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: Radius of Circle Inscribed in a Triangle of Sides $a, b, c$: $4.15$
  • 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 7$: Geometric Formulas: Radius of Circle Inscribed in a Triangle of Sides $a, b, c$: $7.15.$
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