Notes
three triangles in a square solution

Solution to the Three Triangles in a Square Puzzle

Solution by Area of a Triangle and Pythagoras' Theorem

In the diagram above, $E F G$ is a straight line parallel to the side of the square.

Triangles $C F B$ and $D C H$ have the same area and the same length base since $C B$ and $D C$ are sides of the square, so by the formula for the area of a triangle their heights must be the same. Hence $F G$ and $D H$ have the same length.

Then since $D H$ and $H A$ make up one side of the square, which is the same length as $E G$, $E F$ must have the same length as $H A$, namely $4$. This means that the area of triangle $A H F$ is $\frac{1}{2} \times 4 \times 4 = 8$, and so this is the area of all the triangles.

Let $x$ be the side length of the square. Then $D H$ has length $x - 4$, so since the orange triangle has area $8$, we have: $8 = \frac{1}{2}x (x - 4)$ so $x^2 - 4x = 16$.

Let $y$ be the length of $C H$. Applying Pythagoras' theorem to triangle $D H C$ shows that:

Hence $y = \sqrt{48} = 4\sqrt{3}$.