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?
2007-07-11 21:24:27 · 8 answers · asked by sunshine 1 in Science & Mathematics ➔ Mathematics
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 solve
n - 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.
2007-07-11 21:29:21 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
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.
2007-07-12 01:32:17 · answer #2 · answered by norie 2 · 0⤊ 0⤋
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 = 0
n(n-3) - 2n=0
n^2 - 5n = 0
n=5
there are 5 sides.
2007-07-11 21:34:11 · answer #3 · answered by Mr. Engr. 3 · 0⤊ 0⤋
the number of diagonls=1/2n(n-3)
0=1/2n(n-3)
n-3=0
n=3
Remark
n is the number of sides
2007-07-11 21:32:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋
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=0
solving this one gets n=5
2007-07-11 21:34:48 · answer #5 · answered by koolriks 2 · 0⤊ 0⤋
n(n-3)/2 - n = 0
n(n-3) - 2n=0
n^2 - 5n = 0
n=0 n=5
there are 5 sides.
2007-07-11 22:48:32 · answer #6 · answered by sanjeewa 4 · 0⤊ 0⤋
wont pentagons, hexagons, octagons etc have the same number of sides and diagnols?
2007-07-11 21:29:16 · answer #7 · answered by Albus Percival Wulfric Brian Dumbledore™ 5 · 0⤊ 0⤋
too drunk to answer this! sorry :)
2007-07-11 21:28:08 · answer #8 · answered by wildflower 2 · 0⤊ 0⤋