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

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Joined: 11 May 2008
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Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD  [#permalink]

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03 Aug 2008, 18:21

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

(2) AB = BC = CD = AD

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Manager
Joined: 15 Jul 2008
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03 Aug 2008, 18:26
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
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03 Aug 2008, 19:02
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
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03 Aug 2008, 19:04
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
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Current Student
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03 Aug 2008, 19:14
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?
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04 Aug 2008, 01:24
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
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Joined: 15 Jul 2008
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04 Aug 2008, 04:08
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.
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04 Aug 2008, 11:49
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.
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Joined: 17 Jun 2008
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04 Aug 2008, 12:06
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?
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04 Aug 2008, 12:08
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.
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04 Aug 2008, 12:21
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
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04 Aug 2008, 12:28
1
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.

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04 Aug 2008, 12:39
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
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Senior Manager
Joined: 31 Jul 2008
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04 Aug 2008, 15:37
Yeh , D should be the answer as for each of the property mentioned only a Rhombus OR Sqaure (a special case Rhombus)
fits.

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Re: SQUARES and rhombus &nbs [#permalink] 04 Aug 2008, 15:37
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# Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD

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