Area of Triangle in Terms of Circumradius/Proof 2
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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
\(\ds 2 R\) | \(=\) | \(\ds \dfrac a {\sin A}\) | 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 |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The circumcircle