Correct Answer: $\frac{(135\sqrt{3})}{2}$

Solution :

Given: $PQRS$ is a quadrilateral, $QR = 18$ cm and $PS = 9$ cm.
Construction: Extend the lines $QP$ and $RS$ as they meet at $T$ and form $\triangle QRT$ an equilateral triangle.
Take the angles as shown in the figure.
The area of an equilateral $\triangle QRT$ = $\frac{\sqrt{3}}{4}×(18)^2=81\sqrt{3}$ cm ²
And in the $\triangle TSP$ is a right-angled triangle.
$\tan 30^{\circ}=\frac{TS}{PS}=\frac{TS}{9}$
⇒ $\frac{1}{\sqrt3}=\frac{TS}{9}$
⇒ $TS=\frac{9}{\sqrt3}$
The area of right-angled $\triangle TSP$ = $\frac{1}{2}×TS×PS$
The area of right-angled triangle = $\frac{1}{2}×\frac{9}{\sqrt3}×9=\frac{27\sqrt3}{2}$ cm ²
The required area of quadrilateral $PQRS$ = The area of an equilateral $\triangle QRT$ – The area of right-angled $\triangle TSP$
The required area of quadrilateral $PQRS$ = $81\sqrt{3} $ – $\frac{27\sqrt3}{2}$ = $\frac{162\sqrt{3}–27\sqrt3}{2}$ = $\frac{135\sqrt{3}}{2}$ cm ²
Hence, the correct answer is $\frac{135\sqrt{3}}{2}$.