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Answer :

\(\because \) ROQ is a diameter

\( \therefore ∠ \ RPQ \ = \ 90° \)

In \( ∆ \ PRQ \) ,

\( RQ^2 \ = \ RP^2 \ + \ PQ^2 \)

\( \Rightarrow \ RQ^2 \ = \ 7^2 \ + \ 24^2 \ \)

\( = \ 49 \ + \ 576 \ \)

\( \Rightarrow RQ^2 = \ 625 \)

\( \Rightarrow \ RQ \ = \ 25 \) cm

\(\therefore \) Radius \( r \ = \ \frac{1}{2} RQ \ = \ \frac{25}{2} \)cm

Area of the semi circle \( = \ \frac{1}{2} \pi r^2 \ \)

\( = \ \frac{1}{2} \ × \ \frac{22}{7} \ × \

( \frac{25}{2})^2 \ \)

\( = \ \frac{6875}{28} cm^2 \)

and area of \( ∆ \ RPQ \ = \ \frac{1}{2} \ × \ RP \ × \ PQ \)

\( = \ 12 \ × \ 7 \ × \ 24 \ = \ 84 \) cm^{2}

Area of the shaded region

= Area of the semi circle - Area ( \(∆ \ RPQ \) )

\( = \ \frac{6875}{28} \ - \ 84 \ = \ \frac{4523}{28} \) cm^{2}

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