# Solution to the Two Overlapping Hexagons and a Rectangle Puzzle +-- {.image} [[TwoOverlappingHexagonsandaRectangle.png:pic]] > One corner of each regular hexagon lies on the diagonal of the rectangle. What fraction of the rectangle is shaded? =-- ## Solution by [[Similarity]] and [[hexagon|Lengths in a Regular Hexagon]] +-- {.image} [[TwoOverlappingHexagonsandaRectangleLabelled.png:pic]] =-- 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.