10.298. Determine the area of the segment if its perimeter is $p$, and the arc contains $120^{\circ}$.
## Solution.

Let the radius of the circle be $R$. Since the arc of the segment contains $120^{\circ}$, its length is $\frac{2 \pi R}{3}$, and the chord spanning this arc (the side of the inscribed regular triangle) is $R \sqrt{3}$. Therefore, the perimeter of the segment is $p=\frac{2 \pi R}{3}+R \sqrt{3}$, from which we get $R=\frac{3 p}{2 \pi+3 \sqrt{3}}$. The area of the segment is:

$$
\begin{gathered}
S=\frac{\pi R^{2}}{3}-\frac{R^{2} \sqrt{3}}{4}=\frac{R^{2}(4 \pi-3 \sqrt{3})}{12}=\frac{3 p^{2}(4 \pi-3 \sqrt{3})}{4(2 \pi+3 \sqrt{3})^{2}} \\
\text { Answer: } \frac{3 p^{2}(4 \pi-3 \sqrt{3})}{4(2 \pi+3 \sqrt{3})^{2}}
\end{gathered}
$$