Notes
three squares iv solution

Solution to the Three Squares IV Puzzle

Solution by Transformation and Congruent Triangles

With the points labelled as above, triangles $D E F$ and $D B C$ are related by a rotation of $90^\circ$ about point $D$. They are therefore congruent, so line segments $D E$ and $D B$ have the same length. This means that $D$ lies on the diagonal $G H$, and this further means that $D$ is the same distance from $F$ as it is from $A$.

Therefore the side length of the blue square is $\sqrt{2}$ times the side length of the green square, and so the area of the blue square is twice the area of the green square, meaning that it has area $2 \times 16 = 32$.

Solution by Invariance Principle

The point $C$ in the above diagram must lie on the base of the black square, and with that constraint there are some configurations that make the solution simple to see.

In this configuration, the blue and black squares coincide, while the diagonal of the green square is clearly one side of both.

In this configuration, the blue and black squares are alongisde and congruent. The diagonal of the green square is one side of the black square, so also one side of the blue.