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

Question

If a polygon of n sides has 275 diagonals, then n is equal to (A) 25 (B) 35 (C) 20 (D) 15

Tardigrade Admin · Accepted Answer

A polygon of n sides has number of diagonals =(n(n-3)/2)=275 (given) ⇒ n2-3 n-550 =0 ⇒ n2-25 n+22 n-550=0 ⇒ n(n-25)+22(n-25)=0 ⇒ (n-25)(n+22)=0 ⇒ n=25,(∵ n=-22 is not possible)

Q. If a polygon of $n$ sides has $275$ diagonals, then $n$ is equal to

25

35

20

15

A polygon of $n$ sides has number of diagonals
$= \frac{n ( n - 3 )}{2} = 275$ (given)
$\Rightarrow n^{2} - 3 n - 550 = 0$
$\Rightarrow n^{2} - 25 n + 22 n - 550 = 0$
$\Rightarrow n (n - 25) + 22 (n - 25) = 0$
$\Rightarrow (n - 25) (n + 22) = 0$
$\Rightarrow n = 25, (∵ n = - 22$ is not possible)

Q. If a polygon of n sides has 275 diagonals, then n is equal to

25

35

20

15

A polygon of n sides has number of diagonals =2n(n−3)​=275 (given) ⇒n2−3n−550=0 ⇒n2−25n+22n−550=0 ⇒n(n−25)+22(n−25)=0 ⇒(n−25)(n+22)=0 ⇒n=25,(∵n=−22 is not possible)

Q. If a polygon of $n$ sides has $275$ diagonals, then $n$ is equal to

A polygon of $n$ sides has number of diagonals
$= \frac{n ( n - 3 )}{2} = 275$ (given)
$\Rightarrow n^{2} - 3 n - 550 = 0$
$\Rightarrow n^{2} - 25 n + 22 n - 550 = 0$
$\Rightarrow n (n - 25) + 22 (n - 25) = 0$
$\Rightarrow (n - 25) (n + 22) = 0$
$\Rightarrow n = 25, (∵ n = - 22$ is not possible)