Areas add easily, don't they?

We find that $\cos \angle ABC=-\frac{1}{4}$ by Law of Cosines on $\angle ABC$ in $\Delta ABC$ , so $\frac{4}{\sin \angle ABC}= \frac{16}{\sqrt{15}} =2R$ by Extended Law of Sines, giving $R=\frac{8\sqrt{15}}{15}$ .

Now the desired area is $[ABC] + 2[AOC]$ , so we use $[ABC]= \frac{abc}{4R} = \frac{3\sqrt{15}}{4}$ and $2[AOC] = 4 \cdot \frac{2}{\sqrt{15}}= \frac{8\sqrt{15}}{15}$ , which adds up to $\frac{77\sqrt{15}}{60}$ .

Thus, the answer is $152$ .

$By~Heron's ~Formula~~Area ~ABC=\dfrac 1 4*\sqrt{9*1*5*3}.\\ Circumradius~R = \dfrac{abc}{4*(area~ ABC)} =\dfrac{8*\sqrt{15}}{15}.\\ The~three~\Delta s~are~isosceles~with~base~a, b, c,~~and~R~as~equal~sides.\\ \therefore~their~base~altitudes~are~L_a, L_b, L_c~~given~by~\sqrt{R^2-(half~base)^2}.\\ \therefore~Areas~are~~~~\Delta_a=\frac1 2 *L_a*a,~~~~\Delta_b=\frac1 2 *L_b*b,~~~~\Delta_c=\frac1 2 *L_c*c,~~~~\\ Sum~of~areas=\frac 1 2* \Big( \sqrt{\frac{64}{15}-(\frac 2 2)^2}*\frac 2 2.~~+~~\sqrt{\frac{64}{15}-(\frac 3 2)^2}*\frac 3 2.~~+~~\sqrt{\frac{64}{15}-(\frac 4 2)^2}*\frac 4 2.\Big )\\ Sum~of~areas=\dfrac{77\sqrt{15}}{60}=\dfrac{a* \sqrt b} c .\\ So~a+b+c=77+15+60=\Large \color{#D61F06}{152}.$

Note:- ABC is obtuse, so O is outside the triangle ABC.