Question 78947
<pre><font size = 5><b>
please show me how to find the area of a regular 
hexagon with a perimeter of 100cm.
   ________
  /        \
 /          \
/            \
\            /
 \          /
  \________/


Since the sum of all 6 sides is 100cm and since all
the sides are equal, each one is 100/6 cm or 50/3 cm.

The hexagon is made up of 6 equilateral tringles
like this:

   _______
  /\     /\
 /  \   /  \
/____\ /____\
\    / \    /
 \  /   \  /
  \/_____\/

So we will find the area of one of those equilateral
triangles and then multiply it by 6.

     / \    
50/3/   \50/3  
   /_____\
    50/3 

To find the area of that equilateral triangle we must
find its height so we can plug in the formula for the
area of a triangle:

A = (base)(height)/2

To do that we draw in the height, hy

     /|\    
50/3/ |h\50/3  
   /__|__\
    50/3
 
To find that height, we notice that the
height divides the equilateral triangle
into two right triangles, and here is the
left one:


     /|    
50/3/ |h  
   /__|
     
where the bottom side is 1/2 of 50/3 or 25/3

     /|    
50/3/ |h  
   /__|
   25/3

Now we use the Pythagorean theorem to find h

{{{(25/3)^2 + h^2}}} = {{{(50/3)^2}}}

   {{{625/9 + h^2}}} = {{{2500/9}}}

           {{{h^2}}} = {{{2500/9 - 625/9}}}

           {{{h^2}}} = {{{1875/9}}}

     {{{sqrt(h^2)}}} = {{{sqrt(1875)/sqrt(9)}}}

             {{{h}}} = {{{sqrt(625*3)/3}}}

             {{{h}}} = {{{(25sqrt(3))/3}}}
                    
Now we can find the area of the equilateral triangle

     /|\    
50/3/ |h\50/3  
   /__|__\
    50/3

A = {{{1/2}}}(base)(height)

A = {{{1/2}}}{{{50/3}}}{{{(25sqrt(3))/3}}}

A ={{{25/3}}}{{{(25sqrt(3))/3}}}

A = {{{(625sqrt(3))/9}}}


and since the area of the regular hexagon is 6 of these,
we multiply that by 6 and get5

Area of hexagon = {{{(6)(625sqrt(3))/9}}}

Area of hexagon = {{{(2)(625sqrt(3))/3}}}

Area of hexagon = {{{(1250sqrt(3))/3}}}

Edwin</pre>