An n-sided polygon wil have n*(n-3)/2 diagonals.
Consider joining each vertex to every other vertex. That gives potentially n-1 vertices. However, two of these will be sides of the polygon and so not diagonals. So each vertex gives rise to (n-3) diagonals. There are n vertices in the polygon and so that gives n*(n-3) diagonals. But, this method counts each diagonal twice: once from each end and so the correct answer is n*(n-3)/2.
An 11 sided polygon has 44 diagonals.
A four-sided polygon has two diagonals.
There is no such polygon that has 95 diagonals because a 15 sided polygon has 90 diagonals and a 16 sided polygon has 104 diagonals.
A 5-sided polygon is called a pentagon. You can draw up to 5 diagonals in a pentagon.
It works out as having 5150 diagonals
An 11 sided polygon has 44 diagonals.
There are 90 diagonals in a 15-sided polygon.
A four-sided polygon has two diagonals.
There are 170 diagonals
It has 35 diagonals
There is no such polygon that has 95 diagonals because a 15 sided polygon has 90 diagonals and a 16 sided polygon has 104 diagonals.
100 sided polygon has = n(n-3)/2 diagonals = 100(100-3)/2 = 4850 diagonals
A 5-sided polygon is called a pentagon. You can draw up to 5 diagonals in a pentagon.
A 20 sided polygon has 170 diagonals
There are 275 diagonals in a 25 sided polygon
It works out as having 5150 diagonals
A 4-sided polygon has 2 diagonals. A 5-sided polygon, 5. 6-sided, 9. By induction, an n-sided polygon has 'n*(n-3)/2' diagonals. "n" is how many sides a shape has. so, (7)*(7-3)/2 now we have, 7*4/2 so, 28/2 = 14 Therefore, a 7-sided polygon would have 14 diagonals.