$[$ Properties of Sections $]$ $[$ Tetrahedron and Pyramid $]$

The base of a quadrilateral pyramid $S A B C D$ is a parallelogram $A B C D$. 1) Construct the section of the pyramid by a plane passing through the midpoint of edge $A B$ and parallel to the plane $S A D$. 2) Find the area of the resulting section if the area of face $S A D$ is 16.
1) Let $M$ be the midpoint of edge $A B$. According to the theorem of the intersection of two parallel planes by a third intersecting plane, the intersecting plane intersects the plane of face $A S B$ along a line parallel to $A S$, i.e., along the midline $M N$ of triangle $A S B$. Similarly, the intersecting plane intersects the faces $A B C D$ and $C S D$ along lines parallel to $A D$ and $S D$ respectively. Therefore, the points $N, L$, and $K$ where the intersecting plane intersects the edges $S B, C D$, and $S C$ are the midpoints of segments $S B, C D$, and $S C$ respectively, and the desired section is the trapezoid $M N K L$. 2) Let the lines $M N$ and $L K$, lying in the intersecting plane, intersect at point $Q$. Then point $Q$ lies on the line of intersection of the planes $A S B$ and $C S D$, drawn through two parallel lines $A B$ and $C D$, i.e., on the line passing through point $S$ parallel to lines $A B$ and $C D$. Triangle $M Q L$ is equal to triangle $A S D$ (by three sides), and $N K$ is the midline of triangle $M Q L$ (since $N K \| M L$ and $N K=\frac{1}{2} B C=\frac{1}{2} M L$). Triangle $N Q K$ is similar to triangle $M Q L$ with a similarity ratio of $\frac{1}{2}$, so

$$
S_{\triangle N Q K}=\frac{1}{\mathbf{4}} S_{\triangle M Q L}=\frac{1}{\mathbf{4}} S_{\triangle A S D}=\frac{16}{4}=4 .
$$

$$
S_{M N K L}=S_{\triangle M Q L}-S_{\triangle N Q K}=16-4=12
$$

## Answer

12.00