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# 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 $94pi$?

## 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 $94pi$?

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%$(pi r_2^2)$
$9pi r_1^2=4pi r_2^2$
$frac{r_1^2}{r_2^2}=frac{4}{9}$ ____________ (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$pi$
$therefore r_1^2+frac{2}{5}r_2^2=94$
$thereforefrac{4}{9}r_2^2+frac{2}{5}r_2^2=94$
$thereforefrac{38}{45}r_2^2=94$
$therefore r_2^2=111.32$
$therefore r_2=10.55$
$therefore r_1^2=frac{4}{9}times 111.32$
$therefore r_1^2=49.48$
$therefore r_1=7.03$
$therefore r_1+r_2=7.03+10.55=17.58$
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