Given a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, where $A B=4, A D=A A 1=14$. Point $M$ is the midpoint of edge $C C 1$. Find the area of the section of the parallelepiped by the plane passing through points $A 1, D$ and $M$.
Let the lines $D M$ and $D_1 C_1$ intersect at point $K$, and the lines $A_1 K$ and $B_1 C_1$ intersect at point $L$. Then the considered section is the quadrilateral $A_1 D M L$ (Fig.1). Its orthogonal projection onto the base plane $A_1 B_1 C_1 D_1$ is the quadrilateral $A_1 L C_1 D_1$. From the equality of triangles $C M D$ and $C_1 M K$ (by side and two adjacent angles), it follows that $K C_1 = C D = 4$, and from the equality of triangles $K L C_1$ and $A_1 L B_1$ (also by side and two adjacent angles), $C_1 L = B_1 L$. Therefore, $L$ is the midpoint of edge $B_1 C_1$. Hence,

$$
S_{A_1 L C_1 D_1} = \frac{3}{4} S_{A_1 B_1 C_1 D_1} = \frac{3}{4} \cdot 14 \cdot 4 = 42.
$$

Let $H$ be the base of the perpendicular dropped from vertex $D_1$ to the line $A_1 K$ (Fig.2). By the theorem of three perpendiculars, $D H \perp A_1 K$, so $D H D_1$ is the linear angle of the dihedral angle between the planes of the section and the base. Denote $\angle D H D_1 = \alpha$. From the right triangles $A_1 K D_1$ and $D D_1 H$, we find that

$$
\begin{aligned}
& A_1 K = \sqrt{A_1 D_1^2 + D_1 K^2} = \sqrt{14^2 + 8^2} = 2 \sqrt{65} \\
& D_1 H = \frac{A_1 D_1 \cdot D_1 K}{A_1 K} = \frac{14 \cdot 8}{2 \sqrt{65}} = \frac{56}{\sqrt{65}}, \\
& \operatorname{tg} \alpha = \operatorname{tg} \angle D H D_1 = \frac{D D_1}{D_1 H} = \frac{14}{\sqrt{65}} = \frac{\sqrt{65}}{4}.
\end{aligned}
$$

Then $\cos \alpha = \frac{4}{\sqrt{65}}$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-43.jpg?height=114&width=478&top_left_y=2007&top_left_x=790)

## Answer