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

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(C) The circular ground has a radius $R = 35 \, m$. $A$ road of width $w = 3.5 \, m$ runs inside the boundary. The inner radius is $r = R - w = 35 - 3

Vedclass · Accepted Answer

$11704$

Vedclass · Answer

(C) The circular ground has a radius $R = 35 \, m$. $A$ road of width $w = 3.5 \, m$ runs inside the boundary. The inner radius is $r = R - w = 35 - 3.5 = 31.5 \, m$.The area to be repaired is the area of the sector of the outer circle minus the area of the corresponding sector of the inner circle,with a central angle $	heta = 72^{\circ}$.Area of the region $= \frac{	heta}{360^{\circ}} 	imes \pi 	imes (R^2 - r^2)$$= \frac{72}{360} 	imes \frac{22}{7} 	imes (35^2 - 31.5^2)$$= \frac{1}{5} 	imes \frac{22}{7} 	imes (35 - 31.5)(35 + 31.5)$$= \frac{22}{35} 	imes 3.5 	imes 66.5$$= \frac{22}{35} 	imes \frac{7}{2} 	imes 66.5$$= 11 	imes 0.2 	imes 66.5 = 2.2 	imes 66.5 = 146.3 \, m^{2}$.The cost of repairing is $146.3 \, m^{2} 	imes ₹ 80 / m^{2} = ₹ 11,704$.

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

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