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Area of a right trapezium

CAT 2020 Slot 3 · Quantitative Ability · Medium · Geometry

This is a medium Quantitative Ability question from the CAT 2020 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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In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD=45∘\angle BAD = 45^\circ. If DC=5 cm\text{DC} = 5\text{ cm}, BC=4 cm\text{BC} = 4\text{ cm}, the area of the trapezium in sq. cm is

Answer

28

Explanation

In trapezium ABCD, AB∥DCAB \parallel DC and BC⊥DCBC \perp DC. Since BC⊥DCBC \perp DC and AB∥DCAB \parallel DC, BC⊥ABBC \perp AB as well. Height h=BC=4 cmh = BC = 4\text{ cm}.

Draw perpendicular from D to AB meeting AB at point P. DP=BC=4 cmDP = BC = 4\text{ cm} and AP=DP/tan⁡45∘=4/1=4 cmAP = DP / \tan 45^\circ = 4 / 1 = 4\text{ cm}. Also PB=DC=5 cmPB = DC = 5\text{ cm}. So AB=AP+PB=4+5=9 cmAB = AP + PB = 4 + 5 = 9\text{ cm}.

Area of trapezium = 12(AB+DC)×h=12(9+5)×4=28 sq. cm\frac{1}{2} (AB + DC) \times h = \frac{1}{2} (9 + 5) \times 4 = 28\text{ sq. cm}.

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