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CAT 2002 — QA Question 38

Basics of Mensuration/PrismEasy
Passage / Data

Answer the following question based on the information given below.

In the diagram below, ∠ABC = 90° = ∠DCH = ∠DOE = ∠EHK = ∠FKL = ∠GLM = ∠LMN, AB = BC = 2CH = 2CD = EH = FK = 2HK = 4KL = 2LM = MN

The ratio of the areas of the two quadrangles ABCD and DEFG is

Answer & solution

  • A

    1 : 2

  • B

    2 : 1

  • 12 : 7

  • D

    None of these

Solution

Area of trapezium ABCD = 12 BC (CD + AB)
Also BC = 2CD, AB = BC = 2CD

Area of trapezium ABCD = 12 × CD (CD + 2DC) = 3CD2

Area of trapezium DEFG = 12 EO(EF + DG)

EO = CD

EF = CD

DG = CH + HK + KL = CD + CD + CD2=52CD

Area of trapezium DEFG = 12×CD(CD+52CD)=12×CD(7CD2)=7CD24...(ii) 

Ratio of ABCD : DEFG = 12 : 7

Hence, option (c).

CAT 2002 QA Q38: The ratio of the areas of the two quadrangles ABCD and DEFG is — Solution | TheCATExam