A polygon with n sides had n*(n-3)/2 diagonals. So a 20 sided polygon would have 20*17/3 = 170
There are: 0.5*(1002-(3*100)) = 4850 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 is no such polygon because a polygon must have at least 3 sides or more.
For a polygon with n sides, there would be n*(n-3)/2
A polygon with n sides had n*(n-3)/2 diagonals. So a 20 sided polygon would have 20*17/3 = 170
There are: 0.5*(1002-(3*100)) = 4850 diagonals
A polygon with n sides has 1/2*n*(n - 3) diagonals. So, for a polygon, with n = 12, there are 54 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 is no such polygon because a polygon must have at least 3 sides or more.
A polygon with n sides has n*(n - 3)/2 diagonals. So n = 23 gives 23*20/10 = 230 diagonals
For a polygon with n sides, there would be n*(n-3)/2
n(n-3)/2
A polygon with n sides has n(n-3)/2 diagonals. A heptagon has 7 sides giving it 7(7-3)/2 = 14 diagonals.
In a polygon with n sides, we have n(n-3)/2 diagonals. In a convex polygon with n sides, you can draw n-3 diagonals from each vertex, but you are counting each one twice you so you need to divide by do. That is why we have n(n-3) divided by 2
You can find the number of diagonals in a polygon using the formula n(n-3)/2, where n is the number of sides. Therefore an 11 sided polygon has 44 diagonals.
It must have 19 sides to comply with the formula:- 0.5*(192-(3*19)) = 152 diagonals