Correct Answer: $150\sqrt{15}\text{ cm}^{2}$

Solution :
In $\triangle$ ABD,
AB = 20 cm. AD = 30 cm. BD = 40 cm
Semi perimeter (s) = $\frac{a+b+c}{2}$ = $\frac{20+30+40}{2}$ = 45 cm
Now, the area of $\triangle$ABD
= $\sqrt{s(s - a)(s - b)(s - c)}$
= $\sqrt{45(45 - 20)(45 - 30)(45 - 40)}$
= $\sqrt{45 \times 25 \times 15 \times 5}$
= $\sqrt{5 \times 3 \times 5 \times 5 \times 5 \times 3 \times 5}$
= $\sqrt{5^2 \times 5^2 \times 5 \times 3^2 \times 3}$
= $5 \times 5 \times 3\sqrt{15}$
= $75\sqrt{15}$
$\therefore$ Area of parallelogram ABCD $=2 \times 75\sqrt{15}=150\sqrt{15}\text{ cm}^2$
Hence, the correct answer is $150\sqrt{15}\text{ cm}^2$.