ABCD is a trapezoid where BC is parallel to AD and perpendicular to AB. Kindly note that BC< AD. P is a point on AD such that CPD is an equilateral triangle. Q is a point on BC such that AQ is parallel to PC. If the area of the triangle CPD is $$4\sqrt{\ 3}$$, find the area of the triangle ABQ.

Given that, CPD is an equilateral triangle

$$\angle\ CPD=\angle\ PDC=\angle\ DCP=60^{\circ\ }$$

As AQ is parallel to PC $$\angle\ CPD=\angle\ QAP=60^{\circ\ }$$

As BC is parallel to AD $$\angle\ QAP=\angle\ AQB=60^{\circ\ }$$ ( alternate interior angles).

Given, Area of equilateral triangle CPD is $$4\sqrt{\ 3}$$

This implies, $$\ \frac{\ \sqrt{3}a^2}{4}=4\sqrt{3}$$

Solving, we get $$a=\ 4$$.

From figure, length of $$AB$$ = Height of $$\triangle\ CPD=\frac{\sqrt{\ 3}a}{2}=2\sqrt{\ 3}$$.

From $$\triangle\ ABQ,\ BQ=\ \frac{\ AB}{\tan\ 60^{\circ\ }}=\frac{2\sqrt{\ 3}}{\sqrt{\ 3}}=2$$

Area of $$\triangle\ ABQ=\frac{1}{2}\times\ AB\times\ BQ=\ \ \frac{\ 2\times\ 2\sqrt{\ 3}}{2}=2\sqrt{\ 3}\ $$

Option A is correct.