10.402 A circle of radius $R$ with center at point $O$ is divided into six equal parts by points $A, B, C, D, E, F$. Determine the area of the figure $C O E$, bounded by the arc $O C$ with center at point $B$, the arc $O E$ with center at point $F$, and the arc $C E$ with center at point $A$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-244.jpg?height=325&width=294&top_left_y=103&top_left_x=159)

The area of the desired figure $\Phi$, bounded by

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-244.jpg?height=47&width=723&top_left_y=156&top_left_x=479)
( $O M E$ - segment of sector FOME ).

$$
\begin{aligned}
& S_{A O E}=\frac{1}{2} O A \cdot O E \cdot \sin 120^{\circ}=\frac{R^{2} \sqrt{3}}{4} \\
& S_{O M E}=S_{F O M E}-S_{F O E}=\frac{\pi R^{2}}{6}-\frac{R^{2} \sqrt{3}}{4}
\end{aligned}
$$

$(O F=O E=E F=R) . S_{A C K E}=\frac{\pi R^{2}}{2}\left(A E=A C=R \sqrt{3}, \angle E A C=60^{\circ}\right) \Rightarrow S_{\Phi}=\frac{\pi R^{2}}{6}$.

Answer: $\frac{\pi R^{2}}{6}$.