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.

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:

E, F, G, and H are the vertices of a polygon. Is polygon [#permalink]
28 Nov 2010, 12:18

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

63% (01:23) correct
37% (00:17) wrong based on 93 sessions

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?

Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]
28 Nov 2010, 12:32

Expert's post

jscott319 wrote:

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: DS Geometry Problem from my MGMAT CAT Test [#permalink]
06 Nov 2012, 02:34

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.....

Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]
06 Nov 2012, 04:59

Expert's post

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°.

Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]
06 Nov 2012, 05:16

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 [#permalink]
11 Jul 2014, 00:34

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. _________________