11.212 The distance between non-intersecting diagonals of two adjacent lateral faces of a cube is $d$. Determine the total surface area of the cube.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-267.jpg?height=363&width=392&top_left_y=128&top_left_x=160)

Let the distance between the diagonals $C N$ and $B D$ of the faces of the given cube $A B C D K L M N$ be $d$. $\triangle B D K \| \triangle C L N$, since $L C\|K D, C N\| B K$.

Therefore, the distance between these triangles is equal to the distance between the skew lines $C N$ and $B D$, i.e., it is equal to $d$. Let $P$ and $Q$ be the centers of the regular $\triangle B D K$ and $\triangle C L N$. Due to the symmetry of the cube, its diagonal $A M$ will pass through $P$ and $Q$, and $A P$ and $M Q$ are the heights of the regular pyramids $A B D K$ and $M L C N \Rightarrow$

$\Rightarrow P Q=d, A P=M Q=\frac{A M-d}{2}=\frac{x \sqrt{3}-d}{2}$, where $x$ is the side length of the cube. On the other hand,

$A P=\sqrt{A B^{2}-B P^{2}}=\frac{x}{\sqrt{3}}\left(B P=\frac{x \sqrt{2}}{\sqrt{3}}\right) \Rightarrow \frac{x}{\sqrt{3}}=\frac{x \sqrt{3}-d}{2} \Rightarrow x=d \sqrt{3}$ and the surface area

of the cube is $18 d^{2}$.

Answer: $18 d^{2}$.