The number of diagonals in a polygon with n sides is n*(n-3)/2
This is double the number of sides so n*(n-3)/2 = 2*n
That is n*(n-3) = 4n
Dividing both sides by n gives (n-3) = 4
So that n = 7.
A heptagon has 7*4/2 = 14 diagonals.
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It is: 0.5*(n2-3n) = diagonals whereas n is the number of sides of the polygon
number of diagonals = n(n-3)/2 n - number of sides of the polygon
The number of triangles formed by the diagonals of a 21 sided polygon is C(21,3) = 21x20x19/3x2x1 = 1,330.
1/2*(n2-3n) = 665 diagonals where n means number of sides
Using the formula (x)(x-3)/2 = Diagonals ; simply replace the diagonals with the number of diagonals you're given. Then, you'll havev (x)(x-3)/2 = Diagonals. Simplify it, and you'll be given x(power of 2) - 3X = (2)(Diagonals). Subtract the amount of diagonals from both sides, and you'll have x(power of 2) - 3X - 2Diagonals = 0. From there, use the quadratic formula to find the number of sides the polygon has.