Question

a polygon of n sides has n(n-3)/2 diagonals. how many sides has a polygon with 54 diagonals?

Answer 1

As per the question we can write that
$\frac{n(n - 3)}{2} = 54$
If you solve this equation you'll get a quadratic equation
$n^{2} - 3n - 108 = 0$
On solving you'll get the answer
n=12.

Hope you liked it

Answer 2

Given:

Sides of polygon = n

Number of diagonals of polygon = n(n-3)/2

To Find:

Sides of polygon with 54 diagonals.

Solution:

Number of diagonals = 54

Also, no. of diagonals = n(n-3)/2

therefore,

n(n-3)/2 = 54

n(n-3) = 108

n²-3n = 108

n²-3n - 108 = 0

n²-12n+9n-108 = 0

n(n-12)+9(n-12) = 0

(n+9)(n-12) = 0

Either, n+9 = 0;

           n = -9 {which is not possible as no side can be negative}

or, n-12 = 0

     n = 12

Hence, the no. of sides of a polygon with 54 diagonals will be 12.