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 $\triangle ABC$ 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:

$\rho_a$ is the radius of $C$

$I_a$ is the center of $C$.

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

\(\ds (ABC)\)

\(=\)

\(\ds \rho_a \paren {s - a}\)

$\blacksquare$

Also presented as

Some sources present this result as:

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

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(55)$