Why is this true that $\left(d(Q)\right)^n=\left(\sqrt{n}\right)^n\mathcal{L}_n(Q)$ for any cube $Q$ in $\mathbb{R}^n$?

Question

I read a roof about Hausdorff measure. It introduces an equation

$$\alpha(n)\left(\frac{d(Q)}{2}\right)^n=\alpha(n)\left(\frac{\sqrt{n}}{2}\right)^n\mathcal{L}_n(Q)$$

for each cube $Q\subset\mathbb{R}^n$, where $\alpha(n)$ is a constant dependent only on $n$, $d(Q)=\sup\{|x-y|: x,y\in Q\}$, i.e., $d(Q)$ denotes the diameter of the subset $Q$. Here it only requires that these cubes are parallel to the coordinate axes.

Frankly I can't believe this is true, because for me it says that the Lebesgue measure of any cube is dependent only on its diameter, which is obviously wrong. Am I mistaken ?

For your interests, this equation appears in page 71, Measure theory and fine properties of functions, 1e, by Evans.

Answer 1

The diameter of an $n$-dimensional cube (the diagonal of the cube) is $\sqrt n$ times the length of the side of the cube.