Length of Circumradius of Triangle

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

Then the length of the circumradius $R$ of $\triangle ABC$ is given by:
 * $R = \dfrac {abc} {4 \sqrt {s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right)}}$

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

Proof

 * CircumradiusLengthProof.png

Let $O$ be the circumcenter of $\triangle ABC$.

Let $\mathcal A$ be the area of $\triangle ABC$.

Let a perpendicular be dropped from $C$ to $AB$ at $E$.

Let $h := CE$.

Then:

Let a diameter $CD$ of the circumcircle be passed through $O$.

By definition of circumradius, $CD = 2 R$.

By Thales' Theorem, $\angle CAD$ is a right angle.

By Angles on Equal Arcs are Equal, $\angle ADC = \angle ABC$.

It follows from Sum of Angles of Triangle Equals Two Right Angles that $\angle ACD = \angle ECB$.

Thus by Equiangular Triangles are Similar $\triangle DAC$ and $\triangle BEC$ are similar.

So:

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