Relation between the radius and the area of tangential polygon I've recently found a book with loads of formulas for triangle area, but unfortunaly the formulas were just listed, there wasn't a proof for them. I've tried to proof them. But I've stopped at one of the more-known formulas $P = rs$, where $P$ is the area of the triangle, $s$ is the semiperimetar, while $r$ is the radius of the inscribed circle.
I've searched through the interner, but the best thing that I found was that $r = \frac{P}{s}$ holds for all tangential polygon and as you can guess I haven't found a proof.
So please can you help or give me any clue, because I'm still at square one at the moment.
 A: 
The radius is $r$ and half the perimeter is $s$. $[XYZ]$ is the area of $\triangle XYZ$. We know that the area of the tangential polygon is the sum of areas of triangles like $AIG$ and $AGB$. We know that $[AIG]=\frac 12 |AI||IG|$, because $AI$ is perpendicular to $IG$, because $EG$ is tangent to the circle. Also, we have $[AGB]=\frac 12|AB||GB|$. We also have $|GI|=|GB|$, because they are both tangent to the same circle from the same point. Now, $[AIGB]=r\frac{|IG|+|GB|}2$. When calculating the total area, we get
$$
A=r\frac{\text{perimeter}}2=rs
$$
A: Join the incentre to each vertex.  
You then have $n$ small triangles (where $n$ is the number of edges) each with a base which is an edge and a perpendicular height which is the inradius.  So half the product of these is the area of each small triangle.
Add up the $n$ areas and factor out the half and the inradius.  You are left with the sum of the edges, which is the perimeter.
So the area of the polygon is half the perimeter times the inradius.
Khan Academy has a 7-minute lecture on this 
