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Answer :
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