Area of parallelogram ABCD

Let $CD = x$. In $\triangle ADC$, using Cosine Rule for $\angle ADC = 30^\circ$: $AC^2 = AD^2 + CD^2 - 2(AD)(CD)\cos 30^\circ$ $20^2 = 10^2 + x^2 - 2(10)(x)\frac{\sqrt{3}}{2}$ $400 = 100 + x^2 - 10\sqrt{3}x \implies x^2 - 10\sqrt{3}x - 300 = 0$

Solving for positive $x$: $x = \frac{10\sqrt{3} + \sqrt{300 + 1200}}{2} = 5\sqrt{3} + 5\sqrt{15}$

Area of parallelogram = $AD \cdot CD \cdot \sin 30^\circ = 10 \cdot x \cdot \frac{1}{2} = 5x$: $\text{Area} = 5(5\sqrt{3} + 5\sqrt{15}) = 25(\sqrt{3} + \sqrt{15})$.