A polygon has 54 diagonals. Then,the number o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.Given that the number of diagonals is $54$,we have:

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.Given that the number of diagonals is $54$,we have:$\frac{n(n-3)}{2} = 54$$n(n-3) = 108$$n^2 - 3n - 108 = 0$Factoring the quadratic equation:$n^2 - 12n + 9n - 108 = 0$$n(n - 12) + 9(n - 12) = 0$$(n + 9)(n - 12) = 0$Since the number of sides $n$ must be positive,we have $n = 12$.Therefore,the number of sides in the polygon is $12$.

$A$ polygon has $54$ diagonals. Then,the number of its sides is

Similar Questions

$64$ small drops of mercury,each of radius $r$ and charge $q$,coalesce to form a big drop. The ratio of the surface charge density of each small drop to that of the big drop is:

What are the products formed when chlorine is passed through an aqueous hypo solution?

Which of the following orders is correct for the heat of hydrogenation of these compounds?

The maximum and minimum melting points of the first and second transition series elements,respectively,are obtained with:

If $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module