# Solution to the Four Squares II Puzzle +-- {.image} [[FourSquaresII.png:pic]] > Four squares. What’s the shaded area? =-- ## Solution by [[Dissection]] and [[Similar Triangles]] +-- {.image} [[FourSquaresIILabelled.png:pic]] =-- 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 $\frac{4}{5} \times 45 = 36$.