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Re: geometry, data sufficiency [#permalink]
05 Apr 2011, 15:28

2

This post received KUDOS

1

This post was BOOKMARKED

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.

Re: geometry, data sufficiency [#permalink]
05 Apr 2011, 16:50

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. _________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]
19 Mar 2014, 10:43

1

This post received KUDOS

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...?\ _________________

What is the area of parallelogram ABCD ? [#permalink]
20 Mar 2014, 00:56

1

This post received KUDOS

Expert's post

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.

Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]
17 Apr 2014, 06:16

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.

Answer: C.

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.

Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]
17 Apr 2014, 06:27

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.

Answer: C.

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 clarify.

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

Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]
17 Apr 2014, 06:32

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.

Answer: C.

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 clarify.

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

Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]
17 Apr 2014, 07:01

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.

Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]
17 Apr 2014, 07:28

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. _________________

Re: What is the area of parallelogram ABCD ? 1. AB = BC = CD = [#permalink]
17 Apr 2014, 09:41

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.

Re: What is the area of parallelogram ABCD ? [#permalink]
04 Feb 2015, 16:42

1

This post received KUDOS

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.

Answer: C.

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. _________________

"Hardwork is the easiest way to success." - Aviram

One more shot at the GMAT...aiming for a more balanced score.

Re: What is the area of parallelogram ABCD ? [#permalink]
04 Feb 2015, 16:47

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.

Answer: C.

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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