Question

The width of the track is 3.5 m., then the radius of the inner circle is : *

Answer 1

Answer:

The area of the inner circle with inner radius of 21m is pi r squared 3.1416*21^2= 1385.44 sq.m.

The area of the whole circle including the track is (21 + 3.5)^2*3.1416= 24.5^2*3.1416= 1885.74 sq.m.

so, the area of track is 1885.74–1385.44=500.3 sq.m.

Answer 2

As per data given in the question,

The width of the track is $3.5$m.

We have to find the radius of the inner circle.

Let inner circle radius is $r_{1}$ and outer circle radius is $r_{2}$

We know that area of circle$=\pi r^{2}$

so,

$r_{1}+3.5=r_{2}$

$\pi r_{1} ^{2}=962.5m^{2} \\ \\ so, r_{1} =17.5m\\ \\ \ r_{2} =21m$

Finally we get the outer radius of circle is $21$m and inner circle radius is $17.5$m.