If a polygon has 44 diagonals, find the number of its sides.

1.	If a polygon has 44 diagonals, find the number of its sides.
Answer» A polygon of n sides has n vertices. By joining any two vertices of a polygon, we obtain either a side or a diagonal of the polygon. A number of line segments obtained by joining the vertices of a n sided polygon taken two at a time = Number of ways of selecting 2 out of n. = ⁿC₂ = \(\frac{n(n-1)}{2}\) Out of these lines, n lines are the sides of the polygon, sides can’t be diagonals. ∴ Number of diagonals of the polygon = \(\frac{n(n-1)}{2}\) - n = \(\frac{n(n-3)}{2}\) Given that a polygon has 44 diagonals. Let n be the number of sides of the polygon. \(\frac{n(n-3)}{2}\) = 44 ⇒ n(n – 3) = 88 ⇒ n² – 3n – 88 = 0 ⇒ (n + 8) (n – 11) ⇒ n = -8 (or) n = 11 n cannot be negative. ∴ n = 11 is number of sides of polygon is 11.

Answer»

A polygon of n sides has n vertices. By joining any two vertices of a polygon, we obtain either a side or a diagonal of the polygon.

A number of line segments obtained by joining the vertices of a n sided polygon taken two at a time = Number of ways of selecting 2 out of n.

= ⁿC₂

= \(\frac{n(n-1)}{2}\)

Out of these lines, n lines are the sides of the polygon, sides can’t be diagonals.

∴ Number of diagonals of the polygon = \(\frac{n(n-1)}{2}\) - n = \(\frac{n(n-3)}{2}\)

Given that a polygon has 44 diagonals.

Let n be the number of sides of the polygon.

\(\frac{n(n-3)}{2}\) = 44

⇒ n(n – 3) = 88

⇒ n² – 3n – 88 = 0

⇒ (n + 8) (n – 11)

⇒ n = -8 (or) n = 11

n cannot be negative.

∴ n = 11 is number of sides of polygon is 11.

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