Notes
two rectangles of equal area solution

Solution to the Two Rectangles of Equal Area Puzzle

Solution by Similarity and Vertically Opposite Angles

With the points labelled as above, let $A B$ have length $a$ and $B C$ have length $b$. The area of triangle $C B A$ is $\frac{1}{2} a b$ so $a b = 6$.

The triangles $E C D$ and $A C B$ are similar since they are both right-angled and angles $B \hat{C} A$ and $D \hat{C} E$ are equal as they are vertically opposite. The area scale factor is $4$, so the length scale factor is $\sqrt{4} = 2$. This means that the length of $C D$ is twice that of $A B$, and of $D E$ is twice that of $A B$. In terms of $a$ and $b$, $C D$ has length $2 b$ and $D E$ has length $2 a$.

Triangle $E F G$ is also right-angled and angles $G \hat{E} F$ and $D \hat{E} C$ are equal as they are vertically opposite. So $G F E$ is similar to $C B A$. Let the scale factor be $c$, so that the length of $E F$ is $c a$ and of $G F$ is $c b$.

Rectangle $G F D I$ has width $c b$ and height $c a + 2 a$ while rectangle $D B A H$ has width $3 b$ and height $a$. As the pink and blue rectangles have the same area, these two rectangles also have the same area. So:

which simplifies to $c(c + 2) = 3$. The solution to this is $c = 1$ (the other possible solution is $c = -3$ but $c$ is a positive scale factor). So $G F E$ is actually congruent to $C B A$ and the length of $F H$ is $4a$. The area of the blue rectangle is therefore $4a \times 3 b = 12 a b = 72$.