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:

Well: (1) True for square or for rhombus but every square is a rhombus, so sufficient (2) Again true for square or for rhombus but every square is a rhombus, so sufficient

Well: (1) True for square or for rhombus but every square is a rhombus, so sufficient (2) Again true for square or for rhombus but every square is a rhombus, so sufficient

D

Hi Bunuel

When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus?

And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus?

Is that the correct line of reasoning? Thanks so much in advance Cheers J

Well: (1) True for square or for rhombus but every square is a rhombus, so sufficient (2) Again true for square or for rhombus but every square is a rhombus, so sufficient

D

Hi Bunuel

When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus?

And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus?

Is that the correct line of reasoning? Thanks so much in advance Cheers J

For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

For (2): AB = BC = CD = AD, means that ABCD is either a rhombus or square (so still a rhombus).

Re: Is quadrilateral ABCD a rhombus? [#permalink]
06 Dec 2013, 10:10

Is quadrilateral ABCD a rhombus?

(1) Line segments AC and BD are perpendicular bisectors of each other.

But doesn't a rhombus also allow for two perpendicular bisectors? A square has two perpendicular bisectors but a square is not a rhombus (because a rhombus does not have all four angles = 90. Oh, but wait, a rhombus is a square in the same way a rectangle is a square but not the other way around. Tricky.

(2) AB = BC = CD = AD

I guess the same logic applies above, ABCD could be a square or rhombus but regardless of which one it is its still a rhombus. If the stem asked if ABCD was a square, then this would be insufficient.

Re: Data Sufficiency - doubts [#permalink]
08 Aug 2014, 20:36

kritim22 wrote:

Hi,

can someone help me with this question please:

Is quadrilateral ABCD a rhombus?

(1) Line segments AC and BD are perpendicular bisectors of each other.

(2) AB = BC = CD = AD

The answer it says is D- each statement is enough on its own.

Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus?

Let say AC and BD meet at O. We have OD= OB, OA= OC. If you use pithagoras theorem, you can easily see that AD= DC=BC=AB. For example, AD^2= OA^2 + OD^2 So (1) and (2) basically say the same thing. The definition of rhombus is a quadrilateral whose four sides have the same length. So D is the answer. _________________

......................................................................... +1 Kudos please, if you like my post

Re: Data Sufficiency - doubts [#permalink]
08 Aug 2014, 23:56

kritim22 wrote:

Hi,

can someone help me with this question please:

Is quadrilateral ABCD a rhombus?

(1) Line segments AC and BD are perpendicular bisectors of each other.

(2) AB = BC = CD = AD

The answer it says is D- each statement is enough on its own.

Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus?

A square is a special case of rhombus. It has all angles of 90 degrees. so proving that quadrilateral ABCD is a square is sufficient to say that it is a rhombus.

Re: Is quadrilateral ABCD a rhombus? [#permalink]
05 Nov 2014, 04:09

Expert's post

honchos wrote:

Bunuel wrote:

Answer D.

(1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...

But Bunuel this would be true even for rectangle: Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.

For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus.

Re: Is quadrilateral ABCD a rhombus? [#permalink]
06 Nov 2014, 01:17

Bunuel,

For (1), can you consider the case of a kite?

Bunuel wrote:

honchos wrote:

Bunuel wrote:

Answer D.

(1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...

But Bunuel this would be true even for rectangle: Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.

For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus.

Re: Is quadrilateral ABCD a rhombus? [#permalink]
10 Mar 2015, 21:58

1

This post received KUDOS

Expert's post

TooLong150 wrote:

This question is poor, because (1) could be right kite, which is not necessarily a rhombus or a rhombus, but (2) will always satisfy a rhombus.

Actually, the question is fine.

A rhombus is a quadrilateral all of whose sides are of the same length - that's all. You could have a rhombus which also has all angles 90 which makes it a square or a rhombus in the shape of a kite. But nevertheless, if it is a quadrilateral and has all sides equal, it IS A RHOMBUS.

(1) Line segments AC and BD are perpendicular bisectors of each other.

Make 2 lines - a vertical and a horizontal - which are perpendicular bisectors of each other. Make them in any way of any length - just that they should be perpendicular bisectors of each other. When you join the end points, you will get all sides equal. Think of it this way - each side you get will be a hypotenuse of a right triangle. The legs of the right triangle will have the same pair of lengths in all 4 cases. So AB = BC = CD = AD. So ABCD must be a rhombus.

(2) AB = BC = CD = AD

This statement directly tells you that all sides are equal so ABCD must be a rhombus.

Originally, I was supposed to have an in-person interview for Yale in New Haven, CT. However, as I mentioned in my last post about how to prepare for b-school interviews...

Hi Starlord, As far as i'm concerned, the assessments look at cognitive and emotional traits - they are not IQ tests or skills tests. The games are actually pulled from...

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...

The 2015 Pan American Games and more than 6,000 athletes have come to Toronto. With them they have brought a great positive vibe and it made me think...