jmc

We draw a horizontal line through $B$ (meeting the $y$-axis at $P$) and a vertical line through $C$ (meeting the $x$-axis at $Q$). Suppose the point of intersection of these two lines is $R$. We know that $P$ has coordinates $(0,3)$ (since $B$ has $y$-coordinate 3) and $Q$ has coordinates $(5,0)$ (since $C$ has $x$-coordinate 5), so $R$ has coordinates $(5,3)$.

Using the given coordinates, $OA=1$, $AP=2$, $PB=1$, $BR=4$, $RC=1$, $CQ=2$, $QD=1$, and $DO=4$.

The area of $ABCD$ equals the area of $PRQO$ minus the areas of triangles $APB$, $BRC$, $CQD$, and $DOA$.

$PRQO$ is a rectangle, so it has area $3\times 5=15$.

Triangles $APB$ and $CQD$ have bases $PB$ and $QD$ of length 1 and heights $AP$ and $CQ$ of length 2, so each has area $$\frac{1}{2}(1)(2)=1.$$Triangles $BRC$ and $DOA$ have bases $BR$ and $DO$ of length 4 and heights $CR$ and $AO$ of length 1, so each has area $$\frac{1}{2}(4)(1)=2.$$Thus, the area of $ABCD$ is $$15-1-1-2-2=\boxed{9}.$$(Alternatively, we could notice that $ABCD$ is a parallelogram. Therefore, if we draw the diagonal $AC$, the area is split into two equal pieces. Dropping a perpendicular from $C$ to $Q$ on the $x$-axis produces a trapezoid $ACQO$ from which only two triangles need to be removed to determine half of the area of $ABCD$.)