Notes
two tilted squares solution

Solution to the Two Tilted Squares Puzzle

Solution by Pythagoras' Theorem and Congruent Triangles

With the points labelled as in the above diagram, let $a$, $b$, and $c$ be the lengths of sides $B C$, $F C$, and $F B$ respectively. Applying Pythagoras' theorem to triangle $F C B$ results in $c^2 = a^2 + b^2$.

Triangles $F C B$, $F G H$, and $B A I$ are all right-angled triangles with hypotenuse a side of the square $I B F H$ and angle $I \hat{B} A$, $B \hat{F} C$, and $H \hat{F} G$ are all equal, so the triangles are all congruent.

Since $F G$ and $E F$ are of length $b$, applying Pythagoras' theorem to triangle $E G H$ results in $6^2 = (2 b)^2 + a^2 = 4 b^2 + a^2$.

The area of the pink region can be found by finding the area of the full region and subtracting the area of the bottom triangle, $I B D$. The squares have areas $c^2$ and $b^2$, the middle triangle has area $\frac{1}{2} a b$. Triangle $I B D$ has base $B D$ of length $a + b$ and height $I A$ of length $a$. So it has area $\frac{1}{2} a(a + b)$. The area of the pink region is therefore:

Solution by Invariance Principle

In this version of the diagram, the two squares are aligned and so the total distance across both is $6$ meaning that each has a side length of $3$. Their area is then $2 \times 3^2 = 18$.