The number of diagonals in a certain regular polygon is equal to $1/2$ times the number of sides. How many sides does this polygon have?

tomtom Jan 7, 2024

#1**+1 **

Recall the formula for the number of diagonals of a polygon

**D= n * (n-3) /2 ** where n is the number of sides and D is the number of diagonals

So you have this:

1/2 n = n * (n-3)/2 <==== solve for 'n'

1/2n = ( n^2 - 3n )/2 multiply through by 2

n = n^2 - 3n

n^2 - 4n = 0

n(n-4) = 0 shows n = 0 (throw out) or **n = 4 sides ( a quadrilateral.....square, rectangle etc) **

ElectricPavlov Jan 7, 2024

#1**+1 **

Best Answer

Recall the formula for the number of diagonals of a polygon

**D= n * (n-3) /2 ** where n is the number of sides and D is the number of diagonals

So you have this:

1/2 n = n * (n-3)/2 <==== solve for 'n'

1/2n = ( n^2 - 3n )/2 multiply through by 2

n = n^2 - 3n

n^2 - 4n = 0

n(n-4) = 0 shows n = 0 (throw out) or **n = 4 sides ( a quadrilateral.....square, rectangle etc) **

ElectricPavlov Jan 7, 2024