Using a few facts that can be taken off the internet:
* Sum of interior angles of a pentagon:

* Area of a regular pentagon (in terms of side length, S):

The geometric proofs of these claims are probably not incredibly difficult, but probably not that easy either, so it's easier just to start from here, and then clean up with simple geometry that we can do.
Interior angle of a pentagon:

Side length of the pentagon (using the area of 800 square centimetres):



The angle that appears twice in the isoceles triangle is:

(180 comes from the straight line that is made by a face of the regular pentagon and a face of a triangle - this also confirms the triangle is isoceles.)
Therefore, the angle at the far end is

Bisect the isoceles triangle to get a triangle with angle:

and an opposite side length of:

Therefore, the adjacent side length is equal to:


The adjacent side length also happens to be the height of the triangle, so the area of the triangle is:



The fraction on the right simplifies to

, confirming Ahmad's answer.
Simplification 
to come in the next post. Hey, I might as well research on how to derive those 2 facts I stated above (but I doubt I'll get
that far).