Notes
three aligned squares solution

Three Aligned Squares

Solution by Pythagoras' Theorem

As the lengths of the sides of the three squares are consecutive integers, they can be written as $n+1$, $n$, and $n-1$ where $n$ is the length of the side of the light blue square.

Since $C D$ has length $n-1$, the length of $A D$ is $2 n$, so $B$ is the midpoint of $A D$. The length of $F B$ is $n+1-(n-1) = 2$. Since $B$ is the midpoint of $A D$, $F$ is the midpoint of $A E$ and so $A F$ has length $2\sqrt{10}$. Applying Pythagoras' theorem to triangle $A B F$ shows that $40 = 4 + n^2$ which means that $n =6$. The squares therefore have sides $5$, $6$, and $7$.

Since $F B$ has length $2$ and $B C$ has length $1$, the total area of the shape is $7^2 + 6^2 + 5^2 - 2 = 108$.