# Area of Triangle in Terms of Circumradius

## Theorem

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

Then the area $\AA$ of $\triangle ABC$ is given by:

- $\AA = \dfrac {a b c} {4 R}$

where $R$ is the circumradius of $\triangle ABC$.

## Proof

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

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

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

Let $h := CE$.

Then:

\(\ds \AA\) | \(=\) | \(\ds \frac {c h} 2\) | Area of Triangle in Terms of Side and Altitude | |||||||||||

\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds h\) | \(=\) | \(\ds \frac {2 \AA} c\) |

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:

\(\ds \frac {CA} {CD}\) | \(=\) | \(\ds \frac {CE} {CB}\) | $\triangle DAC$ and $\triangle BEC$ are similar | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac b {2 R}\) | \(=\) | \(\ds \frac h a\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {2 \AA} {a c}\) | substituting for $h$ from $(1)$ above | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \AA\) | \(=\) | \(\ds \frac {a b c} {4 R}\) |

$\blacksquare$

## Sources

- For a video presentation of the contents of this page, visit the Khan Academy.