10.132. A circle is inscribed in a circular sector, the arc of which contains $60^{\circ}$. Find the ratio of the area of this circle to the area of the sector.
## Solution.

Let $B, D, E$ be the points of tangency. Denote the radius of the circle from which the sector is cut as $R$, and the radius of the inscribed circle as $r$ (Fig. 10.119). Thus, $B O=D O=E O=r$, and $B O=R . O O=R-r$. $D O=O O \sin 30^{\circ}$. This means that $r=\frac{1}{2}(R-r)$, from which $R=3 r$. The area of the sector $S_{1}=\frac{\pi R^{2} 60^{\circ}}{360^{\circ}}=\frac{\pi R^{2}}{6}=\frac{3 \pi r^{2}}{2}$. The area of the inscribed circle $S_{2}=\pi r^{2}$; $S_{2}: S_{1}=\pi r^{2}: \frac{3 \pi r^{2}}{2}=2: 3$.

Answer: $2: 3$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-618.jpg?height=452&width=300&top_left_y=94&top_left_x=220)

Fig. 10.119

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-618.jpg?height=362&width=516&top_left_y=193&top_left_x=732)

Fig. 10.120