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02 Jun 2020, 09:06
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C
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Difficulty:
45%
(medium)
Question Stats:
70% (02:17) correct
30%
(02:44)
wrong
based on 91
sessions
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In the given figure, ABC is an equilateral triangle such that AD= DB and DF is parallel to EG. If a point is chosen at random inside the triangle, what is the probability that the point would lie inside the quadrilateral DEGF?
A. 0.20
B. 0.33
C. 0.50
D. 0.60
E. 0.75
Are You Up For the Challenge: 700 Level Questions
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02 Jun 2020, 09:13
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Bunuel
Given, AD = DB i.e. D is midpoint of equilateral ABC
i.e. ADE is an equilateral triangle with each side = (1/2)*AB
Let, AB = BC = AC = 4
i.e. DE = (1/2)*AB = (1/2)*4 = 2
∆BDF is a 30º-60º-90º (Ratio of sides 1:√3:2) triangle because angle B = 60º and angle BFD = 90º
SInce, BD = 2, therefore, BF = 1 and DF = √3
i.e. Area of quadrilaterla DEGF = DE(DF = 2*√3 = 2√3
Probability of [point in the quadrilateral = Area of Quadrilateral / ARea of the ∆ABC\( = 2√3 / (√3/4)4^2 = 1/2\)
Answer: Option C
02 Jun 2020, 12:15
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C.
Probability = Area of quadrilateral DFGE/ Area of triangle ABC
= (\sqrt{3}/8) / (\sqrt{3}/4)
= 1/2
Probability = Area of quadrilateral DFGE/ Area of triangle ABC
= (\sqrt{3}/8) / (\sqrt{3}/4)
= 1/2
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