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CAT 2002 — QA Question 37

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 magnitude of ∠FGO =

Answer & solution

  • A

    30°

  • B

    45°

  • C

    60°

  • None of these

Solution

Given that AB = BC = 2CH = 2CD = EH = EK = 2HK = 4KL = 2LM = MN

And EO = FP

Also,

2CD = EH

EO = FP = CD

∴ KL = PG = CD2

FP = CD : PG = CD2 : ∠FPG = 90°

∵ The angle are proportionate to the sides opposite to the angles.

∴ ∠FGO = ∠FGP = tan–1 2

Hence, option (d).

CAT 2002 QA Q37: The magnitude of ∠FGO = — Solution | TheCATExam