Answer: $\frac{8}{3}$
Solution: The areas of the left and right triangles are both one-fourth the area of the area of the rectangle $A B N M$ which gives an area of $\frac{1}{2}$. The area of the lower triangle is $\frac{1}{2}(2)\left(\frac{3}{2}\right)=\frac{3}{2}$. The area of the upper triangle is $\frac{1}{9}$ th the lower triangle (compare their altitudes). Hence, the area is $\frac{3}{2}\left(\frac{1}{9}\right)$.
Therefore, the area of the shaded region is
$$
\frac{1}{2}+\frac{1}{2}+\frac{3}{2}+\frac{1}{6}=\frac{8}{3} .
$$
5. Refer to the figure on the right. The quadrilateral $A B C D$ is a square with a side of length 2 units while $M$ and $N$ are the midpoints of $A D$ and $B C$, respectively. Determine the area of the shaded region.