Area of Triangle in Terms of Circumradius/Proof 1
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
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.