Area of Triangle in Terms of Inradius

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Then the area $\AA$ of $\triangle ABC$ is given by:

$\AA = r s$

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:

$\AA = \map \Area {\triangle AIB} + \map \Area {\triangle BIC} + \map \Area {\triangle CIA}$


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:

\(\ds \map \Area {\triangle AIB}\) \(=\) \(\ds \frac {c r} 2\)
\(\ds \map \Area {\triangle BIC}\) \(=\) \(\ds \frac {a r} 2\)
\(\ds \map \Area {\triangle CIA}\) \(=\) \(\ds \frac {b r} 2\)

Thus:

$\AA = r \dfrac {a + b + c} 2$


That is:

$\AA = r s$

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

$\blacksquare$


Sources