10.170. Determine the area of a circle inscribed in a sector of a circle of radius $R$ with a chord of $2 a$.
Solution.

Let's drop the radii of the inscribed circle to the points of tangency (Fig. 10.141). Then, from the right triangles $\triangle A B D$ and $\triangle O K B$, we get

$$
\sin \alpha=\frac{O K}{O B}=\frac{A D}{A B} \Leftrightarrow \frac{r}{R-r}=\frac{a}{R} \Leftrightarrow R r=R a-r a \Leftrightarrow r=\frac{R a}{R+a}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-635.jpg?height=500&width=596&top_left_y=78&top_left_x=96)

Fig. 10.141

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-635.jpg?height=416&width=550&top_left_y=162&top_left_x=681)

Fig. 10.142

From this, $\pi\left(\frac{R a}{R+a}\right)^{2}$ is the area of the inscribed circle.

Answer: $\pi\left(\frac{R a}{R+a}\right)^{2}$.