Is ABCD a Rombus 1 ABCD is a Square 2. ABCD diagonals bisect

Answer is D.

Statement 1 says that ABCD is a square. A square is just a special rhombus with all angles equal to \(90^o\)

Statement 2: Diagonals bisect at \(90^o\)

Let's draw a picture to visualize this: (Let's consider the center point to be O, I forgot to label this)

So as you can see from the image, since the diagonals are bisected, we have AO = OC. And we have OB to be common for triangles AOB and BOC. So consider right triangle AOB:
\(AB^2 = AO^2 + OB ^2 = OC^2 + OB^2 = BC^2.\)

So we get that AB = BC. Similarly we can prove that all the sides are equal using this method. Hence we get a quadrilateral where all sides are equal, and diagonals bisecting at \(90^o\). Hence it's a rhombus. Sufficient.

Hey whiplash2411,
I agree A is Correct choice..
But not D!
I don't think B also leads to solution here..
B just says bisect each other. It doesn't mean that both should be of equal length!

Guys pls correct me if im wrong

If the diagonals of a quadrilateral are perpendicular bisectors of each other (so if the diagonal bisect each other at a right angle), the quadrilateral is a rhombus.
_________________

I believe the answer here is A.
What is rombus - it is quadrilateral with the following properties:
* Opposite angles of a rhombus have equal measure
* The two diagonals of a rhombus are perpendicular;
* Its diagonals bisect opposite angles (that is why we have first property)
In choice B we do not see all the properties, it could be a rombus but at the same time it could be an quadrilateral with different angles and opposite sides - not sufficient.

OA is D, not A.

There is a property which says if the diagonals of a quadrilateral are perpendicular bisectors of each other (so if the diagonal bisect each other at a right angle), the quadrilateral is a rhombus.
_________________

Most probably i missed the word bisects, which means divides into equal sides, and understood it as intersects. In this case obviously that will be a rombus and no other figure could be drawn. Yes once again i confirm that GMAT has so many small tricky parts.
Thank you Bunuel for clarification.

The answer is D.
All Square are Rombus, so first is that A Satisfies the condition.

Now, if any two diagonals bisects each other, then also it will always be rombus, so B also satisfies the condition.

Since both A and B alone satisfies the condition the answer is D

D

Rule for Rhombus:
Diagonals should bisect each other and angle between diagonals should be 90degrees.
If all sides of a Rhombus are equal then it is a square.

Any square is a Rhombus
If the diagonal bisects at 90 degrees then it is a Rhombus.

So either of the statement alone is sufficient to answer the question.

Hi Bunuel,

I have a query here.

in case of Kite also diagonal are perpendicular bisector. so in st2 cant we consider this as a kite?

Please clarify.

Thanks

The diagonals of a kite may be perpendicular, but they do not both bisect each other. 'Bisect' means 'cuts perfectly in half', and if you draw a skewed kite, and draw its diagonals, you'll see that one of the two diagonals is not cut perfectly in half at their intersection point.

That said, the question in the original post above is really not the kind of question you see on the GMAT. The GMAT does not test if you know the definition of specialized figures like rhombuses or kites, nor will it test if you know obscure facts about their diagonals.
_________________

Never late to study. Bisecting is a trip, eliminating the kite

D

Once into parallelograms, there are two types. Imagine two sets.

Set 1: All angles are 90 degrees and only opposite sides are equal
Set 2: All sides are equal but angles are not equal to 90 degrees
Also, a set 3, which is the intersection of these two sets



Set 1 Contains a Rectangle

Set 2 Contains a Rhombus

Set 3, the intersection of the two which has all sides equal as well as all angles equal, is a square.

So.
A square is a rectangle.
But a rectangle is not a square.

A square is a rhombus.
But a rhombus is not a square

_________________

Isn't it true that, there could be irregular quadrilaterals which have perpendicular diagonals. Not necessarily has to be rhombus. However, if it’s a rhombus, then the diagonal has to be perpendicular?

If the two diagonals of a four-sided figure are perpendicular bisectors of each other then the figure must be a rhombus.

Bunuel

But It could also be a kite according to option B.

Does diagonals BISECT each other for kite?
_________________

Imo D as from statement one we know ABCD is a square and a square is also a rhombus so sufficient.
statement 2 if diagonal bisect at 90 degree then this property will tell us that the figure can be rhombus or square in both case its a rhombus. so sufficient .

The second statement does not indicate whether the two diagonals bisect each other (which is a Rhombus), or whether the diagonals bisect at one side (which is a kite), thus it is not safe to assume that it is a rhombus.

Posted from my mobile device

But it could be a square or a kite?