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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the given figure, ABC is an equilateral triangle such that AD= DB

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Math Expert V
Joined: 02 Sep 2009
Posts: 64891
In the given figure, ABC is an equilateral triangle such that AD= DB  [#permalink]

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1
2 00:00

Difficulty:   65% (hard)

Question Stats: 65% (02:14) correct 35% (02:50) wrong based on 60 sessions

### HideShow timer Statistics 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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Attachment: 1.png [ 9.12 KiB | Viewed 663 times ]

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

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Bunuel wrote: 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

Attachment:
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$$= 2√3 / (√3/4)4^2 = 1/2$$

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

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C.

Probability = Area of quadrilateral DFGE/ Area of triangle ABC
= (\sqrt{3}/8) / (\sqrt{3}/4)
= 1/2
Attachments image.jpg [ 165.77 KiB | Viewed 518 times ]

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Target Test Prep Representative V
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In the given figure, ABC is an equilateral triangle such that AD= DB  [#permalink]

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1
Bunuel wrote: 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.

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# Scott Woodbury-Stewart

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See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews 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   