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Area of parallelogram ABCD

CAT 2021 Slot 3 · Quantitative Ability · Hard · Geometry

This is a hard Quantitative Ability question from the CAT 2021 Slot 3 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10 cm and 20 cm, respectively. If the angle ∠ADC\angle ADC is equal to 30∘30^\circ then the area of the parallelogram, in sq. cm, is

  1. A.

    25(3+15)2\frac{25(\sqrt{3}+\sqrt{15})}{2}

  2. B.

    25(5+15)25(\sqrt{5}+\sqrt{15})

  3. C.

    25(5+15)2\frac{25(\sqrt{5}+\sqrt{15})}{2}

  4. D.

    25(3+15)25(\sqrt{3}+\sqrt{15})

Answer

D

Explanation

Let CD=xCD = x. In △ADC\triangle ADC, using Cosine Rule for ∠ADC=30∘\angle ADC = 30^\circ: AC2=AD2+CD2−2(AD)(CD)cos⁡30∘AC^2 = AD^2 + CD^2 - 2(AD)(CD)\cos 30^\circ 202=102+x2−2(10)(x)3220^2 = 10^2 + x^2 - 2(10)(x)\frac{\sqrt{3}}{2} 400=100+x2−103x  ⟹  x2−103x−300=0400 = 100 + x^2 - 10\sqrt{3}x \implies x^2 - 10\sqrt{3}x - 300 = 0

Solving for positive xx: x=103+300+12002=53+515x = \frac{10\sqrt{3} + \sqrt{300 + 1200}}{2} = 5\sqrt{3} + 5\sqrt{15}

Area of parallelogram = AD⋅CD⋅sin⁡30∘=10⋅x⋅12=5xAD \cdot CD \cdot \sin 30^\circ = 10 \cdot x \cdot \frac{1}{2} = 5x: Area=5(53+515)=25(3+15)\text{Area} = 5(5\sqrt{3} + 5\sqrt{15}) = 25(\sqrt{3} + \sqrt{15}).

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