Autumn tournament

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 $\frac{8+4}{2}\times 2=12$.