Formula for sides of a triangle where the Perimeter equals to the Area I was wondering if there is a formula that could generate the values of the sides of a triangle where his area equals to his perimeter. I only found that if the triangle is equilateral then 
$$l=\frac{12}{√3}$$
where $l$ is the side of the triangle.
Thanks for support
Peterix
P.S. There is a similar problem here

Given any triangle $T$ there is a triangle $T'$ similar to $T$ such that the area of $T'$ is the same as the perimeter of $T'$. 
Suppose the perimeter of $T$ is $P$ and the area of $T$ is $A$. Then dilate $T$ by a factor of $\frac{P}{A}$ to produce the triangle $T'$. (That is, multiply all the lengths by $\frac{P}{A}$.)

The perimeter of $T'$ will be the perimeter $T$ multiplied by $\frac{P}{A}$ to give:
$$
\text{Perimeter of } T' = \frac{P}{A}\cdot P = \frac{P^2}{A}
$$
The ara of $T'$ will be the area of $T$ multiplied by $(\frac{P}{A})^2$ to give:
$$
\text{Area of } T' = \left(\frac{P}{A}\right)^2 A = \frac{P^2}{A}
$$
For example, an equilateral triangle with all edges equal to 1 has area $\sqrt3/4$ and perimeter 3. The equilateral triangle with perimeter equal to its area is dilated by $3 \div (\sqrt3/4) = 12/ \sqrt3$, so its side lengths are all $12/\sqrt3$.
Another example: a 3-4-5 triangle has perimeter 12 and area 6. The similar triangle with the area equal to its perimeter is dilated by $12/6 = 2$. Thus the new triangle is a 6-8-10 triangle and has area and perimeter 24.
A final example: a right-angled triangle with short sides $a$ and $b$ has perimeter $P = a+ b+ \sqrt{a^2+b^2}$ and area $A = \frac12 ab$. Dilate this by $P/A$ to give sides of:
$$
\begin{align}
a \cdot \frac{a + b + \sqrt{a^2 + b^2}}{\frac12 ab} &= 2\left( \frac{a}{b} + 1 + \sqrt{\left(\frac{a}{b}\right)^2 + 1}\right)\\
\text{and } \quad b \cdot \frac{a + b + \sqrt{a^2 + b^2}}{\frac12 ab} &= 2\left( \frac{b}{a} + 1 + \sqrt{\left(\frac{b}{a}\right)^2 + 1}\right)\\
\text{and }\quad\sqrt{a^2+b^2} \cdot \frac{a + b + \sqrt{a^2 + b^2}}{\frac12 ab} &= 2\left(\sqrt{\left(\frac{a}{b}\right)^2 + 1} + \sqrt{\left(\frac{b}{a}\right)^2 + 1} + \frac{a}{b} + \frac{b}{a}\right)
\end{align}
$$ 
