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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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