If a polygon of n sides has 275 diagonals,the | vedclass

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

Question

(A) 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 $275$,we have

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) 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 $275$,we have:$\frac{n(n-3)}{2} = 275$$n(n-3) = 550$$n^2 - 3n - 550 = 0$Solving the quadratic equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-550)}}{2(1)}$$n = \frac{3 \pm \sqrt{9 + 2200}}{2}$$n = \frac{3 \pm \sqrt{2209}}{2}$$n = \frac{3 \pm 47}{2}$Since $n$ must be positive,$n = \frac{3 + 47}{2} = \frac{50}{2} = 25$.Thus,the number of sides $n$ is $25$.

If a polygon of $n$ sides has $275$ diagonals,then $n$ is

Similar Questions

There are $20$ straight lines in a plane such that no two of them are parallel and no three of them are concurrent. If their points of intersection are joined,then the number of new line segments formed is

The number of straight lines that can be formed by joining $8$ points on a circle is:

Three parallel straight lines $L_1, L_2$ and $L_3$ lie on the same plane. Consider $5$ points on $L_1, 7$ points on $L_2$ and $9$ points on $L_3$. Then the maximum possible number of triangles formed with vertices at these points is:

Given six line segments of lengths $2, 3, 4, 5, 6, 7$ units,the number of triangles that can be formed by these lines is

$p$ points are chosen on each of the three coplanar lines. The maximum number of triangles formed with vertices at these points is

Vedclass Test Series

Exam Paper Generator

Online Exam Module