Assuming that the base of the circle lies along the longest base, here is a way to find the length of the shorter base (which along with the length of the longer base, 14, and the height of the trapezoid, 6, readily yields its area):
Label the vertices of the trapezoid ABCD, with AD being the longer base of length 14. Let the smaller of the bases have length $x$, and let the center of the circle be at $O$.
Area of trapezoid= $$\frac{6}{2}(x+14)=3x+42$$ by the area of a trapezoid formula. We can calculate this in a different manner by doing area=
$$A_{OAB}+A_{OBC}+A_{OCA}=\frac{1}{2}(6AB+6*x+6*CD)=6AB+3x$$
Therefore, $$3x+42=6AB+3x\Leftrightarrow AB=7$$
Drop the perpendicular Q from B to AD and applying Pythagorean Theorem, we get $$AB^2=AQ^2+BQ^2 \Leftrightarrow 49=AQ^2+36 \Leftrightarrow AQ=\sqrt{13}$$
By symmetry, $$AD=2AQ+BC$$, which gives $$BC=14-2\sqrt{13}$$
Cheers,
Rofler
I'm using the same notation as in the link mentioned by David Mitra. By similarity we have
$$
\frac{r}{OA}=\frac{r_a}{OA-r-r_a}=\sin\left(\frac{\alpha}{2}\right)
$$
where $\alpha=\angle A$. Then
$$
\frac{r_a}{r} = \frac{1-\sin(\alpha/2)}{1+\sin(\alpha/2)}
= \frac{1-\cos((\beta+\gamma)/2)}{1+\cos((\beta+\gamma)/2)}
= \tan\left(\frac{\beta+\gamma}{4}\right)^2
$$
where we used that $\alpha+\beta+\gamma=\pi$. Similarly
$$
\frac{r_b}{r} = \tan\left(\frac{\alpha+\gamma}{4}\right)^2
\qquad\qquad
\frac{r_c}{r} = \tan\left(\frac{\alpha+\beta}{4}\right)^2
$$
Therefore, $\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}$ can be written as
$$
\begin{split}
\frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r}
= \frac{\sin(\rho)\sin(\sigma)\cos(\tau)+\sin(\rho)\cos(\sigma)\sin(\tau)
+\cos(\rho)\sin(\sigma)\sin(\tau)}{\cos(\rho)\cos(\sigma)\cos(\tau)}
\end{split}
$$
where $\rho=(\beta+\gamma)/4$, $\sigma=(\alpha+\gamma)/4$, and $\tau=(\alpha+\beta)/4$. Note that $\rho+\sigma+\gamma=\pi/2$.
Consider $x$, $y$, $z$ such that $x+y+z=\pi/2$. Then
$$
\begin{split}
\sin(x)\sin(y)\cos(z) &= \frac{1}{2}(\cos(x-y)-\cos(x+y))\cos(z)
\\&=
\frac{1}{4}(\cos(x-y+z)+\cos(x-y-z) -\cos(x+y+z)-\cos(x+y-z))
\\&= \frac{1}{4}( \cos\left(\frac{\pi}{2}-2y\right) + \cos\left(-\frac{\pi}{2}+2x\right) - \cos\left(\frac{\pi}{2}\right)-\cos\left(\frac{\pi}{2}-2z\right)
\\&=\frac{1}{4}\left( \sin(2y)+\sin(2x)-\sin(2z)\right)
\end{split}
$$
And similarly
$$
\begin{split}
\cos(x)\cos(y)\cos(z) &= \frac{1}{2}(\cos(x-y)+\cos(x+y))\cos(z)
\\&=
\frac{1}{4}(\cos(x-y+z)+\cos(x-y-z) +\cos(x+y+z)+\cos(x+y-z))
\\&=\frac{1}{4}\left( \sin(2y)+\sin(2x)+\sin(2z)\right)
\end{split}
$$
This means that $(\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c})/r$ is equal to
$$
\frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r} = \frac{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}=1
$$
Best Answer
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 $\frac{3}{2}(A-B)$.