jmc

Draw a rectangle with vertices $(-3,7),(-3,-1),(6,-1),(6,7)$ about the pentagon, as shown below: The area of the pentagon is the difference between the area of the rectangle and the four right triangles formed at the vertices of the rectangle. We find that $$\begin{array}{rl}{A}_1& =\frac{1}{2}\cdot 5\cdot 2=5,\\ A_2& =\frac{1}{2}\cdot 4\cdot 3=6,\\ A_3& =\frac{1}{2}\cdot 2\cdot 5=5,\\ A_4& =\frac{1}{2}\cdot 3\cdot 6=9,\end{array}$$ while the area of the entire rectangle is $9\times 8=72$. Thus, the area of the pentagon is equal to $72-5-6-5-9=\boxed{47}$ square units.