Inradius and area, r·s = S

Dependencies: Incircle exists (the incircle exists + the incenter $I$ is at distance $r$ from each side), Triangle area = ½ × base × height ($S = \tfrac12 \cdot$ base $\cdot$ height).

Statement

Let $\triangle ABC$ have side lengths $a = BC$, $b = CA$, $c = AB$, area $S$, inradius $r$, and semiperimeter

Then the area, semiperimeter, and inradius of the triangle satisfy

The equivalent form $r = \dfrac{S}{s}$ reduces "inradius" problems to the two familiar quantities "area + perimeter"; chained with Heron's formula $S = \sqrt{s(s-a)(s-b)(s-c)}$, it immediately yields a purely algebraic expression of $r$ in the three side lengths.