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SVP
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0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
Please view the attachment for the problem.
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Senior Manager
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The sum of al the internal angles in anpolygon =(n-2)*180
in the polygon mentioned in the question, all sides are angles are same
that means x+y = 80, or x=y=40
so [(n-2)*180]/n = 80/2
n= 18/7 ???????
none of the answer choices are there!
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CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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SunShine wrote: The sum of al the internal angles in anpolygon =(n-2)*180
in the polygon mentioned in the question, all sides are angles are same
that means x+y = 80, or x=y=40
so [(n-2)*180]/n = 80/2
n= 18/7 ???????
none of the answer choices are there!
X and Y are not the interior angles of the ploygon. But they are angles with the paper.
Let sum of two shown angles of the polygon = a
Then a + x+ + y = 360
So a = 360-80 = 280
i.e each angle of polygon is 140 degrees.
So (n-2) * 180 = n * 140
which yields n = 9
Am I right laxie???
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
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Director
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may be i am dumb , but i can not figure out which angles are X and Y. Are the ones next to the paper or the angles X and Y are opposite
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Director
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x,y should be the angles of the polygon, otherwise the question wouldn't be solvable with given information.
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Director
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SunShine wrote: How do you say that? the red part??? allabout wrote: x+y=80
all angles being equal
x=40
A polygon needs a total of 360 degrees
360/40=9 Sorry, I'll erase my point because it's just a mess.
Only quadrangles have a angle sum of 360°, according to the formula (n-2)180. But it's funny that, while calculating with the angle sum of a quadrangle, I haven't even noticed that this supposition contradicts the result with n=9.
Sorry that I confused you
Last edited by allabout on 01 Feb 2006, 09:04, edited 1 time in total.
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VP
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I agree with ps_dahiya's solution. X and Y have to be angles made with the paper, otherwise we will get n = 18/7.
2*a+X+Y = 360 (we know that it is a quadrilateral)
X+Y = 80
=> a = 140.
Sum of all angles = (n-2)*180
Each angle = (n-2)*180/n
(n-2)*180/n = 140
Hence n = 9
_________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
- Bernard Edmonds
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VP
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X+Y = 80.
Since all the angles of the polygon as same, the visible angles amount to
360 - ( X + Y ) = 280 => Each angle is 140, where 360 is the sum of angles of the visible polygon (4-2)180 = 360.
Now for the entire polygon including the hiddent part we get,
(n-2)180 = 140 * number of angles.
Only n = 9 gives 7 as factor for 140...SUFF.
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SVP
Joined: 24 Sep 2005
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ps_dahiya wrote: SunShine wrote: The sum of al the internal angles in anpolygon =(n-2)*180
in the polygon mentioned in the question, all sides are angles are same
that means x+y = 80, or x=y=40
so [(n-2)*180]/n = 80/2
n= 18/7 ???????
none of the answer choices are there! X and Y are not the interior angles of the ploygon. But they are angles with the paper. Let sum of two shown angles of the polygon = a Then a + x+ + y = 360 So a = 360-80 = 280 i.e each angle of polygon is 140 degrees. So (n-2) * 180 = n * 140 which yields n = 9 Am I right laxie??? Good job, buddy!!! 
9 is OA
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SVP
Joined: 24 Sep 2005
Posts: 1913
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55
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ps_dahiya wrote: SunShine wrote: The sum of al the internal angles in anpolygon =(n-2)*180
in the polygon mentioned in the question, all sides are angles are same
that means x+y = 80, or x=y=40
so [(n-2)*180]/n = 80/2
n= 18/7 ???????
none of the answer choices are there! X and Y are not the interior angles of the ploygon. But they are angles with the paper. Let sum of two shown angles of the polygon = a Then a + x+ + y = 360 So a = 360-80 = 280 i.e each angle of polygon is 140 degrees. So (n-2) * 180 = n * 140 which yields n = 9 Am I right laxie??? Good job, buddy!!!
9 is OA
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