# Area of Triangle in Terms of Inradius/Proof

## Theorem

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

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

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

- 1953: L. Harwood Clarke:
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