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**A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the major segment of the circle. (Use π = 3.14 and √3 = 1.73)**

Answer:

Given,

Radius = 15 cm

θ = 60°

So,

Area of sector OAPB = (60°/360°)×πr^{2} cm^{2}

= 225/6 πcm^{2}

Now, ΔAOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60°

So, Area of ΔAOB = (√3/4) ×a^{2}

Or, (√3/4) ×15^{2}

∴ Area of ΔAOB = 97.31 cm^{2}

Now, area of minor segment APB = Area of OAPB – Area of ΔAOB

Or, area of minor segment APB = ((225/6)π – 97.31) cm^{2} = 20.43 cm^{2}

And,

Area of major segment = Area of circle – Area of segment APB

Or, area of major segment = (π×15^{2}) – 20.4 = 686.06 cm^{2}

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