Area of Triangle in Terms of Inradius
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 $\triangle ABC$.
$\blacksquare$
Also presented as
This can also been seen in some sources presented as:
- $r = \dfrac \AA s$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(53)$