Notes
quadrilateral from two squares in a square solution

Solution to the Quadrilateral from Two Squares in a Square Puzzle

Solution by Area of a Trapezium and Square

In the above diagram, the quadrilateral formed is $A B C D$ which is a trapezium. Let $a$ be the side length of the smaller square and $b$ of the larger. The area of the trapezium is $\frac{1}{2}(a + b) \times (a + b) = \frac{1}{2} (a + b)^2$.

Consider the diagonal $E I$ of the outer square. The triangle $E F D$ is an isosceles right-angled triangle so $E F$ has the same length as $F D$, which is $\frac{1}{2} a$. Therefore $E G$ has length $\frac{3}{2} a$. Similarly, $G I$ has length $\frac{3}{2} b$. The diagonal $E I$ then has length $\frac{3}{2} (a + b)$.

The area of a square can be calculated as half the square of the length of a diagonal, so $\frac{1}{2} \left( \frac{3}{2} (a + b) \right)^2 = 12^2 = 144$. Rearranging this shows that $\frac{1}{2} (a + b)^2 = \frac{4}{9} \times 144 = 64$.