Notes
three squares and a rectangle solution

Solution to the Three Squares and a Rectangle Puzzle

Solution by Similar Triangles and Lengths in a Square

Consider the diagram as labelled above. In this diagram, triangles $A B C$ and $E D C$ are right angled and angle $A \hat{C} B$ is equal to angle $D \hat{C} E$ since each together with angle $A \hat{C} E$ make a right-angle. Therefore, triangles $A B C$ and $E D C$ are similar.

This means that the ratios $A C : B C$ and $E C : D C$ are equal. So the lengths satisfy:

Multiplying up gives:

Then $E C$ is the diagonal of a square of side length $3$, and $B C$ of $4$, so $E C$ has length $3 \sqrt{2}$ while $B C$ has length $4 \sqrt{2}$. Hence

Solution by Invariance Principle

The tilt of the shaded rectangle is not specified, allowing a range of angles with two special cases.

In this configuration, the small square is shrunken to a point meaning that the sides of the rectangle are the diagonals of the squares. So the area is $4\sqrt{2} \times 3\sqrt{2} = 24$.

In this configuration, the small square is the same size as the square of side length $4$, meaning that the shaded rectangle has sides $3$ and $8$ so area $24$.