Area of Triangle in Terms of Exradius

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 $\triangle ABC$ 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

 * Area-of-Triangle-by-Exradius.png

Let $C$ be the excircle of $\triangle ABC$ which is tangent to $a$.

By definition:
 * $\rho_a$ is the radius of $C$
 * $I_a$ is the center of $C$.

Then we have:

Also presented as
Some sources present this result as:


 * $\rho_a = \dfrac \AA {s - a}$