Square in a triangle.

We note that if we remove the square and move the right bottom triangle to the left, it forms a triangle similar to the large triangle. If the length of the side of the square be $x$ . Then, we have:

$\Rightarrow \dfrac {30-x}{x} = \dfrac {30}{20}\quad \Rightarrow 60-2x = 3x \quad \Rightarrow x = \dfrac {60}{5} = \boxed{12}$

It can be easily shown that the side of this square is given by [b*h/(b+h)] where b is the length of the base and h is the height of the triangle