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New post 21 Feb 2005, 16:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

see attached
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New post 21 Feb 2005, 20:11
C. explanation later.

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New post 22 Feb 2005, 06:40
C) !

1) it is a parallelogramm, but not enough

2) it is a rhombus or square, but not enough

1) + 2) it is a rhombus and this information is sufficient

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New post 22 Feb 2005, 09:06
christoph wrote:
C) !

1) it is a parallelogramm, but not enough

2) it is a rhombus or square, but not enough

1) + 2) it is a rhombus and this information is sufficient


SO IN RHOMBUS DIAGONAL OPPOSITE TO LARGER ANGLE IS LARGER THAN DIAGONAL OPPOSITE TO SMALLER ANGLE?

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New post 22 Feb 2005, 10:20
christoph wrote:
C) !

1) it is a parallelogramm, but not enough



Not clear why the figure has to be parallelogram.

My explanation:

1) Gives nothing...except that atleast two angles are not same.

2) It is either Rhombus or Square

Combining (1) and (2), it can not be square (since all angles equal in square ).. hence ABCD is rhoumbus...and AC > BD.

Hence (C)

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New post 22 Feb 2005, 10:36
Agree with C.

Quadrilateral only tells that is has 4 sides.That is all it tells.
Now (1) is not sufficient to tell if AC>BD because you can streatch D as far as you want still not disturbing the angle ABC and angle BCD.This makes BD sometimes can be longer than AC and sometimes shorter than AC(Assuming you fix the points (A,B,C).All you need is one scenario to prove given info is not sufficient.

I chose B initially thinking in haste (2) is sufficient by itself as I thought of only square meeting (2) requirements (bad me) but realized later rhombus also fits the criteria.All sides are same.For square,diagnals are equal but for rhomus dignals are differrent in lenght.So (2) is not sufficient by it self.

Now combinig (1) and (2) leaves only rhombus to fit the criteria where AC>BD. So C is correct.

Last edited by 700Plus on 22 Feb 2005, 10:38, edited 1 time in total.

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New post 22 Feb 2005, 10:38
ketanm wrote:
christoph wrote:
C) !

1) it is a parallelogramm, but not enough



Not clear why the figure has to be parallelogram.

My explanation:

1) Gives nothing...except that atleast two angles are not same.

2) It is either Rhombus or Square

Combining (1) and (2), it can not be square (since all angles equal in square ).. hence ABCD is rhoumbus...and AC > BD.

Hence (C)

Ketan


you are right...only by the fact that abc > bcd it could be a parallelogram, a rhombus or a trapezium.

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New post 22 Feb 2005, 17:04
Why not A?

From i) we get angle ABC > BCD

Split the quadrilateral into 2 triangles

ABC and BCD

Since angle B is greater than angle C side AC is greater than BD.(Sides opposite to greatest angles have greater sides in length)

ii) From 2 it may be a square or a rhombus since both have sides equal.


I am not sure if my explanation has a serious flaw. Please explain.

Thanks

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New post 22 Feb 2005, 17:16
My answer is C...from B you don't get jack...from B you get that it's a parallelogram but that could mean either AC>BD or BD>AC so then you have to find out where the larger and smaller angles are A provides that enabling you to figure out which one is bigger
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New post 22 Feb 2005, 20:00
gmat2me2 wrote:
Why not A?

From i) we get angle ABC > BCD

Split the quadrilateral into 2 triangles

ABC and BCD

Since angle B is greater than angle C side AC is greater than BD.(Sides opposite to greatest angles have greater sides in length)



Tried to describe initially, but then though it would be better if I simply draw a figure.

In the figure (see attachement)... angle B =90, angle C < 90 [Took angle 90 so that you can easily see that angle C less than 90]

Still BD > AC
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New post 12 Mar 2005, 13:39
can somebody finalize this please?

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New post 13 Mar 2005, 09:41
B says that all the sides are equal and that means it would have been a square if the angle would have been mentioned as 90. But the most important thing is it doesn't say anything about the angles. I don't know if a lot of people had that issue but when i first looked at the question i was like its a square then i realized that the angle wasn't mentioned. well anyways, (A) gives us the angle.. so together sufficient..

C

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New post 13 Mar 2005, 20:07
I'm also getting a C

OA?

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New post 14 Mar 2005, 11:51
Diagonal drawn from the point of the larger angle is < diagonal drawn from smaller angle.
or diagonal opposite larger angle is larger than diag opp. smaller angle
C

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  [#permalink] 14 Mar 2005, 11:51
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