The equation is: 0.5*(n2-3n) = diagonals whereas n is the number of sides of the polygon
A 7 sided polygon has 14 diagonals
11
There can be 14 lines in a seven side shape * * * * * That is the total number of diagonals from ALL vertices. Not what the question asked, though. From one vertex, there can be 4. One to every other vertex except for itself and one each on either side.
They are equal.
yes a polygon does have the most number and sides because they have the most number of shapes with the most number of sides
The number of triangles formed by the diagonals of a 21 sided polygon is C(21,3) = 21x20x19/3x2x1 = 1,330.
65
The five sided Polygon has 5 diagonals
A 7 sided polygon has 14 diagonals
nope.aviImproved Answer:-33 because 1/2*(332-99) = 495
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
To find the number of sides ( n ) of a polygon given the number of diagonals, we can use the formula for the number of diagonals in a polygon: ( D = \frac{n(n-3)}{2} ). Setting ( D = 1175 ), we get the equation ( n(n-3) = 2350 ). Solving the quadratic equation ( n^2 - 3n - 2350 = 0 ) using the quadratic formula yields ( n \approx 50 ). Thus, the polygon has 50 sides.
Number of diagonals = 1/2*(n2-3n) where n = the number of sides of the polygon.
A polygon with 14 diagonals will have 7 sides
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.
By using the polygon diagonal formula or the quadratic equation formula in which in both formulae they work out that the polygon in question has 21 sides.