8,9

All edges of a regular triangular prism are equal to $a$. Find the area of the section of the prism by a plane passing through an edge of the base and the midpoint of the non-parallel edge of the other base.
Let $M$ be the midpoint of side $B_1C_1$ of the base $A_1B_1C_1$ of the given regular prism $ABC A_1B_1C_1$. Since the intersecting plane passes through the line $AB$, which is parallel to the plane of the face $A_1B_1C_1$, it intersects the plane of the upper base $A_1B_1C_1$ along a line parallel to $AB$, and thus to $A_1B_1$. Therefore, the line of intersection of these planes also passes through the midpoint $N$ of the segment $A_1C_1$. Hence, $MN$ is the midline of triangle $A_1B_1C_1$. The desired section is an isosceles trapezoid $ANMB$ with bases $AB$ and $MN$, where $AB = a$ and $MN = \frac{1}{2}a$. Let $K$ and $L$ be the midpoints of $MN$ and $AB$ respectively, and $P$ be the orthogonal projection of point $L$ onto the plane of the base $A_1B_1C_1$. Then $P$ is the midpoint of $A_1B_1$. We find the height $KL$ of the trapezoid $ANMB$ from the right triangle $PKL$:

$$
\begin{gathered}
LP = A_1A = a, \quad KP = \frac{1}{2}C_1P = \frac{a\sqrt{3}}{4}, \\
KL = \sqrt{PL^2 + KP^2} = \sqrt{a^2 + \frac{3a^2}{16}} = \frac{a\sqrt{19}}{4}.
\end{gathered}
$$

Therefore,

$$
S_{ANMB} = \frac{1}{2}(AB + MN) \cdot KL = \frac{1}{2} \cdot \frac{3}{2}a \cdot \frac{a\sqrt{19}}{4} = \frac{3a^2\sqrt{19}}{16}
$$

## Answer

$\frac{3a^2\sqrt{19}}{16}$