It is not possible to give a formula. This is because poyhedra are not unique shapes. For example, there are 257 topologically different convex octahedra!
A hexahedron (six faces) can be a triangular bipyramid (two triangular pyramids stuck base-to-base). This has only one diagonal in its interion. A hexahedron can also be a pentagon based pyramid which has 5 diagonals, on the surface of the base. Or it can be a cube (or variant) and that has 16 diagonals (12 on the faces and 4 in the interior). There are 4 more convex hexahedra topologically different from these.
The formula is: 0.5*(n2-3n) = diagonals whereas n is the number of sides of the polygon
1/2*(n2-3n) = number of diagonals Rearranging the formula: n2-3n-(2*diagonals) = 0 Solve as a quadratic equation and taking the positive value of n as the number of sides.
For a 9 sided polygon it is: 0.5*(92-(3*9)) = 27 diagonals
1/2*(n2-3n) = diagonals when n equals number of sides.
The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon
Number of sides - 2
The formula is V-E+F=2 and it tells us that if we take the number of vertices a polyhedron has and subtract the number of edges and then add the number of faces, that result will always be 2.
The formula is: 0.5*(n2-3n) = number of diagonals when n is the number of sides of the polygon
It is: 0.5*(n2-3n) = diagonals whereas n is the number of sides of the polygon
The formula is: 0.5*(n2-3n) = diagonals whereas n is the number of sides of the polygon
Total diagonals formula: 0.5*(n^2 -3n) whereas n is the number of sides of the polygon
n side polygon has n(n-3)/2 diagonals therefore a hexagon has 9 diagonals
1/2*(n2-3n) = number of diagonals Rearranging the formula: n2-3n-(2*diagonals) = 0 Solve as a quadratic equation and taking the positive value of n as the number of sides.
It is: 0.5*(n2-3n) = diagonals whereas 'n' is the number of sides of the polygon
For a 9 sided polygon it is: 0.5*(92-(3*9)) = 27 diagonals
1/2*(n2-3n) = diagonals when n equals 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.