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Re: Two "triangles" in a circle [#permalink]
17 Sep 2009, 09:02

1

This post received KUDOS

I wud go with option B.

Line joining the midpoint and parallel to the third side will be half the length of the third side.So statement 2 states it is parallel to the third side.So DE = 7 cm.

Re: Two "triangles" in a circle [#permalink]
17 Sep 2009, 13:36

They responded and indicated that I was right, they were wrong... E is correct, unless they indicate it is a straight line. They're changing the problem so that B is correct.

Re: Two "triangles" in a circle [#permalink]
20 Sep 2009, 13:20

Just because DE is parallel to CA does not force DEB to be a triangle. Notice that line EB is segmented by point A.

We don't know the length of DE, and we don't know for sure that EB is a straight line (as the question is written). Segment EA could be at a slight obtuse angle to segment AB, which would form a quadrilateral: DEAB. If that were the case, we would not have two similar triangles to create a ratio and determine the length of DE.

This has been confirmed by the folks at Manhattan GMAT. Hope this helps.

Re: Two "triangles" in a circle [#permalink]
16 Oct 2009, 10:12

I am at loss with this Manhattan question.

If EB is assumed to be a straight line, then statement 2 is sufficient.

What about statement 1?

OA says: "(1) INSUFFICIENT: Just knowing that x = 60° tells us nothing about triangle EDB. To illustrate, note that the exact location of point E is still unknown. Point E could be very close to the circle, making DE relatively short in length. However, point E could be quite far away from the circle, making DE relatively long in length. We cannot determine the length of DE with certainty. "

But, given that X=60, if point E would be "very close to the circle", then EB would not go through A. But EB does go through A. Therefore, point E can't be "very close to the circle" => EDB is an equilateral triangle, which makes statement 1 sufficient.

Re: Two "triangles" in a circle [#permalink]
05 Feb 2011, 11:18

guys, i have a question :

if we know for sure that 2 triangles have one same angle and that a side of one is twice that of the other, would not that info suffice to conclude that they are similar?

its like identical angle and sides in proportion. do we need any more info?

Re: Two "triangles" in a circle [#permalink]
05 Feb 2011, 12:44

1

This post received KUDOS

Expert's post

tinki wrote:

guys, i have a question :

if we know for sure that 2 triangles have one same angle and that a side of one is twice that of the other, would not that info suffice to conclude that they are similar?

its like identical angle and sides in proportion. do we need any more info?

Re: Two "triangles" in a circle [#permalink]
05 Feb 2011, 15:25

this might sound crazy, but why cant it be A? I mean x=60, and from the diagram we see that AC = AB because both are radius of the same circle.So, angle A should be equal to angle B. And x=C=60. So, angle A+B = 180-60=120. 120 divided by 2 is 60 each. so, A=60,B=60,C=60 and ABC is equilateral. DBE is proportional to ABC.So DBE must also be an equilateral triangle, therefore DE=DB=7.

Re: Two "triangles" in a circle [#permalink]
05 Feb 2011, 15:43

Expert's post

Sednima wrote:

this might sound crazy, but why cant it be A? I mean x=60, and from the diagram we see that AC = AB because both are radius of the same circle.So, angle A should be equal to angle B. And x=C=60. So, angle A+B = 180-60=120. 120 divided by 2 is 60 each. so, A=60,B=60,C=60 and ABC is equilateral. DBE is proportional to ABC.So DBE must also be an equilateral triangle, therefore DE=DB=7.

Thats just my opinion.Tell me why I may be wrong.

Read the first spoiler in initial post. _________________

stmt 1 does not show how the smaller triangle relates to the larger one. rather we only find out that the smaller triangle is an equilateral triangle. Stmt 2 is sufficient because we can see that angle cab is equal to deb. also angle edb and acb are the same angle and thus we can conclude these are similar triangles. after knowing that we can setup a proportion for these similar triangles.... DE/DB = CA/CB... DE is equal to the diameter. _________________

Re: In the figure to the right, if point C is the center of the [#permalink]
20 Apr 2012, 19:18

1

This post received KUDOS

For disproving A in the spoiler, I was confused myself, I was thinking if the E changes positions it will no longer intersect at A, since x is given and the line CA has to meet BA at that point forcing line BE position to be fixed.

What can change though is angle BDE can get bigger or smaller moving E either down the original BE line or farther away from the path that BE line will take since triangle EDB is not given to be a equilateral in the question stem.

Took me a while to visualize, I need lots more practice with DS

Re: DS Question from Manhattan CAT Code [#permalink]
05 May 2013, 01:52

bharatdasaka wrote:

Even though DE is not parallel to BC, dont you think we can come to the conclusion that Angle B=60 ? if not kindly explain why.

And thanks a lot for the reply!

Even if you come correctly to the conclusion that B is 60° (using statement A), this is not enough to answer the question! As I showed you in my previous post where B=60°, we cannot say anything about the lenght of DE

Hope it's clear _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: In the figure to the right, if point C is the center of the [#permalink]
05 May 2013, 22:50

I dont see how the answer is B.

I think for triangle similarity you need two things. 1- C is midpoint of DB = given, A is midpoint of EB, this we cannot find. 2- DE II BA, given.

With the above, the correct answer should be C. Can someone explain.

EDIT: now i see that if DE ll BA and c is midpoint of DB, then this is sufficient and A being a midpoint of EB is actually a conclusion of those two assumptions. DE is equal to diameter of circle.

For my Cambridge essay I have to write down by short and long term career objectives as a part of the personal statement. Easy enough I said, done it...