[ Cross-sectional Area ]

All edges of a regular quadrilateral pyramid are equal to $a$. A plane is drawn through one side of the base and the midpoint of one of the opposite lateral edges. Find the area of the resulting cross-section.

#
In the cross-section, an isosceles trapezoid will be obtained, with bases equal to $a$ and $\frac{1}{2} a$.

## Solution

Assume that the plane passes through the side $A B$ of the base $A B C D$ of a regular quadrilateral pyramid $S A B C D$ and the midpoint $N$ of the lateral edge $2 C$.

Since the intersecting plane and the plane of the lateral face $D S C$ are drawn through parallel lines $A B$ and $C D$ respectively and have a common point $N$, these planes intersect along a line parallel to the lines $A B$ and $C D$ and passing through the point $N$.

Let $M$ be the point of intersection of this line with the lateral edge $S D$. Then point $M$ is the midpoint of $S D$, so $M N$ is the midline of triangle $D S C$, and $M N=\frac{1}{2} D C=\frac{1}{2} a$.

The quadrilateral $A B N M$ obtained in the cross-section is an isosceles trapezoid (the equality $A M=B N$ follows from the equality of triangles $B N C$ and $A M D$). Let's find the height of this trapezoid.

Let $Q$ and $L$ be the midpoints of segments $C D$ and $A B$, $R$ be the point of intersection of the apothem $S Q$ with segment $M N$, $O$ be the center of the base of the given pyramid, and $P$ be the orthogonal projection of point $R$ onto the plane of the base $A B C D$. Then $R P$ is the midline of the right triangle $S O Q$, and $S O$ is the height of the isosceles right triangle $A S C (A S=S C=a, A C=a \sqrt{2})$. Therefore,

$$
\begin{gathered}
S O=O C=\frac{1}{2} a \sqrt{2}, R P=\frac{1}{2} S O=\frac{1}{4} a \sqrt{2} \\
O P=\frac{1}{2} O Q=\frac{1}{4} a, L P=L O+O P=\frac{1}{2} a+\frac{1}{4} a=\frac{3}{4} a
\end{gathered}
$$

The height $L R$ of the trapezoid $A B N M$ is found using the Pythagorean theorem in the right triangle $L P R$:

$$
L R=\sqrt{L P^{2}+R P^{2}}=\sqrt{9 a^{2} / 16+2 a^{2} / 16}=\frac{1}{4} a \sqrt{11}
$$

Therefore,

$$
S(A B N M)=\frac{1}{2}(A B+M N) \cdot L R=\frac{3}{2} a \cdot \frac{1}{8} a \sqrt{11}=3 a^{2} \sqrt{11} / 16
$$

Answer

$$
3 a^{2} \sqrt{11} / 16
$$