Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 3453 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.

Re: Quadrlateral DS [#permalink]
10 Mar 2013, 15:43

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 _________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: Quadrlateral DS [#permalink]
10 Mar 2013, 21:43

1

This post received KUDOS

Expert's post

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. _________________

Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]
12 Mar 2013, 20:28

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. _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]
13 Mar 2013, 02:37

Expert's post

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. _________________

Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]
13 Mar 2013, 22:17

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. _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]
06 Aug 2013, 11:06

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. _________________

Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]
27 Sep 2014, 10:59

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?

Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]
27 Sep 2014, 11:17

Expert's post

PathFinder007 wrote:

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?

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. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...