Identify the given information.

We have two circles with the same radius, $r = 8$ cm.
The distance between their centers is $d = 12.8$ cm.
The common chord AB connects the two points where the circles intersect.
The line connecting the centers of the circles (let's call them $C_1$ and $C_2$) is perpendicular to the common chord AB and bisects it.
Since the radii of both circles are equal, the midpoint of the chord AB (let's call it M) is also the midpoint of the line segment $C_1C_2$.
Therefore, the distance from one center to M is $h = \frac{d}{2} = \frac{12.8}{2} = 6.4$ cm.

Consider the right-angled triangle $C_1MA$, where $C_1$ is the center of one circle, M is the midpoint of the chord AB, and A is one of the intersection points.
$C_1A$ is the radius, $r = 8$ cm (this is the hypotenuse).
$C_1M$ is the distance from the center to the chord, $h = 6.4$ cm.
$AM$ is half the length of the chord AB.
According to the Pythagorean theorem:

The length of the common chord AB is twice the length of AM.

The common region is formed by two identical circular segments, one from each circle.
The area of a circular segment is given by the formula:

Where $r = 8$ cm and $h = 6.4$ cm.
First, calculate the angle $\theta = \cos^{-1}\left(\frac{h}{r}\right)$ in radians:

Next, calculate $\sqrt{r^2-h^2}$:

Now substitute these values into the segment area formula:

The total common area is twice the area of one circular segment.

Rounding to two decimal places: