# Solution to the Two Rectangles of Equal Area Puzzle +-- {.image} [[TwoRectanglesofEqualArea.png:pic]] > The pink and blue rectangles have the same area. What is it? =-- ## Solution by [[Similarity]] and [[Vertically Opposite Angles]] +-- {.image} [[TwoRectanglesofEqualAreaLabelled.png:pic]] =-- 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 triangle|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 triangle|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: $$ c b( c a + 2 a) = 3 b a $$ 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 $4a \times 3 b = 12 a b = 72$.