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# What is the are of parallelogram ABCD? AB = BC = CD = DA = 1

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What is the are of parallelogram ABCD? AB = BC = CD = DA = 1 [#permalink]

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25 Sep 2007, 04:18
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What is the are of parallelogram ABCD?

[1] AB = BC = CD = DA = 1
[2] AC = BD = sqrt(2)

I think the answer is B, because we know the diagonal of a square is s*sqrt(2), where s = the length of a side of the square, and if we know the two diagonals of this parallelogram are equal then we know it is a square, right? Can we not assume that the side is 1 given that we know the diagonal?

The OA is: C
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25 Sep 2007, 09:03
I think you right since I don't think there is any parallelogram with equal diagonals except square.
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25 Sep 2007, 11:04
I agree.

Stmt 1: tells us that all sides are equal so this could be a square or a rhombus. Insuff

Stmt 2: tells us that both diagonals are equal so this must be a square. The diagonals of a square equal S=Sqrt(2) where S is the side, so the sides are 1 and the area is 1.
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25 Sep 2007, 11:39
The answer has to be C.

Option A - Tell you that all sides are equal, this means it could be a rhombus or a square, the only decidiing factor is angle. In square all angles are 90 Degrees

Option B - It tells you the diagnal is Sqrt( 2), We have to remember that only if all sides are equal the diagonal formulae of ASqrt(2) is applicable
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25 Sep 2007, 11:48
Bluebird wrote:
What is the are of parallelogram ABCD?

[1] AB = BC = CD = DA = 1
[2] AC = BD = sqrt(2)

I think the answer is B, because we know the diagonal of a square is s*sqrt(2), where s = the length of a side of the square, and if we know the two diagonals of this parallelogram are equal then we know it is a square, right? Can we not assume that the side is 1 given that we know the diagonal?

The OA is: C

C is correct.
It can be rhombus from stmnt 2. Only by combining stmnt 1 , we can prove its a sqaure.
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25 Sep 2007, 11:51
Thank you for all the replies. Beatgmat, your logic makes sense, but is it possible for the diagonal 1 = diagonal 2 = sqrt(2) and the sides NOT all be equal to 1?
25 Sep 2007, 11:51
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