10.155. In a circle of radius $R$, two parallel chords are drawn on opposite sides of the center, one of which is equal to the side of an inscribed regular triangle, and the other to the side of an inscribed regular hexagon. Determine the area of the part of the circle contained between the chords.
Solution.

The length of the chord equal to the side of the inscribed regular hexagon, $a_{6}=R$, and the triangle $a_{3}=\frac{3 R}{\sqrt{3}}=\sqrt{3} R$ (Fig. 10.133). The area of the sector $O A B S_{1}$ is equal to the area of the segment plus the area of the triangle

$$
\begin{aligned}
& A O B: S_{1}=S_{\text {segm }}+S_{A O B}=\frac{\pi R^{2} \cdot 60^{\circ}}{360^{\circ}}=\frac{\pi R^{2}}{6} . \text { From which } \\
& S_{\text {segm } 1}=\frac{\pi R^{2}}{6}-\frac{1}{2} R^{2} \sin 60^{\circ}=\frac{\pi R^{2}}{6}-\frac{R^{2} \sqrt{3}}{4}
\end{aligned}
$$

The area of the sector $O C D \quad S_{2}$ is also equal to the sum of the areas of the segment and the triangle $O D C: S_{2}=S_{\text {segm2 }}+S_{O D C}=\frac{\pi R^{2} \cdot 120^{\circ}}{360^{\circ}}=\frac{\pi R^{2}}{3}$, from which $S_{\text {segm2 }}=\frac{\pi R^{2}}{3}-\frac{1}{2} R^{2} \sin 120^{\circ}=\frac{\pi R^{2}}{3}-\frac{\sqrt{3} R^{2}}{4}$. The area of the circle contained between the chords:

$$
\begin{aligned}
S= & \pi R^{2}-S_{\mathrm{segm} 1}-S_{\mathrm{segm} 2}=\pi R^{2}-\frac{\pi R^{2}}{6}+\frac{R^{2} \sqrt{3}}{4}-\frac{\pi R^{2}}{3}+\frac{\sqrt{3} R^{2}}{4}=\frac{R^{2}(\pi+\sqrt{3})}{2} . \\
& \text { Answer: } \frac{R^{2}(\pi+\sqrt{3})}{2} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-629.jpg?height=316&width=490&top_left_y=94&top_left_x=105)

Fig. 10.134