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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

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

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).

Re: Is quadrilateral ABCD a rhombus? [#permalink]
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.

Re: Data Sufficiency - doubts [#permalink]
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. _________________

......................................................................... +1 Kudos please, if you like my post

Re: Data Sufficiency - doubts [#permalink]
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.

Re: Is quadrilateral ABCD a rhombus? [#permalink]
05 Nov 2014, 04:09

Expert's post

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.

Re: Is quadrilateral ABCD a rhombus? [#permalink]
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.

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