Notes
three squares and a triangle solution

Solution to the Three Squares and a Triangle Puzzle

Solution by Similar Triangles and Area of Rectangles and Triangles

Label the points as above, where $I$ is such that $F I$ is perpendicular to $H D$. As each square has area $4$, the side lengths are $2$.

Since the pink triangle is right-angled, angles $A \hat{H} B$ and $F \hat{H} G$ add up to $180^\circ$, meaning that triangles $B A H$ and $H G F$ are similar. This shows that the lengths of the line segements $H I$ and $F I$ are in the ratio $1 : 3$.

Since $B D$ is the diagonal of a square, and it continues through to $F$ then $D E F I$ is also a square. Therefore, line segments $I D$ and $F I$ are the same length. Therefore, $I$ splits $H D$ into the ratio $1 : 3$. Since $H D$ has length $4$, this means that $H I$ has length $1$ and $I D$ has length $3$.

The original three squares have area $3 \times 4 = 12$. Rectangle $D E G H$ has area $4 \times 3 = 12$. So the outer shape has area $24$. The four white triangles have the following areas:

• $H A B$: $\frac{1}{2} \times 2 \times 8 = 8$
• $D C B$: $\frac{1}{2} \times 2 \times 2 = 2$
• $D E F$: $\frac{1}{2} \times 3 \times 3 = \frac{9}{2}$
• $F G H$: $\frac{1}{2} \times 1 \times 3 = \frac{3}{2}$

So the pink triangle has area: