Question

Does the formula pr = bh apply to inscribed circles in quadrilaterals? or only triangles?

Answer 1

It depends. But if `r` is the inradius, `p` is the semiperimeter and `S` is the area,

`pr=S`

is true for any tangential polygon.

Explanation:

Before proving this, note that a tangential polygon is any convex polygon which can have an incircle.

This is an example of a tangential polygon:

![https://en.wikipedia.org/wiki/Tangential_polygon]()

Because the sides of the polygon are tangent to the circle the line that goes through the incenter, `I`, and through the point where the side is tangent, is perpendicular to the respective side.

Now, what we have visualised is a particular case, where the polygon is four sided. Still, let's solve this first.

Let `M`, `N`, `P` and `Q` be the points of tangency.

Notice how the area, `S`, can be written as the sum of the areas of the triangles:

`S=S_(DeltaAIB) + S_(DeltaBIC) + S_(DeltaCID)+S_(DeltaDIA)`

The simplest formula of the area of a triangle is:

`"Area"=("base"xx"height")/2`

If we take the bases to be the sides of the polygon, then the heights will be the inradii:

`S=(ABxxIM)/2+(BCxxIN)/2+(CDxxIP)/2+(DAxxIQ)/2`

`S=(AB*r + BC*r+CD*r+DA*r)/2=r(AB+BC+CD+DA)/2`

We know that, by definition,

`p = (AB+BC+CD+DA)/2`

`:. S = rp`

The general case is solved analogously. If `A_1`, `A_2`, ..., `A_n` are the vertices of a tangential polygon with `n` sides and `T_1`, `T_2`, ..., `T_n` are the points of tangency, then the area of the polygon can be written as:

`S=S_(DeltaA_1IA_2)+S_(DeltaA_2IA_3)+...+S_(DeltaA_(n-1)IA_n)`

`S=(A_1A_2xxIT_1)/2+(A_2A_3xxIT_2)/2+...+(A_(n-1)A_nxxIT_n)/2`

`S=(A_1A_2*r+A_2A_3*r+...+A_(n-1)A_n*r)/2`

`S=r(A_1A_2+A_3A_4+...+A_(n-1)A_n)/2=rp`