Notes
overlapping squares solution

Solution to the Overlapping Squares Puzzle

Solution by Pythagoras' Theorem, Area of a Square, and Crossed Trapezium

With the points labelled as in the above diagram, let $a$ be the length of $A B$, $b$ of $B C$, $c$ of $A C$, $d$ of $A D$, and $e$ of $D B$. As several of these are sides of squares, then $a^2 = 16$, $c^2 = 20$, and $e^2 = 17$. In particular, $a = 4$.

Triangle $D A B$ has a right-angle at $A$, so Pythagoras' theorem applies and shows that $e^2 = d^2 + a^2$ which rearranges to $d^2 = e^2 -a^2 = 17 - 16 = 1$, so $d = 1$. A similar argument applied to triangle $A B C$ shows that $b^2 = c^2 - a^2 = 20 - 16 = 4$ so $b = 2$.

Triangles $A E D$ and $C B E$ are part of a crossed trapezium and so are similar. The length of $C B$ is twice that of $A D$, so the height of $E$ above $C B$ is twice that of $E$ above $A D$. Since these heights must add up to the length of $A B$, which is $4$, then $E$ is $\frac{4}{3}$ above $A D$ and $\frac{8}{3}$ above $C B$. The areas of the triangles are therefore $\frac{1}{2} \times 1 \times \frac{4}{3} = \frac{2}{3}$ and $\frac{1}{2} \times 2 \times \frac{8}{3} = \frac{8}{3}$ respectively. Their difference is therefore $\frac{8}{3} - \frac{2}{3} = 2$.