if ABCD is a quadrilateral, is AB = BC=CD=DA?

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if ABCD is a quadrilateral, is AB = BC=CD=DA?

Statement I . AC is perpendicular to BD

Statement II . AB+CD= BC+ AD

OA E

If someone could throw a few pointers on these properties then twud be great! :D

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by ajith » Sun Jan 24, 2010 7:35 am
bhumika.k.shah wrote:if ABCD is a quadrilateral, is AB = BC=CD=DA?

Statement I . AC is perpendicular to BD

Statement II . AB+CD= BC+ AD

OA E

If someone could throw a few pointers on these properties then twud be great! :D
The question asks whether the quadrilateral ABCD is a regular parallelogram.
Statement 1 and 2 together are not sufficient

I will prove it by an example ( This image follows statement 1 and 2)

Image
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by bhumika.k.shah » Sun Jan 24, 2010 7:37 am
thanks for the drawing... but how do i prove that neither of the statements are sufficient ?

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by bhumika.k.shah » Sun Jan 24, 2010 7:38 am
o! and also that what properties of the quadrilateral are similar for a parallelogram ???

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by bhumika.k.shah » Sun Jan 24, 2010 7:38 am
Doesnt this look like a rhombus???
ajith wrote:
bhumika.k.shah wrote:if ABCD is a quadrilateral, is AB = BC=CD=DA?

Statement I . AC is perpendicular to BD

Statement II . AB+CD= BC+ AD

OA E

If someone could throw a few pointers on these properties then twud be great! :D
The question asks whether the quadrilateral ABCD is a regular parallelogram.
Statement 1 and 2 together are not sufficient

I will prove it by an example ( This image follows statement 1 and 2)

Image

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by ksundar » Sun Jan 24, 2010 7:39 am
Statement I is not sufficient because a quadrilateral will have four sides and the sides to be perpendicular should have a point in common. i.e AB can be perpendicular to BC or AD.

Statement II can be true but will not prove that all the sides are equal.

Statement I and II together will not prove that the quadrilateral is a square. So the answer is E.

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by ajith » Sun Jan 24, 2010 7:40 am
bhumika.k.shah wrote:thanks for the drawing... but how do i prove that neither of the statements are sufficient ?
Name the diagram ABCD it follows statement 1 and statement2 ( And clearly all sides are not equal)

To disprove a claim only an example against it will do (the example is the kite in the picture)
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by bhumika.k.shah » Sun Jan 24, 2010 7:41 am
i thought ajith said we needed to prove whether its a parallelogram...
u r saying a square???

i think its a rhombus! :P

now i am totally confused :(

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by bhumika.k.shah » Sun Jan 24, 2010 7:43 am
ajith ,
if u r saying its kite shaped , isnt it a regular rhombus then ???
ajith wrote:
bhumika.k.shah wrote:thanks for the drawing... but how do i prove that neither of the statements are sufficient ?
Name the diagram ABCD it follows statement 1 and statement2 ( And clearly all sides are not equal)

To disprove a claim only an example against it will do (the example is the kite in the picture)

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by ksundar » Sun Jan 24, 2010 7:43 am
bhumika.k.shah wrote:i thought ajith said we needed to prove whether its a parallelogram...
u r saying a square???

i think its a rhombus! :P

now i am totally confused :(
Bhoomika, question is, are all the sides equal, which is a square!

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by ajith » Sun Jan 24, 2010 7:47 am
bhumika.k.shah wrote:ajith ,
if u r saying its kite shaped , isnt it a regular rhombus then ???
ajith wrote:
bhumika.k.shah wrote:thanks for the drawing... but how do i prove that neither of the statements are sufficient ?
Name the diagram ABCD it follows statement 1 and statement2 ( And clearly all sides are not equal)

To disprove a claim only an example against it will do (the example is the kite in the picture)
it is not a Rhombus

Rhombus has to have equal sides and opposite sides parallel and its diagonals bisect at 90 degrees
Kite has adjacent sides equal and Diagonals intersect at 90 degrees
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by ajith » Sun Jan 24, 2010 7:49 am
ksundar wrote: Bhoomika, question is, are all the sides equal, which is a square!
If all sides are equal it becomes a Rhombus not square. Square is a special rhombus which has all sides equal and sides perpendicular
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by ksundar » Sun Jan 24, 2010 7:50 am
ksundar wrote:
bhumika.k.shah wrote:i thought ajith said we needed to prove whether its a parallelogram...
u r saying a square???

i think its a rhombus! :P

now i am totally confused :(
Bhoomika, question is, are all the sides equal, which is a square!
Rhombus: parallelogram with four equal sides; an oblique-angled equilateral parallelogram

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by luiscarlos59 » Sat Feb 11, 2012 3:23 pm
Isnt with statement 1) rephrased as all sides are equal? so its a square?

