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Hamish
science forum beginner
Joined: 02 Jul 2005
Posts: 2
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 Posted: Sat Jul 02, 2005 1:29 am Post subject:
Internal angles of a polygon
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If polygon has > 2 sides, and is simple and closed, then the sum of it's
internal angles = (sides - 2) * 180 degree. Correct?
If it is just closed, then if the sum of the internal angles is > (sides -
2) * 180 degrees then the polygon is self-intersecting?
If it is closed, is there anyway possible that the sum of the internal
angles < (sides - 2) * 180 degrees? |
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Jim Heckman
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 121
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| Posted: Sat Jul 02, 2005 9:40 pm Post subject:
Re: Internal angles of a polygon
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On 1-Jul-2005, "Hamish" <h.dean@xtra.co.nz>
wrote in message <WSnxe.12003$U4.1506176@news.xtra.co.nz>:
| Quote: | If polygon has > 2 sides, and is simple and closed, then the sum of it's
internal angles = (sides - 2) * 180 degree. Correct?
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Yes.
| Quote: | If it is just closed, then if the sum of the internal angles is > (sides -
2) * 180 degrees then the polygon is self-intersecting?
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I think so.
| Quote: | If it is closed, is there anyway possible that the sum of the internal
angles < (sides - 2) * 180 degrees?
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No. To see this, note that if you 'walk' around the 'outside', the
'turn' angles, i.e., the supplements of the corresponding internal
angles, must sum to 360 degrees to leave you facing in the same
direction as when you started.
--
Jim Heckman |
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Henry
science forum addict
Joined: 15 May 2005
Posts: 58
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| Posted: Sat Jul 02, 2005 10:43 pm Post subject:
Re: Internal angles of a polygon
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On Sat, 2 Jul 2005 15:29:56 +1200, "Hamish" <h.dean@xtra.co.nz> wrote:
| Quote: | If polygon has > 2 sides, and is simple and closed, then the sum of it's
internal angles = (sides - 2) * 180 degree. Correct?
If it is just closed, then if the sum of the internal angles is > (sides -
2) * 180 degrees then the polygon is self-intersecting?
If it is closed, is there anyway possible that the sum of the internal
angles < (sides - 2) * 180 degrees?
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Just out of curiosity, what do you think is the sum of the internal
angles of the "quadrilateral" with sides starting at (0,0) going to
(1,0) then to (0,1) then (1,1) and finally back to (0,0)? |
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Hamish
science forum beginner
Joined: 02 Jul 2005
Posts: 2
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| Posted: Sun Jul 03, 2005 3:14 am Post subject:
Re: Internal angles of a polygon
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| Quote: | If polygon has > 2 sides, and is simple and closed, then the sum of it's
internal angles = (sides - 2) * 180 degree. Correct?
If it is just closed, then if the sum of the internal angles is
(sides -
2) * 180 degrees then the polygon is self-intersecting?
If it is closed, is there anyway possible that the sum of the internal
angles < (sides - 2) * 180 degrees?
Just out of curiosity, what do you think is the sum of the internal
angles of the "quadrilateral" with sides starting at (0,0) going to
(1,0) then to (0,1) then (1,1) and finally back to (0,0)?
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720. |
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Henry
science forum addict
Joined: 15 May 2005
Posts: 58
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| Posted: Sun Jul 03, 2005 11:57 am Post subject:
Re: Internal angles of a polygon
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On Sun, 3 Jul 2005 17:14:15 +1200, "Hamish" <h.dean@xtra.co.nz> wrote:
| Quote: | If it is closed, is there anyway possible that the sum of the internal
angles < (sides - 2) * 180 degrees?
Just out of curiosity, what do you think is the sum of the internal
angles of the "quadrilateral" with sides starting at (0,0) going to
(1,0) then to (0,1) then (1,1) and finally back to (0,0)?
720.
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Then the answer is No |
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