# The base of a regular pyramid is a hexagon. What is the area of the base of the pyramid? Express your answer in radical form.

May 24, 2017

Area of hexagon is $216 \sqrt{3}$ $c {m}^{2}$

#### Explanation:

We know that the angle at $C$ is 30 degrees, so half of the base must be $\frac{12}{2}$ which is $6$ as $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.
Finding $a$ is a matter of Pythagorean,

as such ${a}^{2} + {6}^{2} = {12}^{2}$

or ${a}^{2} = 144 - 36 = 108$ and

$a = \sqrt{108} = \sqrt{6 \times 6 \times 3} = 6 \sqrt{3}$.

Now, we have the height and half of the base, which means the area of $\frac{1}{6}$th of the hexagon will be $\frac{1}{2} b h$ where $b = 12$ and $h = 6 \sqrt{3}$.

The area of the hexagon therefore will be $6 \times \frac{1}{2} b h$.

$= 6 \times \frac{1}{2} \times 12 \times 6 \sqrt{3} = 216 \sqrt{3}$ $c {m}^{2}$