Length of Inradius of Triangle

Theorem

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

Then the length of the inradius $r$ of $\triangle ABC$ is given by:

$r = \dfrac {\sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } } s$

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

Proof

Let $\AA$ be the area of $\triangle ABC$.

From Area of Triangle in Terms of Inradius:

$\AA = r s$

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

From Heron's Formula:

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

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

Hence the result:

$r = \dfrac {\sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } } s$

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 4$: Geometric Formulas: $4.15$