In a regular quadrilateral pyramid $P A B C D$, the side of the base is equal to $a$, and the lateral edge is equal to $\frac{5}{2} a$. One base of the cylinder lies in the plane $P A B$, and the other is inscribed in the section of the pyramid. Find the lateral surface area of the cylinder.
Let the plane of section intersect the edges $AD, PD, PC$ and $BC$ of the regular pyramid $PABCD$ at points $K, L, M$ and $N$ respectively (Fig.1). Then, by the theorem of the intersection of two parallel planes by a third, $KN \| AB, KL \| AP, MN \| BP$. Let $\frac{DK}{AD}=x$. Then

$$
\frac{PL}{PD}=\frac{PM}{PC}=1-x
$$

Therefore, $LM \| CD$ and the sides of the isosceles trapezoid $KLMN$ are:

$$
KN=AB=a, \quad MN=KL=x \cdot AP=\frac{5}{2} ax, \quad LM=(1-x) CD=(1-x) a
$$

Since a circle (the base of the cylinder) can be inscribed in the trapezoid $KLMN$,

$$
KL + MN = KN + LM, \text{ or } 5ax = a + (1-x)a
$$

From this, we find that $x=\frac{1}{3}$. Let $r$ be the radius of the base of the cylinder. Then $r$ is the radius of the circle inscribed in the trapezoid $KLMN$ with bases $a, \frac{2}{3}a$ and lateral sides equal to $\frac{5}{6}a$. If $MF$ is the height of the trapezoid, then

$$
\begin{gathered}
FN=\frac{1}{2}(KN-LM)=\frac{1}{2}\left(a-\frac{2}{3}a\right)=\frac{1}{6}a, \\
MF=\sqrt{MN^2 - FN^2}=\sqrt{\frac{25a^2}{36}-\frac{a^2}{36}}=a\sqrt{\frac{2}{3}}
\end{gathered}
$$

Therefore, $r=\frac{1}{2}MF=\frac{1}{2}a\sqrt{\frac{2}{3}}$. To find the height of the cylinder, consider the section of the pyramid by a plane passing through the vertex $P$ and the midpoints $G$ and $H$ of the edges $AB$ and $CD$ (Fig.2). From the triangle $APB$, we find that

$$
PG=\sqrt{AP^2 - AG^2}=\sqrt{\frac{25a^2}{4}-\frac{a^2}{4}}=a\sqrt{6}
$$

If $E$ is the point of tangency of the circle of the base of the cylinder with the line $KN$, then the perpendicular $ET$ dropped from point $E$ to the line $PG$ is equal to the height of the cylinder. Let $HR$ be the height of the isosceles triangle $PGH$, drawn to the lateral side, and $PQ$ be the height drawn to its base. Then

$$
\begin{aligned}
& PQ=\sqrt{PG^2 - QG^2}=\sqrt{6a^2 - \frac{a^2}{4}}=\frac{a\sqrt{23}}{2},
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_55a07599adfd5314dc75g-22.jpg?height=128&width=824&top_left_y=2194&top_left_x=617)

Therefore, the lateral surface area of the cylinder is

$$
2\pi r \cdot ET = 2\pi \cdot \frac{1}{2}a\sqrt{\frac{2}{3}} \cdot \frac{a\sqrt{23}}{3\sqrt{6}} = \frac{\pi a^2 \sqrt{23}}{9}
$$

## Answer

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