Notes
rectangle on a hexagon solution

Solution to the Rectangle on a Hexagon Puzzle

Solution by Dissection

In the above diagram, triangles $G H F$ and $E B D$ are congruent so the rectangle has the same area as parallelogram $H B D F$. Then triangles $F C D$ and $H A B$ are congruent, so the rectangle has the same area as the rectangle $H A C F$. From the properties of a regular hexagon, this rectangle comprises four of the six equilateral triangles that make up the hexagon, so its area is $\frac{4}{6} \times 30 = 20$.

Solution by Shearing

Shear the rectangle parallel to $H B$ so that $G$ ends up at $F$, then shear the resulting parallelogram parallel to $H F$ so that $B$ ends up at $A$. This shows that the shaded rectangle has the same area as $H A C F$.

Solution by Invariance Principle

The shaded rectangle can be drawn with $B$ at any point between $A$ and $C$. In particular, it can be drawn with $B$ at $A$ which results in the rectangle $H A C F$.