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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 ​$$94\pi$$​?

## 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 ​$$94\pi$$​?

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​$$\pi$$​
$$\therefore r_1^2+\frac{2}{5}r_2^2=94$$
$$\therefore\frac{4}{9}r_2^2+\frac{2}{5}r_2^2=94$$
$$\therefore\frac{38}{45}r_2^2=94$$
$$\therefore r_2^2=111.32$$
$$\therefore r_2=10.55$$
$$\therefore r_1^2=\frac{4}{9}\times 111.32$$​
$$\therefore r_1^2=49.48$$
$$\therefore r_1=7.03$$
$$\therefore r_1+r_2=7.03+10.55=17.58$$
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