Triangle perimeter from angle bisectors - 1

We note that $G$ is the incenter of $\triangle ABC$ . Let the inradius be $r$ . The side lengths of $\triangle ABC$ are:

$\begin{cases} a = \sqrt{4-r^2} + \sqrt{9-r^2} \\ b = \sqrt{9-r^2} + \sqrt{36-r^2} \\ c = \sqrt{4-r^2} + \sqrt{36-r^2} \end{cases} \implies s = \dfrac {a+b+c}2 = \sqrt{4-r^2} + \sqrt{9-r^2} + \sqrt{36-r^2}$

We note that area of $\triangle ABC$ is $sr$ and by Heron's formula :

$\begin{aligned} sr & = \sqrt{s(s-a)(s-b)(s-c)} \\ \implies sr^2 & = (s-a)(s-b)(s-c) \\ r^2 \left(\sqrt{4-r^2} + \sqrt{9-r^2} + \sqrt{36-r^2} \right) & = \sqrt{4-r^2} \cdot \sqrt{9-r^2} \cdot \sqrt{36-r^2} \end{aligned}$

Solving for $r$ (I did it numerically), we have $r \approx 1.458757077$ and perimeter $2s = 2 \left(\sqrt{4-r^2} + \sqrt{9-r^2} + \sqrt{36-r^2} \right) \approx \boxed{19.6}$ .

Use the angle bisector theorem to write down the following equations; bc((b+c)²-a²)=[6(a+b+c)]² ab((a+b)²-c²)=[3(a+b+c)]² ca((a+c)²-b²)=[2(a+b+c)]² p=a+b+c These yield: a = 3.989, b = 8.441 & c = 7.188 and p = 19.6193