Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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]

Show Tags

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

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]

Show Tags

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.

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]

Show Tags

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!

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...