Notes
toppled square solution

Toppled Square

Solution by Pythagoras' Theorem

The heights of the three squares on the right-hand side are $3 \sqrt{3}$, $2 \sqrt{3}$, and $\sqrt{3}$. Therefore the total height of $F$ above the base is $6\sqrt{3}$.

The length of $C D$ is $3\sqrt{3} - 2 \sqrt{3} = \sqrt{3}$ and of $E D$ is $2\sqrt{3}$. Let $a$ be the length of $E C$ then applying Pythagoras' theorem to triangle $E C D$ shows that $x^2 = 3 + 12 = 15$ so $x = \sqrt{15}$. Since the height of $F$ above the base is $3$ times the length of $E D$, the length of $F A$ is $3\sqrt{15}$. Therefore the area of the large square is $135$.