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Answer :

Here \(r \ = \ 25 \) cm, \( \theta \ = \ 115º \)

Total area cleaned at each sweep of the blades

\( = \ 2 \ × \) Area of the sector

(having \( r \ = \ 25 \) and \( \theta \ = \ 115º \) )

\( = \ 2 \ × \ \frac{ \theta }{360} \ × \ \pi r^2 \)

\(= 2 \ × \ \frac{115}{360} \ × \ \frac{22}{7} \ × \ (25)^2 \)

\( = \ \frac{23 \ × \ 11 \ × \ 625}{18 \ × \ 7} \ = \ \frac{158125}{126} \) cm^{2}

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