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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 \( ∆ \ RPQ \ = \ \frac{1}{2} \ × \ RP \ × \ PQ \)
\( = \ 12 \ × \ 7 \ × \ 24 \ = \ 84 \) cm2
Area of the shaded region
= Area of the semi circle - Area ( \(∆ \ RPQ \) )
\( = \ \frac{6875}{28} \ - \ 84 \ = \ \frac{4523}{28} \) cm2