Sum of areas

Draw $XC$ . Since triangle $XCD$ and parallelogram $ABCD$ share the same base $(DC)$ and a common altitude (from $X$ to $DC$ ), the area of triangle $XCD$ is equal to one-half the area of parallelogram $ABCD$ .

Similarly, triangle $XCD$ and parallelogram $XYZD$ share the same base $XD$ and the same altitude to that base (from $C$ to $XD)$ ; thus the area of triangle $XCD$ is equal to one-half the area of parallelogram $XYZD$ . Therefore, the area of the two parallelograms are equal.

$Area_{(ABCD)}=Area_{(XYZD)}=3(6)=18$

Therefore,

$Area_{(XYC)}+Area_{(DCZ)}=\dfrac{1}{2}Area_{(XYZD)}=\dfrac{1}{2}(18)=$ $\boxed{9}$ $\large\boxed{\color{#D61F06}\text{answer}}$

Original parallelogram: $Area_{(ABCD)}=3\times6=18$

Triangle within that parallelogram with the same base and height: $Area_{(DXC)}=\frac{1}{2}\times Area_{(ABCD)}=9$

Same triangle sharing base and height with another parallelogram: $Area_{(DXC)}=\frac{1}{2}\times Area_{(DXYZ)}\implies Area_{(DXYZ)}=2\times9=18$

The area we are looking for as the area of the parallelogram minus the area of the triangle:

$Area_{(XYC)}+Area_{(DCZ)}=Area_{(DXYZ)}- Area_{(DXC)}=9$