Q2 of 26 Page 242

A chord of a circle of radius 12 cm. subtends an angle of 120o at the centre. Find the area of the corresponding minor segment of the circle

(use π = 3.14 and √3 = 1.732)


Let Major segment A1 minor segment be A2







area of sector ACD = 150.72 sq.cm


Sin 30° =



AE = 6cm


Cos 30° =



DE = 1.732 × 6


DE = 10.392cm


CD = 2 × 10.392 = 20.784cm


Area of minor segment = Area of sector ACD – Area ΔACD


= 150.72 –20.784


= 1550.72-62.352


= 88.44 cm2


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9

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Hint : The distance of two points to the point of contact is the same.

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(use π = 3.14)


i. Minor segment


ii. Major segment

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(use π = 22/7)

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(use π = 3.14)