Is Quadrilateral ABCD a Rectangle? 1) Diagonals of ABCD Bisect each o

Here is an example -->

Check out this trapezoid with equal diagonals




It clearly isn't a rectangle
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Main statement tells us that ABCD is a Quadrilateral..

Let's see the statements..

1) Diagonals of ABCD Bisect each other at 90 degrees
this tells us that the diagonals cut each other at middle of each other at 90°
The diagonals can be of different sizes..
So it is surely a RHOMBUS but cannot say if it is rectangle/square
Insufficient

2) Diagonals of ABCD are equal in dimension
Diagonals are equal cannot alone suffice if it's a rectangle..

Combined..
It is a rhombus from statement I and diagonals are EQUAL from statement II..
It is a square and each square is also a rectangle..
Sufficient
C
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Hi,

Surely if diagonals are EQUAL and BISECT each other, it would be rectangle or square..
But here it is just diagonals are EQUAL. Say both are 5 m but they bisect at 1 m from top, it will not be rectangle or square..
Once you draw, you will see various Quadrilateral can be made .
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Yes, it is not a parallelogram.
The reason is that the diagonals have to be equal and bisect each other and if you do the same here, it will become a rhombus[/quote]

Sorry, I meant Rectangle. So the OA has to be E.[/quote]

NO, the OA will be C.

you have not read the complete info of statement I.
1) Diagonals of ABCD Bisect each other at 90 degrees
so statement I tells us that the diagonals bisect each other. Bisect means they intersect at the centre.

so combined it has to be a rectangle - equal diagonals and bisecting each other at 90 would be a square and square is also a rectangle
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Hello chetan2u / Bunuel,
How do we distinguish when to treat a square as A SQUARE and not a rectangle.
It's true all squares are rectangles. I did reach at the conclusion that from both A and B the quadrilateral is a square. But marked "E".
Regards

hi chetan2u ...

stat2 states diagnols are equal...in a quadrialteral, only square and rectangle have equal diagnols,,,
i cant think of any other figure where diagnols ar equal
please help

ok..makes sense now,,,,thanks

i guess that shud be intersect??? bisect means dividing each other into two equal parts...
correct me if im rong,,
thanks

Statement1 is not surely a rhombus. The diagonals of a kite are also perpendicular to each other.

Posted from my mobile device

.


I think there is a difference between "bisect" and "intersect"
In a quadrilateral, if diagonals intersect each other at 90 degrees, then it could be a rhombus or a square or a kite. In first two cases we can call it a rhombus only (because square is a special kind of rhombus) but if its a kite, it cannot be called a rhombus. .
(Please note that the statement does NOT say 'intersect each other at 90' but it says bisect each other at 90'.
Diagonals of a square, a rectangle and an isosceles trapezoid are equal in dimension. In the first case (if its a square) we can call it a rhombus (square is a special kind of rhombus) but in other two cases we cannot.

Hi chetan2u

Please correct me if I'm wrong. I'm attaching a file with a figure in which diagonals are equal and cross at 90 degree yet the figure is a kite, not a parallelogram.

Yes, it is not a parallelogram.
The reason is that the diagonals have to be equal and bisect each other and if you do the same here, it will become a rhombus
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Hi chetan2u

Please correct me if I'm wrong. I'm attaching a file with a figure in which diagonals are equal and cross at 90 degree yet the figure is a kite, not a parallelogram.[/quote]

Yes, it is not a parallelogram.
The reason is that the diagonals have to be equal and bisect each other and if you do the same here, it will become a rhombus[/quote]

Sorry, I meant Rectangle. So the OA has to be E?

Why statement 1 alone is not sufficient??
Diagonals of a rectangle do not bisect at 90 deg. So statement 1 should be sufficient to decide that its not a rectangle but any other quadrilateral (may be a square)
Please help.

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A square is always a rectangle, although it may not be necessary that a rectangle is a square.
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Bunuel, GMATinsight
Should not the answer be E, as both Rhombus and Square satisfy both the conditions of 1 and 2 ?

abhishekmayank

After combining the statements, we get only a square (which is a rhombus too but is also a rectangle) hence Sufficient

You can't get any rhombus that is not a square after combining the two statements

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