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# What is the area of parallelogram ABCD ?

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What is the area of parallelogram ABCD ? [#permalink]

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05 Apr 2011, 15:41
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What is the area of parallelogram ABCD ?

(1) AB = BC = CD = DA = 1
(2) AC = BD = $$\sqrt{2}$$

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Re: geometry, data sufficiency [#permalink]

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05 Apr 2011, 16:28
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Knesl wrote:
What is the area of parallelogram $$ABCD$$ ?

1. $$AB = BC = CD = DA = 1$$
2. $$AC = BD = \sqrt{2}$$

(C) 2008 GMAT Club - s10#1

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

why is statement 2 not sufficient? when the two diagonals are to be the same then it is possible only in case of square. Therefore, the sides are defined as well. Or am I wrong?

1. Can be square or rhombus.

2.
Diagonals are same for rectangle and square.

For square the area will be:
Area = 1*1 as the side will be 1. Diagonal is $$sqrt{2}$$, Diagonal=hypotenuse of 45-90-45 right triangle. Side= 1.

For rectangle the sides can be:
0.5, 1.12; Area = 0.56
OR
0.75, 1.2; Area = 0.9

Basically, all combination of l and w that satisfies:
l^2+w^2=2. And there are infinite such possibilities.

Combining;
We know it's a square and area is 1.

Ans: "C"
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Re: geometry, data sufficiency [#permalink]

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05 Apr 2011, 17:50
The answer is C as fluke has explained. To add a bit more, it were a square then there is no need to calculate the sides, the area can be simply 1/2 * d1 * d2.
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Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]

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19 Mar 2014, 11:43
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I have a doubt in the explanation of this question. The official ans says that all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus but this is the property of square(a parallelogram) as well...?\
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What is the area of parallelogram ABCD ? [#permalink]

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20 Mar 2014, 01:56
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swati007 wrote:
I have a doubt in the explanation of this question. The official ans says that all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus but this is the property of square(a parallelogram) as well...?

Yes, both a rhombus and a square have equal sides. From (1) we know that ABCD is a rhombus. A square is a special type of a rhombus, so from (1) ABCD is a rhombus and can be a square.

What is the area of parallelogram $$ABCD$$?

Notice that we are told that ABCD is a parallelogram.

(1) $$AB = BC =CD = DA = 1$$ --> all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to $$\frac{d_1*d_2}{2}$$ (where $$d_1$$ and $$d_2$$ are the lengths of the diagonals) or $$bh$$ (where $$b$$ is the length of the base and $$h$$ is the altitude), so we don't have enough data to calculate the area. Not sufficient.

(2) $$AC = BD = \sqrt{2}$$ --> the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.

(1)+(2) ABCD is a rectangle and a rhombus, so it's a square --> area=side^2=1^2=1. Sufficient.

Hope it's clear.
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Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]

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17 Apr 2014, 07:16
Bunuel wrote:
swati007 wrote:
I have a doubt in the explanation of this question. The official ans says that all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus but this is the property of square(a parallelogram) as well...?

Yes, both a rhombus and a square have equal sides. From (1) we know that ABCD is a rhombus. A rhombus is a special type of a square, so from (1) ABCD is a rhombus and can be a square.

What is the area of parallelogram $$ABCD$$?

Notice that we are told that ABCD is a parallelogram.

(1) $$AB = BC =CD = DA = 1$$ --> all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to $$\frac{d_1*d_2}{2}$$ (where $$d_1$$ and $$d_2$$ are the lengths of the diagonals) or $$bh$$ (where $$b$$ is the length of the base and $$h$$ is the altitude), so we don't have enough data to calculate the area. Not sufficient.

(2) $$AC = BD = \sqrt{2}$$ --> the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.

(1)+(2) ABCD is a rectangle and a rhombus, so it's a square --> area=side^2=1^2=1. Sufficient.

Hope it's clear.

HI Bunnel,

Diagonal of a square is also equals. then if both the diagonals are equal and root 2 then we have side as 1 and we can calculate the area.

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Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]

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17 Apr 2014, 07:27
Expert's post
PathFinder007 wrote:
Bunuel wrote:
swati007 wrote:
I have a doubt in the explanation of this question. The official ans says that all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus but this is the property of square(a parallelogram) as well...?

Yes, both a rhombus and a square have equal sides. From (1) we know that ABCD is a rhombus. A rhombus is a special type of a square, so from (1) ABCD is a rhombus and can be a square.

What is the area of parallelogram $$ABCD$$?

