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

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 ajith
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The question asks whether the quadrilateral ABCD is a regular parallelogram.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!
Statement 1 and 2 together are not sufficient
I will prove it by an example ( This image follows statement 1 and 2)
Always borrow money from a pessimist, he doesn't expect to be paid back.

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Doesnt this look like a rhombus???
ajith wrote:The question asks whether the quadrilateral ABCD is a regular parallelogram.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!
Statement 1 and 2 together are not sufficient
I will prove it by an example ( This image follows statement 1 and 2)
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.
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.
 ajith
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Name the diagram ABCD it follows statement 1 and statement2 ( And clearly all sides are not equal)bhumika.k.shah wrote:thanks for the drawing... but how do i prove that neither of the statements are sufficient ?
To disprove a claim only an example against it will do (the example is the kite in the picture)
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i thought ajith said we needed to prove whether its a parallelogram...
u r saying a square???
i think its a rhombus!
now i am totally confused
u r saying a square???
i think its a rhombus!
now i am totally confused

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ajith ,
if u r saying its kite shaped , isnt it a regular rhombus then ???
if u r saying its kite shaped , isnt it a regular rhombus then ???
ajith wrote:Name the diagram ABCD it follows statement 1 and statement2 ( And clearly all sides are not equal)bhumika.k.shah wrote:thanks for the drawing... but how do i prove that neither of the statements are sufficient ?
To disprove a claim only an example against it will do (the example is the kite in the picture)
Bhoomika, question is, are all the sides equal, which is a square!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!
now i am totally confused
 ajith
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it is not a Rhombusbhumika.k.shah wrote:ajith ,
if u r saying its kite shaped , isnt it a regular rhombus then ???
ajith wrote:Name the diagram ABCD it follows statement 1 and statement2 ( And clearly all sides are not equal)bhumika.k.shah wrote:thanks for the drawing... but how do i prove that neither of the statements are sufficient ?
To disprove a claim only an example against it will do (the example is the kite in the picture)
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
Always borrow money from a pessimist, he doesn't expect to be paid back.
 ajith
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If all sides are equal it becomes a Rhombus not square. Square is a special rhombus which has all sides equal and sides perpendicularksundar wrote: Bhoomika, question is, are all the sides equal, which is a square!
Always borrow money from a pessimist, he doesn't expect to be paid back.
Rhombus: parallelogram with four equal sides; an obliqueangled equilateral parallelogramksundar wrote:Bhoomika, question is, are all the sides equal, which is a square!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!
now i am totally confused

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 Mike@Magoosh
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Hi, there. I'll throw in my 2Â¢.
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 testwriter 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 willynilly.
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 nextdoor neighbor is the President. To leap to that conclusion would be supremely gullible.
Much in the same way, be hyperreluctant 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 nextdoor 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
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 testwriter 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 willynilly.
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 nextdoor neighbor is the President. To leap to that conclusion would be supremely gullible.
Much in the same way, be hyperreluctant 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 nextdoor 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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