Inradius, you say?

Note that the area of a triangle is

$A = rs$

where $r$ and $s$ are the inradius and semi-perimeter of the triangle. But we are given that the the perimeter is equal to the area. Thus, the semi-perimeter must be half of the area of the triangle. From this, we get that the inradius must be of length $2$ .

We can prove this formula easily. Let $D$ , $E$ and $F$ be points of tangency from the incircle to $\Delta ABC$ at $BC$ , $CA$ and $AB$ respectively. Also, let $I$ be the incentre. We have $AE=AF=x$ , $BF=BD=y$ and $CD=CE=z$ . Thus, $AB=x+y$ , $BC=y+z$ and $CA=z+x$ , $ID=IE=IF=r$ ( $r$ is the inradius), so the perimeter is $2x+2y+2z$ . The area of $\Delta AIF=\Delta AIE= \dfrac{1}{2}rx$ Similarly, we can find the area of the other 4 triangles to get that the area of $\Delta ABC=(x+y+z)r=rs$ , where $s$ is the semi-perimeter.

Thus, the formula is proven.