ABCP is a quadrant of a circle of radius 14 cm With AC as a diameter, a semicircle is drawn. Find the area of the shaded region. (Take Pi=22/7)

The figure consists of a rt. angled triangle ABC, with AC as the hypotenuse. A semicircle APC is drawn with AC as the diameter. Over this, another the arc of the quadrant extends. The shaded region is between the arc of the semicircle and the arc of the quadrant.

**Given That:** ABCP is a quadrant of a circle of radius 14 cm and ACQA is a semicircle

We are asked to find area of shaded region.

Required Area = Area APCQA

= Area ACQA – Area ACPA

= Area of semicircle ACQA – (Area of quadrant ABCPA – Area of ∆ ABC) …… (1)

Now, in right angled triangle ABC,

Thus, from equation (1),

Hence, area of shaded region is 98 cm^{2}.

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