Paper

The intersection of the rectangles is a rhombus, so area is half the product of diagonals.

The diagonal $DB=\sqrt{6^2+2^2}=2\sqrt{10}$

$\triangle DBC$ is similar to $\triangle DHG$ therefore $\frac{GH}{GD}=\frac{2}{6}=\frac{1}{3}$

$GH=\frac{1}{3}\times GD=\frac{1}{3}\times \frac{DB}{2}=\frac{\sqrt{10}}{3}$

$Area=2\sqrt{10}\times \frac{\sqrt{10}}{3}=\frac{20}{3}=6.667$

Let $DC$ and $A'B$ intersect at $H$ .
Let $AB$ intersect $C'D$ at $I$ .

Then the figure is symmetric about the line $IH$ . Hence, we have $DH = HB$ . Set this length to be $x$ .

Applying Pythagoras theorem on triangle $BHC$ , we have $BC ^2 + BH^2 = BH^2$ , or that $2^2 + (6-x)^2 = x^2$ . Expanding and simplifying, we get $4 + 36 -12x + x^2 = x^2$ , or that $x = \frac{ 40}{12} = \frac{108}{3}$ .

Hence, the blue area, which can be calculated as base DH * height BC, is $2 \times \frac{10}{3} \approx 6.67$ .

Note that all the triangles around the shaded region are congruent. This can be easily proved using AAS congruency as all triangles have AD=BC=A'D=BC'=2 and one angle is 90 degrees while the other can be shown equal by some angle chasing.

Let us say that AB intersect C'D at X and let us call AX = x and thus XB = 6-x . Now as ADX is congruent to C'BX we get C'X = AX = x . Now in triangle C'BX, by Pythagoras' theorem we have

C'B^2 + C'X^2 = XB^2

or 2^2 + x ^2 = (6- x )^2

which solves to x = 8/3

The triangles' area become = 1/2 * 2 * 8/3 = 8/3

Now the shaded region's area is = area of rectangle - 2 * area of triangle

= 2 * 6 - 2 *8/3

=12 - 16/3

= 20/3

=6.667 or 6.67

If the side length of the rhombus is x, then we see right-angle triangles with sides 2, x, and (6-x).

Pythagoras:

$x² = 2² + (6-x)²$

$x² = 40 - 12x + x²$

$x = 40/12 = 10/3$

The area of a parallellogram is base × perpendicular height

$= 2 × 10/3 = 6⅔ = 6.67$

I will give a general solution for rectangles with width $a$ and height $b$ .

The shaded region is a parallelogram. To find its area, we must multiply the base $b$ by the height $y$ . The challenge is to find the value of $y$ .

(In this drawing, the $x$ s and $y$ s are equal because of the obvious symmetry of the situation.)

On one hand, $a = x + y$ . On the other hand, applying the Pythagorean Theorem to the green right triangle, $y^2 = x^2 + b^2.$ Substitution of $x = a - y$ yields $y^2 = (a - y)^2 + b^2 \ \ \ \ \therefore \ \ \ \ y^2 = a^2 - 2ay + y^2 + b^2;$ $2ay = a^2 + b^2;$ $y = \frac{a^2 + b^2}{2a}.$ Thus the shaded area is $A = by = \frac b{2a} (a^2 + b^2).$ In this case, $a = 6$ and $b = 2$ , so that $A = \frac 2{2\cdot 6}(6^2 + 2^2) = \frac16\cdot 40 = \frac{20}3 \approx \boxed{6.67}.$