The radius of a circular ground is 90, m. Ins | vedclass

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(D) The radius of the outer circle is $R = 90\, m$.The width of the road is $10\, m$,so the radius of the inner circle is $r = 90 - 10 = 80\, m$.The a

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$5338$

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(D) The radius of the outer circle is $R = 90\, m$.The width of the road is $10\, m$,so the radius of the inner circle is $r = 90 - 10 = 80\, m$.The area of the road is the difference between the area of the outer circle and the area of the inner circle.Area of the road $= \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$.Area $= 3.14 	imes (90^2 - 80^2)$.Area $= 3.14 	imes (8100 - 6400)$.Area $= 3.14 	imes 1700$.Area $= 5338\, m^2$.

The radius of a circular ground is $90\, m$. Inside it,a road of width $10\, m$ runs around its boundary. Find the area of the road. $(\pi=3.14)$ (in $m^2$)

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