Perpendicular medians

Let $Q$ the centroid and $CF$ a median of $\triangle ABC$ .

Focus in $\triangle AQB$ : as $\triangle AQB$ is rectangle $AB$ , $F$ circumcircle's center thus $FQ$ is radii this circumcircle and $FQ=AF=FB=2.5$

By medians propierties $QC=2FQ$ so $FC=7.5$

$\triangle ABF$ area is $9$ and its base is $2.5 \rightarrow \frac{2.5h}{2}=9 \rightarrow h=7.2$

Let $G$ the interseccion of $AB$ with alture since $C$ , apply pythagoras:

$GF^2=7.5^2-7.2^2$ $GF=\sqrt{7.5^2-7.2^2}$

$GF=2.1$

$AG=2.5-2.1=0.4$

To find $AC$ ad $BC$ apply pythagoras again:

$AC^2=0.4^2+7.2^2=52$ $AC=\sqrt{52}$

$BC^2=4.6^2+7.2^2=73$ $BC=\sqrt{73}$

$\triangle ABC$ perimeter is $\boxed{5+\sqrt{73}+\sqrt{52} \approx 20.755}$