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Statement 1 says that ABCD is a square. A square is just a special rhombus with all angles equal to \(90^o\)

Statement 2: Diagonals bisect at \(90^o\)

Let's draw a picture to visualize this:

Attachment:

c73058.jpg [ 10.66 KiB | Viewed 4681 times ]

(Let's consider the center point to be O, I forgot to label this)

So as you can see from the image, since the diagonals are bisected, we have AO = OC. And we have OB to be common for triangles AOB and BOC. So consider right triangle AOB: \(AB^2 = AO^2 + OB ^2 = OC^2 + OB^2 = BC^2.\)

So we get that AB = BC. Similarly we can prove that all the sides are equal using this method. Hence we get a quadrilateral where all sides are equal, and diagonals bisecting at \(90^o\). Hence it's a rhombus. Sufficient.

Statement 1 says that ABCD is a square. A square is just a special rhombus with all angles equal to \(90^o\)

Statement 2: Diagonals bisect at \(90^o\)

Let's draw a picture to visualize this:

Attachment:

c73058.jpg

(Let's consider the center point to be O, I forgot to label this)

So as you can see from the image, since the diagonals are bisected, we have AO = OC. And we have OB to be common for triangles AOB and BOC. So consider right triangle AOB: \(AB^2 = AO^2 + OB ^2 = OC^2 + OB^2 = BC^2.\)

So we get that AB = BC. Similarly we can prove that all the sides are equal using this method. Hence we get a quadrilateral where all sides are equal, and diagonals bisecting at \(90^o\). Hence it's a rhombus. Sufficient.

Hey whiplash2411, I agree A is Correct choice.. But not D! I don't think B also leads to solution here.. B just says bisect each other. It doesn't mean that both should be of equal length!

Guys pls correct me if im wrong
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Statement 1 says that ABCD is a square. A square is just a special rhombus with all angles equal to \(90^o\)

Statement 2: Diagonals bisect at \(90^o\)

Let's draw a picture to visualize this:

Attachment:

c73058.jpg

(Let's consider the center point to be O, I forgot to label this)

So as you can see from the image, since the diagonals are bisected, we have AO = OC. And we have OB to be common for triangles AOB and BOC. So consider right triangle AOB: \(AB^2 = AO^2 + OB ^2 = OC^2 + OB^2 = BC^2.\)

So we get that AB = BC. Similarly we can prove that all the sides are equal using this method. Hence we get a quadrilateral where all sides are equal, and diagonals bisecting at \(90^o\). Hence it's a rhombus. Sufficient.

Hey whiplash2411, I agree A is Correct choice.. But not D! I don't think B also leads to solution here.. B just says bisect each other. It doesn't mean that both should be of equal length!

Guys pls correct me if im wrong

If the diagonals of a quadrilateral are perpendicular bisectors of each other (so if the diagonal bisect each other at a right angle), the quadrilateral is a rhombus.
_________________

Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]

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12 Mar 2013, 21:28

I believe the answer here is A. What is rombus - it is quadrilateral with the following properties: * Opposite angles of a rhombus have equal measure * The two diagonals of a rhombus are perpendicular; * Its diagonals bisect opposite angles (that is why we have first property) In choice B we do not see all the properties, it could be a rombus but at the same time it could be an quadrilateral with different angles and opposite sides - not sufficient.
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I believe the answer here is A. What is rombus - it is quadrilateral with the following properties: * Opposite angles of a rhombus have equal measure * The two diagonals of a rhombus are perpendicular; * Its diagonals bisect opposite angles (that is why we have first property) In choice B we do not see all the properties, it could be a rombus but at the same time it could be an quadrilateral with different angles and opposite sides - not sufficient.

OA is D, not A.

There is a property which says if the diagonals of a quadrilateral are perpendicular bisectors of each other (so if the diagonal bisect each other at a right angle), the quadrilateral is a rhombus.
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Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]

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13 Mar 2013, 23:17

Bunuel wrote:

ziko wrote:

I believe the answer here is A. What is rombus - it is quadrilateral with the following properties: * Opposite angles of a rhombus have equal measure * The two diagonals of a rhombus are perpendicular; * Its diagonals bisect opposite angles (that is why we have first property) In choice B we do not see all the properties, it could be a rombus but at the same time it could be an quadrilateral with different angles and opposite sides - not sufficient.

OA is D, not A.

