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 is to broad. A parallelogram can be several things

Statement 2 correctly identifies this polygon as a rhombus. If this polygon is a rhombus then it IS NOT a square. By identifying the polygon as a rhombus doesn't this prove that the polygon EFGH IS NOT a square?

E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square?

(1) EFGH is a parallelogram.

(2) The diagonals of EFGH are perpendicular bisectors of one another.

I answered B to this question.

Statement 1 is to broad. A parallelogram can be several things

Statement 2 correctly identifies this polygon as a rhombus. If this polygon is a rhombus then it IS NOT a square. By identifying the polygon as a rhombus doesn't this prove that the polygon EFGH IS NOT a square?

I am providing the official answer. Please help!

Rhombus is a quadrilateral with all four sides equal in length. So, all squares are rhombuses but not vise-versa.

(1) EFGH is a parallelogram --> all squares are parallelograms but not vise-versa. Not sufficient.

(2) The diagonals of EFGH are perpendicular bisectors of one another --> EFGH is a rhombus --> all squares are rhombuses but not vise-versa. Not sufficient.

(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not vise-versa. Not sufficient.

Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square [#permalink]

Show Tags

06 Nov 2012, 03:34

Bunuel wrote:

(2) The diagonals of EFGH are perpendicular bisectors of one another --> EFGH is a rhombus --> all squares are rhombuses but not vise-versa. Not sufficient.

(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not vise-versa. Not sufficient.

Bunuel - One doubt, If the diagonals of EFGH are perpendicular bisectors of one another then the 2 diagonals are of the same measure right? then it could be rectangle also because diagonals of rectangle are equal and bisect each other.....

(2) The diagonals of EFGH are perpendicular bisectors of one another --> EFGH is a rhombus --> all squares are rhombuses but not vise-versa. Not sufficient.

(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not vise-versa. Not sufficient.

Bunuel - One doubt, If the diagonals of EFGH are perpendicular bisectors of one another then the 2 diagonals are of the same measure right? then it could be rectangle also because diagonals of rectangle are equal and bisect each other.....

Cheers

The diagonals of EFGH are perpendicular bisectors of one another, means that the diagonals cut one another into two equal parts at 90°. If a rectangle is not a square, then its diagonals do not cut each other at 90°.

Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square [#permalink]

Show Tags

06 Nov 2012, 06:16

Bunuel wrote:

Jp27 wrote:

Bunuel wrote:

(2) The diagonals of EFGH are perpendicular bisectors of one another --> EFGH is a rhombus --> all squares are rhombuses but not vise-versa. Not sufficient.

(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not vise-versa. Not sufficient.

Bunuel - One doubt, If the diagonals of EFGH are perpendicular bisectors of one another then the 2 diagonals are of the same measure right? then it could be rectangle also because diagonals of rectangle are equal and bisect each other.....

Cheers

The diagonals of EFGH are perpendicular bisectors of one another, means that the diagonals cut one another into two equal parts at 90°. If a rectangle is not a square, then its diagonals do not cut each other at 90°.

Hope it's clear.

yes Bunuel totally clear! many thanks for all your responses.

Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square [#permalink]

Show Tags

01 Jul 2015, 23:07

Bunuel wrote:

Rhombus is a quadrilateral with all four sides equal in length. So, all squares are rhombuses but not vise-versa.

(1) EFGH is a parallelogram --> all squares are parallelograms but not vise-versa. Not sufficient.

(2) The diagonals of EFGH are perpendicular bisectors of one another --> EFGH is a rhombus --> all squares are rhombuses but not vise-versa. Not sufficient.

(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not vise-versa. Not sufficient.

In 2, the diagonals are perpendicular bisectors could happen for a rhombus as well as a square right? Is that why we're saying it's Insufficient? As far as I know,diagonals are perpendicular bisectors of each other for square ,rhombus and even rectangles.Please correct me if I'm wrong.

Rhombus is a quadrilateral with all four sides equal in length. So, all squares are rhombuses but not vise-versa.

(1) EFGH is a parallelogram --> all squares are parallelograms but not vise-versa. Not sufficient.

(2) The diagonals of EFGH are perpendicular bisectors of one another --> EFGH is a rhombus --> all squares are rhombuses but not vise-versa. Not sufficient.

(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not vise-versa. Not sufficient.

In 2, the diagonals are perpendicular bisectors could happen for a rhombus as well as a square right? Is that why we're saying it's Insufficient? As far as I know,diagonals are perpendicular bisectors of each other for square ,rhombus and even rectangles.Please correct me if I'm wrong.

From (2) the figure can be a square or rhombus, yes.

Diagonals of a rectangle are bisectors of each other but not perpendicular to each other, unless of course it's a square.
_________________

Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square [#permalink]

Show Tags

31 Dec 2016, 02:50

To prove that a quadrilateral is a square, you must prove that it is both a rhombus (all sides are equal) and a rectangle (all angles are equal).

(1) INSUFFICIENT: Not all parallelograms are squares (however all squares are parallelograms).

(2) INSUFFICIENT: If a quadrilateral has diagonals that are perpendicular bisectors of one another, that quadrilateral is a rhombus. Not all rhombuses are squares (however all squares are rhombuses).

If we look at the two statements together, they are still insufficient. Statement (2) tells us that ABCD is a rhombus, so statement one adds no more information (all rhombuses are parallelograms). To prove that a rhombus is a square, you need to know that one of its angles is a right angle or that its diagonals are equal (i.e. that it is also a rectangle).

Statement 1 is to broad. A parallelogram can be several things

Statement 2 correctly identifies this polygon as a rhombus. If this polygon is a rhombus then it IS NOT a square. By identifying the polygon as a rhombus doesn't this prove that the polygon EFGH IS NOT a square?

I am providing the official answer. Please help!

The GMAT uses the more controversial definition of a rhombus; in other words, all squares are technically rhombuses.

Statement 1

No because we could have a standard rectangle, rhombus or kite.

Statement 2

Too broad because we could have a rhombus- but not all rhombuses are squares; however, all squares are rhombuses.

Statement 1 and 2

The most that we can infer is that this shape is a rhombus- but we cannot exact any further details