What is the area of a hexagon with an apothem of 9?

Nov 27, 2015

$162 \sqrt{3}$ square units

Explanation:

The apothem is the length from the center of a regular polygon to the midpoint of one of its sides. It is perpendicular (${90}^{\circ}$) to the side.

You can use the apothem as the height for the whole triangle:

To find the area of the whole triangle, we first need to find the length of the base, since the base length is unknown.

To find the base length, we can use the formula:

$b a s e = a p o t h e m \cdot 2 \cdot \tan \left(\frac{\pi}{n}\right)$

where:
$\pi = \pi$ radians
$n$ = number of whole triangles formed in a hexagon

$b a s e = a p o t h e m \cdot 2 \cdot \tan \left(\frac{\pi}{n}\right)$
$b a s e = 9 \cdot 2 \cdot \tan \left(\frac{\pi}{6}\right)$
$b a s e = 18 \cdot \tan \left(\frac{\pi}{6}\right)$
$b a s e = 18 \cdot \frac{\sqrt{3}}{3}$
$b a s e = \frac{18 \sqrt{3}}{3}$
$b a s e = \frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{18}}}}^{6} \sqrt{3}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}$
$b a s e = 6 \sqrt{3}$

To find the area of the hexagon, find the area of the whole triangle and multiply the value by $6$, since $6$ triangles can be formed in a hexagon:

$A r e a = \left(\frac{b a s e \cdot a p o t h e m}{2}\right) \cdot 6$
$A r e a = \left(\frac{b a s e \cdot a p o t h e m}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right) \cdot {\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}}}^{3}$
$A r e a = b a s e \cdot a p o t h e m \cdot 3$
$A r e a = 6 \sqrt{3} \cdot 9 \cdot 3$
$A r e a = 54 \sqrt{3} \cdot 3$
$A r e a = 162 \sqrt{3}$