Through the vertex of a regular quadrilateral pyramid and the midpoints of two adjacent sides of the base, a plane is drawn. Find the area of the resulting section if the side of the base of the pyramid is $a$, and the lateral edge is $2a$.

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Let $M$ and $N$ be the midpoints of sides $AB$ and $BC$ of the base $ABCD$ of a regular quadrilateral pyramid $PABCD$, respectively, with $AB = a$ and $AP = 2a$. Let $O$ be the center of the square $ABCD$, and $K$ be the intersection point of $MN$ and $BD$. Then $OK$ is the orthogonal projection of the inclined line $PK$ on the plane $ABCD$, and $OK \perp MN$. By the theorem of three perpendiculars, $PK \perp MN$, i.e., $PK$ is the height of triangle $PMN$. Since $MN$ is the midline of triangle $ABC$,

$$
MN = \frac{1}{2} AC = \frac{1}{2} a \sqrt{2}, \quad OK = \frac{1}{2} OB = \frac{1}{4} BD = \frac{1}{4} a \sqrt{2}
$$

By the Pythagorean theorem in right triangles $AOP$ and $ROK$, we find that

$$
OP = \sqrt{AP^2 - AO^2} = \sqrt{4a^2 - \frac{a^2}{2}} = \frac{1}{2} a \sqrt{14}
$$

$$
PK = \sqrt{OP^2 + OK^2} = \sqrt{\frac{7a^2}{2} + \frac{a^2}{8}} = \frac{1}{4} a \sqrt{58}
$$

Therefore,

$$
S_{\triangle PMN} = \frac{1}{2} MN \cdot PK = \frac{1}{2} \cdot \frac{1}{2} a \sqrt{2} \cdot \frac{1}{4} a \sqrt{58} = \frac{1}{8} a^2 \sqrt{29}
$$

## Answer

$\frac{a^2 \sqrt{29}}{8}$.