 Parallelogram $ABCD$ is given such that $\angle ABC$ equals $30^o$ . Let $X$ be the foot of the perpendicular from $A$ onto $BC$, and $Y$ the foot of the perpendicular from $C$ to $AB$. If $AX = 20$ and $CY = 22$, find the area of the parallelogram.

1. Given that $\angle ABC = 30^\circ$ in parallelogram $ABCD$, we need to find the area of the parallelogram.
2. Let $X$ be the foot of the perpendicular from $A$ onto $BC$, and $Y$ be the foot of the perpendicular from $C$ onto $AB$.
3. We are given $AX = 20$ and $CY = 22$.
4. In $\triangle CYB$, we know that $\angle CBY = 30^\circ$ and $\triangle CYB$ is a right triangle at $Y$. Therefore, $\angle BCY = 60^\circ$.
5. Using the cosine of $60^\circ$, we have:
   \[
   \cos 60^\circ = \frac{YC}{BC}
   \]
   Since $\cos 60^\circ = \frac{1}{2}$, we get:
   \[
   \frac{1}{2} = \frac{22}{BC}
   \]
6. Solving for $BC$, we get:
   \[
   BC = 2 \times 22 = 44
   \]
7. The area of the parallelogram $ABCD$ can be calculated using the base $BC$ and the height $AX$:
   \[
   \text{Area} = BC \times AX = 44 \times 20 = 880
   \]

The final answer is $\boxed{880}$