Question

If a polygon of ‘n’ sides has 1/2 n (n−3) diagonals. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?

Answer 1

Number of diagonals = n(n - 3)/2
n(n - 3) = 2 × 65
⇒ n² - 3n = 130
⇒ n² - 3n - 130 = 0
Performing factorization we get:
⇒ n² - 13n + 10n - 130 = 0
⇒ n(n - 13) + 10(n - 13) = 0
⇒ (n + 10)(n - 13) = 0
n = 13, - 10
Since no. of sides cannot be negative so
Number of Sides = 13


When No. of Diagonals is 50
n(n - 3) = 50 × 2
⇒ n² - 3n - 150 = 0
Discriminant = b² - 4ac = (9 - 4 × 1 × ( - 150)) = 609
Since 609 is not a perfect square so n can never be a whole number.
Hence 50 diagonals are not possible.

Answer 2

In the attachments I have answered   this problem----------------------------------     Concept:--------------------------------------- The number of diagonals of n sided    polygon is n(n-3)/2------------------------      ---------------------------------------------------- See the attachment for detailed   solution.