E, F, G, and H are the vertices of a polygon. Is polygon : GMAT Data Sufficiency (DS)
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# E, F, G, and H are the vertices of a polygon. Is polygon

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

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28 Nov 2010, 12:18
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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?

[Reveal] Spoiler: OA
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28 Nov 2010, 12:32
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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?

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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Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]

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28 Nov 2010, 12:41
Isnt a square BOTH a Rhombus and a Rectangle? I though it can not separately be one or the other?
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Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]

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28 Nov 2010, 12:43
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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: DS Geometry Problem from my MGMAT CAT Test [#permalink]

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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.....

Cheers
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Re: DS Geometry Problem from my MGMAT CAT Test [#permalink]

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

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

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11 Jul 2014, 00:34
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Re: E, F, G, and H are the vertices of a polygon. Is polygon [#permalink]

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01 Jul 2015, 22:07
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.

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

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01 Jul 2015, 23:52
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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 [#permalink]

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01 Jul 2015, 23:57
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Re: E, F, G, and H are the vertices of a polygon. Is polygon [#permalink]

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15 Jul 2016, 00:10
Hello from the GMAT Club BumpBot!

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Re: E, F, G, and H are the vertices of a polygon. Is polygon   [#permalink] 15 Jul 2016, 00:10
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