There is a property which says if the diagonals of a quadrilateral are perpendicular bisectors of each other (so if the diagonal bisect each other at a right angle), the quadrilateral is a rhombus.

Most probably i missed the word bisects, which means divides into equal sides, and understood it as intersects. In this case obviously that will be a rombus and no other figure could be drawn. Yes once again i confirm that GMAT has so many small tricky parts. Thank you Bunuel for clarification.
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Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]

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06 Aug 2013, 12:06

rxs0005 wrote:

Is ABCD a rhombus?

(1) ABCD is a square (2) ABCD diagonals bisect at 90 degrees

D

Rule for Rhombus: Diagonals should bisect each other and angle between diagonals should be 90degrees. If all sides of a Rhombus are equal then it is a square.

Any square is a Rhombus If the diagonal bisects at 90 degrees then it is a Rhombus.

So either of the statement alone is sufficient to answer the question.
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Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]

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27 Sep 2014, 11:59

Bunuel wrote:

shanmugamgsn wrote:

whiplash2411 wrote:

Answer is D.

Statement 1 says that ABCD is a square. A square is just a special rhombus with all angles equal to \(90^o\)

Statement 2: Diagonals bisect at \(90^o\)

Let's draw a picture to visualize this:

Attachment:

c73058.jpg

(Let's consider the center point to be O, I forgot to label this)

So as you can see from the image, since the diagonals are bisected, we have AO = OC. And we have OB to be common for triangles AOB and BOC. So consider right triangle AOB: \(AB^2 = AO^2 + OB ^2 = OC^2 + OB^2 = BC^2.\)

So we get that AB = BC. Similarly we can prove that all the sides are equal using this method. Hence we get a quadrilateral where all sides are equal, and diagonals bisecting at \(90^o\). Hence it's a rhombus. Sufficient.

Hey whiplash2411, I agree A is Correct choice.. But not D! I don't think B also leads to solution here.. B just says bisect each other. It doesn't mean that both should be of equal length!

Guys pls correct me if im wrong

If the diagonals of a quadrilateral are perpendicular bisectors of each other (so if the diagonal bisect each other at a right angle), the quadrilateral is a rhombus.

Hi Bunuel,

I have a query here.

in case of Kite also diagonal are perpendicular bisector. so in st2 cant we consider this as a kite?

in case of Kite also diagonal are perpendicular bisector. so in st2 cant we consider this as a kite?

Please clarify.

Thanks

The diagonals of a kite may be perpendicular, but they do not both bisect each other. 'Bisect' means 'cuts perfectly in half', and if you draw a skewed kite, and draw its diagonals, you'll see that one of the two diagonals is not cut perfectly in half at their intersection point.

That said, the question in the original post above is really not the kind of question you see on the GMAT. The GMAT does not test if you know the definition of specialized figures like rhombuses or kites, nor will it test if you know obscure facts about their diagonals.
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Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]

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20 Oct 2015, 03:56

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Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]

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19 Dec 2015, 21:36

Once into parallelograms, there are two types. Imagine two sets.

Set 1: All angles are 90 degrees and only opposite sides are equal Set 2: All sides are equal but angles are not equal to 90 degrees Also, a set 3, which is the intersection of these two sets

Set 1 Contains a Rectangle

Set 2 Contains a Rhombus

Set 3, the intersection of the two which has all sides equal as well as all angles equal, is a square.

So. A square is a rectangle. But a rectangle is not a square.

A square is a rhombus. But a rhombus is not a square

Attachments

Screen Shot 2015-12-20 at 10.04.55 AM.png [ 31.42 KiB | Viewed 913 times ]

Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]

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13 Mar 2017, 01:54

Bunuel wrote:

ziko wrote:

I believe the answer here is A. What is rombus - it is quadrilateral with the following properties: * Opposite angles of a rhombus have equal measure * The two diagonals of a rhombus are perpendicular; * Its diagonals bisect opposite angles (that is why we have first property) In choice B we do not see all the properties, it could be a rombus but at the same time it could be an quadrilateral with different angles and opposite sides - not sufficient.

OA is D, not A.

There is a property which says if the diagonals of a quadrilateral are perpendicular bisectors of each other (so if the diagonal bisect each other at a right angle), the quadrilateral is a rhombus.

Isn't it true that, there could be irregular quadrilaterals which have perpendicular diagonals. Not necessarily has to be rhombus. However, if it’s a rhombus, then the diagonal has to be perpendicular?

gmatclubot

Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect
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13 Mar 2017, 01:54

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