10.336. An equilateral triangle $ABC$ with side $a=$ $=2$ cm has a circle inscribed in it; point $A$ is the center of a second circle with radius 1 cm. Find the area of the intersection of these circles.
## Solution.

The desired area is the sum of the areas of segment $D m E$ of a circle with radius $R=\sqrt{3} / 3$ cm centered at $O$, inscribed in $A B C$, and segment of a circle with radius $r=1$ cm (Fig. 10.121),

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0800.jpg?height=545&width=600&top_left_y=824&top_left_x=661)

Fig. 10.121 $\angle D O E=120^{\circ}$. Then

$$
S_{D m E}=\frac{1}{3} \pi R^{2}-S_{\triangle D O E}=\frac{\pi R^{2}}{3}-\frac{R^{2}}{2} \sin 120^{\circ}=\frac{\pi}{9}-\frac{\sqrt{3}}{2} \text{ cm}^{2}
$$

$S_{D n E}=\frac{1}{6} \pi r^{2}-S_{\triangle D A E}==\frac{\pi}{6}-\frac{\sqrt{3}}{4}$ cm $^{2}$. Therefore, the desired area $S=\frac{5 \pi-6 \sqrt{3}}{18} \text{ cm}^{2}$.

Answer: $\frac{5 \pi-6 \sqrt{3}}{18} \text{ cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0801.jpg?height=484&width=508&top_left_y=79&top_left_x=176)

Fig. 10.122

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0801.jpg?height=482&width=507&top_left_y=78&top_left_x=671)

Fig. 10.123