Mensuration-36

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ABCP is a quadrant of a circle of radius 14 cm. With AC as diameter, semicircle is drawn. Find the area of the shaded portion APCQA .

Solution: In right angle triangle ABC,

AC2 = AB2 + BC2 (Pythagoras theorem)

=> AC2 = 142 + 142 = 2 x 142

=> AC = √(2 x 14) = 14 √2

=> AC / 2 = 14 √2 / 2 cm = 7 √2 cm

Area of the portion between the quadrant and the semi-circle

= (Area of region ABCQA) - Area of quadrant ABCPA

Area of region ABCQA = Area of triangle ABC + Area of semicircle ACQA

= (1/2 x 14 x 14 + 1/2 x 22/7 x 7 √2 x 7 √2) cm2 = (98 + 154) cm2

= 252 cm2

Area of quadrant ABCPA = (1/4 x 22/7 x 14 x 14) cm2 = 154 cm2

Therefore, Area of shaded region APCQA = 252 cm2 - 154 cm2 = 98 cm2

image:tri36.png

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