The side of the base of a regular quadrilateral pyramid is $a$, and the apothem of the pyramid is $\frac{3}{2} a$. The orthogonal projection of the pyramid onto a plane perpendicular to one of its lateral faces is an isosceles trapezoid. Find the area of this trapezoid.
Let $M$ and $N$ be the midpoints of sides $BC$ and $AD$ of the base $ABCD$ of a regular quadrilateral pyramid $PABCD$ (Fig.1). Then $BC = a$, $PM = \frac{3}{2}a$. From point $N$, drop a perpendicular $AE$ to the apothem $PM$. Then the line $NE$ is perpendicular to the plane of the face $PBC$, since this line is perpendicular to two intersecting lines $PM$ and $BC$ of this plane. Any plane $\gamma$, passing through the line $NE$, is perpendicular to the plane of the face $PBC$. Let the orthogonal projection of the pyramid onto the plane $y$ be an isosceles trapezoid. In parallel projection, parallel lines $AD$ and $BC$ are mapped to parallel lines, so if $A', B', C'$, and $D'$ are the orthogonal projections of points $A, B, C$, and $D$ respectively, then points $A'$ and $D'$ lie on one base of the trapezoid, and points $B'$ and $C'$ lie on the other. The projection $P'$ of the vertex $P$ of the pyramid lies either on the line $B'C'$ or on the line $A'D'$, and one of the segments $B'C'$ or $A'D'$ is the base of the trapezoid, while the point $P'$ lies on the extension of the other (otherwise, the projection of the pyramid would be a parallelogram). Suppose that $P'$ lies on the extension of the base $B'C'$ beyond point $C'$ (Fig.3). Since the planes $y$ and $PBC$ are perpendicular, the line $B'P'$ lies in the plane $PBC$. In this case, $B'P'$ is the base of the trapezoid.

Let $\angle BPM = \angle CPM = \phi$ (Fig.2). From the right triangle $CMP$, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-38.jpg?height=120&width=1218&top_left_y=2621&top_left_x=420)

Let $PH$ be the height of the pyramid. Denote $\angle PMH = \beta$. From the right triangle $MPN$, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-38.jpg?height=117&width=788&top_left_y=2810&top_left_x=635)

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-39.jpg?height=128&width=932&top_left_y=39&top_left_x=563)

The line $NE$, lying in the plane $\gamma$ and perpendicular to the inclined line $AD$ to this plane, is perpendicular to the base $A'D'$ of the trapezoid $A'D'P'C'$ by the theorem of three perpendiculars. Since $N$ is the midpoint of $AD$ (and thus the midpoint of $A'D'$), $NE$ is the perpendicular bisector of the base $A'D'$ of the isosceles trapezoid $A'D'P'C'$. Therefore, $NE$ is also the perpendicular bisector of the base $B'P'$ of this trapezoid. Since $B'P'$ is the orthogonal projection of the segment $BP$ onto the plane $y$, the midpoint $E$ of its projection $B'P'$ on this plane is the projection of the midpoint $K$ of the segment $BP$. Let the lines $B'C'$ and $BC$ intersect at point $F$. Denote $\angle P'FC = \alpha$. Then $\angle PEK = \angle EPP' = \angle P'FC = \alpha$. By the cosine and sine theorems from triangle $PKE$, we find that

$$
\begin{gathered}
KE = \sqrt{PK^2 + PE^2 - 2PK \cdot PE \cos \varphi} = \\
= \sqrt{\left(\frac{a \sqrt{10}}{4}\right)^2 + \left(\frac{7}{6}a\right)^2 - 2 \cdot \frac{a \sqrt{10}}{4} \cdot \frac{7}{6}a \cdot \frac{3}{\sqrt{10}}} = \frac{a \sqrt{34}}{12} \\
\sin \alpha = \frac{KP' \sin \varphi \varphi}{KE} = \frac{\frac{a \sqrt{10}}{1} \cdot \frac{1}{\sqrt{10}}}{\frac{a \sqrt{31}}{12}} = \frac{3}{\sqrt{34}}. \\
\text{Then} \\
\cos \alpha = \frac{5}{\sqrt{34}}, A'D' = AD \cos \alpha = \frac{5a}{\sqrt{34}}
\end{gathered}
$$

Note that the lines $B'P'$ and $BP$ intersect at an angle $\beta - \alpha$. Then

$$
\begin{gathered}
B'P' = BP \cos (\beta - \alpha) = \frac{a \sqrt{10}}{2}(\cos \beta \cos \alpha - \sin \beta \sin \alpha) = \\
= \frac{a \sqrt{10}}{2}\left(\frac{1}{3} \cdot \frac{5}{\sqrt{34}} - \frac{2 \sqrt{2}}{3} \cdot \frac{3}{\sqrt{34}}\right) = \frac{a \sqrt{10}}{2} \cdot \frac{7}{\sqrt{35}} = \frac{7a}{\sqrt{34}} \\
\text{Therefore,} \\
S_{A'D'P'C'} = \frac{1}{2}(A'D' + C'P') NE = \frac{1}{2}\left(\sqrt{34} + \frac{7a}{\sqrt{34}}\right) \cdot \frac{2a \sqrt{2}}{3} = \frac{4a^2}{\sqrt{17}}
\end{gathered}
$$

## Answer

$\frac{4}{\sqrt{17}} a^2$.

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