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

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

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.

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.

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

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

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

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?

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?

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

Thanks Bunuel.Really appreciate it.

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