10.113. The side of an equilateral triangle inscribed in a circle is equal to $a$. Calculate the area of the segment cut off by it.
## Solution.

The area of segment $A n B$ is the difference between the areas of sector $A O B$ and $\triangle A O B$ (Fig. 10.107). We find $S_{\text {sect } A O B}=\frac{\pi R^{2}}{3}, S_{\triangle A O B}=\frac{1}{2} a \cdot \frac{R}{2}=\frac{a R}{4}$, hence $S=\frac{\pi R^{2}}{3}-\frac{a R}{4}$. Since $R=\frac{a}{\sqrt{3}}$, we finally get $S=\frac{a^{2}(4 \pi-3 \sqrt{3})}{36}$.

Answer: $\quad \frac{a^{2}(4 \pi-3 \sqrt{3})}{36}$.