Notes
equilateral triangle in a hexagon solution

Solution to the Equilateral Triangle in a Hexagon Puzzle

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

With the points labelled as above, let $h$ be half the height of the hexagon. The side length is $3 a$, so the area of the hexagon is $6 \times \frac{1}{2} \times 3 a \times h = 9 a h$.

Consider the region $G B C I$. By symmetry, $G$ is one third of the way between $B$ and $A$, so the distance between the lines $B C$ and $G H$ is $\frac{1}{3} h$. Similarly, $I$ is one third of the way from $D$ to $C$, so the height of $I$ above $G H$ is also $\frac{1}{3} h$. Since $B C$ has length $3 a$ and $A D$ has length $6 a$, the length of $G H$ is $4 a$. So the area of $G B C I$ is:

There are three such regions, meaning that the total unshaded area is $\frac{11}{2} a h$ which means that the shaded region has area $\frac{7}{2} a h$ and this is $\frac{7}{18}$ths of the hexagon.