Revision as of 17:03, 22 November 2016 by First (talk | contribs) (Proof)

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]

A Property

  • 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.


Add in the incircle and drop the altitudes from the incenter to the sides of the triangle. Also draw the lines $\overline{AI}, \overline{BI}$, and $\overline{CI}$. After this AB, AC, and BC are the bases of $\triangle{AIB}, {AIC}$, and ${BIC}$ respectively. But they all have the same height(the inradius), so $[ABC]=\frac{(a+b+c) \times r}{2} =rs$.


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