Notice that we are told that ABCD is a parallelogram.

(1) $$AB = BC =CD = DA = 1$$ --> all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to $$\frac{d_1*d_2}{2}$$ (where $$d_1$$ and $$d_2$$ are the lengths of the diagonals) or $$bh$$ (where $$b$$ is the length of the base and $$h$$ is the altitude), so we don't have enough data to calculate the area. Not sufficient.

(2) $$AC = BD = \sqrt{2}$$ --> the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.

(1)+(2) ABCD is a rectangle and a rhombus, so it's a square --> area=side^2=1^2=1. Sufficient.

Hope it's clear.

HI Bunnel,

Diagonal of a square is also equals. then if both the diagonals are equal and root 2 then we have side as 1 and we can calculate the area.

Please read the red part in my solution. Why should the sides equal to 1? Why cannot they be any numbers satisfying $$x^2+y^2=2$$?
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Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]

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17 Apr 2014, 07:32
Expert's post
Bunuel wrote:
PathFinder007 wrote:
Bunuel wrote:

Yes, both a rhombus and a square have equal sides. From (1) we know that ABCD is a rhombus. A rhombus is a special type of a square, so from (1) ABCD is a rhombus and can be a square.

What is the area of parallelogram $$ABCD$$?

Notice that we are told that ABCD is a parallelogram.

(1) $$AB = BC =CD = DA = 1$$ --> all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to $$\frac{d_1*d_2}{2}$$ (where $$d_1$$ and $$d_2$$ are the lengths of the diagonals) or $$bh$$ (where $$b$$ is the length of the base and $$h$$ is the altitude), so we don't have enough data to calculate the area. Not sufficient.

(2) $$AC = BD = \sqrt{2}$$ --> the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.

(1)+(2) ABCD is a rectangle and a rhombus, so it's a square --> area=side^2=1^2=1. Sufficient.

Hope it's clear.

HI Bunnel,

Diagonal of a square is also equals. then if both the diagonals are equal and root 2 then we have side as 1 and we can calculate the area.

Please read the red part in my solution. Why should the sides equal to 1? Why cannot they be any numbers satisfying $$x^2+y^2=2$$?

For more on this trap check the following questions:
the-circular-base-of-an-above-ground-swimming-pool-lies-in-a-167645.html
figure-abcd-is-a-rectangle-with-sides-of-length-x-centimete-48899.html
in-right-triangle-abc-bc-is-the-hypotenuse-if-bc-is-13-and-163591.html
m22-73309-20.html
if-vertices-of-a-triangle-have-coordinates-2-2-3-2-and-82159-20.html
if-p-is-the-perimeter-of-rectangle-q-what-is-the-value-of-p-135832.html
if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html
what-is-the-area-of-rectangular-region-r-105414.html
what-is-the-perimeter-of-rectangle-r-96381.html
pythagorean-triples-131161.html
given-that-abcd-is-a-rectangle-is-the-area-of-triangle-abe-127051.html
m13-q5-69732-20.html#p1176059
m20-07-triangle-inside-a-circle-71559.html
what-is-the-perimeter-of-rectangle-r-96381.html
what-is-the-area-of-rectangular-region-r-166186.html
if-distinct-points-a-b-c-and-d-form-a-right-triangle-abc-129328.html

Hope this helps.
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Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]

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17 Apr 2014, 08:01
Because diagonal of a square = site root2

now as it is given diagonals are equal and this is also property of a square . so if diagonal is root 2 then my site will be 1. and area of a square would be one.

Thanks
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Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]

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17 Apr 2014, 08:28
Expert's post
PathFinder007 wrote:
Because diagonal of a square = site root2

now as it is given diagonals are equal and this is also property of a square . so if diagonal is root 2 then my site will be 1. and area of a square would be one.

Thanks

First of all from (2) we know that ABCD is a rectangle, not necessarily a square.

Next, the fact that the diagonals equals to $$\sqrt{2}$$ does not mean that the sides must be equal to 1. The sides can be:

$$\frac{1}{2}$$ and $$\frac{\sqrt{7}}{2}$$;
$$\frac{1}{3}$$ and $$\frac{\sqrt{7}}{\sqrt{3}}$$;
...

Basically the lengths of the sides can be any positive (x, y) satisfying $$x^2+y^2=(\sqrt{2})^2$$.

Please follow the links in my post above for questions which use the same trap.
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Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]

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17 Apr 2014, 10:41
Bunuel wrote:
PathFinder007 wrote:
Because diagonal of a square = site root2

now as it is given diagonals are equal and this is also property of a square . so if diagonal is root 2 then my site will be 1. and area of a square would be one.

