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The area of a circle is $ 616 \mathrm{~cm}^{2} $. There is a $ 2 \mathrm{~m} $ wide road on its edge. What will be area of that road?
Given:
The area of a circle is \( 616 \mathrm{~cm}^{2} \). There is a \( 2 \mathrm{~m} \) wide road on its edge.
To do:
We have to find the area of the road.
Solution:
Let the radius of the inner circle be $r$.
Width of the road $=2\ m$.
This implies,
Radius of the outer circle$=r+2\ m$.
Area of the inner circle$=616\ m^2$
Therefore,
$\pi r^2=616$
$\frac{22}{7}\times r^2=616$
$r^2=616\times\frac{7}{22}$
$r^2=28\times7$
$r^2=7\times4\times7$
$r^2=(2\times7)^2$
$r=14\ m$
Radius of the outer circle$=14+2\ m=16\ m$.
Area of the outer circle$=\pi (16)^2$
$=\frac{22}{7}\times256$
$=804.57\ m^2$
Area of the road$=$Area of the outer circle$-$ Area of the inner circle
$=804.57-616\ m^2$
$=188.57\ m^2$
The area of the road is $188.57\ m^2$.
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