15.7. If $A B C D E F$ is a regular hexagon with each side of length 6 units, what is the area of $\triangle A C E$ ?
15.7. $27 \sqrt{3}$ square units
Note that $\triangle A C E$ is equilateral. Each interior angle of $A B C D E F$ measures $\frac{1}{6}(6-2)\left(180^{\circ}\right)=120^{\circ}$. Using a property of a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, we have
$$
\frac{1}{2} A C=\frac{\sqrt{3}}{2} \cdot 6 \quad \text { or } \quad A C=6 \sqrt{3} .
$$

The height of $\triangle A C E$ is $\frac{\sqrt{3}}{2} \cdot 6 \sqrt{3}=9$, so that area of $\triangle A C E=\frac{1}{2} \cdot 6 \sqrt{3} \cdot 9=27 \sqrt{3}$.