Find the area of a regular hexagon with a 48-inch perimeter

If the perimeter is 48 inches then each side is 8 inches.

The angle at the centre that subtends to adjacent vertices is 360/6 = 60degrees.

Therefore the hexagon consists of 6 equilateral triangles of sidelength 8 units.

Using Heron's formula

$$\\Area = 6 \;\; *\;\; \sqrt{s(s-a)(s-b)(s-c)}\qquad where\quad s=(a+b+c)/2\\\\ s=24/2=12\\\\ Area = 6 \;\; * \;\; \sqrt{12(12-8)(12-8)(12-8)}\\\\ Area = 6 \;\; * \;\; \sqrt{12*64}\\\\ Area = 6 \;\; * \;\; 8\sqrt{4*3}\\\\ Area = 6 \;\; * \;\; 16\sqrt{3}\\\\ Area = 96\sqrt{3}\;\;inches^2\\\\$$

Heron's Formula strikes, again  !!!!!!  ........ LOL!!!

You showed me Heron's formula.       I still had to look it up though.   

I was going to do it the usual way which would not have been very difficult but then I thought it was the ideal situation for ME to use Heron's formula :))