geometry. i think? - XP Math - Forums

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 07-24-2007 #1 sunshine Guest   Posts: n/a geometry. i think? the number of diagonals of a regular polygon is subtracted from the number of sides of the polygon, and the result is zero. what is the number of sides of this polygon?
 07-24-2007 #2 wildflower Guest   Posts: n/a too drunk to answer this! sorry
 07-24-2007 #3 ..~*@\$*~.. Guest   Posts: n/a wont pentagons, hexagons, octagons etc have the same number of sides and diagnols?
 07-24-2007 #4 Scarlet Manuka Guest   Posts: n/a If there are n sides, each vertex has (n-3) diagonals leading from it (there are n-1 other vertices, two of which are its neighbours; a line to any other is a diagonal). So the number of diagonals is n (n-3) / 2 (each diagonal can be started from either end). So we have to solven - n (n-3) / 2 = 0 n (n-3) - 2n = 0 n (n - 5) = 0 n = 0 or 5, obviously 0 doesn't count. So the answer is 5.
 07-24-2007 #5 salah h Guest   Posts: n/a the number of diagonls=1/2n(n-3)0=1/2n(n-3)n-3=0n=3Remarkn is the number of sides
 07-24-2007 #6 ebayberto Guest   Posts: n/a the number of diagonals is determined by the equation n(n-3)/2 where n equals the number of sides.. then in your problem, we will be having an equation;n(n-3)/2 - n = 0n(n-3) - 2n=0n^2 - 5n = 0n=5there are 5 sides.
 07-24-2007 #7 koolriks Guest   Posts: n/a no of diagnols is n(n-1)/2-n if the number of sides is 'n'so the equation becomes n(n-1)/2-2*n=0solving this one gets n=5
 07-24-2007 #8 sanjeewa Guest   Posts: n/a n(n-3)/2 - n = 0n(n-3) - 2n=0n^2 - 5n = 0n=0 n=5there are 5 sides.
 07-24-2007 #9 norie Guest   Posts: n/a i think in every polygon the maximum diagonals you can have is 1/2 of the number of its sides. for example a square has 4 sides and its has a max. of 2 diagonals. so i think a zero answer is impossible.

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