The radius of a circular ground is 35, m. Ins | vedclass

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Question

Here, $\odot( O , OA )$ is the circular ground and the measure of the angle between radif $\overline{ OA }$ and $\overline{ OC }$ is $72$.

For mino

Vedclass · Accepted Answer

$11704$

Vedclass · Answer

Here, $\odot( O , OA )$ is the circular ground and the measure of the angle between radif $\overline{ OA }$ and $\overline{ OC }$ is $72$.

For minor sector OABC, radius $r_{1}=35 m$

and $	heta=m \angle AOC =72$

For minor sector ODEF,

radius $r_{2}=35-3.5=31.5 m$ and

$	heta=m \angle DOF =72$

Area of the region to be repaired

$=$ Area of minor sector $OABC$ $-$ Area of minor sector $ODEF$

$=\frac{\pi r_{1}^{2} 	heta}{360}-\frac{\pi r_{2}^{2} 	heta}{360}$

$=\frac{\pi 	heta}{360}\left(r_{1}^{2}-r_{2}^{2}ight)$

$=\frac{\pi 	heta}{360}\left(r_{1}+r_{2}ight)\left(r_{1}-r_{2}ight)$

$=\frac{22}{7} 	imes \frac{72}{360}(35+31.5)(35-31.5)$

$=\frac{22}{7 	imes 5}(66.5)(3.5)$

$=11 	imes 13.3$

$=146.3 m ^{2}$

cost of repairing $1 m ^{2}$ region $=₹ 80$ $	herefore$ cost of repairing $146.3 m ^{2}$ region

$=₹(146.3 	imes 80)$

$=₹ 11,704$

Thus, the cost of repairing is $₹ 11,704$.

The radius of a circular ground is $35\, m$. Inside it, $3.5 \,m$ broad road runs around its boundary. A part of the road between two radii forming an angle of measure $72$ at the centre is to be repaired. Find the cost of repairing at the rate of ₹ $80 / m ^{2}$. (in ₹)

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