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Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]
11 Oct 2013, 02:56

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

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

Answer: C.

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? _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]
14 Oct 2013, 03:44

1

This post received 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.

Answer: C.

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. _________________

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

Re: ABC and AED are triangles with BC parallel to ED. Find the a [#permalink]
09 Dec 2014, 09:36

Expert's post

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.

Please explain.

Thanks

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

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

Please explain.

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

gmatclubot

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

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