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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- Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is : (a) \( \frac{p}{180} \ × \ 2\pi R \) (b) \( \frac{p}{180} \ × \ \pi R^2 \) (c) \( \frac{p}{360} \ × \ 2\pi R \) (d) \( \frac{p}{720} \ × \ 2\pi R^2 \)

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- NCERT solutions for class 10 maths chapter 3 Pair of linear equations in two variables
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- NCERT solutions for class 10 science chapter 2 Acids, Bases and Salts
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