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

Question

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!

Bunuel · Accepted Answer

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.

Answer: E.

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

Hope it helps.

jscott319 · Answer

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

Bunuel · Answer

A square is a special type of: 
Rhombus;
Parallelogram;
Rectangle.

Jp27 · Answer

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

Bunuel · Answer

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.

Jp27 · Answer

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

davesinger786 · Answer

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.

Bunuel · Answer

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.

davesinger786 · Answer

Oh ,right.My bad.It can't be a rectangle.Thanks for your explanation.Kudos

anairamitch1804 · Answer

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

Nunuboy1994 · Answer

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

SchruteDwight · Answer

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?

Leonfranco · Answer

Hi Bunuel-a query,
If statement two read diagonals of EFGH are EQUAL,perpendicular bisectors of each other .Would that statement be sufficient by itself?

Bunuel · Answer

Yes, because in this case EFGH has to be a square.

Leonfranco · Answer

Thanks Bunuel.Really appreciate it.

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

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

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