Inscribed kissing circles in an isosceles trapezoid 5 equal circles in an isosceles trapezoid. Radius of circle is 4. Find black colored area.

I don't have any ideas, could you give me a hand? Thanks.
 A: Clearly the area of the colored region is 3 times the area of one of the curved triangles.
Draw a regular hexagon around one of the circles, in such a way that all of its sides are tangent to the circle (i.e., the circle is inscribed in the hexagon). There are 6 regions of the hexagon that are not contained in the circle, one at each vertex of the hexagon:

Note that any one of the curved triangles in your picture consists of 3 of these hexagon corners. 
Let $A$ be the area of a hexagon with inradius 4. Let $B$ be the area of a circle with radius 4. The total area of 6 hexagon corner regions is $A-B$. The region we want to find is made of 3 curved triangles, each of which is made of 3 hexagon corners, for a total of 9 hexagon corners. Thus, the area of the colored region is $\frac{3}{2}(A-B)$.
A: It's pretty easy to figure out the area of a hexagon and easier to look it up, but I don't know it off the top of my head and I think the problem can be done without it. The area formulas for trapezoids and circles would be enough considering this figure:

A: The hexagonal solution is lovely, and the trapezoid picture is pretty nice.  But there are cruder ways that perhaps require less insight.
Let's find the area of one of the three black chunks.  To do this, join the centers of the circles it fits between.  We get an equilateral triangle of side $8$.
From the area of this, you must subtract three $60^\circ$ sectors, which together have half the area of a circle of radius $4$, that is, $8\pi$.
With basic trigonometry, or otherwise, one finds that an equilateral triangle of side $a$ has area $a^2\sqrt{3}/4$.  Here $a=8$, so the area is $16\sqrt{3}$.  Thus one little black chunk has area $16\sqrt{3}-8\pi$, and all three have combined area
$$48\sqrt{3}-24\pi.$$
