Question

Find the area enclosed between two cocentric circles of radii $3.5$ cm and $7$ cm circle, such that the area enclosed between it and the $7$ cm circle is same as that between the $2$ inner circles. Find the radius of the third circle correct to one decimal place.

Answer 1

The area enclosed by two concentric of radii $r_1$ and $r_2$ $(r_1>r_2)=\pi (r_1^2-r_2^2)$

$\Rightarrow$ Thus, the area enclosed by two inner concentric circles $=\pi \left[\right(7{)}^2-\left(3.5{)}^2\right]$

$\Rightarrow$ The area enclosed by two inner concentric circles $=\frac{22}{7}\times (49-12.25)$

$\Rightarrow$ The area enclosed by two inner concentric circles $=\frac{22}{7}\times 36.75=115.5c{m}^2$

According to the given information,

$\Rightarrow$ $\pi \left[\right(r^2-\left(7{)}^2\right]=\pi \left[\right(7{)}^2-\left(3.5{)}^2\right]$

$\Rightarrow$ $r^2=[7^2+7^2-(3.5{)}^2]$

$\Rightarrow$ $r^2=(3.5{)}^2\times (4+4-1)$

$\Rightarrow$ $r^2=7\times (3.5{)}^2$

$\Rightarrow$ $r^2=85.75$

$\therefore$ $r=9.26cm$