total area, in m covered by the shaded region if the /PAQ in radians.
2
dimension of the room is 6 m × 5 m. (ii) perimeter rantau berlorek.
the perimter of the shaded region.
(iii) luas rantau berlorek.
Tip Penting
the area of the shaded region.
Luas rantau berlorek/Area of the shaded region
= 2(luas sektor BAD – luas segi tiga BAD) P
2(area of sector BAD – area of triangle BAD)
12
(i) Luas rantau berlorek θ O 12 cm
Area of the shaded region A B
1
2 π
4
2
= 2 3 1 (20 )  2 – (20 ) sin 90°
2
2
2
= 228.32 cm 2 Q
(i) sin q = 8
(ii) Jumlah jubin diperlukan = 30 × 25 = 750 12
Total tiles needed = 41.81° = 0.73 rad
Jumlah luas diliput oleh rantau berlorek
Total area covered by shaded region ∠PAQ = 1.46 rad
= 228.32 × 750 = 171 240 cm 2 (ii) Perimeter rantau berlorek
= 17.124 m 2 Perimeter of the shaded region
= 2[16 + 12(1.46)]
= 67.04 cm

(iii) Luas rantau berlorek
Area of shaded region
1
= 2 3 1 (12) (1.46) – (12) sin 83.62° 4
2
2
2
2
= 67.13 cm 2

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