(Q30)If a polygon of 'n' sides has
1
2
 n(n-3)diagonls. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?

Answer:

Given: Number of diagonals of a polygon with n-sides =
n(n-3)
2

Number of diagonals of a given polygon = 65

i.e,
n(n-3)
2
 = 65

Where n is the number of sides of the polygon

n2 - 3n = 2× 65

n2 - 3n - 130 = 0

n2 - n + 10n - 130= 0

n(n - ) + 10(n-) = 0

(n - )(n + 10) = 0

n - = 0 or n + 10 = 0

n = or n = -10

But n can't be negative.

∴n = number of sides =

Also to check 50 as the number of diagonals of a polygon =
n(n-3)
2
 =50

n2 - 3n = 100

n2 - 3n - 100 = 0

There is no real value of n for which the above equation is satisified.

∴There cant be a polygon with 50 diagonals