10.138. Three equal circles of radius $r$ touch each other pairwise. Calculate the area of the figure located outside the circles and bounded by their arcs between the points of tangency.
## Solution.

The segments connecting the centers of the circles (Fig. 10.124) form an equilateral triangle with side length $2r$. Its area is: $S_{1}=\frac{4 r^{2} \sqrt{3}}{4}=r^{2} \sqrt{3}$. The area of one sector $S_{2}=\frac{\pi r^{2} \cdot 60^{\circ}}{360^{\circ}}=\frac{\pi r^{2}}{6}$. Since there are three sectors, the area of the figure located outside the circles and bounded by their arcs between the points of tangency is

$$
\begin{aligned}
& S=r^{2} \sqrt{3}-\frac{\pi r^{2}}{2}=\frac{r^{2}(2 \sqrt{3}-\pi)}{2} \\
& \text { Answer: } \frac{r^{2}(2 \sqrt{3}-\pi)}{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-621.jpg?height=324&width=436&top_left_y=94&top_left_x=96)

Fig. 10.125