(35*(35-3))/2
(35*32)/2
1120/2
560 diagonals
The number of diagonals is n(n-1)/2 - n substitute n=35 35(34)/2 - 35 = 560
100 diagonals * * * * * No, it is 0.5*10*(10-3) = 35
35 sides
It has 35 sides.
octagon
It will have ten sides
It has 10 sides because using the formula 0.5*(102-30) = 35 diagonals
if a polygon has 560 diagnose how many vertices does it have
It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
It has 35 diagonals
There are 560 diagonals by using the diagonal formula
A diagonal of a polygon is a segment drawn from one vertex to another non-adjacent vertex in a polygon. This leaves 32 diagonals that can be drawn from one vertex in a 35 sided polygon.
54 Diagonals. * * * * * A polygon with n sides has 1/2*n*(n-3) diagonals. So a decagon would have 1/2*10*7 = 35 diagonals. How the Community answer got 54 is anybody's guess!
The number of diagonals is n(n-1)/2 - n substitute n=35 35(34)/2 - 35 = 560
100 diagonals * * * * * No, it is 0.5*10*(10-3) = 35
1oo diagonals, because 1 side has 10 diagonals so 10 sides (10x10) = 100 diagonals! This is the correct answer..... a decagon has 35 diagonals. The equation for the this is D=n(n-3)/2 meaning diagonals = number of sides (N) multiplied by number of sides minus 3. and that number over 2. D=10(10-3)/2
35 diagonals