3475.11 – The Unknown Edge

Let $x$ be the unknown edge length. In the figure, the diagonal of the cube, which we are given is 20, is $AG$. It is also the hypotenuse of $\triangle ACG$. By the Pythagorean Theorem, using that $AC = x\sqrt{2}$,

$$20 = AG = \sqrt{(x\sqrt{2})^2 + x^2} = \sqrt{3x^2}.$$

Solving for $x$,

$$\begin{aligned}3 x^2 &= 400 \\[0.5em]x^2 &= \dfrac{400}{3} \\[1em]x &= \dfrac{20}{\sqrt{3}} \\[1em]&\approx 11.5 cm\end{aligned}$$

This is the Pythagorean Theorem in three dimensions:

$$AG^2 = x^2 + x^2 + x^2.$$

It works for a rectangular solid (not just a cube) and actually works in any number of dimensions as well!