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.

Let 14 cm be the radius of the quadrant with A as the centre.

Then the area of the quadrant ABC

\(= \ \frac{1}{4} \ \pi r^2 \ = \ \frac{1}{4} \ × \ \frac{22}{7} \ × \ 196 \)

\(= \ 154 \)cm²

Area of \(\triangle \) ABC

\( = \ \frac{1}{2} \ × \ AC \ × \ AB \)

\( = \ \frac{1}{2} \ × \ 14 \ × \ 14 \ = \ 98 \) cm²

Now, since AC = AB = 14 cm and \( ∠ \ BAC \ = \ 90º \)

\(\therefore \) By Pyrthagoras Theorem,

\( BC \ = \ \sqrt{AC^2 \ + \ AB^2} \ \)
\( = \ \sqrt{14^2 \ + \ 14^2} \ \)
\( = \ 14 \sqrt{2} \)

\(\therefore \) Radius of the semicircle BNC \( = \ 7\sqrt{2} \)cm

Area of the semicircle BNC \( = \ \frac{ \pi}{2} (7 \sqrt{2})^2 \)

\(= \ \frac{1}{2} \ × \ \frac{22}{7} \ × \ 98 \ = \ 154 \) cm²

\(\therefore \) The area of the shaded region

= The area of semicircle with diameter BC - ( Area of the quadrant ABC - Area of ∆ ABC)

\(= \ 154 \ - \ ( 154 \ - \ 98) = \ 98 \) cm²