Question

a polygon of n sides has n(n-3)/2 diagonals. how many sides has a polygon with 324 diagonals

Answer 1

n (n -3) is equals to 324 x 2
n^2-3n-648=0
n^2-27n+24n-648=0
n(n-27)+24(n-7)=0
n=27&-24
so n=27

Answer 2

Concept

A polygon is  a simple figure bounded by straight lines. Poly means many in Greek and gon means angle. The simplest polygon is a triangle whose three sides and three angles add up to 180 degrees.

Now, for an 'n' sided polygon, the number of diagonals can be obtained with the  formula:

number of diagonals = n(n-3)/2

Given

We have been given that a polygon of n sides has $\frac{ n(n-3)}{2}$ diagonals.

Find

We are asked to determine the sides of the polygon with 324 diagonals .

Solution

It is given diagonals of the polygon 324.

According to the formula of diagonal

$\frac{ n(n-3)}{2}=324\\n(n-3)=324\times2\\n^2-3n=648\\n^2-3n-648=0\\n^2-(27-24)n-648=0\\n^2-27n+24n-648=0\\n(n-27)+24(n-27)=0\\(n+24)(n-27)=0$

Hence, the two values of n are -24 and 27 .

As sides can't be negative so the sides of polygon with 324 diagonals is  27 .

Therefore , 27 sides has a polygon with 324 diagonals .

#SPJ3