A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. [Take π = 22/7 and √3 = 1.732.]

∠ACB = 60°

Chord AB subtends an angle of 60° at the centre

Radius = 30 cm

Let Radius be R

In triangle ABC, AC = BC

So, ∠CAB = ∠CBA

∠ ACB + ∠ CAB + ∠ CBA = 180°

60° + 2∠CAB = 180°

2∠CAB = 180° - 60° = 120°

∠CAB = 120°/2 = 60°

∠CAB = ∠CBA = 60°

∴ ΔABC is a equilateral triangle

Length of side of an equilateral triangle = radius of circle = 30 cm

Area of equilateral triangle = √3/4 × side^{2} = 1.732/4 × 30 × 30 cm^{2}

= 389.7 cm^{2}

Area of sector ACB = = 3.14 × 30 × 30 × 60°/360° = 471.45 cm^{2}

Area of minor Segment = Area of sector ACB – Area of ΔABC

= 471.45 cm^{2} – 389.7 cm^{2} = 81.75 cm^{2}

Area of circle = R^{2} = 3.14 × 30 × 30 cm^{2} = 2828.57 cm^{2}

Area of major segment = Area of circle – Area of minor segment

= 2826 cm^{2} – 81.75 cm^{2}

= 2744.25 cm^{2}

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