Area stacked between common tangent and circles Is there any way to find area of shaded region?

The radii of circles are $4$ and $12$ units.
 A: 
By similar triangles, $CD=8$.
By Pythagoras' Theorem, $CG=EF=8\sqrt{3}$
Area of trapezoid $ACEF=\frac{(4+12)\times 8\sqrt{3}}{2}=64\sqrt{3}$
$\angle BAF=\cos^{-1} \left(  \frac{1}{2} \right)=60^{\circ}$ and 
$\angle BCE=180^{\circ}-60^{\circ}=120^{\circ}$
$\therefore$ area of sector $BAF
=\pi (12)^{2} \times \frac{60^{\circ}}{360^{\circ}}
=24\pi$
$\therefore$ area of sector $BCE
=\pi (4)^{2} \times \frac{120^{\circ}}{360^{\circ}}
=\frac{16\pi}{3}$
The required area $=64\sqrt{3}-24\pi-\frac{16\pi}{3}
                   =64\sqrt{3}-\frac{88\pi}{3}$
A: 
Hint: BC // DE
Hint: Trapezoid - Sectors
A: You need three things: Thales, Pythagoras and trigonometry. 
Let $O_1$ the center of the circle of radius $R_1$ and $O_2$ similarly. 
Let's call $T$ the tangent. 
Let's call $O$ the intersection between $O_1O_2$ and $T$. 
Then Thales will give you $OO_1$ distance.
Pythagoras will give you then the lenght between $O$ and the circle $1$
Thales again for the height of the trapezoid...
Then you need to calculate the areas of the circles sections. Trigonometry is your friend. 
