Question

In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region (in $c{m}^2$).

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Answer 1

Area of quadrant = $\frac{90}{360}\times \pi \times 14\times 14$ = 153.94 $c{m}^2$

Area of $△ABC$ = 0.5 x ${14}^2$ = 98 $c{m}^2$

The diameter = 14 × $\sqrt{2}$ = 19.8 cm.

Area of semi-circle = 0.5 $\pi \times {9.9}^2$ = 153.95 $c{m}^2$

The area of segment = 153.94 - 98 = 55.94$c{m}^2$

The required area = 153.95 - 55.94 = 98.01$c{m}^2$.