Problem 5. $DABC$ is a triangular pyramid (Fig. 5). $\overline{AK}=\overline{KD}, \overline{BP}=\overline{PC}, \overline{DM}=0.4 \overline{DC}$. Find the area $S$ of the section of the pyramid by the plane $KMP$, if the vertex $A$ is at a distance $h=1$ from the plane $KMP$ and the volume of the pyramid $DABC$ is 5.
Solution. We construct the section of the pyramid $DABC$ using the method of parallel projections. We choose the line $KP$ as the direction of projection. In this case, the pyramid $\dot{D}ABC$ is represented by the parallelogram $ABDC$ (Fig. 6), and the cutting plane is represented by the line $KM$, which intersects the edge $AB$ at point $E$. Point $K=P$ is the center of symmetry of the parallelogram $ABDC$, so $AE: EB = DM: MC = 2: 3$.

Thus, the section of the pyramid $DABC$ by the plane $KMP$ is the quadrilateral $KMPE$ (Fig. 5).

The quadrilateral pyramids $AEKMP$ and $DEKMP$ have the same base $KMPE$ and $AK = KD$. Therefore, their volumes are equal. The area of triangle $AMD$ is 0.4 of the area of triangle $ACD$, and the distance from point $P$ to the plane $ACD$ is half the distance from point $B$ to this plane. Therefore, the volumes of pyramids $PAMC$ and $DEPB$ constitute 0.6 of the volume of pyramid $DABC$. This means that the volume of pyramid $AKMPE$ is 0.2 of the volume of pyramid $DABC$, i.e., it is equal to 1.

Thus, $\frac{1}{3} S h = 1$. But $h = 1$, so $S = 3$.