Notes
three squares and an equilateral triangle solution

Solution to the Three Squares and an Equilateral Triangle Puzzle

There is a hidden assumption here that the lower edge is a straight line.

Solution by Angles in a Triangle, Isosceles Triangles, and Lengths in a Square

With the points labelled as above, angle $A \hat{C} D = 45^\circ$ and angle $A \hat{C} B = 60^\circ$, so angle $D \hat{C} B = 15^\circ$. Then since angle $D \hat{C} E = 90^\circ$, angle $B \hat{C} E = 75^\circ$.

Angle $A \hat{B} C = 60^\circ$ and angle $A \hat{B} E = 90^\circ$, so angle $C \hat{B} E = 30^\circ$. Then angle $C \hat{E} B = 180^\circ - 30^\circ - 75^\circ = 75^\circ$. Thus triangle $B C E$ is isosceles.

This means that $B E$ and $B C$ are the same length, then also $B E$ and $A C$ are the same length. So the side length of the right-hand square is the same length as the diagonal of the pink square.

The area of the pink square is therefore half the area of the right-hand square, thus is $4$.