Thanks

First of all from (2) we know that ABCD is a rectangle, not necessarily a square.

Next, the fact that the diagonals equals to $$\sqrt{2}$$ does not mean that the sides must be equal to 1. The sides can be:

$$\frac{1}{2}$$ and $$\frac{\sqrt{7}}{2}$$;
$$\frac{1}{3}$$ and $$\frac{\sqrt{7}}{\sqrt{3}}$$;
...

Basically the lengths of the sides can be any positive (x, y) satisfying $$x^2+y^2=(\sqrt{2})^2$$.

Please follow the links in my post above for questions which use the same trap.

Clear. Thanks for your valuable input.
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What is the area of parallelogram ABCD ? [#permalink]

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02 Feb 2015, 16:27
Knesl wrote:
What is the area of parallelogram ABCD ?

(1) AB = BC = CD = DA = 1
(2) AC = BD = $$\sqrt{2}$$

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Dear Bunuel, can we say - based on statement 2- that the parallelogram could be rhombus? If the answer is not can

you tell me why?
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Re: What is the area of parallelogram ABCD ? [#permalink]

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03 Feb 2015, 03:35
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23a2012 wrote:
Knesl wrote:
What is the area of parallelogram ABCD ?

(1) AB = BC = CD = DA = 1
(2) AC = BD = $$\sqrt{2}$$

(C) 2008 GMAT Club - M13-05

Dear Bunuel, can we say - based on statement 2- that the parallelogram could be rhombus? If the answer is not can

you tell me why?

From (2) we have that ABCD is a rectangle, and if it's a square, then it becomes a rhombus too.
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Re: What is the area of parallelogram ABCD ? [#permalink]

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04 Feb 2015, 17:42
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Bunuel wrote:
swati007 wrote:
I have a doubt in the explanation of this question. The official ans says that all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus but this is the property of square(a parallelogram) as well...?

Yes, both a rhombus and a square have equal sides. From (1) we know that ABCD is a rhombus. A rhombus is a special type of a square, so from (1) ABCD is a rhombus and can be a square.

What is the area of parallelogram $$ABCD$$?

Notice that we are told that ABCD is a parallelogram.

(1) $$AB = BC =CD = DA = 1$$ --> all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to $$\frac{d_1*d_2}{2}$$ (where $$d_1$$ and $$d_2$$ are the lengths of the diagonals) or $$bh$$ (where $$b$$ is the length of the base and $$h$$ is the altitude), so we don't have enough data to calculate the area. Not sufficient.

(2) $$AC = BD = \sqrt{2}$$ --> the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.

(1)+(2) ABCD is a rectangle and a rhombus, so it's a square --> area=side^2=1^2=1. Sufficient.

Hope it's clear.

Thanks as always for your valuable and detailed explanations. However, you've mentioned that a 'rhombus is a special type of square', where as a square is a special type of rhombus. Parallelogram->Rectangle/Rhombus->Square.
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Re: What is the area of parallelogram ABCD ? [#permalink]

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04 Feb 2015, 17:47
Expert's post
aviram wrote:
Bunuel wrote:
swati007 wrote:
I have a doubt in the explanation of this question. The official ans says that all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus but this is the property of square(a parallelogram) as well...?

Yes, both a rhombus and a square have equal sides. From (1) we know that ABCD is a rhombus. A rhombus is a special type of a square, so from (1) ABCD is a rhombus and can be a square.

What is the area of parallelogram $$ABCD$$?

Notice that we are told that ABCD is a parallelogram.

(1) $$AB = BC =CD = DA = 1$$ --> all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to $$\frac{d_1*d_2}{2}$$ (where $$d_1$$ and $$d_2$$ are the lengths of the diagonals) or $$bh$$ (where $$b$$ is the length of the base and $$h$$ is the altitude), so we don't have enough data to calculate the area. Not sufficient.

(2) $$AC = BD = \sqrt{2}$$ --> the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.

(1)+(2) ABCD is a rectangle and a rhombus, so it's a square --> area=side^2=1^2=1. Sufficient.

Hope it's clear.

Thanks as always for your valuable and detailed explanations. However, you've mentioned that a 'rhombus is a special type of square', where as a square is a special type of rhombus. Parallelogram->Rectangle/Rhombus->Square.

Typo edited. Thank you.
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Re: What is the area of parallelogram ABCD ?   [#permalink] 04 Feb 2015, 17:47
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