Area of Triangle in Terms of Circumradius

Theorem

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

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 1

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 \ \$ $\ds \AA$ $=$ $\ds \frac {a b c} {4 R}$

$\blacksquare$

Proof 2

 $\ds 2 R$ $=$ $\ds \dfrac a {\sin A}$ Law of Sines $\ds$ $=$ $\ds \dfrac {a b c} {b c \sin A}$ $\ds$ $=$ $\ds \dfrac {a b c} {2 \AA}$ Area of Triangle in Terms of Two Sides and Angle $\ds \leadsto \ \$ $\ds R$ $=$ $\ds \dfrac {a b c} {4 \AA}$ Area of Triangle in Terms of Two Sides and Angle

$\blacksquare$

Also presented as

This can also been seen in some sources presented as:

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