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# In the figure, AC = 3, CE = x, and BC is parallel to DE. If

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In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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Updated on: 01 Jan 2019, 23:58
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In the figure, AC = 3, CE = x, and BC is parallel to DE. If the area of triangle ABC is 1/12 of the area of triangle ADE, then x = ?

A. $$3\sqrt{3}-3$$
B. $$3$$
C. $$6$$
D. $$6\sqrt{3}-3$$
E. $$6\sqrt{3}$$

Guys - as I neither have an OA nor the answer choices can someone please let me know how to solve this question?

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Originally posted by enigma123 on 04 Feb 2012, 16:42.
Last edited by Gladiator59 on 01 Jan 2019, 23:58, edited 1 time in total.
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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04 Feb 2012, 17:09
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In the figure, AC = 3, CE = x, and BC is parallel to DE. If the area of triangle ABC is 1/12 of the area of triangle ADE, then x = ?

Another useful property: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{S^2}{s^2}$$.

Next, note that ABC and ADE are similar triangles, thus the ratio of their areas is the square of the ratio of their sides: $$\frac{AE^2}{AC^2}=\frac{12}{1}$$ --> $$\frac{AE}{AC}=\sqrt{12}=2\sqrt{3}$$ --> $$AE=2\sqrt{3}*AC=6\sqrt{3}$$ --> $$x=AE-AC=6\sqrt{3}-3$$.

For more on triangles check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
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##### General Discussion
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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04 Feb 2012, 17:14
Thanks Bunuel. But how do we know that they are similar triangles?
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MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 438
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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04 Feb 2012, 17:15
Is this because Angle b = Angle D and Angle c = Angle E?
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Best Regards,
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MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Math Expert
Joined: 02 Sep 2009
Posts: 65809
Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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04 Feb 2012, 17:19
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enigma123 wrote:
Is this because Angle b = Angle D and Angle c = Angle E?

Similar triangles are triangles in which the three angles are identical. Trianlges ABC and ADE share one common angle D and also since BC is parallel to DE, <ABC=<ADE and <ACB=<AED.

Notice that it is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.

Another useful property of similar triangles: in similar triangles, the sides of the triangles are in some proportion to one another. For example, a triangle with lengths 3, 4, and 5 has the same angle measures as a triangle with lengths 6, 8, and 10. The two triangles are similar, and all of the sides of the larger triangle are twice the size of the corresponding legs on the smaller triangle.

Hope it helps.
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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05 Feb 2012, 02:47
on solving we get AE = 6\sqrt{3}
so value of x = 6\sqrt{3} - 3
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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05 Feb 2012, 03:48
Bunuel in this question how did you get AE^2/AC^2 = 12/1?
_________________
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MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
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Joined: 02 Sep 2009
Posts: 65809
Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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05 Feb 2012, 03:50
enigma123 wrote:
Bunuel in this question how did you get AE^2/AC^2 = 12/1?

Given: the area of triangle ABC is 1/12 of the area of triangle ADE --> so the ratio of the area ADE to the area of ABC is 12 (12/1).
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If  [#permalink]

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13 Jul 2020, 07:43
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If   [#permalink] 13 Jul 2020, 07:43