how do I figure out area of a polygon with perimeter of 108

Question

anonymous · Answer

It depends.  Is the polygon regular (that is, does it have equal internal angles and equal sides)?

xkehaulanix · Answer

Is it a regular polygon?

anonymous · Answer

It's a 9 sided polygon

anonymous · Answer

Okay, it sounds like it's regular, since 9 divides 108.  If it's regular, you can look at the shape and break it up into 9 isosceles triangles.  

You can find the area using a combination of cosine rule (to find the other sides of the isosceles triangle) and then formula for area of a triangle (1/2 ab sinC).  The angle C will be 360degrees/(number of sides).

The formula for the area will be

$$A=\frac{na^2\sin (360^o/n)}{4(1-\cos(360^o/n))}$$where n is the number of sides, a is the side length.  Here, $$a=\frac{108}{9}=12$$ and n = 9, so$$A=\frac{9 \times 12^2.\sin(40^o)}{4 \times (1-\cos(40^o))} \approx 890 $$square units.

If you want a proper derivation of the formula, let me know.  I have to rush right now.  Also, the formula only holds for regular polygons; that is, polygons that have the same side length and equal internal angles.