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In figure AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Area of the sector $$= \ \frac{90}{360} \ × \ \frac{22}{7} \ × \ 7 \ × \ 7 \$$
$$=\ \frac{77}{2}$$ cm2

Area of $$∆ \ OCB \ = \ \frac{1}{2} \ × \ OC \ × \ OB \$$
$$= \ \frac{1}{2} \ ×\ 7 \ × \ 7 \$$
$$= \ \frac{49}{2}$$cm2

$$\therefore$$ The area of the segment BPC

$$= \ \frac{77}{2} \ - \ \frac{49}{2} \ = \ \frac{28}{2} \ = \ 14$$cm2

Similarly the area of the segment AQC = 14 cm2

Also, the area of the circle with DO as diameter

$$= \ \frac{22}{7} \ × \ \frac{7}{2} \ × \ \frac{7}{2} \ = \ \frac{77}{2}$$ cm2

$$\therefore$$ The total area of the shaded region

$$= \ 14 \ + \ 14 \ + \ \frac{77}{2}$$

$$= \ \frac{28 \ + \ 28 \ + \ 77}{2} \ = \ \frac{133}{2} \ = \ 66.5$$ cm2