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° + 2CAB = 180°


2CAB = 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 × side2 = 1.732/4 × 30 × 30 cm2


= 389.7 cm2


Area of sector ACB = = 3.14 × 30 × 30 × 60°/360° = 471.45 cm2


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


= 471.45 cm2 – 389.7 cm2 = 81.75 cm2


Area of circle = R2 = 3.14 × 30 × 30 cm2 = 2828.57 cm2


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


= 2826 cm2 – 81.75 cm2


= 2744.25 cm2


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