4.9. The circumference of each of two equal circles of radius $R$ passes through the center of the other. Find the area of the common part of these circles.
4.9. Since in $\triangle O A O_{1}$ all sides are equal to $R$, then $\angle A O B=120^{\circ}$ (Fig. 4.8). We find

$$
S_{\triangle A O B}=\frac{1}{2} R^{2} \sin 120^{\circ}=\frac{1}{2} R^{2} \sin 60^{\circ}=\frac{R^{2} \sqrt{3}}{4}, S_{\mathrm{ceKT} A O B}=\frac{1}{3} \pi R^{2}
$$

from which $S_{\text {segm }}=\frac{1}{3} \pi R^{2}-\frac{R^{2} \sqrt{3}}{4}$. Thus, the area of the common part of the two circles is

$$
\begin{aligned}
& \quad \frac{2}{3} \pi R^{2}-\frac{R^{2} \sqrt{3}}{2}=\frac{R^{2}(4 \pi-3 \sqrt{3})}{6} \\
& \text { Answer: } \frac{R^{2}(4 \pi-3 \sqrt{3})}{6}
\end{aligned}
$$

Fig. 4.8

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-577.jpg?height=285&width=327&top_left_y=1140&top_left_x=46)