Question

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.

Accepted Answer

A. Area of equilateral $	riangle CPD = \frac{\sqrt{3}}{4} a^2 = 4\sqrt{3} \implies a = 4$. Height of $	riangle CPD = \frac{\sqrt{3}}{2} 	imes 4 = 2\sqrt{3}$. Since $BC \parallel AD$ and $AB \perp AD$, $AB = 	ext{height} = 2\sqrt{3}$. AQ $\parallel$ PC $\implies \angle QAB = 30^\circ \implies BQ = AB 	an 30^\circ = \frac{2\sqrt{3}}{\sqrt{3}} = 2$. Area of $	riangle ABQ = \frac{1}{2} 	imes AB 	imes BQ = \frac{1}{2} 	imes 2\sqrt{3} 	imes 2 = 2\sqrt{3}$.

Trapezoid Area with Equilateral Triangle

Explanation

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