jmc

The graph of the two inequalities is shown below:



The shaded region is the solution set to the two given inequalities. Angle $ADC$ is a right angle because $\overline{AD}$ has slope -1 and $\overline{DC}$ has slope 1, and the two slopes are negative reciprocals. Similarly, the other three angles between sides bounding the shaded region are also right angles. Since $AD=DC$ by symmetry, $ABCD$ is a square. A diagonal of the square is $BD$, which measures 3 units. So a side of the square measures $3/\sqrt{2}$ units and the area is $(3/\sqrt{2}{)}^2=\boxed{4.5}$ square units.