Is this possible? Assume a polygon that has all equal angles and lengths
Sure!
A regular polygon has "n" sides.
Therefore it may be subdivided into "n" isosceles triangles, all meeting at the centre.
The angle at the central apex of each triangle is 360/n degrees.
Subtract that from 180 degrees and divide by 2 to get the other angles "A" of each isosceles triangle.
Take half the length of a side of the polygon.
Use TanA multiplied by half a side length squared to get a triangle area.
"n" of these is the polygon area.
Here is the situation....
"x" is half the length of a side.
"h" is the perpendicular height of each isosceles triangle.
$\displaystyle TanA=\frac{h}{x}\Rightarrow\ h=xTanA\Rightarrow\ xh=x^2TanA$
That is the area of each isosceles triangle
(twice the area of the right-angled triangle as 2 of these form the isosceles triangle).
Multiply by "n" for the polygon area.