Radius of the inscribed circle

The circle inscribed in a triangle is tangent to all three sides. Its radius has a very simple expression in terms of the area and the semiperimeter.

Theorem — Radius of the inscribed circle

The radius $r$ of the inscribed circle of a triangle with area $S$ and perimeter $2p$ is: $r = \frac{2S}{2p} = \frac{S}{p}$ where $p$ is the semiperimeter.

The centre $O$ (incentre) is equidistant from the three sides: $OH\cong OK\cong OR=r$.

Proof — $r = S/p$

1. The inscribed circle is tangent to all the sides, so $OH\perp AB$, $OK\perp BC$, $OR\perp AC$, with $OH\cong OK\cong OR = r$.
2. We split triangle $ABC$ into three sub-triangles — $\triangle ABO$, $\triangle BCO$, $\triangle ACO$ — each with height equal to the radius $r$: $S_{ABC} = \frac{AB\cdot r}{2}+\frac{BC\cdot r}{2}+\frac{AC\cdot r}{2} = \frac{(AB+BC+AC)\cdot r}{2} = \frac{2p\cdot r}{2}.$
3. Hence $S = p\cdot r$, that is $r = \dfrac{S}{p}$.

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Topics: Euclidean circle
Concepts: Inscribed circle
Skills: Proving · Using formulae