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# 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