ABC and AED are triangles with BC parallel to ED. Find the a : GMAT Data Sufficiency (DS)
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# ABC and AED are triangles with BC parallel to ED. Find the a

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ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]

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11 Oct 2013, 02:42
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ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.
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Triangle_Barrons.png [ 45.21 KiB | Viewed 1754 times ]

(1) BC = 8
(2) ED = 5
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Oct 2013, 02:43, edited 1 time in total.
Renamed the topic and edited the question.
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Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]

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11 Oct 2013, 02:56
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ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.

(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.

(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.

Property:
If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Since BC/ED=8/5, then $$\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}$$. As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.

Hope it's clear.
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Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]

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14 Oct 2013, 03:41
Bunuel wrote:
ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.

(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.

(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.

Property:
If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Since BC/ED=8/5, then $$\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}$$. As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.

Hope it's clear.

Wonderfully done - one question please: "the area of ABC is 16" (in the stimulus) - is there anything we can infer from this information? Or does this information help us in any way before looking at S1 and S2?
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There are times when I do not mind kudos...I do enjoy giving some for help

Math Expert
Joined: 02 Sep 2009
Posts: 35950
Followers: 6863

Kudos [?]: 90115 [1] , given: 10418

Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]

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14 Oct 2013, 03:44
1
KUDOS
Expert's post
obs23 wrote:
Bunuel wrote:
ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.

(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.

(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.

Property:
If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Since BC/ED=8/5, then $$\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}$$. As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.

Hope it's clear.

Wonderfully done - one question please: "the area of ABC is 16" (in the stimulus) - is there anything we can infer from this information? Or does this information help us in any way before looking at S1 and S2?

We just know that the area is 16, we can get nothing more from it.
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Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]

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09 Dec 2014, 09:11
Hi Bunuel,

Without combining (1) and (2), can't we say both the triangles are similar? i am asking this because i did not understand the reason for insufficiency "We don't know where is ED positioned. Not sufficient."

I marked the answer D because i thought two triangles are similar. Hence, knowing BC=8 will give me the height of the triangle ABC because the area is given. As we know, for similar triangles their ratios of heights and bases are equal i will be able to find the area of triangle AED.

Thanks
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Math Expert
Joined: 02 Sep 2009
Posts: 35950
Followers: 6863

Kudos [?]: 90115 [0], given: 10418

Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]

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09 Dec 2014, 09:36
HKD1710 wrote:
Hi Bunuel,

Without combining (1) and (2), can't we say both the triangles are similar? i am asking this because i did not understand the reason for insufficiency "We don't know where is ED positioned. Not sufficient."

I marked the answer D because i thought two triangles are similar. Hence, knowing BC=8 will give me the height of the triangle ABC because the area is given. As we know, for similar triangles their ratios of heights and bases are equal i will be able to find the area of triangle AED.

Thanks

Yes, we know that they are similar from the stem but this does not help. Check the image below:
Attachment:

Untitled.png [ 49.29 KiB | Viewed 859 times ]

_________________
Senior Manager
Joined: 21 Jun 2014
Posts: 458
Concentration: General Management, Technology
GMAT 1: 540 Q45 V20
GPA: 2.49
WE: Information Technology (Computer Software)
Followers: 12

Kudos [?]: 155 [0], given: 91

Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]

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09 Dec 2014, 10:12
Bunuel wrote:
HKD1710 wrote:
Hi Bunuel,

Without combining (1) and (2), can't we say both the triangles are similar? i am asking this because i did not understand the reason for insufficiency "We don't know where is ED positioned. Not sufficient."

I marked the answer D because i thought two triangles are similar. Hence, knowing BC=8 will give me the height of the triangle ABC because the area is given. As we know, for similar triangles their ratios of heights and bases are equal i will be able to find the area of triangle AED.

Thanks

Yes, we know that they are similar from the stem but this does not help. Check the image below:
Attachment:
Untitled.png

By looking at the image attached, i understood what you meant by "We don't know where is ED positioned. Not sufficient."

I conclude that in case of similar triangles when the figure is not drawn to scale and value of base for both the triangles is not known then knowing the value of base for only one triangle will always lead to this situation and will be insufficient. Please confirm!

Thank you
_________________

---------------------------------------------------------------
Target - 720-740
helpful post means press '+1' for Kudos!
http://gmatclub.com/forum/information-on-new-gmat-esr-report-beta-221111.html
http://gmatclub.com/forum/list-of-one-year-full-time-mba-programs-222103.html

Re: ABC and AED are triangles with BC parallel to ED. Find the a   [#permalink] 09 Dec 2014, 10:12
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# ABC and AED are triangles with BC parallel to ED. Find the a

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