Notes
two overlapping hexagons and a rectangle solution

Solution to the Two Overlapping Hexagons and a Rectangle Puzzle

Solution by Similarity and Lengths in a Regular Hexagon

The two triangles at the top of the shaded region are congruent to the two white triangles below it, showing that the area of the shaded region is the same as the area of the rectangle with height parallel to $K H$ and width parallel to $E F$.

The height of the outer rectangle, which is the length of $C B$, is the larger diameter of the hexagons, which is twice a side length. Thus the length of $C B$ is twice that of $H G$. The points $E$ and $F$ correspond to $H$ and $G$ along $A B$, so $E F$ is half of $A B$ and is located centrally along that edge.

So along $A B$, $J$ is at the midpoint, while $E$ is at the midpoint of $A J$ and $F$ of $J B$. Point $I$ is at the midpoint of $A F$. So the points along $A B$ lie at the following locations (as proportions of $A B$): $E$ is at $\frac{1}{4}$, $I$ at $\frac{3}{8}$, $J$ at $\frac{1}{2}$, and $F$ at $\frac{3}{4}$. This means that $J$ is one third of the way between $I$ and $F$, so $K$ is one third of the way from $B$ to $G$. Since $B G$ is one quarter of $B C$, $K G$ is two thirds of one quarter of $B C$, which is $\frac{1}{6}$.

As a proportion of $B G$ then $K H$ is $\frac{1}{2} + \frac{1}{6} = \frac{2}{3}$ while $E F$ is $\frac{1}{2}$ of $A B$. So the shaded region has area equal to $\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$ of the area of the outer rectangle.