The area of triangle formed by the common tangents of the circles  $x^2+y^2+12x=0$ and  $x^2+y^2-4x=0$ is

 $\hspace{0.17em}\left.\begin{array}{r}{r}_1=2\\ r_2=6\end{array}\right\}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{r}_2=3\left(r_1\right)\therefore \Delta$ is equilateral
          
In radius  $=2\Rightarrow side=4\sqrt{3}Area=\frac{\sqrt{3}}{4}\times 48=12\sqrt{3}$