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# Is quadrilateral ABCD a rectangle?

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Manager
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Difficulty:

35% (medium)

Question Stats:

48% (01:38) correct 52% (00:44) wrong based on 70 sessions

(1) Line segments AC and BD bisect one another.
(2) Angle ABC is a right angle.

[Reveal] Spoiler:
Rephrase: does ABCD have 1)opposite sides that are parallel and equal, 2) bisecting diagonals and 3) opposite and equal angles of 90; adjacent angles that sum to 180.

(1) Line segments AC and BD bisect one another.
Bisecting diagonals make quadrilateral ABCD a parallelogram but not necessarily a rectangle.

(2) Angle ABC is a right angle.
One right angle isn't enough to establish that quadrilateral ABCD is a rectangle.

Answer C: A paralleogram with one right angle has all right angles because opposite angles are equal. Therefore ABCD is a rectangle

Source: MGMAT question bank
[Reveal] Spoiler: OA

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Manager
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Are there diagrams to these questions?
Manager
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gottabwise wrote:
[Reveal] Spoiler:
Rephrase: does ABCD have 1)opposite sides that are parallel and equal, 2) bisecting diagonals and 3) opposite and equal angles of 90; adjacent angles that sum to 180.

(1) Line segments AC and BD bisect one another.
[Reveal] Spoiler:
Bisecting diagonals make quadrilateral ABCD a parallelogram but not necessarily a rectangle.

(2) Angle ABC is a right angle.
[Reveal] Spoiler:
One right angle isn't enough to establish that quadrilateral ABCD is a rectangle.

[Reveal] Spoiler:
Answer C: A paralleogram with one right angle has all right angles because opposite angles are equal. Therefore ABCD is a rectangle.

Source: MGMAT question bank

I posted this because of how it means you recall the properties of parallelograms.

Senior Manager
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no diagrams????
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Manager
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Statement 1: Line segments AC and BD bisect one another. This is the case for rectangle, square and rhombus.
Not suff.
Stetement 2: Angle ABC is a right angle.
Any quadrilateral can have one right angle and other different angle. Not Suff.

Combining both statements could give us either rectangle or a square. Not suff.
Manager
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Sorry for the delay folks. Diagrams were not provided with the original question.
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Statement 1: Line segments AC and BD bisect one another. This is the case for rectangle, square and rhombus.
Not suff.
Stetement 2: Angle ABC is a right angle.
Any quadrilateral can have one right angle and other different angle. Not Suff.

Combining both statements could give us either rectangle or a square. Not suff.

But isn't square a specific case of rectangle. That way i would say answer is C
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I agree with arundas, squares are technically rectangles. Rectangles aren't always squares
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OK! What if it is a right trapezoid? I mean it will have one (even two) right angles, the segments AC and BD will bisect one another and at the same time it will not be a rectagle. What in this case?
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@ mossovet814: If it is a trapezoid then the diagonals (AC and BD) won't bisect each other.

Senior Manager
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It's "С"

Last edited by Financier on 22 Aug 2010, 22:23, edited 1 time in total.
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pankajattri wrote:
@ mossovet814: If it is a trapezoid then the diagonals (AC and BD) won't bisect each other.

OK! This is a right trapezoid. Two angles are right and diagonals bisect one another. Still it is not a rectangle. I know the question is stupid... but I just want to find a flaw in my reasoning.
Attachments

trapezoid.jpg [ 10.88 KiB | Viewed 4525 times ]

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Financier
Ok I must admit I am confused. Diagonals of a trapezoid bisect each other?
a). Diagonals of quadrangle allways bisect each other.
b). Trapezoid is a specific case of quadrangle .
I don't think that the diagonals of a trapezoid bisect each other.

Financier wrote:
pankajattri wrote:
@ mossovet814: If it is a trapezoid then the diagonals (AC and BD) won't bisect each other.

Dear Pankajattri,

First of all, I strongly suggest that you review the basics of Geometry so that you could recall that:
a). Diagonals of quadrangle allways bisect each other.
b). Trapezoid is a specific case of quadrangle .

Second, the Rules of GMATclub suppose that folks do not just utter statements without any foundation,
but rather unveil what rules they are based on. Above are the statements that prove that diagonals of Trapezoid intersect each other.

Third, the correct answer is E.

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Dear Financier,

Even If I was wrong, which by the way I am not, I don't know why couldn't you find a more suitable way of posting your comment.

As for as the question is concerned; the diagonals of a PARALLELOGRAM bisect each other and not of each QUADRANGLE. All parallelograms are quadrangles but not vice versa.

I really hope that you know what does "bisection", "quadrangle" and "parallelogram" mean.

And next time please do not utter statements without any foundation.
Senior Manager
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Pankajattri,

I was wrong, excuse me and my tone. I'm soooo sorry. I messed up words "bisect" and "intersect". This happens when people study a lot. The correct answer is "C".
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Guys just wanted to add . as I had little confusion when I first saw this question --

From (1) ----> The diagonals bisect each other , ----> its a ||gm , for it to be a recrangle both diagonals should be equal.
From(2) ----> |_ ABC= 90 ----> clearly not sufficient.

When we combine both 1 & 2 ----> Its a ||gm with one angle 90. ----> that has to be a rectangle and both diagonals become automatically equivalent.
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