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# M11-05

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:43
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Difficulty:

35% (medium)

Question Stats:

62% (00:52) correct 38% (00:52) wrong based on 169 sessions

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Is quadrilateral $$ABCD$$ a rhombus?

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

(2) $$AB + CD = BC + AD$$

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Joined: 02 Sep 2009
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15 Sep 2014, 23:43
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, 21:53
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
Math Expert
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15 Jun 2016, 01:39
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, 02:26
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?
Intern
Joined: 29 Jun 2016
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17 Aug 2016, 07:49
A kite-shaped figure is a funny explanation.
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Joined: 09 Oct 2015
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27 Sep 2016, 07:00
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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Joined: 10 Aug 2014
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07 Oct 2016, 03:54
This is a HIGH quality question,i agree with the explanation.
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Location: India
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09 Nov 2016, 03:23
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, 06:20
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, 04:51
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, 06:06
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, 04:42
Is it sufficient for statement 1 IF IT SAYS AC and BD are perpendicular bisectors of each other?
Intern
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08 Jun 2017, 09:04
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, 14:08
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, 07:12
Those haven't understood the solution. Pls refer to the below link

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14 Nov 2017, 18:22
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 06 Apr 2016
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14 Dec 2017, 01:27
The existing explanations doesn't seem sufficient to me!

From statement 1,

We know that the diagram may be a rectangle/rhombus/square, hence insufficient.

Statement 2 also says the same since

AB+CD=BC+AD applies for all the three.

Is my reasoning correct?

Bunuel can you pls help me with this one!
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Joined: 02 Sep 2009
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14 Dec 2017, 02:57
3
Darknight2 wrote:
The existing explanations doesn't seem sufficient to me!

From statement 1,

We know that the diagram may be a rectangle/rhombus/square, hence insufficient.

Statement 2 also says the same since

AB+CD=BC+AD applies for all the three.

Is my reasoning correct?

Bunuel can you pls help me with this one!

No. All squares and rectangle, are rhombi. This is not the reason the answer is E. The point is that ABCD could be a kite, so NOT a rhombus.

Both kite and rhombus have the diagonals perpendicular to each other and the sum of the opposite sides equal to each other. So, ABCD can be a kite or a rhombus. Look at the diagram below:

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Math Expert
Joined: 02 Sep 2009
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14 Dec 2017, 02:58
Bunuel wrote:
Darknight2 wrote:
The existing explanations doesn't seem sufficient to me!

From statement 1,

We know that the diagram may be a rectangle/rhombus/square, hence insufficient.

Statement 2 also says the same since

AB+CD=BC+AD applies for all the three.

Is my reasoning correct?

Bunuel can you pls help me with this one!

No. All squares and rectangle, are rhombi. This is not the reason the answer is E. The point is that ABCD could be a kite, so NOT a rhombus.

Both kite and rhombus have the diagonals perpendicular to each other and the sum of the opposite sides equal to each other. So, ABCD can be a kite or a rhombus. Look at the diagram below:

For more practice Properties of Polygons Questions.
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Re: M11-05 &nbs [#permalink] 14 Dec 2017, 02:58

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# M11-05

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