jmc

This triangle's vertices are the intersection points of each pair of lines. The intersection point of $y=0$ and $y=x+4$ is (-4,0). The intersection point of $y=0$ and $x+3y=12$ is (12,0). To find the intersection of the last 2 lines, we substitute the first equation for $y$ and then solve for $x$. Doing so, we get $$\begin{array}{rl}x+3y& =12\\ x+3(x+4)& =12\\ x+3x+12& =12\\ 4x& =0\\ x& =0\end{array}$$ Thus, $y=4$, and the intersection point is (0,4). Let the base of the triangle be the side of the triangle on the $x$-axis. Because this side is between the points (-4,0) and (12,0), its length is $12-(-4)=12+4=16$. The height must be perpendicular to this side and through the final vertex. This is along the $y$-axis. Thus, the height of the triangle is just the $y$-coordinate of the other point, which is 4. Thus, the area of the triangle is $\frac{1}{2}\cdot 16\cdot 4=8\cdot 4=\boxed{32}\text{sq units}$.