Answer :

Clearly from the figure,

Area of one design

= Area of the sector AOB - Area ( \(∆ \ AOB \) )

\( = \ \frac{ \theta }{360} \ × \ \pi r^2 \ - \frac{1}{2} \ × \ (r)^2 \ × \ sin60° \), where \(r \ = \ 28 \) cm

\( = \ \frac{60}{360} \ × \) \( \frac{22}{7} \ × \ (28)^2 \) \( - \ \frac{1}{2} \ × \ (28)^2 \ × \ \frac{ \sqrt{3}}{2} \)

\( = \ \frac{1232}{3} \ - \ 333.2 \)

\( = \ \frac{1232 \ - \ 999.6}{3} \)

\( = \ \frac{232.4}{3} \)cm^{2}

Area of 6 such design \( = \ 6 \ × \ \frac{232.4}{3} \ = \ 464.8 \)cm^{2}

Cost of making such designs @ Rs. 0.35 per cm^{2}

= Rs \(0.35 \ × \ 464.8 \)

= Rs 162.68

- Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°
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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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