Xavier Aptitude Test 2023 Paper

Given that, CPD is an equilateral triangle$\angle \mathit{C}\mathit{P}\mathit{D}=\angle \mathit{P}\mathit{D}\mathit{C}=\angle \mathit{D}\mathit{C}\mathit{P}={60}^{\circ }$
As AQ is parallel to $\mathrm{PC}\angle \mathit{C}\mathit{P}\mathit{D}=\angle \mathit{Q}\mathit{A}\mathit{P}={60}^{\circ }$
As BC is parallel to $\mathrm{AD}\angle \mathit{Q}\mathit{A}\mathit{P}=\angle \mathit{A}\mathit{Q}\mathit{B}={60}^{\circ }$ (alternate interior angles).
Given, Area of equilateral triangle CPD is $4\sqrt{3}$
This implies, $\frac{\sqrt{3}{\mathit{a}}^2}{4}=4\sqrt{3}$
Solving, we get $\mathit{a}=4$.
From figure, length of $\mathit{A}\mathit{B}=$ Height of $△\mathit{C}\mathit{P}\mathit{D}=\frac{\sqrt{3}\mathit{a}}{2}=2\sqrt{3}$.
From $△\mathit{A}\mathit{B}\mathit{Q},\mathit{B}\mathit{Q}=\frac{\mathit{A}\mathit{B}}{\tan {60}^{\circ }}=\frac{2\sqrt{3}}{\sqrt{3}}=2$
Area of $△\mathit{A}\mathit{B}\mathit{Q}=\frac{1}{2}\times \mathit{A}\mathit{B}\times \mathit{B}\mathit{Q}=\frac{2\times 2\sqrt{3}}{2}=2\sqrt{3}$
Option A is correct.