Area equals perimeter

Denote the inradius as $r$ . In the above image, we can divide $\triangle ABC$ into three triangles: $\triangle AOB, \triangle BOC,$ and $\triangle COA$ .

Since the three radii are perpendicular to the sides of $\triangle ABC$ , we can see that the areas of each of the smaller triangles are, respectively, $\dfrac{\overline{AB}*r}{2}, \dfrac{\overline{BC}*r}{2}, \dfrac{\overline{CA}*r}{2}$ .

Thus, the total area of $\triangle ABC$ is $a = \dfrac{\overline{AB}*r+\overline{BC}*r+\overline{CA}*r}{2} = \dfrac{p*r}{2}$

Since $a = p$ , this simplifies to $r = \boxed{2}$ .