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In the figure, AC = 3, CE = x, and BC is parallel to DE. If
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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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MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730
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.
Added options and OA.




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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If
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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: \(\frac{AE^2}{AC^2}=\frac{12}{1}\) > \(\frac{AE}{AC}=\sqrt{12}=2\sqrt{3}\) > \(AE=2\sqrt{3}*AC=6\sqrt{3}\) > \(x=AEAC=6\sqrt{3}3\). For more on triangles check Triangles chapter of Math Book: mathtriangles87197.htmlHope it helps.
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If
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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
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If
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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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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If
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04 Feb 2012, 17:19
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
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05 Feb 2012, 02:47
ar(ABC)/ar(ADE)= (AC/AE)^2 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
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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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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If
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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
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10 Jul 2019, 04:06
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Re: In the figure, AC = 3, CE = x, and BC is parallel to DE. If
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10 Jul 2019, 04:06




