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Trapezoid Area with Equilateral Triangle

XAT 2023 · Quantitative Ability & Data Interpretation · Hard · Geometry

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ABCD is a trapezoid where BC is parallel to AD and perpendicular to AB. Kindly note that BC<ADBC < 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 434\sqrt{3}, find the area of the triangle ABQ.

  1. A.

    232\sqrt{3}

  2. B.

    434\sqrt{3}

  3. C.

    4

  4. D.

    838\sqrt{3}

  5. E.

    None of the above

Answer

A

Explanation

Area of equilateral △CPD=34a2=43  ⟹  a=4\triangle CPD = \frac{\sqrt{3}}{4} a^2 = 4\sqrt{3} \implies a = 4. Height of △CPD=32×4=23\triangle CPD = \frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3}. Since BC∥ADBC \parallel AD and AB⊥ADAB \perp AD, AB=height=23AB = \text{height} = 2\sqrt{3}. AQ ∥\parallel PC   ⟹  ∠QAB=30∘  ⟹  BQ=ABtan⁡30∘=233=2\implies \angle QAB = 30^\circ \implies BQ = AB \tan 30^\circ = \frac{2\sqrt{3}}{\sqrt{3}} = 2. Area of △ABQ=12×AB×BQ=12×23×2=23\triangle ABQ = \frac{1}{2} \times AB \times BQ = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3}.

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