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)

Radius (r) of circle = 15 cm


Area of sector OPRQ = (60/360)×Πr2


 


=  ×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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