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M14, Q32 : quadrilateral

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M14, Q32 : quadrilateral [#permalink] New post 18 Mar 2009, 14:12
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If points A , B , C , and D form a quadrilateral, is AC longer than BD ?

1. \angle ABC \gt \angle BCD
2. AB = BC = CD = DA

[Reveal] Spoiler: OA
C

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Re: DS : gmatclub test question M14, Q32 : quadrilateral [#permalink] New post 20 Mar 2009, 23:33
What is the OA?

My answer:
Both Statements are required to answer this question.

Statement 1 alone doesnot say anything about the diagonals AC & BD.
Statement 2 says that the quadrilateral can be a square or a trapezoid so we still can't answer whether AC>BD.

But if we combine both the statements, this quadrilateral should be a trapezoid where angle ABC > angle BCD.
And Now we can say that AC>BD.
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Re: DS : gmatclub test question M14, Q32 : quadrilateral [#permalink] New post 26 Mar 2009, 04:44
I go with B.
1) We dont have any data about angle DAB.
2) It can be a square or a rhombus ( not trapezoid ).
So, diagonals will be equal.
What is OA?


abhishekik wrote:
What is the OA?

My answer:
Both Statements are required to answer this question.

Statement 1 alone doesnot say anything about the diagonals AC & BD.
Statement 2 says that the quadrilateral can be a square or a trapezoid so we still can't answer whether AC>BD.

But if we combine both the statements, this quadrilateral should be a trapezoid where angle ABC > angle BCD.
And Now we can say that AC>BD.
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Re: DS : gmatclub test question M14, Q32 : quadrilateral [#permalink] New post 05 Aug 2009, 17:13
Economist wrote:
I go with B.
1) We dont have any data about angle DAB.
2) It can be a square or a rhombus ( not trapezoid ).
So, diagonals will be equal.
What is OA?


abhishekik wrote:
What is the OA?

My answer:
Both Statements are required to answer this question.

Statement 1 alone doesnot say anything about the diagonals AC & BD.
Statement 2 says that the quadrilateral can be a square or a trapezoid so we still can't answer whether AC>BD.

But if we combine both the statements, this quadrilateral should be a trapezoid where angle ABC > angle BCD.
And Now we can say that AC>BD.


what kind of trapezoid has all sides equal??
shouldnt the answer be B?
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Re: DS : gmatclub test question M14, Q32 : quadrilateral [#permalink] New post 05 Aug 2009, 17:19
viperm5 wrote:
Economist wrote:
I go with B.
1) We dont have any data about angle DAB.
2) It can be a square or a rhombus ( not trapezoid ).
So, diagonals will be equal.
What is OA?


abhishekik wrote:
What is the OA?

My answer:
Both Statements are required to answer this question.

Statement 1 alone doesnot say anything about the diagonals AC & BD.
Statement 2 says that the quadrilateral can be a square or a trapezoid so we still can't answer whether AC>BD.

But if we combine both the statements, this quadrilateral should be a trapezoid where angle ABC > angle BCD.
And Now we can say that AC>BD.


what kind of trapezoid has all sides equal??
shouldnt the answer be B?


oh wait i figured it out.
S2 basically says it is a rhombus.
however you dont know what the internal angles of the rhombus are, hence you dont know which diagonal of the rhombus is longer....
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Re: M14, Q32 : quadrilateral [#permalink] New post 16 Jun 2010, 05:15
Can somebody exlain why Answer A is wrong.

I did the following reasoning:

If angle ABC is bigger than angle BCD, according to properties of triangle side AC will be bigger than side DB.
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Re: M14, Q32 : quadrilateral [#permalink] New post 16 Jun 2010, 05:27
Why can't we apply rule " side opposite to greater angle is greater" here. Since here as per option A we have angle ABC> angle BCD. so doesn't it apply AC>BD?? Please post the OE
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Re: M14, Q32 : quadrilateral [#permalink] New post 16 Jun 2010, 05:52
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Its not stated that the shape is a parallelogram, so the shape could be something like a trapezoid. Knowing something about two of the interior angles doesn't tell us anything about the rest, and we therefore do not know anything about the lengths of the diagonals; 1 does not hold.

Based on 2, we know that the shape must be a rhombus, but we don't know anything about the interior angles; 2 does not hold.

Together, we know that the same is a rhombus and which of the angles are large/small -- we can therefore answer the question; C.
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Re: M14, Q32 : quadrilateral [#permalink] New post 16 Jun 2010, 07:34
Moss wrote:
Its not stated that the shape is a parallelogram, so the shape could be something like a trapezoid. Knowing something about two of the interior angles doesn't tell us anything about the rest, and we therefore do not know anything about the lengths of the diagonals; 1 does not hold.

Based on 2, we know that the shape must be a rhombus, but we don't know anything about the interior angles; 2 does not hold.

Together, we know that the same is a rhombus and which of the angles are large/small -- we can therefore answer the question; C.


Thanks for the explanation Moss. Now it is clear to me :)
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Re: DS : gmatclub test question M14, Q32 : quadrilateral [#permalink] New post 16 Jun 2010, 11:15
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Economist wrote:
I go with B.
1) We dont have any data about angle DAB.
2) It can be a square or a rhombus ( not trapezoid ).
So, diagonals will be equal.
What is OA?


