Question 78947
It would be easier to draw this, but here goes - I will type it

<pre>

 P ___ Q
U /___\ R
  \___/
T      S
</pre>

Draw diagonals QT and PS too.
Label the center O.
This divides your regular hexagon into six equilateral triangles.
These lines all have equal length:
 PQ, QR, RS, ST, TU,UP,PO,QO,RO,SO,TO,UO
The perimiter is 100 (all lengths in cm here) so if s is the length of those lines then 6 s = 100, or s = 100/6 (cm)


There is a point halfway between S and T on line ST (the midpoint).
Label this as M.  Consider the triangle OSM:

<pre>
   O
   |\
 h | \s
   |  \
 M ---- S
   s/2
</pre>

Use the pythagorean theorem (or trig) to find out the missing height h:

{{{ h^2 = s^2 - (s/2)^2 }}}
{{{ h^2 = 3 s^2 / 4  }}}
{{{ h = (sqrt(3)/2) s }}}

Now you know the area of triangle OMS is
{{{ 1/2 * (sqrt(3)/2) s * (s/2) }}} (one half times the base times the height)

simplified, {{{ (sqrt(3) * s^2) / 8 }}}

The area of the hexagon is 12 times this because the hexagon can be decomposed exactly into 12 triangles each congruent to OSM.

{{{ A = 12 * ((sqrt(3) * s^2) / 8) = ((3 * sqrt(3)) / 2) s^2 }}}
{{{ A = ((3 * sqrt(3)) / 2) * (100/6)^2 = 10000 * sqrt(3) / 24 }}}
This is approximately 721.688

As a check, compare to the area of the circumscribed circle of the hexagon
pi * (100/6)^2 = 872.665

I would expect these two areas to be a little closer together.  You may want to check my work above.  But I think it is pretty much right.