Answer
ABCD is a trapezoid.
Area = $9$.
Work Step by Step
From the graph, the quadrilateral $ABCD$ has a pair of parallel sides, so $ABCD$ is a trapezoid.
The area is $\displaystyle \frac{1}{2}$(sum of bases lengths)$\times$(height)
$A=(\displaystyle \frac{b_{1}+b_{2}}{2})h$.
$b_{1}=d(A, B)=\sqrt{(1-5)^{2}+(0-0)^{2}}=\sqrt{4^{2}}=4$.
$b_{2}=d(C, D)=\sqrt{(4-2)^{2}+(3-3)^{2}}=\sqrt{2^{2}}=2$.
The bases are parallel to the x-axis, so
$h$ = difference in y-coordinates $= |3-0|=3$.
Area = $(\displaystyle \frac{4+2}{2})3=9$.
