Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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]

Show Tags

19 Mar 2014, 11:43

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...?\
_________________

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]

Show Tags

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.

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.

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 \(x^2+y^2=2\)?
_________________

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]

Show Tags

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.

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]

Show Tags

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.

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

Show Tags

04 Feb 2015, 17:42

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.

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.

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

Show Tags

23 Aug 2017, 09:10

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

07 Sep 2017, 20:19

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

Statement 1

Knowing htat the paralleogram has four equal sides is enough to determine that the figure is a rhombus since a rhombus is essentially an equilateral paralleogram. But we do not know the diagonal length of this rhombus and we cannot calculate the lengths of the diagonals with just the length of the sides. Insuff

Statement 2

Knowing that the parallelogram has two equal diagonals implies that the parallelogram is a rectangle... But what kind of rectangle? A square? If it we knew that that the parallelogram was a square then we could calculate the area of it by simply squaring the diagonal lengths and dividing by 2. Logically, a square IS a rhombus because the diagonals of a square can be squared and then divided by 2 in order to find the area.

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...