Q. 64.0( 368 Votes )

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle

(Use π = 3.14 and = 1.73)

Answer :


Radius (r) of circle = 15 cm

Area of sector OPRQ = (60/360)×Πr2

= 1/6 × 3.14 × (15)2

= 117.75 cm2

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2 ∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.

Area of ΔOPQ = ×(Side)2

=  × (15)2

=

= 56.25 × 

= 97.3125 cm2

Area of segment PRQ = Area of sector OPRQ - Area of ΔOPQ

= 117.75 - 97.3125

= 20.4375 cm2

Area of major segment PSQ = Area of circle - Area of segment PRQ

= Π(15)2 – 20.4375

= 3.14 ×225 – 20.4375

= 706.5 – 20.4375

= 686.0625 cm2

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