Given a cube $A B C D A 1 B 1 C 1 D 1$ with edge $a$. Let $M$ be the midpoint of edge $D 1 C 1$. Find the perimeter of triangle $A 1 D M$, as well as the distance from vertex $D 1$ to the plane passing through the vertices of this triangle.
From the right triangles $A 1 D 1 M, D D 1 M$ and $D D 1 A 1$, by the Pythagorean theorem, we find that

$$
A 1 M=\sqrt{A_{1} D_{1}^{2}+D_{1} M^{2}}=\sqrt{\mathfrak{a}^{2}+\frac{\alpha^{2}}{1}}=\frac{a \sqrt{5}}{2}
$$

$$
\begin{aligned}
& D M=\sqrt{D_{1} D^{2}+D_{1} M^{2}}=\sqrt{\boldsymbol{a}^{2}+\frac{\alpha^{2}}{1}}=\frac{a \sqrt{5}}{2} \\
& A 1 D=\sqrt{A_{1} D_{1}^{2}+D_{1} D^{2}}=\sqrt{\boldsymbol{a}^{2}+\pi^{2}}=a \sqrt{2}
\end{aligned}
$$

Therefore, the perimeter of triangle $A 1 D M$ is

$$
A 1 M+D M+A 1 D=\frac{a \sqrt{5}}{2}+\frac{a \sqrt{5}}{2}+a \sqrt{2}=a \sqrt{5}+a \sqrt{2}=a(\sqrt{5}+\sqrt{2}) .
$$

Now let's find the distance from vertex $D 1$ to the plane passing through the vertices of this triangle.

![](https://cdn.mathpix.com/cropped/2024_05_06_acc13965d0370bfc03dcg-09.jpg?height=48&width=263&top_left_y=565&top_left_x=14)

Let $K$ be the midpoint of the base $A 1 D$ of the isosceles triangle $A 1 M D$. Then $M K$ is the height of triangle $A 1 M D$. By the Pythagorean theorem from the right triangle $M K D$, we find that

$$
M K=\sqrt{M D^{2}-K D^{2}}=\sqrt{\frac{\mathrm{k}^{2}}{1}-\frac{\frac{0}{2}^{2}}{2}}=\frac{a \sqrt{3}}{2}
$$

Let $V$ be the volume of the pyramid $A 1 D 1 M D$. Then

$$
V=\frac{1}{3} S_{\Delta A} 1 D 1 M \cdot D D 1=\frac{1}{3} \cdot \frac{1}{2} A 1 D 1 \cdot D 1 M \cdot D D 1=\frac{1}{3} \cdot \frac{1}{2} a \cdot \frac{a}{2} \cdot a=\frac{a^{3}}{12}
$$

On the other hand,

$$
V=\frac{1}{3} S_{\Delta A} 1 M D \cdot h
$$

where $h$ is the desired distance from vertex $D 1$ to the plane passing through the vertices of triangle $A 1 M D$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_acc13965d0370bfc03dcg-09.jpg?height=117&width=648&top_left_y=1228&top_left_x=705)

![](https://cdn.mathpix.com/cropped/2024_05_06_acc13965d0370bfc03dcg-09.jpg?height=49&width=263&top_left_y=1416&top_left_x=14)

Let's take the origin of the coordinate system at vertex $D 1$, and direct the coordinate axes along the rays $D 1 A 1, D 1 C 1$ and $D 1 D$. Then the equation of the plane passing through points $A 1, D$ and $M$ is

$$
\frac{\pi}{a}+\frac{y}{2}+\frac{z}{a}=1
$$

(the equation of the plane in intercept form), or $x+2 y+z-a=0$. Therefore,

$$
h=\frac{|0+0+0-a|}{\sqrt{1+1+1}}=\frac{a}{\sqrt{6}}
$$

## Answer

$a(\sqrt{5}+\sqrt{2}) ; \frac{a}{\sqrt{6}}$.