117. Given a regular tetrahedron with edge $a$. A sphere touches the three edges of the tetrahedron emanating from one vertex at their ends. Find the area of the part of the spherical surface located inside the tetrahedron.
117. Consider a tetrahedron with edge $2a$. The surface of the sphere that touches all its edges is divided by the surface of the tetrahedron into 4 equal segments and 4 equal curved triangles, each of which is equal to the desired triangle. The radius of the sphere is $\frac{a \sqrt{2}}{2}$, the height of each segment is $a\left(\frac{\sqrt{2}}{2}-\frac{1}{2} \sqrt{\frac{2}{3}}\right)$, therefore, the area of the desired curved triangle is

$$
\begin{aligned}
\frac{1}{4}\left[4 \pi a^{2}\left(\frac{\sqrt{2}}{2}\right)^{2}-4 \cdot 2 \pi a^{2} \frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}-\frac{1}{2} \sqrt{\frac{2}{3}}\right)\right] & = \\
& =\frac{\pi a^{2}}{6}(2 \sqrt{3}-3)
\end{aligned}
$$