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In the figure above, P and Q are centers of two identical circles. Is
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25 Jun 2013, 04:15
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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. (2) The area of one circle is 16 pi. 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 ??
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Re: In the figure above, P and Q are centers of two identical circles. Is
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25 Jun 2013, 13:09
stunn3r wrote: 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. (2) The area of one circle is 16 pi. 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.
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Re: In the figure above, P and Q are centers of two identical circles. Is
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25 Jun 2013, 13:14
shaileshmishra wrote: stunn3r wrote: 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. (2) The area of one circle is 16 pi. 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?
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Re: In the figure above, P and Q are centers of two identical circles. Is
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25 Jun 2013, 13:19
Bunuel wrote: shaileshmishra wrote: stunn3r wrote: 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. (2) The area of one circle is 16 pi. 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
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Re: In the figure above, P and Q are centers of two identical circles. Is
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Re: In the figure above, P and Q are centers of two identical circles. Is
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25 Jun 2013, 14:40
shaileshmishra wrote: stunn3r wrote: 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. (2) The area of one circle is 16 pi. 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 ..
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Re: In the figure above, P and Q are centers of two identical circles. Is
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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
Please correct me if wrong .



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Re: In the figure above, P and Q are centers of two identical circles. Is
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10 Jul 2013, 01:45
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. Hope it's clear.
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Re: In the figure above, P and Q are centers of two identical circles. Is
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16 Apr 2018, 09:08
Bunuel wrote: 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. Answer: C. Bunuel, I am able to follow to the point where we know the arc is 1/4 of the circles circumference, but how do you know the angles of the quadrilateral are 90 and that the sides are equal?



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Re: In the figure above, P and Q are centers of two identical circles. Is
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16 Apr 2018, 10:06
lostnumber wrote: Bunuel wrote: 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. Answer: C. Bunuel, I am able to follow to the point where we know the arc is 1/4 of the circles circumference, but how do you know the angles of the quadrilateral are 90 and that the sides are equal? Hello These are some basics of geometry. As we are given that the two circles are identical, their radii have to be same. And if you observe in the diagram, PM and PN are radii of first circle, while QM and QN are radii of second circle  so these 4 PM, PN, QM, QN must be equal. This makes this quadrilateral a rhombus. Now if you observe, QM is a tangent to the left circle at the point of contact M. And since P is centre of that circle, P must be perpendicular to PM at point M (again basics). So angle PMQ = 90. similarly angle PNQ is also = 90 and other two angles are also 90. Thus its a rhombus with all angles 90 degrees, hence its a square.



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In the figure above, P and Q are centers of two identical circles. Is
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16 Apr 2018, 12:08
amanvermagmat wrote: Hello
These are some basics of geometry. As we are given that the two circles are identical, their radii have to be same. And if you observe in the diagram, PM and PN are radii of first circle, while QM and QN are radii of second circle  so these 4 PM, PN, QM, QN must be equal. This makes this quadrilateral a rhombus.
Now if you observe, QM is a tangent to the left circle at the point of contact M. And since P is centre of that circle, P must be perpendicular to PM at point M (again basics). So angle PMQ = 90. similarly angle PNQ is also = 90 and other two angles are also 90.
Thus its a rhombus with all angles 90 degrees, hence its a square.
Thank you for the answer Aman, I'm trying to wrap my head around some of these geometry properties and struggling Let me try to walk through your explanation and see if I can follow: So, we know that P and Q are the center of the circles. Are we assuming that M and N are on the circumference of the circle based on the diagrams being drawn to scale? If so then I understand makes PM and PN radius of circle 1, and QM and QN radius of circle 2. Since the circles are identical these line segments are all identical in length, meaning we have a rhombus. Ok, I'm following so far now that you've explained it. But I'm still not able to figure out how we know that the angles are 90. Are you saying any line that is a tangent to the circle will always be perpendicular to the center? If that's true than why do we need our two additional pieces of information at all? Couldn't we establish the shape is a square by knowing the circles are equal size and that M & N are the points where the circles intersect? Thank you for helping me understand these concepts



