Notes
three stacked squares solution

Solution to the Three Stacked Squares Puzzle

Solution by Pythagoras' Theorem

Let $a$, $b$, $c$ be the side lengths of the three squares in increasing order. Since the smaller squares stack against the larger, $c = a + b$. Since the two larger squares stretch across the semi-circle, $b + c = 5$. So $c = 5 - b$ and $a = 5 - 2 b$.

With the points labelled as in the diagram, $O E$ is a radius of the semi-circle so has length $2.5$, $D E$ has length $b$, and $C D$ has length $a$. The length of $O D$ is the length of $B D$ minus the radius, so is $a + c - 2.5 = 7.5 - 3 b$.

Pythagoras' theorem applies to triangle $O D E$, meaning that

This simplifies to $0 = 2 b^2 - 9 b + 10$ which factors as $(b - 2)(2 b - 5) = 0$ so $b = 2$ or $b = 2.5$. But if $b = 2.5$ then $a = 0$ and so there wouldn’t be three squares.

With $b = 2$, then $a = 1$ and $c = 3$, so the total area is $14$.