Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 23:38

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD

Author Message
Current Student
Joined: 11 May 2008
Posts: 556
Followers: 8

Kudos [?]: 174 [0], given: 0

Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD [#permalink]

### Show Tags

03 Aug 2008, 17:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

(2) AB = BC = CD = AD
Manager
Joined: 15 Jul 2008
Posts: 207
Followers: 3

Kudos [?]: 57 [0], given: 0

### Show Tags

03 Aug 2008, 17:26
arjtryarjtry wrote:

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

(2) AB = BC = CD = AD

1) is necessary and sufficient condition for rhombus
2) is also same.

guess D
VP
Joined: 17 Jun 2008
Posts: 1397
Followers: 8

Kudos [?]: 290 [0], given: 0

### Show Tags

03 Aug 2008, 18:02
arjtryarjtry wrote:

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

(2) AB = BC = CD = AD

fior a figure to be rhombus necesary and sufficient condn is that its sides are equal and diagnals bisect each other at rt angles .

(1) and (2) both are reqd
IMO C
_________________

cheers
Its Now Or Never

VP
Joined: 17 Jun 2008
Posts: 1397
Followers: 8

Kudos [?]: 290 [0], given: 0

### Show Tags

03 Aug 2008, 18:04
bhushangiri wrote:
arjtryarjtry wrote:

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

(2) AB = BC = CD = AD

1) is necessary and sufficient condition for rhombus
2) is also same.

guess D

its not necessary that a figure with diagonals bisecting each other at rt angles is rhombus,may be its sides are unequal.we can try this drawing on a paper.
hence (2) is important
_________________

cheers
Its Now Or Never

Current Student
Joined: 11 May 2008
Posts: 556
Followers: 8

Kudos [?]: 174 [0], given: 0

### Show Tags

03 Aug 2008, 18:14
i have a diagram for 1 not to be suff... but what bout 2..? can we have a figure with four sides equal , but not parallel?
Attachments

kite.doc [19 KiB]

Director
Joined: 27 May 2008
Posts: 549
Followers: 8

Kudos [?]: 312 [0], given: 0

### Show Tags

04 Aug 2008, 00:24
arjtryarjtry wrote:
i have a diagram for 1 not to be suff... but what bout 2..? can we have a figure with four sides equal , but not parallel?

the figure is wrong... both lines should bisect eachother... in your figure.. the perpendicular line is not divided in half...

IMO D
Manager
Joined: 15 Jul 2008
Posts: 207
Followers: 3

Kudos [?]: 57 [0], given: 0

### Show Tags

04 Aug 2008, 03:08
spriya wrote:
bhushangiri wrote:
arjtryarjtry wrote:

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

(2) AB = BC = CD = AD

1) is necessary and sufficient condition for rhombus
2) is also same.

guess D

its not necessary that a figure with diagonals bisecting each other at rt angles is rhombus,may be its sides are unequal.we can try this drawing on a paper.
hence (2) is important

No.. if the diagonals are perp and bisect each other, then it has to be a rhombus. Distance from any vertex to the adjoining vertex will beI same. Can be proved using the congruence of the 4 triangles using SAS (side angle side) approach.
SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 34

Kudos [?]: 867 [0], given: 5

### Show Tags

04 Aug 2008, 10:49
arjtryarjtry wrote:

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

(2) AB = BC = CD = AD

must be D .

statement 1 and 2 both are properties of rhombus.
_________________

Smiling wins more friends than frowning

VP
Joined: 17 Jun 2008
Posts: 1397
Followers: 8

Kudos [?]: 290 [0], given: 0

### Show Tags

04 Aug 2008, 11:06
durgesh79 wrote:
arjtryarjtry wrote:
i have a diagram for 1 not to be suff... but what bout 2..? can we have a figure with four sides equal , but not parallel?

the figure is wrong... both lines should bisect eachother... in your figure.. the perpendicular line is not divided in half...

IMO D

Oh ok if the diagonals are perpendicular and bisect too

then lets say abcd is a quad. with AC and BD perpen and bisecting

AC=x,BD=y pt of intersection of diagonals be O.
AD= ((x/2)^2 + (y/2)^2 )^1/2 =CD=BC=AB
oops (2) is redundant (1) itself is sufficient

hence IMO A
Whats the OA?
_________________

cheers
Its Now Or Never

SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 34

Kudos [?]: 867 [0], given: 5

### Show Tags

04 Aug 2008, 11:08
spriya wrote:
durgesh79 wrote:
arjtryarjtry wrote:
i have a diagram for 1 not to be suff... but what bout 2..? can we have a figure with four sides equal , but not parallel?

the figure is wrong... both lines should bisect eachother... in your figure.. the perpendicular line is not divided in half...

IMO D

Oh ok if the diagonals are perpendicular and bisect too

then lets say abcd is a quad. with AC and BD perpen and bisecting

AC=x,BD=y pt of intersection of diagonals be O.
AD= ((x/2)^2 + (y/2)^2 )^1/2 =CD=BC=AB
oops (2) is redundant (1) itself is sufficient

hence IMO A
Whats the OA?

Your reasoning is correct. but Why A.. it should D right
each statement itself answers the question.
_________________

Smiling wins more friends than frowning

VP
Joined: 17 Jun 2008
Posts: 1397
Followers: 8

Kudos [?]: 290 [0], given: 0

### Show Tags

04 Aug 2008, 11:21
spriya wrote:
durgesh79 wrote:
arjtryarjtry wrote:

the figure is wrong... both lines should bisect eachother... in your figure.. the perpendicular line is not divided in half...

IMO D

Oh ok if the diagonals are perpendicular and bisect too

then lets say abcd is a quad. with AC and BD perpen and bisecting

AC=x,BD=y pt of intersection of diagonals be O.
AD= ((x/2)^2 + (y/2)^2 )^1/2 =CD=BC=AB
oops (2) is redundant (1) itself is sufficient

hence IMO A
Whats the OA?

Your reasoning is correct. but Why A.. it should D right
each statement itself answers the question.

can we call a parallelogram with 4 side equal a rhombus?this is my doubt
_________________

cheers
Its Now Or Never

SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 34

Kudos [?]: 867 [1] , given: 5

### Show Tags

04 Aug 2008, 11:28
1
KUDOS
spriya wrote:
can we call a parallelogram with 4 side equal a rhombus?this is my doubt

yes thats correct

A parallelogram with equal sides (a = b) is called a rhombus, a parallelogram whose angles are all right angles is called a rectangle.

_________________

Smiling wins more friends than frowning

VP
Joined: 17 Jun 2008
Posts: 1397
Followers: 8

Kudos [?]: 290 [0], given: 0

### Show Tags

04 Aug 2008, 11:39
x2suresh wrote:
spriya wrote:
can we call a parallelogram with 4 side equal a rhombus?this is my doubt

yes thats correct

A parallelogram with equal sides (a = b) is called a rhombus, a parallelogram whose angles are all right angles is called a rectangle.

u get a kudo for this

Correcting my analysis
IMO D
_________________

cheers
Its Now Or Never

Senior Manager
Joined: 31 Jul 2008
Posts: 306
Followers: 1

Kudos [?]: 46 [0], given: 0

### Show Tags

04 Aug 2008, 14:37
Yeh , D should be the answer as for each of the property mentioned only a Rhombus OR Sqaure (a special case Rhombus)
fits.
Re: SQUARES and rhombus   [#permalink] 04 Aug 2008, 14:37
Display posts from previous: Sort by