Central Region Area

Label the points as follows:

Let the radius of the five congruent circles be $r$ , so that $RT = RS = r$ .

By symmetry, $\angle QPT = \frac{360°}{10} = 36°$ , and from right $\triangle PRT$ we have $\angle PRT = 54°$ , $PR = r \sec 54°$ , and $PT = \tan 54°$ .

From $PR + RS = PS$ we have $r \sec 54° + r = 1$ , which solves to $r = \frac{1}{\sec 54° + 1}$ .

The central yellow region is $A = 10A_{PQT} = 10(A_{\triangle PRT} - A_{RQT}) = 10(\frac{1}{2}r^2 \tan 54° - \frac{54}{360} \pi r^2) = (5 \tan 54° - \frac{3}{2}\pi)r^2$ .

Substituting $r = \frac{1}{\sec 54° + 1}$ into $A = (5 \tan 54° - \frac{3}{2}\pi)r^2$ gives $\lfloor 10^4 A \rfloor = \boxed{2973}$ .

Let the radius of the five congruent circles be $r$ . Join the centers of the five congruent circles with straight lines, we get a regular pentagon with side length of $2r$ .

From the figure we note that $OP+PQ = OQ \implies r \sec 54^\circ + r = 1 \implies r = \dfrac 1{1+\sec 54^\circ}$ .

Now consider one of the five congruent isosceles triangles that make up the pentagon. We note that one fifth of the yellow are is the area of $\triangle OPR$ minus $\dfrac {108^\circ}{360^\circ}$ of the area of circle of radius $r$ . Therefore the area of the yellow region:

$A = 5\left(r^2 \tan 54^\circ - \frac {108}{360}\pi r^2\right) = \frac {5\left(\tan 54^\circ - \frac {108}{360}\pi \right)}{(1+\sec 54^\circ)^2} \approx 0.297315551092$

Therefore $\lfloor 10^4A\rfloor = \boxed{2973}$ .