Exercise 90

Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(9,3)$, respectively. What is its area?

Solution

For regular hexagon $ABCDEF$ with $AB=a$, $AC=\sqrt3\,a$, so $[ABCDEF]=6\cdot\frac{\sqrt3}4a^2=\frac{\sqrt3}2\,AC^2$.
$AC^2=9^2+3^2=90$.
$[ABCDEF]=\frac{\sqrt3}2(90)=\boxed{45\sqrt3}$.