Find all integer sided triangles whose area equals their perimeter. Find all integer sided triangles whose area equals their perimeter?
Taking sides to be $e,f,g$ we have,
$$\sqrt{(e+f+g)(e+f-g)(e+g-f)(g+f-e)}=4(e+f+g)$$
how to proceed next ?

Let $a,b,c$ be the lengths of the sides and $s=(a+b+c)/2$.
By Heron's formula, We have
$$\sqrt{s(s-a)(s-b)(s-c)}=2s,$$
i.e.
$$(-a+b+c)(a-b+c)(a+b-c)=16(a+b+c)\tag1$$
Here, let $x=-a+b+c,y=a-b+c,z=a+b-c$, then we have
$$a=\frac{y+z}{2},\quad b=\frac{z+x}{2},\quad c=\frac{x+y}{2}\tag2$$
So, $(1)$ can be written as
$$xyz=16(x+y+z)\tag3$$
Here, note that $x,y,z$ are positive integers and all even from $(2)(3)$. So, let $x=2X,y=2Y,z=2Z$, then $(3)$ can be written as
$$XYZ=4(X+Y+Z).$$
We may suppose that $X\le Y\le Z$. So,
$$XYZ\le 4\cdot 3Z=12Z\Rightarrow XY\le 12.$$
$$X^2\le XY\le 12\Rightarrow X^2\le 12\Rightarrow X\le 3.$$


*

*For $X=1$, we have $(Y-4)(Z-4)=20\Rightarrow (Y,Z)=(5,24),(6,14),(8,9)$.

*For $X=2$, we have $(Y-2)(Z-2)=8\Rightarrow (Y,Z)=(3,10),(4,6)$.

*For $X=3$, we have $3YZ\le 4\cdot 3Z\Rightarrow Y\le 4$. If $Y=3$, then $9Z=4(6+Z)$. If $Y=4$, then $12Z=4(7+Z)$. Hence, there is no solution.
Thus, the answer is the followings : 
$$(6,25,29),(7,15,20),(9,10,17),(5,12,13),(6,8,10).$$
