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In the given figure, ABC is an equilateral triangle such that AD= DB

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In the given figure, ABC is an equilateral triangle such that AD= DB  [#permalink]

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New post 02 Jun 2020, 08:06
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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


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Re: In the given figure, ABC is an equilateral triangle such that AD= DB  [#permalink]

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New post 02 Jun 2020, 08:13
Bunuel wrote:
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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


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1.png


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
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In the given figure, ABC is an equilateral triangle such that AD= DB  [#permalink]

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New post 02 Jun 2020, 11:15
C.

Probability = Area of quadrilateral DFGE/ Area of triangle ABC
= (\sqrt{3}/8) / (\sqrt{3}/4)
= 1/2
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In the given figure, ABC is an equilateral triangle such that AD= DB  [#permalink]

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New post 06 Jun 2020, 04:37
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Bunuel wrote:
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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



Solution:

Since DF is parallel to EG, the remaining two interior angles of the quadrilateral are right angles as well. Then, DEGF is actually a rectangle and DE is parallel to FG as well.

Since DE is parallel to FG, angle ABC is equal to angle ADE and angle ACB is equal to angle AED. Thus, the triangle ADE is also an equilateral triangle.

Let x be the length of a side of ADE. Then, a side of ABC has length 2x. Further, since the triangles ADE and ABC are similar and the length of a side of ABC is twice that of ADE, the height of ABC is twice that of ADE as well. Let h be the height of ADE and thus, the height of ABE is 2h.

Finally, notice that the length of DF is exactly the difference between the heights of ABC and ADE; i.e. h.

Now, the area of ABC is (2x * 2h)/2 = 2xh and the area of DEGF is xh. Thus, the probability that a randomly chosen point inside ABC is located within DEGF is xh/2xh = 1/2.

Answer: C
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In the given figure, ABC is an equilateral triangle such that AD= DB   [#permalink] 06 Jun 2020, 04:37

In the given figure, ABC is an equilateral triangle such that AD= DB

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