Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 31 Aug 2010
Posts: 44

E, F, G, and H are the vertices of a polygon. Is polygon [#permalink]
Show Tags
28 Nov 2010, 13:18
12
This post was BOOKMARKED
Question Stats:
59% (01:26) correct
41% (00:24) wrong based on 412 sessions
HideShow timer Statistics
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!
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 39609

Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]
Show Tags
28 Nov 2010, 13:32
1
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
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 viseversa.(1) EFGH is a parallelogram > all squares are parallelograms but not viseversa. Not sufficient. (2) The diagonals of EFGH are perpendicular bisectors of one another > EFGH is a rhombus > all squares are rhombuses but not viseversa. Not sufficient. (1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not viseversa. Not sufficient. Answer: E. For more on this topic check Polygons chapter of Math Book: mathpolygons87336.htmlHope it helps.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 31 Aug 2010
Posts: 44

Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]
Show Tags
28 Nov 2010, 13:41
Isnt a square BOTH a Rhombus and a Rectangle? I though it can not separately be one or the other?



Math Expert
Joined: 02 Sep 2009
Posts: 39609

Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]
Show Tags
28 Nov 2010, 13:43



Senior Manager
Joined: 22 Dec 2011
Posts: 294

Re: DS Geometry Problem from my MGMAT CAT Test [#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 viseversa. Not sufficient.
(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not viseversa. 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



Math Expert
Joined: 02 Sep 2009
Posts: 39609

Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]
Show Tags
06 Nov 2012, 05:59
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 viseversa. Not sufficient.
(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not viseversa. 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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 22 Dec 2011
Posts: 294

Re: DS Geometry Problem from my MGMAT CAT Test [#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 viseversa. Not sufficient.
(1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not viseversa. 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.



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15926

Re: E, F, G, and H are the vertices of a polygon. Is polygon [#permalink]
Show Tags
11 Jul 2014, 01:34
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Intern
Joined: 10 May 2015
Posts: 30

Re: E, F, G, and H are the vertices of a polygon. Is polygon [#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 viseversa.(1) EFGH is a parallelogram > all squares are parallelograms but not viseversa. Not sufficient. (2) The diagonals of EFGH are perpendicular bisectors of one another > EFGH is a rhombus > all squares are rhombuses but not viseversa. Not sufficient. (1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not viseversa. Not sufficient. Answer: E. For more on this topic check Polygons chapter of Math Book: mathpolygons87336.htmlHope it helps. Hi Bunuel, 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.



Math Expert
Joined: 02 Sep 2009
Posts: 39609

Re: E, F, G, and H are the vertices of a polygon. Is polygon [#permalink]
Show Tags
02 Jul 2015, 00:52
davesinger786 wrote: Bunuel wrote: Rhombus is a quadrilateral with all four sides equal in length. So, all squares are rhombuses but not viseversa.(1) EFGH is a parallelogram > all squares are parallelograms but not viseversa. Not sufficient. (2) The diagonals of EFGH are perpendicular bisectors of one another > EFGH is a rhombus > all squares are rhombuses but not viseversa. Not sufficient. (1)+(2) EFGH is a rhombus (all rhombus are parallelogram). Again all squares are rhombuses but not viseversa. Not sufficient. Answer: E. For more on this topic check Polygons chapter of Math Book: mathpolygons87336.htmlHope it helps. Hi Bunuel, 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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 10 May 2015
Posts: 30

Re: E, F, G, and H are the vertices of a polygon. Is polygon [#permalink]
Show Tags
02 Jul 2015, 00:57
Oh ,right.My bad.It can't be a rectangle.Thanks for your explanation.Kudos



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15926

Re: E, F, G, and H are the vertices of a polygon. Is polygon [#permalink]
Show Tags
15 Jul 2016, 01:10
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Senior Manager
Joined: 26 Oct 2016
Posts: 460
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: E, F, G, and H are the vertices of a polygon. Is polygon [#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). The correct answer is E
_________________
Thanks & Regards, Anaira Mitch




Re: E, F, G, and H are the vertices of a polygon. Is polygon
[#permalink]
31 Dec 2016, 02:50







