Notes
three squares solution

Solution to the Three Squares Puzzle

Solution by Properties of Right-Angled Triangles and Congruent Triangles

With the points labelled as above, the side lengths of the squares are $4\sqrt{2}$, $3\sqrt{2}$, and $5.

Point $B$ is such that triangle $F B A$ is right-angled, and so also point $E$ is such that $F E D$ is also right-angled. Angles $D \hat{F} E$ and $A \hat{F} B$ are equal, as both add to angle $B \hat{F} D$ to make a right-angle. Therefore, triangles $F B A$ and $F E D$ are congruent.

Let line segment $F E$ have length $a$ and $E D$ have length $b$. Then these are also the lengths of line segments $F B$ and $A B$, respectively. Translating line segment $F E$ along $F B$ shows that the lengths of $A B$ and $F E$ add up to $4 \sqrt{2}$, similarly the difference of the length of $F B$ and of $E D$ is $3 \sqrt{2}$. That is, $a + b = 4 \sqrt{2}$ and $a - b = 3 \sqrt{2}$. Hence $a = \frac{7\sqrt{2}}{2}$.

Triangle $F E C$ is right-angled and isosceles with shorter side length $a = \frac{7\sqrt{2}}{2}$. Its diagonal is therefore $\sqrt{2} a = \frac{7 \sqrt{2}}{2} \times \sqrt{2} = 7$.

Hence line segment $F C$ has length $7$.