jmc

Let I, II, and III denote the areas of the triangles as shown on the diagram. The area of $\Delta ABC$ can be obtained by subtracting I+II+III from the area of the rectangle.

I $=4\times 2/2=4$, II $=5\times 2/2=5$, III = $1\times 4/2=2$; I+II+III $=4+5+2=11$.

Subtracting these areas from the area of the large rectangle tells us that the area of $ABC$ is $4\cdot 5-4-5-2=\boxed{9}$ square units.