Two circles $k_1$ and $k_2$ intersect in such a way that outside their intersection is 10% of the area of the circle $k_1$ and 60% the area of the circle $k_2$. What is the ratio of the radii of the circles $k_1$ and $k_2$? What is the sum of the radii if the area of the figure is $94\pi$?

Question:

Two circles $k_1$ and $k_2$ intersect in such a way that outside their intersection is 10% of the area of the circle $k_1$ and 60% the area of the circle $k_2$. What is the ratio of the radii of the circles $k_1$ and $k_2$? What is the sum of the radii if the area of the figure is $94\pi$?

Answer:

Since, outside their intersection is 10% of the area of the circle $k_1$ and 60% the area of the circle $k_2$, 

$\therefore$ 90%$(\pi r_1^2)$=40%$(\pi r_2^2)$

$9\pi r_1^2=4\pi r_2^2$

$\frac{r_1^2}{r_2^2}=\frac{4}{9}$ ____________ (1)

$\frac{r_1}{r_2}=\frac{2}{3}$

$r_1:r_2=2:3$

From (1),

$r_1^2=\frac{4}{9}r_2^2$ _________________ (2)

Now, the area of figure is 94$\pi$.

$\therefore \pi r_1^2$+40% $(\pi r_2^2)$=94$\pi$ 

$\therefore r_1^2+\frac{2}{5}r_2^2=94$ 

$\therefore\frac{4}{9}r_2^2+\frac{2}{5}r_2^2=94$ 

$\therefore\frac{38}{45}r_2^2=94$

$\therefore r_2^2=111.32$ 

$\therefore r_2=10.55$ 

$\therefore r_1^2=\frac{4}{9}\times 111.32$ 

$\therefore r_1^2=49.48$ 

$\therefore r_1=7.03$ 

$\therefore r_1+r_2=7.03+10.55=17.58$