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polygon with n sides is 180(n – 2) degrees

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by arora007 » Wed Dec 29, 2010 3:20 am
The sum of the interior angles of any polygon with n sides is 180(n - 2) degrees. If the sum of the interior angles of polygon P is three times the sum of the interior angles of quadrilateral Q, how many sides does P have?
(A) 6
(B) 8
(C) 10
(D) 12
(E) 14

[spoiler]the answer could be both B & E, but why only B is given the answer ?[/spoiler]
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by shovan85 » Wed Dec 29, 2010 3:26 am
Just put the formula and get the answer... :)

180(p-2) = 3*360 => p = 8, where p is number of sides of the polygon

Can u plz tell how did u get 14?
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by arora007 » Wed Dec 29, 2010 3:31 am
shovan85 wrote:Just put the formula and get the answer... :)

180(p-2) = 3*360 => p = 8, where p is number of sides of the polygon

Can u plz tell how did u get 14?
180(14-2) = 180*12

now when we divide by 3 we get

180 * 4 = 180(6-2) thus n=6, so I believe E also fits the list.
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by shovan85 » Wed Dec 29, 2010 4:06 am
arora007 wrote:
180 * 4 = 180(6-2) thus n=6, so I believe E also fits the list.
But when n=6 then it is a Hexagon not a Quadrilateral.
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by diebeatsthegmat » Wed Dec 29, 2010 6:54 pm
arora007 wrote:
shovan85 wrote:Just put the formula and get the answer... :)

180(p-2) = 3*360 => p = 8, where p is number of sides of the polygon

Can u plz tell how did u get 14?
180(14-2) = 180*12

now when we divide by 3 we get

180 * 4 = 180(6-2) thus n=6, so I believe E also fits the list.

nope, babe, it should be B
180(n-2)=3*360
n-2=6 so n = 8
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by arora007 » Thu Dec 30, 2010 1:41 am
shovan85 wrote:
arora007 wrote:
180 * 4 = 180(6-2) thus n=6, so I believe E also fits the list.
But when n=6 then it is a Hexagon not a Quadrilateral.
Thanks dude... quadrilateral(4 sided polygon). How much more silly can it get ?
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