Page 25 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat

(c) 11.7 cm (d) (e) Diberi AB ialah diameter.
13.1 cm 2 θ Given that AB is a diameter.
θ
O
O
A 2 θ B
8.9 cm
O
26.8 cm
2πj = 11.7 + 2(8.9) 26.8
π
j = 4.695 cm 2π – 2q = 13.1 2q = × 3
q = 11.7 q = 2.12 rad 5
4.695 q = 3π rad
= 2.492 rad 10
= 0.942 rad
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6. Tentukan perimeter tembereng yang berlorek bagi setiap bulatan berpusat O yang berikut. SP 1.2.2 TP4
Determine the perimeter of the shaded segment of each of the following circles with centre O.

(a)
P P Q
Tip Penting 1.3 rad
Perentas PQ dapat diperolehi dengan 7.5 cm
petua kosinus, iaitu O
0.4π rad PQ = √
j + j – 2j kos q, dengan q
2
2
2
9.6 cm dalam darjah.
Q The chord PQ can be obtained by using the 1.3 rad = 1.3 × 180° = 74.48°
cosine rule, that is π
PQ = √
j + j – 2j cos q, such that q is in
2
2
2
7.5 + 7.5 – 2(7.5) kos 74.48°
2
2
2
degrees. Maka, PQ = √
180° Hence = 9.08 cm
0.4p rad = 0.4p × = 72°
π Panjang lengkok PQ = jq = 7.5(1.3)
Maka, PQ = √ Arc length of PQ
2
2
2
9.6 + 9.6 – 2(9.6) kos 72°
Hence = 11.29 cm = 9.75 cm
Panjang lengkok PQ = jq = 9.6(0.4π) Perimeter tembereng berlorek = 9.08 + 9.75
Arc length of PQ = 12.06 cm Perimeter of the shaded segment = 18.83 cm
Perimeter tembereng berlorek = 12.06 + 11.29
Perimeter of the shaded segment = 23.35 cm
(c) P
(b) P
2.8 cm
7.2 cm
O 5.25 cm
O
4.2 rad
Q
Q
Panjang lengkok minor PQ Panjang lengkok major PQ
Minor arc length of PQ Major arc length of PQ
= (2π – 4.2) × 7.2 = 2π(2.8) – 5.25 = 12.34 cm.
= 15 cm ∠POQ = 5.25
/POQ = 2π – 4.2 = 2.08 rad 2.8
= 119.36° = 1.875 rad 180
7.2 + 7.2 – 2(7.2) kos 119.36°
Maka, PQ = √ = 1.875 × p
2
2
2
Hence = 12.43 cm = 107.43°
2.8 + 2.8 – 2(2.8) kos 107.43°
2
2
2
Perimeter tembereng berlorek = 15 + 12.43 Maka, PQ = √
Perimeter of the shaded segment = 27.43 cm Hence = 4.51 cm
Perimeter tembereng berlorek = 12.34 + 4.51
Perimeter of the shaded segment = 16.85 cm
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01 Hybrid PBD Mate Tamb Tg5.indd 4 09/11/2021 9:20 AM