Notes
two squares overlapping a third square solution

Solution to the Two Squares Overlapping a Third Square Puzzle

Solution by Lengths in Equilateral Triangles and Lengths in Squares

In the above diagram, the point labelled $C$ is directly across from $B$, so that angle $A \hat{C} B$ is a right-angle. As the red squares have area $4$, they have side length $2$ and diagonal $2\sqrt{2}$.

The length of line segment $B C$ is half of a diagonal of the red squares, namely $\sqrt{2}$, so triangle $A B C$ is a right-angled triangle with one side half of the hypotenuse. This means that it is half of an equilateral triangle and so $A C$ has length $\sqrt{3}$ times the length of $B C$, so has length $\sqrt{6}$. Since $C D$ also has length $\sqrt{2}$, the full length of the side of the outer square is $\sqrt{6} + \sqrt{2}$.

The length of $E F$ is then $\sqrt{6} + \sqrt{2} - 2\sqrt{2} = \sqrt{6} - \sqrt{2}$. The area of triangle $E G F$ is therefore: