Let’s draw the parallelogram.
Let $\text{AE}$ be the height of the parallelogram.
In $\triangle \text{AED},$
$\Rightarrow \text{sin} \; 30^{\circ} = \frac{\text{AE}}{\text{AD}}$
$\Rightarrow \frac{1}{2} = \frac{\text{AE}}{10}$
$\Rightarrow \boxed{\text{AE} = 5 \;\text{cm}}$
In $\triangle \text{AED},$ apply the Pythagorean theorem.
$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2}$
$\Rightarrow 10^{2} = 5^{2} + \text{(DE)}^{2}$
$\Rightarrow \text{DE} = \sqrt{100-25} = \sqrt{75}$
$\Rightarrow \boxed{\text{DE} = 5 \sqrt{3} \; \text{cm}}$
In $\triangle \text{AEC},$ apply the Pythagorean theorem.
$\Rightarrow 20^{2} = 5^{2} + \text{(EC)}^{2}$
$\Rightarrow \text{(EC)}^{2} = 400-25 = 375$
$\Rightarrow \text{EC} = \sqrt{375}$
$\Rightarrow \boxed{\text{EC} = 5\sqrt{15} \; \text{cm}}$
So, the length of $\text{DC} = \text{DE} + \text{EC}$
$\Rightarrow \boxed{\text{DC} = (5\sqrt{3} + 5\sqrt{15}) \; \text{cm}}$
The area of parallelogram $\text{ABCD} = \text{Base} \times \text{Height} = \text{DC} \times \text{AE}$
$ \qquad \qquad \qquad = (5\sqrt{3} + 5\sqrt{15}) \times 5 = 25(\sqrt{3} + \sqrt15) \; \text{cm}^{2}$
$\therefore$ The area of parallelogram $\text{ABCD}$ is $25(\sqrt{3} + \sqrt15) \; \text{cm}^{2}.$
Correct Answer $:\text{D}$