Question

The radii of two circles are in the ratio $3:8$. If the difference between their areas is $2695\pi c{m}^2$, find the area of the smaller circle.

• A. $441\pi c{m}^2$
• B. $64\pi c{m}^2$
• C. $400\pi c{m}^2$
• D. $482\pi c{m}^2$

Answer 1

The correct option is A $441\pi c{m}^2$

$\frac{r_1}{r_2}=\frac{3}{8}$

$\Rightarrow r_2=\frac{8r_1}{3}$

Difference between Area $=2695\pi c{m}^2$

$\Rightarrow |\pi r_1^2-\pi r_2^2|=2695\pi$

$\Rightarrow \left|\pi \right(r_1^2-r_2^2\left)\right|=2695\pi$

$\Rightarrow \left|\pi (r_1^2-\frac{64r_1^2}{9})\right|=2695\pi$

$\left|\pi (+\frac{55}{9}{r}_1^2)\right|=2695\pi$

$\Rightarrow r_1^2=\frac{2695}{\pi }\times \frac{9}{55}\times \pi$

$r_1^2=49\times 9$

$r_1=7\times 3$

$=21cm$

The area of the smaller circle of radius $r_1\left(21cm\right)$

Area $=\pi r^2$

$=441\times \pi$

$=441\pi$