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# Is quadrilateral ABCD a rhombus?

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14 Oct 2009, 20:09
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(1) Line segments AC and BD are perpendicular bisectors of each other.

(2) AB = BC = CD = AD
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14 Oct 2009, 20:40
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(1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...
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14 Oct 2009, 20:45
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Well:
(1) True for square or for rhombus but every square is a rhombus, so sufficient
(2) Again true for square or for rhombus but every square is a rhombus, so sufficient

D
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17 Jun 2011, 17:45
1. Sufficient

diagnols are perpendicular bisectors. can only happen in a square or rhombus. Square is a kind of rhombus.

2. Sufficient

As we know its a square or rhombus. Square is a kind of rhombus.

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10 Nov 2011, 00:16
Hi guys
I'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:

(1) Line segments AC and BD are perpendicular bisectors of each other
(2) AB = BC = CD = AD

The official answer is as follows:
Statement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus
Statement 2 - A quadrilateral with four equal sides is by definition a rhombus

What I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?
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10 Nov 2011, 02:54
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mk87 wrote:
Hi guys
I'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:

(1) Line segments AC and BD are perpendicular bisectors of each other
(2) AB = BC = CD = AD

The official answer is as follows:
Statement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus
Statement 2 - A quadrilateral with four equal sides is by definition a rhombus

What I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?

Statement 1 says the diagonals are perpendicular bisectors of each other which means that they intersect each other at mid points and form 90 degree angles. It is not possible that they are in the shape of an asymmetric kite. Figure 1 is not possible. Only figure 2 is possible. In figure 2, all sides will be equal.
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Ques4.jpg [ 10.18 KiB | Viewed 10371 times ]

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

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29 Jan 2013, 06:59
gijoedude wrote:

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

(2) AB = BC = CD = AD

A rhombus is a quadrilateral with all its sides equal to each other.

1.
Imagine that the point where the bisectors meet is called X.
Then, BX = DX and AX = CX.

Using the Pythagorean Theorem you could imagine getting the hypotenuse of sides BX and CX.
Since AX = CX, then the hypotenuse of BX and AX is also as long as the previous.
As you could imagine, all the hypotenuse formed are equal, thus forming equal 4 sides. Hence, a rhombus.

SUFFICIENT.

2.

All sides are equal.

SUFFICIENT.

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13 Oct 2013, 15:10
Bunuel wrote:

Well:
(1) True for square or for rhombus but every square is a rhombus, so sufficient
(2) Again true for square or for rhombus but every square is a rhombus, so sufficient

D

Hi Bunuel

When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus?

And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus?

Is that the correct line of reasoning?
Cheers
J

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13 Oct 2013, 15:20
jlgdr wrote:
Bunuel wrote:

Well:
(1) True for square or for rhombus but every square is a rhombus, so sufficient
(2) Again true for square or for rhombus but every square is a rhombus, so sufficient

D

Hi Bunuel

When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus?

And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus?

Is that the correct line of reasoning?
Cheers
J

For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

For (2): AB = BC = CD = AD, means that ABCD is either a rhombus or square (so still a rhombus).

Hope it helps.
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06 Dec 2013, 10:10

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

But doesn't a rhombus also allow for two perpendicular bisectors? A square has two perpendicular bisectors but a square is not a rhombus (because a rhombus does not have all four angles = 90. Oh, but wait, a rhombus is a square in the same way a rectangle is a square but not the other way around. Tricky.

(2) AB = BC = CD = AD

I guess the same logic applies above, ABCD could be a square or rhombus but regardless of which one it is its still a rhombus. If the stem asked if ABCD was a square, then this would be insufficient.

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Re: Data Sufficiency - doubts [#permalink]

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08 Aug 2014, 20:36
kritim22 wrote:
Hi,

can someone help me with this question please:

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

(2) AB = BC = CD = AD

The answer it says is D- each statement is enough on its own.

Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus?

Let say AC and BD meet at O. We have OD= OB, OA= OC. If you use pithagoras theorem, you can easily see that AD= DC=BC=AB.
For example, AD^2= OA^2 + OD^2
So (1) and (2) basically say the same thing.
The definition of rhombus is a quadrilateral whose four sides have the same length. So D is the answer.
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Re: Data Sufficiency - doubts [#permalink]

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08 Aug 2014, 23:56
kritim22 wrote:
Hi,

can someone help me with this question please:

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

(2) AB = BC = CD = AD

The answer it says is D- each statement is enough on its own.

Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus?

A square is a special case of rhombus. It has all angles of 90 degrees. so proving that quadrilateral ABCD is a square is sufficient to say that it is a rhombus.

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05 Nov 2014, 02:31
Bunuel wrote:

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

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...

But Bunuel this would be true even for rectangle:
Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.
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05 Nov 2014, 04:09
honchos wrote:
Bunuel wrote:

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

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...

But Bunuel this would be true even for rectangle:
Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.

For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus.

Hope it's clear.
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06 Nov 2014, 01:17
Bunuel,

For (1), can you consider the case of a kite?

Bunuel wrote:
honchos wrote:
Bunuel wrote:

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

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...

But Bunuel this would be true even for rectangle:
Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.

For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus.

Hope it's clear.

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06 Nov 2014, 06:22
TARGET730 wrote:
Bunuel,

For (1), can you consider the case of a kite?

Yes. In the special case where all 4 sides are the same length, the kite satisfies the definition of a rhombus.
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10 Mar 2015, 19:32
This question is poor, because (1) could be right kite, which is not necessarily a rhombus or a rhombus, but (2) will always satisfy a rhombus.

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10 Mar 2015, 21:58
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TooLong150 wrote:
This question is poor, because (1) could be right kite, which is not necessarily a rhombus or a rhombus, but (2) will always satisfy a rhombus.

Actually, the question is fine.

A rhombus is a quadrilateral all of whose sides are of the same length - that's all. You could have a rhombus which also has all angles 90 which makes it a square or a rhombus in the shape of a kite. But nevertheless, if it is a quadrilateral and has all sides equal, it IS A RHOMBUS.

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

Make 2 lines - a vertical and a horizontal - which are perpendicular bisectors of each other. Make them in any way of any length - just that they should be perpendicular bisectors of each other. When you join the end points, you will get all sides equal. Think of it this way - each side you get will be a hypotenuse of a right triangle. The legs of the right triangle will have the same pair of lengths in all 4 cases. So AB = BC = CD = AD. So ABCD must be a rhombus.

(2) AB = BC = CD = AD

This statement directly tells you that all sides are equal so ABCD must be a rhombus.

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

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22 Aug 2015, 10:10
While somewhat unusual, isn't the fact that it asks if the quadrilateral is a rhombus sufficient in itself? Aren't all quadrilaterals also rhombuses and vise versa because they have 4 sides?

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

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22 Aug 2015, 10:18
gmatninja2016 wrote:
While somewhat unusual, isn't the fact that it asks if the quadrilateral is a rhombus sufficient in itself? Aren't all quadrilaterals also rhombuses and vise versa because they have 4 sides?

No. Quadrilaterals are polygons with 4 sides. A rhombus is a special quadrilateral that has all 4 sides equal.

All rhombuses are quadrilaterals but not all quadrilaterals are rhombuses. Some can even be a trapezoid or any random 4 sides structure/polygon.

Hope this helps.

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Re: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD   [#permalink] 22 Aug 2015, 10:18

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