Area of Triangle in Terms of Inradius and Exradii

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:

$\AA = \sqrt {r \rho_a \rho_b \rho_c}$

where:

$r$ is the inradius

$\rho_a$, $\rho_b$ and $\rho_c$ are the exradii of $\triangle ABC$ with respect to $a$, $b$ and $c$ respectively.

Proof

\(\ds \AA\)

\(=\)

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

Area of Triangle in Terms of Exradius

\(\ds \)

\(=\)

\(\ds \rho_b \paren {s - b}\)

Area of Triangle in Terms of Exradius

\(\ds \)

\(=\)

\(\ds \rho_c \paren {s - c}\)

Area of Triangle in Terms of Exradius

\(\ds \)

\(=\)

\(\ds r s\)

Area of Triangle in Terms of Inradius

\(\ds \leadsto \ \ \)

\(\ds \AA^4\)

\(=\)

\(\ds \rho_a \paren {s - a} \rho_b \paren {s - b} \rho_c \paren {s - c} r s\)

multiplying them all together

\(\ds \)

\(=\)

\(\ds s \paren {s - a} \paren {s - b} \paren {s - c} r \rho_a \rho_b \rho_c\)

rearranging

\(\ds \)

\(=\)

\(\ds \AA^2 r \rho_a \rho_b \rho_c\)

Heron's Formula

\(\ds \leadsto \ \ \)

\(\ds \AA^2\)

\(=\)

\(\ds r \rho_a \rho_b \rho_c\)

\(\ds \AA\)

\(=\)

\(\ds \sqrt {r \rho_a \rho_b \rho_c}\)

$\blacksquare$

Also see

• Heron's Formula, to which this is closely related.