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In the figure above, P and Q are centers of two identical circles. Is
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16 Apr 2018, 21:26
lostnumber wrote: amanvermagmat wrote: Hello
These are some basics of geometry. As we are given that the two circles are identical, their radii have to be same. And if you observe in the diagram, PM and PN are radii of first circle, while QM and QN are radii of second circle  so these 4 PM, PN, QM, QN must be equal. This makes this quadrilateral a rhombus.
Now if you observe, QM is a tangent to the left circle at the point of contact M. And since P is centre of that circle, P must be perpendicular to PM at point M (again basics). So angle PMQ = 90. similarly angle PNQ is also = 90 and other two angles are also 90.
Thus its a rhombus with all angles 90 degrees, hence its a square.
Thank you for the answer Aman, I'm trying to wrap my head around some of these geometry properties and struggling Let me try to walk through your explanation and see if I can follow: So, we know that P and Q are the center of the circles. Are we assuming that M and N are on the circumference of the circle based on the diagrams being drawn to scale? If so then I understand makes PM and PN radius of circle 1, and QM and QN radius of circle 2. Since the circles are identical these line segments are all identical in length, meaning we have a rhombus. Ok, I'm following so far now that you've explained it. But I'm still not able to figure out how we know that the angles are 90. Are you saying any line that is a tangent to the circle will always be perpendicular to the center? If that's true than why do we need our two additional pieces of information at all? Couldn't we establish the shape is a square by knowing the circles are equal size and that M & N are the points where the circles intersect? Thank you for helping me understand these concepts Hello Yes, your first para is completely correct. So you are perfectly clear till the point that we have a rhombus here. As per your second para, yes, you have raised a wonderful point. Since we already know that 4 sides of this quadrilateral are equal (both circles being identical so radii equal) and two angles are already 90 degrees (radius being perpendicular to tangent at point of contact), so we can already establish that this quadrilateral is a square. Infact, we dont need the two statements given Kudos for this point. (either that or I am missing something here)



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In the figure above, P and Q are centers of two identical circles. Is
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17 Apr 2018, 01:48
amanvermagmat wrote: lostnumber wrote: amanvermagmat wrote: Hello
These are some basics of geometry. As we are given that the two circles are identical, their radii have to be same. And if you observe in the diagram, PM and PN are radii of first circle, while QM and QN are radii of second circle  so these 4 PM, PN, QM, QN must be equal. This makes this quadrilateral a rhombus.
Now if you observe, QM is a tangent to the left circle at the point of contact M. And since P is centre of that circle, P must be perpendicular to PM at point M (again basics). So angle PMQ = 90. similarly angle PNQ is also = 90 and other two angles are also 90.
Thus its a rhombus with all angles 90 degrees, hence its a square.
Thank you for the answer Aman, I'm trying to wrap my head around some of these geometry properties and struggling Let me try to walk through your explanation and see if I can follow: So, we know that P and Q are the center of the circles. Are we assuming that M and N are on the circumference of the circle based on the diagrams being drawn to scale? If so then I understand makes PM and PN radius of circle 1, and QM and QN radius of circle 2. Since the circles are identical these line segments are all identical in length, meaning we have a rhombus. Ok, I'm following so far now that you've explained it. But I'm still not able to figure out how we know that the angles are 90. Are you saying any line that is a tangent to the circle will always be perpendicular to the center? If that's true than why do we need our two additional pieces of information at all? Couldn't we establish the shape is a square by knowing the circles are equal size and that M & N are the points where the circles intersect? Thank you for helping me understand these concepts Hello Yes, your first para is completely correct. So you are perfectly clear till the point that we have a rhombus here. As per your second para, yes, you have raised a wonderful point. Since we already know that 4 sides of this quadrilateral are equal (both circles being identical so radii equal) and two angles are already 90 degrees (radius being perpendicular to tangent at point of contact), so we can already establish that this quadrilateral is a square. Infact, we dont need the two statements given Kudos for this point. (either that or I am missing something here) amanvermagmat lostnumberwe cannot assume that MQ is perpendicular to PM. MQ will be perpendicular to PM only when distance between both identical circle's centre is \(\sqrt{2}\) times their radius. At only PQ = \(\sqrt{2}.r\) Quad will be square, otherwise it will be rhombus.
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In the figure above, P and Q are centers of two identical circles. Is
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17 Apr 2018, 10:21
Princ wrote: Hello Yes, your first para is completely correct. So you are perfectly clear till the point that we have a rhombus here. As per your second para, yes, you have raised a wonderful point. Since we already know that 4 sides of this quadrilateral are equal (both circles being identical so radii equal) and two angles are already 90 degrees (radius being perpendicular to tangent at point of contact), so we can already establish that this quadrilateral is a square. Infact, we dont need the two statements given Kudos for this point. (either that or I am missing something here) we cannot assume that MQ is perpendicular to PM. MQ will be perpendicular to PM only when distance between both identical circle's centre is \(\sqrt{2}\) times their radius. At only PQ = \(\sqrt{2}.r\) Quad will be square, otherwise it will be rhombus.[/quote] Hello PrincThis is a very basic theorem of tangent, that a tangent is perpendicular to the radius at point of contact.



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In the figure above, P and Q are centers of two identical circles. Is
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17 Apr 2018, 10:41
Actually problem is coming due to "reveal Spoiler" section under question . In question stem , it is not specified that QM is tangent to circle or not. Please check below image for case when PQ>√2.r. Sent from my XT1068 using GMAT Club Forum mobile app
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Re: In the figure above, P and Q are centers of two identical circles. Is
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18 Apr 2018, 03:53
Princ wrote: Actually problem is coming due to "reveal Spoiler" section under question . In question stem , it is not specified that QM is tangent to circle or not. Please check below image for case when PQ>√2.r. Sent from my XT1068 using GMAT Club Forum mobile appThank you Princ I did not look at the 'reveal spoiler' but I just assumed that QM will be tangent. Thats why my perspective of the statements not needed.




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