A sector representing a quarter of a circle has been cut out. From this sector and the remaining part of the circle, the lateral surfaces of two cones have been made. Find the ratio of the heights of these cones.
Let $R$ be the radius of the given circle, $r_1$ and $r_2$ be the radii of the bases of the obtained cones ($r_1 < r_2$), and $h_1$ and $h_2$ be the heights of the cones. Then the slant height of each of the obtained cones is equal to $R$. The lengths of the circumferences of the bases of the cones are respectively one quarter and three quarters of the circumference of the circle with radius $R$. From the equations

$$
2 \pi r_1 = \frac{1}{1} \cdot 2 \pi R, \quad 2 \pi r_2 = \frac{3}{1} \cdot 2 \pi R
$$

we find that

$$
r_1 = \frac{1}{1} R, \quad r_2 = \frac{3}{1} R.
$$

Consider the axial sections of the cones. By the Pythagorean theorem, we find that

$$
\begin{aligned}
& h_1 = \sqrt{R^2 - \tau_1^2} = \sqrt{R^2 - \frac{R^2}{16}} = \frac{R \sqrt{15}}{1}, \\
& h_2 = \sqrt{R^2 - \tau_2^2} = \sqrt{R^2 - \frac{9 R^2}{16}} = \frac{R \sqrt{7}}{1}.
\end{aligned}
$$

Therefore, $\frac{h_1}{h_2} = \frac{\sqrt{15}}{\sqrt{7}}$.

## Answer

$\frac{\sqrt{15}}{\sqrt{7}}$

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