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Intern  Joined: 31 Aug 2010
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E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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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?

Math Expert V
Joined: 02 Sep 2009
Posts: 58428
Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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2
6
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?

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.

For more on this topic check Polygons chapter of Math Book: math-polygons-87336.html

Hope it helps.
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Intern  Joined: 31 Aug 2010
Posts: 38
Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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Isnt a square BOTH a Rhombus and a Rectangle? I though it can not separately be one or the other?
Math Expert V
Joined: 02 Sep 2009
Posts: 58428
Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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1
jscott319 wrote:
Isnt a square BOTH a Rhombus and a Rectangle? I though it can not separately be one or the other?

A square is a special type of:
Rhombus;
Parallelogram;
Rectangle.
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Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58428
Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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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.
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Manager  Joined: 22 Dec 2011
Posts: 211
Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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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.
Intern  Joined: 10 May 2015
Posts: 24
Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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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.

For more on this topic check Polygons chapter of Math Book: math-polygons-87336.html

Hope 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 V
Joined: 02 Sep 2009
Posts: 58428
Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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davesinger786 wrote:
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.

For more on this topic check Polygons chapter of Math Book: math-polygons-87336.html

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

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Oh ,right.My bad.It can't be a rectangle.Thanks for your explanation.Kudos
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Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square  [#permalink]

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

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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?

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

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

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Hey guys,

What we need for a square is

- The diagonals bisect each other at 90° and they are equal

Since question only gives that they bisect, it could also be a kite, right? Re: E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square   [#permalink] 23 Oct 2018, 20:05
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