Notes
divided hexagon solution

Solution to the Divided Hexagon Puzzle

Solution by Properties of a Regular Hexagon, Area of a Trapezium, and Area of a Triangle

Consider the points labelled as above. Let $a$ be the side length of the hexagon, and let $h$ be the half-height, so $A C$ has length $2 h$. The area of the hexagon is then $6 \times \frac{1}{2} a h = 3 a h$.

The length of $L B$ is $\frac{1}{2} a$ so triangle $A B C$ has area $\frac{1}{2} a h$ which is one sixth of the area of the hexagon, so triangle $A C G$ also has area $\frac{1}{2} a h$. As its height is $2 h$, the length of $C G$ must be $\frac{1}{2} a$ and so $G$ is the midpoint of $C D$.

Then as $H$ is half way along $A G$, $L H$ has length $\frac{1}{4} a$ (since $B C$ has length $2 a$). Therefore $E H$ has length $\frac{5}{4} a$, so trapezium $E H G D$ has area $\frac{1}{2} (\frac{5}{4} a + \frac{1}{2} a) h = \frac{7}{8} a h$. This is $\frac{7}{24}$ths of the area of the hexagon, so the area of triangle $E I H$ is $\frac{1}{24} \times 3 a h = \frac{1}{8} a h$. Since $E H$ has length $\frac{5}{4} a$, this means that $I J$ (the height of $I$ above $H E$) has length $\frac{1}{5} h$. So $K I$ has length $\frac{4}{5} h$ and the area of the green triangle is $\frac{1}{2} \times a \times \frac{4}{5} h = \frac{2}{5} a h$.

This is $\frac{2}{15}$ths of the area of the hexagon.