Area of Triangle in Terms of Inradius

Theorem
Let $\triangle ABC$ be a triangle whose sides are of lengths $a, b, c$.

Then the area $\mathcal A$ of $\triangle ABC$ is given by:
 * $\mathcal A = r s$

where:
 * $r$ is the inradius of $\triangle ABC$
 * $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.

Proof

 * IncenterLengthProof.png

Let $I$ be the incenter of $\triangle ABC$.

Let $r$ be the inradius of $\triangle ABC$.

The total area of $\triangle ABC$ is equal to the sum of the areas of the triangle formed by the vertices of $\triangle ABC$ and its incenter:
 * $\mathcal A = \operatorname {Area} \left({\triangle AIB}\right) + \operatorname {Area} \left({\triangle BIC}\right) + \operatorname {Area} \left({\triangle CIA}\right)$

Let $AB$, $BC$ and $CA$ be the bases of $\triangle AIB, \triangle BIC, \triangle CIA$ respectively.

The lengths of $AB$, $BC$ and $CA$ respectively are $c, a, b$.

The altitude of each of these triangles is $r$.

Thus from Area of Triangle in Terms of Side and Altitude:

Thus:
 * $\displaystyle \mathcal A = r \frac {a + b + c} 2$

That is:
 * $\mathcal A = r s$

where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.