What is the area of a regular hexagon with side #4sqrt3# and apothem 6?

1 Answer
Nov 26, 2015

#72sqrt(3)#

Explanation:

First of all, the problem has more information than needed to solve it. If the side of a regular hexagon equals to #4sqrt(3)#, its apothem can be calculated and will indeed be equal to #6#.
The calculation is simple. We can use Pythagorean Theorem. If the side is #a# and apothem is #h#, the following is true:
#a^2 - (a/2)^2 = h^2#
from which follows that
#h = sqrt(a^2 - (a/2)^2) = (a*sqrt(3))/2#

So, if side is #4sqrt(3)#, apothem is
#h = [4sqrt(3)sqrt(3)]/2 = 6#

The area of a regular hexagon is #6# areas of equilateral triangles with a side equal to a side of a hexagon.
Each such triangle has base #a=4sqrt(3)# and altitude (apothem of a hexagon) #h=(a*sqrt(3))/2=6#.

The area of a hexagon is, therefore,
#S = 6*(1/2)*a*h = 6*(1/2)*4sqrt(3)*6 = 72sqrt(3)#