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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]
25 Jun 2013, 05:10

1

This post received KUDOS

Expert's post

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Attachment:

circle.jpg [ 7.48 KiB | Viewed 1631 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]
25 Jun 2013, 13:22

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]
10 Jul 2013, 01:40

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]
10 Jul 2013, 01:45

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]
31 Jul 2014, 08:33

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Re: In the figure above, P and Q are centers of two identical [#permalink]
10 Sep 2015, 07:22

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).

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