+1911 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?
+37146 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)
+37146 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)
