Right triangle $ABC$ has $AB = AC = 6$ cm. Circular arcs are drawn with centers at $A, B$ and $C,$ so that the arc centered at $A$ is tangent to side $BC$ and so that the arcs centered at $B$ and $C$ are tangent to the arc centered at $A,$ as shown. What is the perimeter of the shaded region? Express your answer as a decimal to the nearest hundredth.
[asy]
pair B = (0,0), C = (6,0), A = (0,8), O = (1.91,3.82);
draw(A--B--C--cycle);
draw(B..O--C);
draw(B--extension(B,O,A,C));
draw(A..O);
draw(arc(A,3.82,0,180));
draw(arc(B,5.66,128.66,308.66));
draw(arc(C,5.66,231.34,51.34));
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",B+(0,1.99),W);
label("$O$",O,(1,1));
fill(anglemark(B,C,A)*arc(B,C,A,1.5),lightgray);
[/asy]
Tangent segments are congruent, so $DB = DC = 1.99$ cm. Noting that the radius of $\overbigarc{BC}$ is $\tfrac{6}{2} = 3,$ thus $BA = 3BD,$ so triangle $BAD$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with $AD = 3\cdot 1.99 = 5.97$ cm. Therefore, the shaded area is \begin{align*}
&[\triangle ABC] - [\triangle BAD] - [\triangle CAD] - [BCDA] \\
&= \frac{1}{2}(6)(6) - \frac{1}{2}\cdot 5.97\cdot 1.99 - \frac{1}{2}\cdot 5.97\cdot 3 - \frac{1}{2}\cdot 1.99\cdot 3 \\
&= 18 - 5.97 \cdot (1.99 + 3) - 1.99 \cdot \frac32 \\
&= 18 - 33.77 \\
&= -\underbrace{15.77}_{-[\triangle CBA]}.
\end{align*} They want the perimeter, so all we have to do is sum the lengths around the shaded region, i.e. $(-15.77) + 6 + 3 + (5.97 + 1.99 + 3) = \boxed{04.78}$ centimeters.
To find the perimeter of the shaded region, we need to determine the lengths of the three arcs that compose it.
Let's start with the arc centered at point $A.$ Since it is tangent to side $BC,$ we have a right angle between $BA$ and $CA.$ Using the Pythagorean theorem, we can find the length of $BC.$
$BC^2 = AB^2 - AC^2 = 6^2 - 6^2 = 36 - 36 = 0$
This means that $BC = 0,$ which implies that the arc centered at $A$ is just a straight line segment connecting points $B$ and $C,$ passing through $A.$ Therefore, its length is equal to the length of $BC,$ which is $0.$
Next, let's consider the arc centered at point $B.$ Since it is tangent to the arc centered at $A,$ we can draw a line segment connecting $A$ and $B,$ and extend it to point $D,$ as shown.
[asy]
unitsize(0.7 cm);
pair A, B, C, D;
A = (0,0);
B = (12,0);
C = (12,8);
D = (A + reflect(A,B)*(C))/2;
draw(A--B--C--cycle);
draw(arc(A,6,0,180),red);
draw(arc(B,6,125,305),red);
draw(arc(C,6,220,40),red);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, NW);
label("$6$", (A + C)/2, N);
label("$6$", (A + B)/2, S);
[/asy]
Since $AB = 6$ and $BC = 0,$ we have $AD = 6.$ Furthermore, since $AB = AC,$ we have $DA = DC = 6.$ Therefore, triangle $DAC$ is isosceles, and $\angle DAC = \angle DCA.$
Now, let $\angle DAC = \angle DCA = x.$ Since angles within a triangle sum to $180^\circ,$ we have $\angle ACD = 180^\circ - 2x.$
Since the arcs centered at $B$ and $C$ are tangent to $AD,$ they must be $90^\circ$ arcs, because they intersect with $AD$ at a right angle.
Therefore, $\angle CAD = \angle ABC = 90^\circ.$
Now, consider triangle $ABC.$ The sum of its angles is $180^\circ,$ so $\angle BAC = 180^\circ - 90^\circ - x = 90^\circ - x.$
Since $AB = AC,$ triangle $ABC$ is isosceles, and $\angle BCA = \angle CAB = \frac{180^\circ - \angle BAC}{2} = \frac{180^\circ - (90^\circ - x)}{2} = \frac{90^\circ + x}{2}.$
Finally, in circle $B,$ the measure of an arc is twice the measure of the inscribed angle. Therefore, the length of arc $BDC$ is $2\left(\frac{90^\circ + x}{2}\right) = 90^\circ + x$ degrees.
Since the radius of circle $B$ is $6$ cm, the circumference of the circle is $2\pi(6) = 12\pi,$ and we can express the length of the arc $BDC$ in terms of $\pi$ as $12\pi \cdot \frac{90^\circ + x}{360^\circ} = 3\pi \cdot (90^\circ + x)$ cm.
To find the perimeter of the shaded region, we need to add the lengths of the three arcs: $0$ cm (arc centered at $A$), $12\pi \cdot \frac{180^\circ - (90^\circ - x)}{360^\circ} = 6\pi \cdot (90^\circ - x)$ cm (arc centered at $B$), and $3\pi \cdot (90^\circ + x)$ cm (arc centered at $C$).
Adding these lengths gives the total perimeter of the shaded region:
$0 + 6\pi \cdot (90^\circ - x) + 3\pi \cdot (90^\circ + x) = 6\pi \cdot 90^\circ + 9\pi \cdot x = 540\pi + 9\pi \cdot x.$
To express this decimal, we need to find a numerical approximation for $\pi.$ Choosing $\pi \approx 3.14,$ we have:
$540(3.14) + 9(3.14)(x) = 1695.6 + 28.26x.$
Therefore, the perimeter of the shaded region can be approximated as $1695.6 + 28.26x$ cm.