8,9

The volume of the pyramid $A B C D$ is 5. A plane is drawn through the midpoints of the edges $A D$ and $B C$, intersecting the edge $C D$ at point $M$. In this case, $D M: M C=2: 3$. Find the area of the section of the pyramid by the specified plane, if the distance from it to the vertex $A$ is 1.
Since the secant plane passes through the midpoints of opposite edges of the tetrahedron, it divides its volume in half. Let the secant plane intersect edge $AB$ at point $K$, and let $P$ and $Q$ be the midpoints of edges $AD$ and $BC$. Then the volume of the polyhedron $PMQKAC$ is $\frac{5}{2}$. On the other hand, it is equal to the sum of the volumes of the quadrilateral pyramid $APMKQ$ and the triangular pyramid $AMCQ$ with a common vertex $A$, and since

$$
V_{AMCQ}=\frac{CM}{CD} \cdot \frac{CQ}{CB} V_{ABCD}=\frac{3}{5} \cdot \frac{1}{2} V_{ABCD}=\frac{3}{5} \cdot \frac{1}{2} \cdot 5=\frac{3}{2}
$$

then

$$
V_{APMQK}=V_{PMQKAC}-V_{AMCQ}=\frac{5}{2}-\frac{3}{2}=1 .
$$

On the other hand, if $h$ is the distance from vertex $A$ to the secant plane, then

$$
V_{APMQK}=\frac{1}{3} S_{PMQK} \cdot h
$$

so $\frac{1}{3} S_{PMQK} \cdot h=1$, from which we find that

$$
S_{PMQK}=\frac{3}{h}=3
$$

## Answer

3.00