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# PS polygon

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01 Feb 2006, 00:33
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Senior Manager
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01 Feb 2006, 01:07
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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01 Feb 2006, 01:51
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???
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01 Feb 2006, 03:31
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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01 Feb 2006, 03:43
x,y should be the angles of the polygon, otherwise the question wouldn't be solvable with given information.
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01 Feb 2006, 06:41
SunShine wrote:
How do you say that? the red part???

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 360Â°, 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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01 Feb 2006, 08:17
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
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01 Feb 2006, 13:08
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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04 Feb 2006, 08:43
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
SVP
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Posts: 1890
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04 Feb 2006, 08:44
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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