10,11

The side of the base $ABCD$ of a regular quadrilateral pyramid $SABCD$ is $a$, and the lateral edge is $b$.

Find the area of the section of the pyramid by a plane passing through the line $BD$ parallel to the line $AS$.
The secant plane intersects the plane $A S C$ along a line passing through the center of the square $A B C D$ and parallel to the lateral edge $A S$. In the section, an isosceles triangle is obtained.

## Solution

The plane of the triangle $A S C$ has a common point $O$ (the intersection point of the diagonals of the square $A B C D$) and passes through the line $A S$, which is parallel to the secant plane. Therefore, it intersects the secant plane along a line parallel to $A S$ and passing through the point $O$.

Let this line intersect the edge $S C$ at point $M$. Since $O$ is the midpoint of $A C$, $O M$ is the midline of the triangle $A S C$, and $O M = \frac{1}{2} A S = \frac{1}{2} b$. The desired section is an isosceles triangle $B M D$ (the equality $B M = D M$ follows from the equality of triangles $B M C$ and $D M C$).

Since $M O$ is the height of the triangle $B M D$,

$$
S(B M D) = \frac{1}{2} B D \cdot O M = \frac{1}{4} a \sqrt{2} \cdot b = \frac{1}{2} a b \sqrt{2}.
$$

## Answer

$$
a b \sqrt{2} / 4
$$