Are you smarter than me? 51

Inradius $r$ is given by: $\space r = \sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}}$ , where $a$ , $b$ and $c$ are the side lengths of the triangle and $s = \frac {a+b+c}{2}$ .

Therefore, the radius of circle centered $C_1$ :

$\quad r_1 = \sqrt{\dfrac{(8-6)(8-5)(8-5)}{8}} = \sqrt{\dfrac{2\times 3 \times 3}{8}} = \dfrac{3}{2}$

We note that $C_1$ , $C_2$ and $C_3$ are aligned on a same straight line. They are incircles of similar triangles.

And we note that: $\space \dfrac {r_2}{r_1} = \dfrac {4-2r_1}{4} = \dfrac {1}{4}$ .

Similarly, $\space \dfrac {r_3}{r_2} = \dfrac {1}{4} \quad \Rightarrow r_2 = \dfrac {3}{8} \quad \Rightarrow r_3 = \dfrac {3}{32}$

The required result:

$\quad \frac {1}{2}\times 6 \times 4 - \pi (r_1^2+r_2^2+r_3^2) + r_2$

$= 12 - \pi \left( (\frac {3}{2})^2 + (\frac {3}{8})^2 + (\frac {3}{32})^2 \right) + \frac {3}{8} = \boxed {4.837}$