The graph of $g(x)$ consists of two rays with a common vertex at $x = 8$. We remove the module and easily calculate its intersection points with the x-axis through the equations $6 - x = 0$ and $x - 10 = 0$ — $A(6, 0)$ and $B(10, 0)$.

After removing the modules in $f(x)$ we see that in the interval $(-\infty, 2)$ $f(x) = -2$, in the interval $[2, 4]$ $f(x) = 2x - 6$ and in the interval $(4, \infty)$ $f(x) = 2$. We solve the equations $f(x) = g(x)$ for each of the three intervals. We get the following intersection points — $D(4, 2)$ and $C(12, 2)$.

It follows that the figure $ABCD$ is a trapezoid with bases $8$ and $4$ and height $2$. Therefore, its area is $\frac{8+4}{2} \times 2 = 12$.