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# M11-05

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:43
Expert's post
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Difficulty:

45% (medium)

Question Stats:

57% (00:48) correct 43% (00:49) wrong based on 108 sessions

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Is quadrilateral $$ABCD$$ a rhombus?

(1) $$AC$$ is perpendicular to $$BD$$

(2) $$AB + CD = BC + AD$$
[Reveal] Spoiler: OA

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16 Sep 2014, 00:43
Official Solution:

Statements (1) and (2) combined are insufficient. $$ABCD$$ can be a rhombus but it can also be a kite-shaped figure. The diagonals of this figure form a cross, and S2 holds because of the symmetry ($$AB = BC$$ and $$CD = AD$$).

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14 Jun 2016, 22:53
Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. $$ABCD$$ can be a rhombus but it can also be a kite-shaped figure. The diagonals of this figure form a cross, and S2 holds because of the symmetry ($$AB = BC$$ and $$CD = AD$$).

Bunuel HI , ABCD can also be a square,correct me if i'm wrong

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15 Jun 2016, 02:39
sharma123 wrote:
Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. $$ABCD$$ can be a rhombus but it can also be a kite-shaped figure. The diagonals of this figure form a cross, and S2 holds because of the symmetry ($$AB = BC$$ and $$CD = AD$$).

Bunuel HI , ABCD can also be a square,correct me if i'm wrong

Yes, it can be a square too.
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06 Jul 2016, 03:26
I think this is a poor-quality question and I don't agree with the explanation. I dont understand how can 2nd statement be true?

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17 Aug 2016, 08:49
A kite-shaped figure is a funny explanation.

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27 Sep 2016, 08:00
Bunuel IF the condition 1 read that the AC is a perpendicular bisector of BD, would that have been enough to conclude that the figure was a rhombus ( considering square is also a special type of rhombus ) ? Will the diagonals of the a kite shaped figure also bisect each other at 90 degrees ?

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07 Oct 2016, 04:54
This is a HIGH quality question,i agree with the explanation.

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Intern
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Location: India
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09 Nov 2016, 04:23
Bunuel wrote:
sharma123 wrote:
Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. $$ABCD$$ can be a rhombus but it can also be a kite-shaped figure. The diagonals of this figure form a cross, and S2 holds because of the symmetry ($$AB = BC$$ and $$CD = AD$$).

Bunuel HI , ABCD can also be a square,correct me if i'm wrong

Yes, it can be a square too.

Isn't it true that every square is a in fact a rhombus?

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09 Nov 2016, 07:20
yeshu_a wrote:
Isn't it true that every square is a in fact a rhombus?

Yes, every square is a rhombus.
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11 Nov 2016, 05:51
I don't understand how the first statement could be true...

How can two sides of a quadrilateral be perpendicular if they don't share any vertex? (AC and BD)

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11 Nov 2016, 07:06
pedropereira wrote:
I don't understand how the first statement could be true...

How can two sides of a quadrilateral be perpendicular if they don't share any vertex? (AC and BD)

AC and BD are diagonals, not sides.
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21 Dec 2016, 05:42
Is it sufficient for statement 1 IF IT SAYS AC and BD are perpendicular bisectors of each other?

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08 Jun 2017, 10:04
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. hi,

In a rhombus diagonals bisect each other at 90 degree. in statement one we are told that the diagonals bisect each other at 90 degree. then it must be a rhombus. even a kite shaped figure is a rhombus. correct me if im wrong.

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08 Jun 2017, 15:08
kanikas1 wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. hi,

In a rhombus diagonals bisect each other at 90 degree. in statement one we are told that the diagonals bisect each other at 90 degree. then it must be a rhombus. even a kite shaped figure is a rhombus. correct me if im wrong.

Yes, you are wrong.

RHOMBUS - a quadrilateral in which all four sides are congruent.
KITE - a quadrilateral in which each pair of consecutive sides are congruent, but opposite sides are not congruent.
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03 Aug 2017, 08:12
Those haven't understood the solution. Pls refer to the below link

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Re: M11-05   [#permalink] 03 Aug 2017, 08:12
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# M11-05

Moderators: Bunuel, Vyshak

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