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by Mike@Magoosh » Sun Feb 12, 2012 10:28 pm
Hi, there. I'll throw in my 2¢. :)

Given ajith's crystal clear and 100% correct explanations, I'm surprised but the level of confusion in this post. I'll do what I can to clear things up.

Prompt: if ABCD is a quadrilateral, is AB = BC=CD=DA?

It's asking whether ABCD is an equilateral quadrilateral, that is to say, a rhombus.

It is not necessarily asking about a square, which is both equilateral and equiangular. That's a trap that the GMAT test-writer has laid for you. For a triangle, equilateral is automatically equiangular as well. For all higher polygons, quadrilaterals and up, equilateral and equiangular an independent of one another, and those shapes that happen to have both are quite special, quite rare, not to be taken for grant, not to be assumed willy-nilly.

Statement #1: AC is perpendicular to BD

The vertices of any quadrilateral are, by convention, named in continuous order as one goes clockwise or counterclockwise around the edge. Here, ABCD means that if we start at A, and walk around the edge, we will come to B, then C, then D, in that order.

A and C are not adjacent, but across from each other, so AC is a diagonal. BD is also a diagonal.

If we know only that the diagonals are perpendicular, the quadrilateral could be a square, or rhombus, or a kite, or an irregular quadrilateral. See p. 1 of the pdf for examples. This statement does not uniquely determine the quadrilateral, so it is insufficient.

Statement #2: AB + CD= BC + AD

This is an unusual constraint --- the sum of one pair of opposite sides equals the sum of the other pair of opposite side. Again, with this constraint alone, the quadrilateral could be a square, or rhombus, or a kite, or an irregular quadrilateral. See p. 2 of the pdf for examples. This statement does not uniquely determine the quadrilateral, so it is insufficient.

Combined Statements #1 & #2: AC is perpendicular to BD, and AB + CD = BC + AD

Here, I believe, no irregular quadrilaterals are possible any more. Nevertheless, as ajith has pointed out, both the kite and rhombus are possible here, so even here, it is not sufficient to determine that the quadrilateral is definitively a rhombus. Therefore, answer = E.

One important thing to keep in mind here are what I call the BIG 4 parallelogram properties:
(a) opposite sides parallel
(b) opposite sides congruent
(c) opposite angles congruent
(d) diagonals bisect each other


Any one of those is sufficient to prove that the quadrilateral is a parallelogram; any one of those four sufficient to prove the other three.

The only way you know you have sufficient information either to prove a quadrilateral is a parallelogram or to prove a quadrilateral is one of the "special" parallelograms (square, rectangle, rhombus) is if you have one of the BIG 4 in the mix some how. Here, even when Statements #1 & #2 are combined, it's not enough to prove one of the BIG 4, so not enough to guarantee the quadrilateral is a rhombus.

The difficulty of this question lies in confusing what could be true with what must be true.

Irregular quadrilaterals are legion, but a rhombus is one of the elite of the quadrilateral world.

The analog of this question:

Prompt: is some randomly chosen American the President of the US?
Statement #1: the randomly chosen American has two arms & two legs
Statement #2: the randomly chosen American's last name begins with a vowel

It's true that if the randomly chosen American were the President, Statements #1 and #2 would both be true, but that is NOT what GMAT DS is asking. GMAT DS is asking: if you know Statements #1 and #2 only, is that enough to guarantee, without a shadow of a doubt, that the randomly chosen American is the President? Obviously, if I tell you the person who lives next door to me has two arms, two legs, and last name that begins with a vowel, you are not going to leap to the conclusion that my next-door neighbor is the President. To leap to that conclusion would be supremely gullible.

Much in the same way, be hyper-reluctant to jump to the conclusion of "elite" quadrilateral (rectangle, rhombus, square) without sufficient evidence (which must include one of the BIG 4). Otherwise, that's just as gullible as leaping to the conclusion that my next-door neighbor is the President when given minimal information.

Does all this make sense? Please let me know if any of you have any questions at all about this.

Mike :)
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