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Answer :
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