2006-09-11 11:45:25 · 6 answers · asked by Lizz 1 in Science & Mathematics ➔ Mathematics
This is a problem where if you remember the formula, it's a snap---if you don't it can take a while.
The formula is n(n-3)/2 which means 35 times 32 divided by 2 = 560 diagonals
2006-09-11 11:52:20 · answer #1 · answered by MollyMAM 6 · 0⤊ 0⤋
Pick a vertex on the polygon, how many diagonals pass from it?There are 34 other vertices 2 of them are neighbors so there passes 32 diagonals, and two edges from that vertex. So diagonals+edges for that vertex equals 34. Pick another vertex, there passes 34 lines from that too but one of the lines connects this vertex to the original vertex so it was counted in the original 34 lines. So there are 34+33 lines passing through these vertices including the edges and diagonals... If you continue this way you get
34+33+32+...+0
for the number of edges and diagonals of a polygon with 35 edges. From the Gauss formula
34+33+32+...+0=(34)(35)/2=595
Among this 595 lines 35 are the edges, so the total number of diagonals is 595-35=560
2006-09-11 12:05:51 · answer #2 · answered by firat c 4 · 0⤊ 0⤋
formula for no of diagonals = n ( n-3) /2 the place n = no of aspects For 35 aspects = 35 X 32 /2 = 35X sixteen = 560 answer 3sides = 3 ( 3-3) /2 = 0 4 = 4( a million) /2 = 2 5 = 5 (2) /2 = 5
2016-12-15 06:22:23 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Depends on the shape of the polygon.
2006-09-11 14:23:01 · answer #4 · answered by · 5 · 0⤊ 0⤋
A pentagon has five diagonals
Hexagon, 9 diagonals
sorry, name of 7 sided too hard to recall!!
Octagon, 20
Just not less than n!!!!!
2006-09-11 12:07:55 · answer #5 · answered by bubsir 4 · 0⤊ 0⤋
Right on, MollyMAM!!! I've forgotten that formula - give her the TEN!!!
2006-09-11 11:55:17 · answer #6 · answered by Anonymous · 0⤊ 0⤋