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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 ​\( 94\pi \)​?

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