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 500,000 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 3911 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
1
This post was BOOKMARKED
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
Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]
20 Oct 2015, 02:56
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect [#permalink]
19 Dec 2015, 20:36
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 144 times ]
_________________
Fais de ta vie un rêve et d'un rêve une réalité
gmatclubot
Re: Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect
[#permalink]
19 Dec 2015, 20:36
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...