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Is quadrilateral ABCD a rhombus?

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Is quadrilateral ABCD a rhombus? [#permalink] New post 14 Oct 2009, 20:09
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Is quadrilateral ABCD a rhombus?

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

(2) AB = BC = CD = AD
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Re: rhombus? [#permalink] New post 14 Oct 2009, 20:15
IMO E.

State 1) It can be a Square.
State 2) It can be a Square again.

Together the statements are not sufficient to answer if this is a rhombus or a square.
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Re: rhombus? [#permalink] New post 14 Oct 2009, 20:24
By definition, square is a special case of rhombus. So all squares can be called rhombi too :)
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Re: rhombus? [#permalink] New post 14 Oct 2009, 20:40
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Re: rhombus? [#permalink] New post 14 Oct 2009, 20:45
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Re: rhombus? [#permalink] New post 25 Oct 2009, 06:00
about the 2nd option. can't it be a parellogram too?
and a pllgram is not necessarily a rhombus.

so the answer shud be A.
(or am i missing the obvious here??)
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Re: rhombus? [#permalink] New post 25 Oct 2009, 09:15
any rhombus has the following two properties:
Opposite angles of a rhombus have equal measure.
The two diagonals of a rhombus are perpendicular.

so any square is a rhombus hence D.
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Re: rhombus? [#permalink] New post 13 Oct 2013, 15:10
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

Hi Bunuel

When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus?

And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus?

Is that the correct line of reasoning?
Thanks so much in advance
Cheers
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Re: rhombus? [#permalink] New post 13 Oct 2013, 15:20
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jlgdr wrote:
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

Hi Bunuel

When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus?

And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus?

Is that the correct line of reasoning?
Thanks so much in advance
Cheers
J :)


For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

For (2): AB = BC = CD = AD, means that ABCD is either a rhombus or square (so still a rhombus).

Hope it helps.
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Re: Is quadrilateral ABCD a rhombus? [#permalink] New post 06 Dec 2013, 10:10
Is quadrilateral ABCD a rhombus?

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

But doesn't a rhombus also allow for two perpendicular bisectors? A square has two perpendicular bisectors but a square is not a rhombus (because a rhombus does not have all four angles = 90. Oh, but wait, a rhombus is a square in the same way a rectangle is a square but not the other way around. Tricky.

(2) AB = BC = CD = AD

I guess the same logic applies above, ABCD could be a square or rhombus but regardless of which one it is its still a rhombus. If the stem asked if ABCD was a square, then this would be insufficient.
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Data Sufficiency - doubts [#permalink] New post 08 Aug 2014, 19:29
Hi,

can someone help me with this question please:

Is quadrilateral ABCD a rhombus?

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

(2) AB = BC = CD = AD

The answer it says is D- each statement is enough on its own.

Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus?
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Re: Data Sufficiency - doubts [#permalink] New post 08 Aug 2014, 20:36
kritim22 wrote:
Hi,

can someone help me with this question please:

Is quadrilateral ABCD a rhombus?

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

(2) AB = BC = CD = AD

The answer it says is D- each statement is enough on its own.

Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus?


Let say AC and BD meet at O. We have OD= OB, OA= OC. If you use pithagoras theorem, you can easily see that AD= DC=BC=AB.
For example, AD^2= OA^2 + OD^2
So (1) and (2) basically say the same thing.
The definition of rhombus is a quadrilateral whose four sides have the same length. So D is the answer.
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Re: Data Sufficiency - doubts [#permalink] New post 08 Aug 2014, 23:56
kritim22 wrote:
Hi,

can someone help me with this question please:

Is quadrilateral ABCD a rhombus?

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

(2) AB = BC = CD = AD

The answer it says is D- each statement is enough on its own.

Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus?


A square is a special case of rhombus. It has all angles of 90 degrees. so proving that quadrilateral ABCD is a square is sufficient to say that it is a rhombus.
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Re: Is quadrilateral ABCD a rhombus? [#permalink] New post 12 Aug 2014, 06:10
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kritim22 wrote:
Hi,

can someone help me with this question please:

Is quadrilateral ABCD a rhombus?

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

(2) AB = BC = CD = AD

The answer it says is D- each statement is enough on its own.

Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus?


Merging topics. Please refer to the discussion above.

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Re: Is quadrilateral ABCD a rhombus? [#permalink] New post 05 Nov 2014, 02:31
Bunuel wrote:
Answer D.

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

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...



But Bunuel this would be true even for rectangle:
Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.
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Re: Is quadrilateral ABCD a rhombus? [#permalink] New post 05 Nov 2014, 04:09
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honchos wrote:
Bunuel wrote:
Answer D.

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

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...



But Bunuel this would be true even for rectangle:
Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.


For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus.

Hope it's clear.
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Re: Is quadrilateral ABCD a rhombus? [#permalink] New post 06 Nov 2014, 01:17
Bunuel,

For (1), can you consider the case of a kite?

Bunuel wrote:
honchos wrote:
Bunuel wrote:
Answer D.

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

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...



But Bunuel this would be true even for rectangle:
Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.


For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus.

Hope it's clear.

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Re: Is quadrilateral ABCD a rhombus? [#permalink] New post 06 Nov 2014, 06:22
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Re: Is quadrilateral ABCD a rhombus?   [#permalink] 06 Nov 2014, 06:22
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