E, F, G, and H are the vertices of a polygon. Rhombus Q

[william.conrad]

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.


(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).

Answer is D.


Statement two - Is this explanation correct? If a quadrilateral has diagnols that are perp bisectors of one another doesn't that mean it could be either a square or a rhombus? [/b]

Still arrive at the same answer regardless...

[rchitta]

I think, there is a small flaw in your approach for the 2nd argument. A quadrilateral whose diagonals are perpendicular bisectors doesn't have to be a rhombus. It is a parallelogram for sure but not a rhombus.

Remember, a rhombus is a quadrilateral with all sides of the same length. It doesn't talk about its diagonals being perpendicular bisectors.

But your answer is right, its insufficient to tell if that quadrilateral is a square.

[esledge]

Answer is D.

Statement two - Is this explanation correct? If a quadrilateral has diagnols that are perp bisectors of one another doesn't that mean it could be either a square or a rhombus? [/b]

Still arrive at the same answer regardless...

Hi William, I think you had a typo above. To clarify: The correct answer is E.

You are correct about statement two. If the diagonals of a quadrilateral are perpendicular bisectors, the quadrilateral must have 4 equal sides, which also implies that the quadrilateral has two pairs of parallel sides. That is, it's a square or a rhombus.

For the diagonals to be just perpendicular, many quadrilaterals are possible. All four sides may be different lengths, and it's possible that no two sides will be parallel. That's not to say that every quadrilateral will have perpendicular diagonals, but if you start by drawing two perpendicular lines, you can obviously draw four lines around the end points and make a wide variety of quadrilaterals.

For the diagonals to be just bisectors, a quadrilateral must have two pairs of parallel sides. The following quadrilaterals have diagonals that are bisectors: square, rhombus, rectangle, parallelogram.

So you can see that it is the "bisector" constraint that is most restrictive. Adding "perpendicular" to it simply rules out the quadrilaterals that have one set of sides that is longer than the other set (draw a long, skinny rectangle or parallelogram to see why). Thus, diagonals that are "perpendicular bisectors" implies rhombus or square.

I think, there is a small flaw in your approach for the 2nd argument. A quadrilateral whose diagonals are perpendicular bisectors doesn't have to be a rhombus. It is a parallelogram for sure but not a rhombus.

Remember, a rhombus is a quadrilateral with all sides of the same length. It doesn't talk about its diagonals being perpendicular bisectors.

rchitta, you are right that a rhombus is usually defined this way, but do you see why diagonals that are perpendicular bisectors are another (related) property?

[lj6871849]

Hi Emily - this might be a silly question pls bear with me.

If I have cord XY is the perpendicular bisector of line AP then XY Bisects AP at equidistant right. If the point of had been O then AO = OP

Properties of diagonals of a square is that they should Bisect @ 90 degree and the 2 diagonals should be of equal length

Now Stat B says "The diagonals of EFGH are perpendicular bisectors of one another"

If the 2 diagonals were AB and CD , and if O was the point of intersection then we knw AO = OB AND CO = OD but we do not know whether AO and CO are equal. Is this understanding correct?

If AO and CO were equal then we for sure know that Diagonals bisect at 90 degree and they are equal.

Even if the stat read "The diagonals of EFGH are perpendicular of one another"

they we can say its a rhombus but not all rhombus are squares hence NS.

Please correct me if Im wrong.

Cheers

[tim]

it sounds like you are correct in concluding that this is a rhombus but not necessarily a square. if the diagonals were only perpendicular but not necessarily bisecting though, we could not even conclude that we had a rhombus. does this address your concern?

[AdhiK295]

Would the rhombus that is also a parallelogram definitely a square?
It seems to me that correct answer is C (both statements are sufficient)

[RonPurewal]

Would the rhombus that is also a parallelogram definitely a square?
It seems to me that correct answer is C (both statements are sufficient)

any rhombus is already a parallelogram. so, nope—still just a rhombus.