$\left[\begin{array}{l}\text { Properties of sections } \\ {[\text { Linear dependence of vectors }]}\end{array}\right]$

The side of the base $A B C D$ of a regular quadrilateral pyramid $S A B C D$ is equal to $a$, and the lateral edge is equal to $b$. Find the area of the section of the pyramid by a plane passing through the midpoint of edge $A B$ and parallel to the lines $B D$ and $A S$.
Let $M$ be the midpoint of $AB$. A secant plane is parallel to the line $BD$ lying in the plane of the base $ABCD$, and has a common point $M$ with the plane of the base. Therefore, the base plane intersects the secant plane along a line passing through point $M$ parallel to $BD$. If $N$ is the point of intersection of this line with the edge $AD$, and $O$ is the center of the base $ABCD$, then $N$ is the midpoint of $AD$, $MN$ is the midline of triangle $ADB$, and the point $R$ of intersection of $MN$ and $AC$ is the midpoint of $AO$. The secant plane intersects the plane of triangle $ASC$ along a line parallel to $AS$ and passing through point $R$. Let this line intersect the lateral edge $SC$ at point $K$. From the similarity of triangles $RKC$ and $ASC$, it follows that $\frac{RK}{AS}=\frac{CR}{AC}=\frac{3}{4}$. Therefore, $RK=\frac{3}{4} AS=\frac{3}{4} b$. Since the secant plane passes through the line $RK$, parallel to $AS$, and thus also to the planes of the faces $ASB$ and $ASD$, it intersects these planes along lines parallel to $AS$ and passing through points $M$ and $N$ respectively, i.e., along the midlines of triangles $ASB$ and $ASD$. Let $F$ and $E$ be the midpoints of $SB$ and $SD$. Then the desired section is the pentagon $MNEKF$, and $MF=NE=\frac{1}{2} b$. Since $NE \| AS$ and $MN \| BD$, and $AS \perp BD$ (by the theorem of three perpendiculars), then $NE \perp MN$. Similarly, $MF \perp MN$. Therefore, $RN$ and $RM$ are the heights of the right trapezoids $RNEK$ and $RMFK$. Then

$$
\begin{aligned}
& S_{MNEKF}=2 S_{RNEK}=2 \cdot \frac{1}{2}(RK+NE) \cdot NR= \\
& =(RK+NE) \cdot NR=\left(\frac{3}{4} b+\frac{1}{2} b\right) \cdot \frac{1}{4} a \sqrt{2}=\frac{5 a b \sqrt{2}}{16}
\end{aligned}
$$

## Answer

$5 a b \sqrt{2}$

16.