Notes
hexagon in a star in a hexagon solution

Hexagon in a Star in a Hexagon

Solution by Similar Triangles and Crossed Trapezium

In the above diagram, the line segments $A B$ and $D E$ are parallel. Angles $E \hat{C} D$ and $A \hat{C} B$ are vertically opposite so are equal. Put together, this means that triangles $D C E$ and $A C B$ are similar.

Triangle $E C F$ is equilateral and $B$ is the midpoint of $C F$, so the length of $C B$ is half of that of $C E$, which has equal length with $C D$. This means that the scale factor from triangle $A C B$ to $D C E$ is $2$. So $D E$ has twice the length of $A B$, and since the area scale factor is the square of the length scale factor, the area of the outer hexagon is $4 \times 18 = 72$.

The quadrilateral $A B E D$ is a trapezium since $A B$ and $D E$ are parallel. Its area is one sixth of the difference between the hexagons, so has area $9$. By considering the areas in a crossed trapezium, triangles $D C A$ and $E C B$ have twice the area of $A C B$, and $D C E$ has four times that area. Putting that together, triangle $A C B$ has area one ninth of the whole trapezium, so the shaded region inside the trapezium has area $5$. The entire shaded region therefore has area $30$.