In a right parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ with bases $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=29, A D=36$, $B D=25, A A_{1}=48$. Find the area of the section $A B_{1} C_{1} D$.
Drop a perpendicular $B K$ from vertex $B$ to line $A D$. Since the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ is a right parallelepiped, $B_{1} B$ is perpendicular to the base plane $A B C D$. Therefore, $B K$ is the orthogonal projection of the slant $B_{1} K$ onto the base plane $A B C D$. By the theorem of three perpendiculars, $B_{1} K \perp A D$. Thus, $B_{1} K$ is the height of the parallelogram $A B_{1} C_{1} D$.

Let $S$ be the area of triangle $A B D$, and $p$ be the semiperimeter of this triangle. Then

$$
\begin{gathered}
p=(36+29+25) / 2=45 \\
S=\sqrt{45(45-36)(45-29)(45-25)}=\sqrt{45 \cdot 9 \cdot 16 \cdot 20}=9 \cdot 10 \cdot 4=360
\end{gathered}
$$

On the other hand, $S=\frac{1}{2} A D \cdot B K=18 \cdot B K$. Therefore, $B K=360 / 18=20$. From the right triangle $B_{1} B K$, we find that

$$
B_{1} K=\sqrt{B B^{2}+B K^{2}}=\sqrt{48^{2}+20^{2}}=\sqrt{2704}=52
$$

Therefore,

$$
S\left(A B_{1} C_{1} D\right)=A D \cdot B_{1} K=36 \cdot 52=1872 .
$$

## Answer

1872.

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