Perimeter from Altitudes

Let $BC = a$ , $CA=b$ , and $AB=c$ . Then the area of $\triangle ABC$ is $[ABC] = \dfrac {3a}2 = \dfrac {5b}2 = \dfrac {7c}2$ . $\implies 3a=5b=7c$ $\implies b = \dfrac 35 a$ and $c = \dfrac 37 a$ . By Heron's formula :

$\begin{aligned} [ABC] & = \sqrt{s(s-a)(s-b)(s-c)} & \small \blue{\text{where }s = \frac {a+b+c}2 = \frac {71}{70}a} \\ \implies \frac {3a}2 & =\frac {a^2}{70^2} \sqrt{71(1)(29)(41)} \approx 0.059295812 a^2 \\ \implies a & \approx 25.29689612 \end{aligned}$

Therefore perimeter of $\triangle ABC$ is $\left(1+ \dfrac 35+\dfrac 37 \right) a \approx \boxed{51.3}$ .

If $D$ is the area of the triangle then the triangle has sides $\tfrac23D$ , $\tfrac25D$ , $\tfrac27D$ . Heron's Formula then tells us that $D \; = \; \sqrt{\tfrac{71}{105}D \times \tfrac{1}{105}D \times \tfrac{29}{105}D \times \tfrac{41}{105}D} \; = \; \frac{D^2}{105^2}\sqrt{84419}$ and so $D \; = \; \frac{105^2}{\sqrt{84419}}$ and hence the triangle has perimeter $2(\tfrac13 + \tfrac15 + \tfrac17)D \; = \; 210 \sqrt{\frac{71}{1189}} \; = \; \boxed{51.31656071}$

Let the position coordinates of $A$ be $(a, 3)$ , of $B$ be $(0,0)$ , and of $C$ be $(b, 0)$ .

Then $9b^2-25a^2=225$

$9b^2-49(b-a)^2=441$

So, let $b=5\cosh α=7\cosh β, a=3\sinh α=7\cosh β-3\sinh β$

Solving we get

$\sinh β\approx 3.47273,\cosh β\approx 3.61384, a\approx 14.87869, b\approx 25.296896$ .

Hence the perimeter of the triangle is $b+\sqrt {a^2+9}+\sqrt {(b-a) ^2+9}\approx \boxed {51.31665}$ .