# Area of a triangle: Incenter (Part 6)

The incenter $I$ of a triangle $\triangle ABC$ is defined by the intersection of the angle bisectors of its three angles. A circle can be inscribed within $\triangle ABC$, as shown in the picture.

This incircle provides a different way of finding the area of $\triangle ABC$ commonly needed for high school math contests like the AMC 10/12. Suppose that the sides $a$, $b$, and $c$ are known and the inradius $r$ is also known. Then $\triangle ABI$ is a right triangle with base $c$ and height $r$. So

$\hbox{Area of ~} \triangle ABI = \displaystyle \frac{1}{2} cr$

Similarly,

$\hbox{Area of ~} \triangle ACI = \displaystyle \frac{1}{2} br$

$\hbox{Area of ~} \triangle BCI = \displaystyle \frac{1}{2} ar$

Since the area of $\triangle ABC$ is the sum of the areas of these three smaller triangles, we conclude that

$\hbox{Area of ~} \triangle ABC = \displaystyle \frac{1}{2} r (a+b+c)$,

or

$\hbox{Area of ~} \triangle ABC = rs$,

where $s = (a+b+c)/2$ is the semiperimeter of $\triangle ABC$.

This also permits the computation of $r$ itself. By Heron’s formula, we know that

$\hbox{Area of ~} \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$

Equating these two expressions for the area of $\triangle ABC$, we can solve for the inradius $r$:

$r = \displaystyle \sqrt{ \frac{(s-a)(s-b)(s-c)}{s} }$

For much more about the inradius and incircle, I’ll refer to the MathWorld website.

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