Notes
four squares ii solution

Solution to the Four Squares II Puzzle

Solution by Dissection and Similar Triangles

With the points labelled as in the above diagram, the shaded region has the same area as that of rectangle $K E F H$. This can be seen by dissection, since triangles $K L E$ and $I J G$ are congruent, and then triangles $I J K$ and $G F E$ are also congruent.

Triangle $G F E$ is also congruent to $L M E$, so the length of $F E$ is the same as that of $M E$. It is also similar to triangle $E D N$. The squares $A B Q P$ and $B C D N$ have area scale factor $4$, so have length scale factor $2$, meaning that $B C$ is twice the length of $A B$. So $D N$ is twice the length of $E D$. By similarity, then, $M E$ is twice the length of $M L$ which is twice the length of $M K$, so $M E$ is $\frac{4}{5}$ths of $K E$.

The area of $K E F H$ is therefore $\frac{4}{5}$ths of the area of a square with side length $K E$, but such as square is $A C E K$ and this has area $20 + 5 + 5 + 5 + 5 + 5 = 45$. So the shaded area is $\frac{4}{5} \times 45 = 36$.