Question

9. The diameter of a circle centre O is 26cm. If a chord PQ is drawn such that it is 5cm from O to the centre of the chord, calculate, correct to the nearest whole number:
(a) ∠POQ;
(b) area of the minor segment formed by the chord PQ. [Take π=22/7]

Accepted Answer

Radius r=13cm (half of diameter 26cm). Perpendicular distance from O to chord PQ = 5cm.
Half-chord length = √(r²−d²) = √(169−25) = √144 = 12cm, so PQ=24cm.

(a) In the right triangle formed by O, the midpoint of PQ, and P: cos(∠POQ/2) = 5/13 → ∠POQ/2 = cos⁻¹(5/13) = 67.38° → ∠POQ = 134.76° ≈ 135° (to the nearest whole number).

(b) Area of sector POQ = (θ/360)×πr² = (134.76/360)×(22/7)×169 ≈ 0.3743×531.14 ≈ 198.8cm².
Area of triangle POQ = (1/2)r²sinθ = 0.5×169×sin(134.76°) = 84.5×0.7106 ≈ 60.1cm².
Area of minor segment = Sector − Triangle = 198.8−60.1 ≈ 138.7 ≈ 139cm² (to the nearest whole number).

NABTEB Mathematics 2024 Theory — Question 9

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