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