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Is quadrilateral ABCD a rhombus?  [#permalink]

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Question Stats: 41% (00:50) correct 59% (00:52) wrong based on 1123 sessions

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Is quadrilateral ABCD a rhombus?

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

(2) AB = BC = CD = AD
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Math Expert V
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Re: rhombus?  [#permalink]

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6
5
Thought about this again: D it is.

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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Re: rhombus?  [#permalink]

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1
6
Answer D.

(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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Rhombus Vs. Kite  [#permalink]

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

Is quadrilateral ABCD a Rhombus?
(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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Re: Rhombus Vs. Kite  [#permalink]

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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:

Is quadrilateral ABCD a Rhombus?
(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.
Attachment: Ques4.jpg [ 10.18 KiB | Viewed 19689 times ]

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

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1
gijoedude wrote:
Is quadrilateral ABCD a rhombus?

(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.

Answer: D
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Re: rhombus?  [#permalink]

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Bunuel wrote:
Thought about this again: D it is.

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?
Thanks so much in advance
Cheers
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Re: rhombus?  [#permalink]

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jlgdr wrote:
Bunuel wrote:
Thought about this again: D it is.

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?
Thanks so much in advance
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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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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Bunuel wrote:
Answer D.

(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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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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honchos wrote:
Bunuel wrote:
Answer D.

(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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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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Bunuel,

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

Bunuel wrote:
honchos wrote:
Bunuel wrote:
Answer D.

(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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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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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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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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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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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.

Answer (D)
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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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Superset
The answer to the question will be either yes or no.

Translation
In order to find the answer, we need:
1# measure of length of sides
2# angle between the diagonals
3# other properties to justify that the quadrilateral is a triangle

Statement analysis
St 1: diagonals are bisectors happens in parallelogram. Diagonals are perpendicular happens in kite. A figure common to both is a rhombus. ANSWER.

St 2: if all sides are equal. The figure will be a rhombus. ANSWER

Option D
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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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Aren't the diagonals of any parallelogram perpendicular bisectors?
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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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cgarmestani wrote:
Aren't the diagonals of any parallelogram perpendicular bisectors?

No.

The diagonals of a parallelogram always bisect (cut in half) each other but they are perpendicular to each other only if a parallelogram is a square.
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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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Bunuel wrote:
Answer D.

(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...

Hi Bunuel,

Thanks for all your help. I have a question. Why is statement 1 a rhombus and not a kite or square? I am guessing it has to do with intersecting at 90 degrees vs. perpendicular bisectors. What is the difference between the two terminologies?
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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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thinkpad18 wrote:
Bunuel wrote:
Answer D.

(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...

Hi Bunuel,

Thanks for all your help. I have a question. Why is statement 1 a rhombus and not a kite or square? I am guessing it has to do with intersecting at 90 degrees vs. perpendicular bisectors. What is the difference between the two terminologies?

A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

A line segment bisector is a line which cuts a line segment into two equal parts.

As for your other doubt, I think it is answered on the previous two pages:
https://gmatclub.com/forum/is-quadrilat ... ml#p999739
https://gmatclub.com/forum/is-quadrilat ... l#p1277833
https://gmatclub.com/forum/is-quadrilat ... l#p1438436
https://gmatclub.com/forum/is-quadrilat ... l#p1497287

Hope it helps.
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Re: Is quadrilateral ABCD a rhombus?  [#permalink]

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Bunuel wrote:
Thought about this again: D it is.

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

Just one question if you could help me.
Are not diagonals of rectangle perpendicular bisectors of each other. I know that they are not equal but they are at 90 degrees to each other and divide the diagonals in equal halves. Re: Is quadrilateral ABCD a rhombus?   [#permalink] 17 Jun 2018, 01:50

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