Rhombus does not have equal diagonals. One of the diagonal is always bigger than the other.
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Re: DS : gmatclub test question M14, Q32 : quadrilateral [#permalink] New post 17 Jun 2010, 05:09
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This is not entirely true because a square is a special case of rhombus. Square is also a rhombus.
gaurav2k101 wrote:
Rhombus does not have equal diagonals. One of the diagonal is always bigger than the other.

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Re: M14, Q32 : quadrilateral [#permalink] New post 29 May 2011, 05:10
agreed Rhombus is there from the second statement.
but could not deduce this fact from the given conversation that AC is longer than BD.By which rule we can deduce this.
Can anyone please elaborate some more how it holding true ?
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Re: M14, Q32 : quadrilateral [#permalink] New post 29 May 2011, 05:23
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pjain01 wrote:
agreed Rhombus is there from the second statement.
but could not deduce this fact from the given conversation that AC is longer than BD.By which rule we can deduce this.
Can anyone please elaborate some more how it holding true ?


According to statements 1 and 2; ABCD is a non-square rhombus.
All non-square rhombuses have unequal diagonals.

Question is:
How do we know that AC is indeed the larger of the two diagonals.

\angle{ABC}>\angle{BCD}
We can infer,
\angle{CDA}>\angle{DAB}
As opposite angles are equal in rhombus.

The diagonal formed by joining vertices of two smaller angles will be greater than the diagonal formed by joining vertices of two greater angles in a rhombus. This may not be a theorem. I am telling this just by observation.
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Re: M14, Q32 : quadrilateral [#permalink] New post 20 Jun 2011, 04:27
I am sorry for a stupid question ) in the st2 it is written,that AB=BC=CD=DA.doesnt it mean that all of a quadrilateral's sides are equal and it is a square? if it were written AB=BC and CD=DA, I would agree wih you.
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Re: M14, Q32 : quadrilateral [#permalink] New post 20 Jun 2011, 04:37
LalaB wrote:
I am sorry for a stupid question ) in the st2 it is written,that AB=BC=CD=DA.doesnt it mean that all of a quadrilateral's sides are equal and it is a square? if it were written AB=BC and CD=DA, I would agree wih you.


AB=BC=CD=DA
Can be rhombus OR square.
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Re: M14, Q32 : quadrilateral [#permalink] New post 20 Jun 2011, 04:44
fluke wrote:
Can be rhombus OR square.

ah ok, I got it .thanks :)
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Re: M14, Q32 : quadrilateral [#permalink] New post 20 Jun 2011, 08:34
In my opinion the answer should be C. The reason should be following. The length of diagonal of a quadrilateral ABCD is given by:
AC=sqrt(AB^2 + BC^2 -2AB*BC*cos( \angle ABC )
BD=sqrt(BC^2 + CD^2 -2BC*CD*cos( \angle BCD )

Note that BC is a common side in both the above equations and we are given that \angle ABC \gt \angle BCD, thus let us plug in some values. Let \angle ABC =30 degree and let \angle BCD = 25 degree. Let BC = 5cm. Now there are three possible combination of AB and CD possible.
1] AB=CD (for example AB = 4 and CD =4)
2] AB>CD (for example AB= 4 and CD =3)
3] AB<CD (for example AB = 3 and CD =4)

So if \angle ABC and \angle BCD are acute angles, then always AC>BD for the above mentioned scenario.

However, if \angle ABC and \angle BCD are obtuse angles then for condition 3, i.e., AB<CD, even if \angle ABC \gt \angle BCD, we can get BD>AC.

Thus, statement 1 alone is not sufficient to answer the condition and we need statement 2 so that we have AB=CD and then no matter if angle is obtuse or acute, AC>BD, when \angle ABC and \angle BCD. I hope it answers any doubt.
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Re: M14, Q32 : quadrilateral [#permalink] New post 20 Jun 2011, 08:58
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My answer is C. The rule that the side facing the bigger angle is longer than the side facing the smaller angle is applied to two triangles with 2 equal couple sides only :)
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Re: M14, Q32 : quadrilateral [#permalink] New post 20 Jun 2011, 09:14
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thuylinh wrote:
My answer is C. The rule that the side facing the bigger angle is longer than the side facing the smaller angle is applied to two triangles with 2 equal couple sides only :)

is it two equal couple sides or only one triangle-specific?

side opposite to the largest angle would be the biggest in that triangle ONLY..right?
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Re: M14, Q32 : quadrilateral [#permalink] New post 24 Jun 2013, 04:42
Answer is Correct, all agree its a rhombus(Square is also a Rhombus with 90 degree angle.)

Larger the angle, its opposite side length is also larger, and smaller the angle its opposite side length is also smaller thats the Property of a Triangle.

Point to be noted here is this-"AB=BC=CD=DA",

Now we have to compare AC and BD, AC>BD, WHY?
Because all other sides being equal for two triangles, AC has bigger opposite angle than BD.

So C is correct combining two we get the answer.
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Re: M14, Q32 : quadrilateral   [#permalink] 24 Jun 2013, 04:42
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