Notes
triangle over two rectangles solution

Solution to the Triangle Over Two Rectangles Puzzle

Solution by Area of a triangle

With the lengths as in the above diagram, the areas are as follows:

• Orange: $\frac{1}{2} d (a + c)$
• Purple: $\frac{1}{2} a b$
• Pink: $a b + c d - \frac{1}{2} d(a + c) - \frac{1}{2} a b$
• White: $\frac{1}{2} c(b - d)$

Simplifying the area of the pink region leads to the expression:

Since the pink and purple regions have the same area, the last two terms in this expression must cancel out and so $a = c$. Putting this into the expression for the orange region gives that as $a d$ and so as this is the same as the area of the purple region it must be that $b = 2 d$. The expression for the area of the white region therefore simplifies to $\frac{1}{4} a b$ which is half the area of the purple region and so the upper triangle has area $6$.

Solution by Dissection

In the above diagram, triangles $B E F$ and $B A F$ are congruent and so the same area. As each coloured region has the same area, this means that the triangles $F E H$ and $C I H$ have the same area. They are also similar since they are right-angled triangles and angles $I \hat{H} C$ and $E \hat{H} F$ are the same as they are vertically opposite angles. They are therefore congruent. This means that $C I$ and $F E$ have the same length and so $I$ is the midpoint of $C G$.

The height of triangle $C D F$ is therefore twice that of $A B F$ so since they have the same area, the base of $A B F$ must be twice the base of $C D F$. That is to say, the length of $C D$ is half that of $A F$. So then the length of $B I$ is half that of $A F$.

Putting that together, the lengths of $C I$ and $B A$ are the same, while that of $B I$ is half that of $A F$, so the area of triangle $I B C$ is half that of triangle $A F B$, and so has area $6$.