The inradius of a polygon is the radius of its incircle (assuming an incircle exists). It is commonly denoted $r$.

$[asy] pathpen = linewidth(0.7); pair A=(0,0),B=(4,0),C=(1.5,2),I=incenter(A,B,C),F=foot(I,A,B); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(CR(D(MP("I",I,SW)),inradius(A,B,C))); D(F--I--foot(I,B,C)--I--foot(I,C,A)); D(rightanglemark(I,F,B,5)); MP("r",(F+I)/2,E); [/asy]$

## Properties

• If $\triangle ABC$ has inradius $r$ and semi-perimeter $s$, then the area of $\triangle ABC$ is $rs$. This formula holds true for other polygons if the incircle exists.
• The inradius satisfies the inequality $2r \le R$, where $R$ is the circumradius (see below).
• If $\triangle ABC$ has inradius $r$ and circumradius $R$, then $\cos{A}+\cos{B}+\cos{C}=\frac{r+R}{R}$.

## Problems

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