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yashikaaggarwal
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20 Jun 2020, 03:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
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Question Stats:
32% (02:25) correct
68%
(02:26)
wrong
based on 22
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If the Area of Triangle ABF is 3 sq. units, that of triangle AFE is 4 sq. units and that of triangle BFD is 2 sq units, Find the area of Quadrilateral FECD?
A. 8
B. 24
C. 72
D. 96
E. 48
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21 Jun 2020, 04:50
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21 Jun 2020, 09:53
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This is not a GMAT question. This is a CAT question.
There is a particular theorem to solve such types of questions.
\(\frac{1}{Area(CAB)} = \frac{1}{Area(AEB)} + \frac{1}{Area(ADB)} - \frac{1}{Area(AFB)}\)
\(\frac{1}{Area(CAB)} = \frac{1}{7} + \frac{1}{5} - \frac{1}{3}\)
\( \frac{1}{Area(CAB)} = \frac{15+21-35}{105}\)
\(\frac{1}{Area(CAB)} = 1/105\)
\(Area(CAB) = 105\)
Area(FECD) = Area(CAB) - Area(AFE) - Area(BFD) - Area(AFB)
\(= 105 - 4 - 3 - 2 = 96\)
OA, D
There is a particular theorem to solve such types of questions.
\(\frac{1}{Area(CAB)} = \frac{1}{Area(AEB)} + \frac{1}{Area(ADB)} - \frac{1}{Area(AFB)}\)
\(\frac{1}{Area(CAB)} = \frac{1}{7} + \frac{1}{5} - \frac{1}{3}\)
\( \frac{1}{Area(CAB)} = \frac{15+21-35}{105}\)
\(\frac{1}{Area(CAB)} = 1/105\)
\(Area(CAB) = 105\)
Area(FECD) = Area(CAB) - Area(AFE) - Area(BFD) - Area(AFB)
\(= 105 - 4 - 3 - 2 = 96\)
OA, D
yashikaaggarwal
