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In the figure to the right, if point C is the center of the.

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by Sprite_TM » Wed May 20, 2009 7:54 pm
this question is from the manhattan gmat CAT test

i don't understand even with the manhattan explanation. i hope someone can explain to me why its an isosoceles triangle. thanks

In the figure to the right, if point C is the center of the circle and DB = 7, what is the length of DE?

(1) x = 60°

(2) DE || CA

a) statement 1 only
b) statement 2 only
c) both statements
d) each alone is sufficient
e) both statements not enough

OA b
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by DanaJ » Wed May 20, 2009 9:51 pm
Ah, geometry! Ever so fascinating... If I were in charge at the GMAC, I'd replace those pesky word problems with geometry!

Anyway, back to our question:

First off, let's look at that picture. C is the center of the circle, so it's half way between D and B. Now, CD, BC and AC are all equal in length, since they're all radii. Since AC = BC, triangle ABC must be an isosceles triangle.

1. tells you that x = 60, meaning that triangle ABC is an isosceles triangle with one 60 degrees angle. This makes it an equilateral triangle, so AB = AC = BC. However, apart from knowing that angle ABC is also 60, this info is not related to the bigger BED triangle. So 1 is not enough.

2. is what we need to establish that triangles BED and BAC are similar triangles. This is because they share angle lengths: angle B is the same and the other two base angles are equal respectively because DE || CA.
Since these two triangles are similar, then the rapport of their sides lengths is the same. This means that:

BC/BD = AC/ED

But remember that BC is a radius and BD is a diameter, so BC/BD will be 1/2. This means that AC/ED = 1/2, or that ED = 2AC. AC is also a radius, so AC = 7/2. This makes DE = 7.
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by lamhe » Wed May 20, 2009 10:03 pm
Yes B for me too....
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by Sprite_TM » Thu May 21, 2009 7:25 am
ahhh i see it now DanaJ. thanks so much. i didnt realize the point from A to C was a radii!!!
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by mauroyb0 » Sat Jun 06, 2009 4:44 am
Hi there,

I perfectly understand the answer and agree with proposition B. However I thought figures were not properly scaled in GMAT questions. As it is not stated anywhere, how can we deduce that point A is really on the circle line?
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by DanaJ » Sat Jun 06, 2009 5:10 am
Well, the figures might not be properly scaled, but that doesn't mean that you can't rely on them at all... Most of the time, they actually ARE scaled, at least that;s what I've noticed... But this scaling actually comes down to lengths rather than position of points: for instance, you can easily notice that triangle ABC is equilateral. However, you need to prove that, you can't just assume that AC = AB - and this is where stmt 1 comes in.
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by Domnu » Sat Jun 06, 2009 2:32 pm
DanaJ wrote:Ah, geometry! Ever so fascinating... If I were in charge at the GMAC, I'd replace those pesky word problems with geometry!

Anyway, back to our question:

First off, let's look at that picture. C is the center of the circle, so it's half way between D and B. Now, CD, BC and AC are all equal in length, since they're all radii. Since AC = BC, triangle ABC must be an isosceles triangle.

1. tells you that x = 60, meaning that triangle ABC is an isosceles triangle with one 60 degrees angle. This makes it an equilateral triangle, so AB = AC = BC. However, apart from knowing that angle ABC is also 60, this info is not related to the bigger BED triangle. So 1 is not enough.

2. is what we need to establish that triangles BED and BAC are similar triangles. This is because they share angle lengths: angle B is the same and the other two base angles are equal respectively because DE || CA.
Since these two triangles are similar, then the rapport of their sides lengths is the same. This means that:

BC/BD = AC/ED

But remember that BC is a radius and BD is a diameter, so BC/BD will be 1/2. This means that AC/ED = 1/2, or that ED = 2AC. AC is also a radius, so AC = 7/2. This makes DE = 7.
Same here ;)

Well, if both are parallel, then we have that the overall triangle is isoceles.
Have you wondered how you could have found such a treasure? -T
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by yogami » Sun Jun 07, 2009 6:43 am
check out basic proportionality theorem.
200 or 800. It don't matter no more.
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by enniguy » Thu Jan 14, 2010 4:08 am
Dana and everyone,

I still feel that the answer should be C. Can you please help me out here? This is how I solved it.
(1) Consider a line DB=7 and a circle around it. Draw the line BE such that BCA=60 degrees. So, without knowing that DE || CA, I can draw a line in many ways. So option 1 is not sufficient alone.

(2) Consider X to be 75 degrees or 80 degrees. You can still find a point A on the circle. Using this point A, draw your new BE line. (Extend your BE line forever). Now, draw a line DE such that DE || CA. This DE will satisfy the figure but with a different length and not 7.

Together, we can fix both the angle X - 60 degrees - (through which we will fix the point A) and length of DE (because DE || CA).

Is there anything wrong in my thinking?
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by DanaJ » Thu Jan 14, 2010 10:31 am
Fact is, you do not need to know that x is 60 degrees. This is because you have similar triangles, as described in the point above.

The construction you describe seems a bit complicated and it's pretty hard for me to follow. This is because not all points are defined (for instance, C in 1 and A in 2). I do not think you need anything else except what you already have in the drawing.
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by pkw209 » Mon Feb 01, 2010 4:41 pm
fantastic. thanks Dana!
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by birgiolasj » Thu Aug 18, 2011 8:27 pm
enniguy wrote:Dana and everyone,

I still feel that the answer should be C. Can you please help me out here? This is how I solved it.
(1) Consider a line DB=7 and a circle around it. Draw the line BE such that BCA=60 degrees. So, without knowing that DE || CA, I can draw a line in many ways. So option 1 is not sufficient alone.

(2) Consider X to be 75 degrees or 80 degrees. You can still find a point A on the circle. Using this point A, draw your new BE line. (Extend your BE line forever). Now, draw a line DE such that DE || CA. This DE will satisfy the figure but with a different length and not 7.

Together, we can fix both the angle X - 60 degrees - (through which we will fix the point A) and length of DE (because DE || CA).

Is there anything wrong in my thinking?
That is what I thought as well initially and got it wrong.

It's true that, under condition (2), you could draw point A anywhere on the circle and that would result in a different DE. However, regardless of where you drew point A (in other words regardless of angle x), the length of DE would not change.

This is illustrated if you imagined angle x to be 1 degree. This would make point A just above point B, line BE almost completely vertical, and DE almost horizontal, but the same length as the diameter.

Alternatively, you could draw out different versions of the diagram with different angles x, and then measure the length of DE. It will be equal to the diameter regardless of the angle (0 & 180 being exceptions as those angles would cause DE cease to exist).

This problem exploits the common assumption that if a figure can be constructed differently, the dimensions of the different constructions would be different as well. Because the second condition makes the triangles similar, the different constructions would still result in the same dimensions.

Definitely a tricky one.

-Justas
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by iburnedmyrice » Tue Aug 04, 2015 6:00 pm
If this is not drawn to scale and they dont say that A in on the circle. Can we just assume it is on? If so then when do we assume and when dont we assume?????
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by Brent@GMATPrepNow » Tue Aug 04, 2015 7:17 pm
iburnedmyrice wrote:If this is not drawn to scale and they dont say that A in on the circle. Can we just assume it is on? If so then when do we assume and when dont we assume?????
Here's a free video that addresses that question: https://www.gmatprepnow.com/module/gmat-geometry?id=863
Basically, if it looks like a point is on a line, then we can assume that it is on the line.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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