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