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# Is the quadrilateral a square?

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11 Feb 2012, 14:08
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ABCD.png [ 10.14 KiB | Viewed 3806 times ]

(1) AB = BC = CD = DA
(2) AC = BD

In this question - How come the answer is C?

In Square the diagonals bisect each other at 90 degrees and are equal. How do we know that diagonals are bisecting each other at 90 degrees?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Feb 2012, 14:49, edited 2 times in total.
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11 Feb 2012, 14:48
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(1) AB = BC = CD = DA --> both square and rhombus have all four sides equal in length. Not sufficient.

(2) AC = BD --> both square and rectangle have diagonals equal in length. Not sufficient.

(1)+(2) ABCD is a square. Sufficient.

Useful property: if the diagonals of a rhombus are equal, then that rhombus must be a square. So you can apply this for (1)+(2) and conclude that ABCD must be a square.

Check Polygons chapter of Math Book for more useful properties and tips: math-polygons-87336.html

Hope it helps.
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05 Aug 2013, 19:07
from 1 it is clear that it is either a rhombus or a square as all sides are equal.

from 2 it is established that it is a square as IN A RHOMBUS DIAGONALS ARE NOT EQUAL.

hence C.

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05 Aug 2013, 23:08
enigma123 wrote:
Attachment:
ABCD.png

(1) AB = BC = CD = DA
(2) AC = BD

In this question - How come the answer is C?

In Square the diagonals bisect each other at 90 degrees and are equal. How do we know that diagonals are bisecting each other at 90 degrees?

square has equal sides and equal diagonals.
rhombus has equal sides but not equal diagonals.
so both together are sufficient...... Answer (C)

About the 2nd question, let p the point of intersection ofthe two diagonal AC and BD.
now angle PAB = angle PBA [because PA=PB]
so angle APB has to be 90 or it will not be possible for the total angle surrounding the point P to be 360.
so diagonals have to bisect at 90 degress
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24 Jul 2015, 02:23
divyarathi15 wrote:

(1) AB = BC = CD = DA
(2) AC = BD

Hi,
1) statement tells us all sides are equal. it could be a rhombus , a square etc.... insuff
2) stat 2 tells us that the diagonals are equal.. it could be a rectangle , a square or any other quad.. insuff

combined we know it can only be a square... suff
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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24 Jul 2015, 02:32
Hi!

We could also say from statement 1 that it could be any irregular quadrilateral (apart from being either a rhombus/sq). Again from statement 2 it can be concluded that the figure could be any irregular quad too. combining both cant it be concluded that the figure could be an irregular quadrilateral too?

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24 Jul 2015, 02:40
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22 Jun 2017, 08:07
enigma123 wrote:
Attachment:
ABCD.png

(1) AB = BC = CD = DA
(2) AC = BD

In this question - How come the answer is C?

In Square the diagonals bisect each other at 90 degrees and are equal. How do we know that diagonals are bisecting each other at 90 degrees?

Solution:

Statement 1: It can be either a square or rhombus. Insufficient.
Statement 2: It can be either a square or a parallelogram or a rectangle.

Combining St 1 and St 2, It has to be a square.

Therefore the answer is Option C.

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27 Jul 2017, 21:42
I did this question in the following method. Correct me if i am wrong . From the statement 1 the quadrilateral can be a Rhombus or a square. Hence not sufficient. Now from statement 2 the quadrilateral is a rhombus or a square or a rectangle.
1 and 2 together C

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27 Jul 2017, 21:47
longhaul123 wrote:
I did this question in the following method. Correct me if i am wrong . From the statement 1 the quadrilateral can be a Rhombus or a square. Hence not sufficient. Now from statement 2 the quadrilateral is a rhombusor a square or a rectangle.
1 and 2 together C

A rhombus with equal diagonal is a square.
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