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:

Here NQ and QN are tangents to left circle, and we know any line from center of a circle to the point where tangent touches the circle will make a 90 degree angle with tangent.

Following this rule angle PNQ and PMQ are 90. and angle NPM and NQM must be equal to each other hence these are also 90. PN=PM=QN=QM=radius .. given. Hence this should be a square without considering any options .. what am I doing wrong ??

Re: In the figure above, P and Q are centers of two identical [#permalink]

Show Tags

25 Jun 2013, 06:10

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Attachment:

circle.jpg [ 7.48 KiB | Viewed 1863 times ]

In the figure above, P and Q are centers of two identical circles. Is quadrilateral a square?

(1) The length of arc MN is 2pi. Clearly insufficient.

(2) The area of one circle is 16 pi --> \(\pi{r^2}=16\pi\) --> \(r=4\). Not sufficient.

(1)+(2) From (2) the circumference of each circle is \(2\pi{r}=8\pi\). So, we have that arc MN (\(2\pi\)) is 1/4th of the circumference which means that angles P and Q are 1/4*360=90 degrees. Since all 4 sides of the quadrilateral are equal and two sides are 90 degrees, then it must be a square. Sufficient.

Here NQ and QN are tangents to left circle, and we know any line from center of a circle to the point where tangent touches the circle will make a 90 degree angle with tangent.

Following this rule angle PNQ and PMQ are 90. and angle NPM and NQM must be equal to each other hence these are also 90. PN=PM=QN=QM=radius .. given. Hence this should be a square without considering any options .. what am I doing wrong ??

hi, this method of yours will not hold in all cases. let us suppose left circle passes through centre of right circle...in that case QM AND QN will not be tangent then you cant make it 90 on the basis of that. might there be other method to prove that a square..but your method will not hold true.... KUDOS if it helped. _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Here NQ and QN are tangents to left circle, and we know any line from center of a circle to the point where tangent touches the circle will make a 90 degree angle with tangent.

Following this rule angle PNQ and PMQ are 90. and angle NPM and NQM must be equal to each other hence these are also 90. PN=PM=QN=QM=radius .. given. Hence this should be a square without considering any options .. what am I doing wrong ??

hi, this method of yours will not hold in all cases. let us suppose left circle passes through centre of right circle...in that case QM AND QN will not be tangent then you cant make it 90 on the basis of that. might there be other method to prove that a square..but your method will not hold true.... KUDOS if it helped.

Where in the solution did you see that 90 degrees were deduced based on tangency? _________________

Here NQ and QN are tangents to left circle, and we know any line from center of a circle to the point where tangent touches the circle will make a 90 degree angle with tangent.

Following this rule angle PNQ and PMQ are 90. and angle NPM and NQM must be equal to each other hence these are also 90. PN=PM=QN=QM=radius .. given. Hence this should be a square without considering any options .. what am I doing wrong ??

hi, this method of yours will not hold in all cases. let us suppose left circle passes through centre of right circle...in that case QM AND QN will not be tangent then you cant make it 90 on the basis of that. might there be other method to prove that a square..but your method will not hold true.... KUDOS if it helped.

Where in the solution did you see that 90 degrees were deduced based on tangency?

hi bunuel, i dint doubted your explanation. actually i repled to the spoiler of the author (stunn3r). if possible please refer to the spoiler. SKM _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: In the figure above, P and Q are centers of two identical [#permalink]

Show Tags

25 Jun 2013, 14:22

Expert's post

shaileshmishra wrote:

Bunuel wrote:

shaileshmishra wrote:

hi, this method of yours will not hold in all cases. let us suppose left circle passes through centre of right circle...in that case QM AND QN will not be tangent then you cant make it 90 on the basis of that. might there be other method to prove that a square..but your method will not hold true.... KUDOS if it helped.

Where in the solution did you see that 90 degrees were deduced based on tangency?

hi bunuel, i dint doubted your explanation. actually i repled to the spoiler of the author (stunn3r). if possible please refer to the spoiler. SKM

My bad. Missed the text under the spoiler. _________________

Here NQ and QN are tangents to left circle, and we know any line from center of a circle to the point where tangent touches the circle will make a 90 degree angle with tangent.

Following this rule angle PNQ and PMQ are 90. and angle NPM and NQM must be equal to each other hence these are also 90. PN=PM=QN=QM=radius .. given. Hence this should be a square without considering any options .. what am I doing wrong ??

hi, this method of yours will not hold in all cases. let us suppose left circle passes through centre of right circle...in that case QM AND QN will not be tangent then you cant make it 90 on the basis of that. might there be other method to prove that a square..but your method will not hold true.... KUDOS if it helped.

That helped .. THank You Shailesh .. and forgive BUNU, he misses things sometimes .. _________________

Re: In the figure above, P and Q are centers of two identical [#permalink]

Show Tags

10 Jul 2013, 02:40

Hi Brunel,

Wont angle PMQ and PNQ be equal to 90 degrees considering the tanget properties.

Also, if we divide the quadilateral by drawing the line PQ, angle MPQ=MQP = 45 degrees as angle PMQ= 90 degree. Similarly, angle QPN will be 45 degrees, so angle P= angle MPQ+ angle QPN = 45+45=90

Re: In the figure above, P and Q are centers of two identical [#permalink]

Show Tags

10 Jul 2013, 02:45

Expert's post

Kriti2013 wrote:

Hi Brunel,

Wont angle PMQ and PNQ be equal to 90 degrees considering the tanget properties.

Also, if we divide the quadilateral by drawing the line PQ, angle MPQ=MQP = 45 degrees as angle PMQ= 90 degree. Similarly, angle QPN will be 45 degrees, so angle P= angle MPQ+ angle QPN = 45+45=90

Please correct me if wrong .

We can derive all these when we combine the statements, but not from the beginning, since we don't know whether radii are the tangents.

Re: In the figure above, P and Q are centers of two identical [#permalink]

Show Tags

31 Jul 2014, 09:33

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: In the figure above, P and Q are centers of two identical [#permalink]

Show Tags

10 Sep 2015, 08:22

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. _________________

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...