20 sides implies 20 corners.
To ask how many ways there are of connecting each corner with _every_ other corner is the same as asking how many ways you can choose two items from 20. There is a formula for calculating that number. However, we must omit 20 of these ways because are the sides of the polygon, not diagonals.
Hence, the number we seek is "20 choose 2" - 20 = 190 - 20 = 170
To calculate "20 choose 2" easily use
* * * * *
Convincing but wrong. Joining a vertex to either of the two adjoining vertices gives a side of the polygon, NOT a diagonal. So the number we seek is NOT 20 choose 2. What we want is 20 seek any of the 20-3 vertices (not itself and not the two adjacent ones). Next, the suggested method would count each diaginal twice - once from each end. So we need to divide the result by 2.
The correct answer for an n-gon is 0.5*n*(n-3)
When n = 20 this gives 0.5*20*17 = 170
A polygon with n sides had n*(n-3)/2 diagonals. So a 20 sided polygon would have 20*17/3 = 170
Such a polygon does not exist. A polygon with n sides has 0.5*n*(n-3) diagonals If there are 10 diagonals then 0.5*n*(n-3) = 10 which requires n2 - 3n - 20 = 0 which has no integer roots.
It has 170 diagonals by using the diagonal formula
20 sides and 170 diagonals
decagon
A polygon with n sides had n*(n-3)/2 diagonals. So a 20 sided polygon would have 20*17/3 = 170
20 sides 1/2*(202-60) = 170 diagonals
It has 20 diagonals
A polygon with n sides has n*(n - 3)/2 diagonals. So n = 23 gives 23*20/10 = 230 diagonals
Such a polygon does not exist. A polygon with n sides has 0.5*n*(n-3) diagonals If there are 10 diagonals then 0.5*n*(n-3) = 10 which requires n2 - 3n - 20 = 0 which has no integer roots.
There are 170 diagonals
A 20 sided polygon has 170 diagonals
A regular octogon (8 sides) has 20 diagonals. An irregular octogon also has 20 diagonals, but only if all its points point outwards.
Sides: 8 Diagonals : 20
170
20
It is 18 diagonals