# Crossed Hexagon +-- {.image} [[CrossedHexagon.png:pic]] > What fraction of the regular hexagon is shaded? =-- ## Solution by [[Similar Triangles]] and [[Vertically Opposite Angles]] +-- {.image} [[CrossedHexagonLabelled.png:pic]] =-- In triangles $B G D$ and $A G E$, the sides $B D$ and $A E$ are [[parallel]], and the angles $B \hat{G} D$ and $E \hat{G} A$ are [[vertically opposite]] and so equal. Therefore, triangles $B G D$ and $A G E$ are [[similar]]. Since $A E$ has twice the length of $B D$, the length [[scale factor]] is $2$ and the [[area scale factor]] is $4$. The length of $F G$ is twice that of $C G$, so $C G$ has length one third of that of $C F$. Triangle $B F D$ is an [[equilateral triangle]] with area one sixth of the area of the whole [[hexagon]]. So the area of $B G D$ is $\frac{1}{18}$th of the area of the whole hexagon. Triangle $A G E$ then has area four times that, so has area $\frac{4}{18}$ths of the whole hexagon. Therefore, the area of the shaded region is $\frac{10}{18} = \frac{5}{9}$ths of the hexagon. ## Solution by [[Crossed Trapezium]] The [[quadrilateral]] $A B D E$ with its diagonal lines is a [[crossed trapezium]]. In such, the areas of the four triangles are in the ratio $1 : s : s : s^2$ where $s$ is the ratio of the parallel sides. Therefore the ratio of the area of the shaded region to the whole trapezium is $\frac{1 + s^2}{1 + 2 s + s^2}$. As $s = 2$, this is $\frac{5}{9}$. Since the other half is a copy of this trapezium, that ratio then holds for the whole hexagon.