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Is quadrilateral ABCD a rhombus? [#permalink]
VeritasKarishma wrote:
mk87 wrote:
Hi guys
I'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:

Is quadrilateral ABCD a Rhombus?
(1) Line segments AC and BD are perpendicular bisectors of each other
(2) AB = BC = CD = AD

The official answer is as follows:
Statement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus
Statement 2 - A quadrilateral with four equal sides is by definition a rhombus

What I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?


Statement 1 says the diagonals are perpendicular bisectors of each other which means that they intersect each other at mid points and form 90 degree angles. It is not possible that they are in the shape of an asymmetric kite. Figure 1 is not possible. Only figure 2 is possible. In figure 2, all sides will be equal.
Attachment:
Ques4.jpg


This is an interesting post. So to be clear, the distinction is this

1. The diagonals of a kite INTERSECT each other, but we don't consider them as bisectors b/c the diagonals are not split evenly
2. The diagonals of a rhombus are PERPENDICULAR BISECTORS (bisection b/c diagonals are split evenly into two)
3. The diagonals of a rectangle are NON-PERPENDICULAR bisectors (b/c the bisection does not form 90 degree angles), but in the case of a square the bisectors ARE perpendicular bisectors.

Bunuel is this accurate?
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Re: Is quadrilateral ABCD a rhombus? [#permalink]
CEdward wrote:
VeritasKarishma wrote:
mk87 wrote:
Hi guys
I'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:

Is quadrilateral ABCD a Rhombus?
(1) Line segments AC and BD are perpendicular bisectors of each other
(2) AB = BC = CD = AD

The official answer is as follows:
Statement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus
Statement 2 - A quadrilateral with four equal sides is by definition a rhombus

What I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?


Statement 1 says the diagonals are perpendicular bisectors of each other which means that they intersect each other at mid points and form 90 degree angles. It is not possible that they are in the shape of an asymmetric kite. Figure 1 is not possible. Only figure 2 is possible. In figure 2, all sides will be equal.
Attachment:
Ques4.jpg


This is an interesting post. So to be clear, the distinction is this

1. The diagonals of a kite INTERSECT each other, but we don't consider them as bisectors b/c the diagonals are not split evenly
2. The diagonals of a rhombus are PERPENDICULAR BISECTORS (bisection b/c diagonals are split evenly into two)
3. The diagonals of a rectangle are NON-PERPENDICULAR bisectors (b/c the bisection does not form 90 degree angles), but in the case of a square the bisectors ARE perpendicular bisectors.

Bunuel is this accurate?


CEdward - this link is pretty awesome for quadrilaterals - https://e-gmat.com/blogs/quadrilateral-properties-formulas-rectangle-square-parallelogram-rhombus-trapezium-trapezoid/
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Re: Is quadrilateral ABCD a rhombus? [#permalink]
Bunuel wrote:
Thought about this again: D it is.

Well:
(1) True for square or for rhombus but every square is a rhombus, so sufficient
(2) Again true for square or for rhombus but every square is a rhombus, so sufficient

D


What about the kite? It has a perpendicular bisector, so 1 is out??
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Re: Is quadrilateral ABCD a rhombus? [#permalink]
I am kind of confused because, in previous DS questions, the answers probably would be E for such a question.

I feel the inference is kind of vague.


According to statement 1, cant the figure be both a Square and a Rhombus?

Same for statement 2...

Can someone please clarify how to infer here thanks.
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Re: Is quadrilateral ABCD a rhombus? [#permalink]
I think the most important thing to understand here is that every square is a rhombus
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Re: Is quadrilateral ABCD a rhombus? [#permalink]
Economist wrote:
Is quadrilateral ABCD a rhombus?

(1) Line segments AC and BD are perpendicular bisectors of each other.

(2) AB = BC = CD = AD


Is it true that square is a kind of rhombus ? Aren't the properties different for both wrt diagonals length ? Can we assume this for all questions where it can be either a square or a rhombus ?
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Re: Is quadrilateral ABCD a rhombus? [#permalink]
Expert Reply
prachisaraf wrote:
Economist wrote:
Is quadrilateral ABCD a rhombus?

(1) Line segments AC and BD are perpendicular bisectors of each other.

(2) AB = BC = CD = AD


Is it true that square is a kind of rhombus ? Aren't the properties different for both wrt diagonals length ? Can we assume this for all questions where it can be either a square or a rhombus ?


Every square is a rhombus, but every rhombus is not a square.
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Re: Is quadrilateral ABCD a rhombus? [#permalink]
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