Question

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

Answer 1

`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)`