Area of Triangle in Terms of Exradius/Proof
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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.
Let $\rho_a$ be the exradius of $\triangle ABC$ with respect to the excircle which is tangent to $a$.
Let $s$ be the semiperimeter of $\triangle ABC$.
Then the area $\AA$ of $\triangle ABC$ is given by:
- $\AA = \rho_a \paren {s - a}$
Proof
Let $C$ be the excircle of $\triangle ABC$ which is tangent to $a$.
By definition:
Then we have:
\(\ds \AA\) | \(=\) | \(\ds \map \Area {\triangle ABI_a} + \map \Area {\triangle ACI_a} - \map \Area {\triangle CBI_a}\) | (see figure above) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {c \rho_a} 2 + \frac {b \rho_a} 2 - \frac {a \rho_a} 2\) | Area of Triangle in Terms of Side and Altitude | |||||||||||
\(\ds \) | \(=\) | \(\ds \rho_a \frac {b + c + a} 2 - \rho_a a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \rho_a s - \rho_a a\) | Definition of Semiperimeter | |||||||||||
\(\ds (ABC)\) | \(=\) | \(\ds \rho_a \paren {s - a}\) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The ex-circles