Question 328662
The number of diagonals in a polygon is given by the equation:


d = s * (s-3) / 2


d is the number of diagonals.
s is the number of sides.


A triangle has 3 * 0 / 2 diagonals = 0
A rectangle has 4 * 1 / 2 = 3
A pentagon has 5 * 2 / 2 = 5
A hexagon has 6 * 3 / 2 = 9


The question is what polygon had 3 times as many diagonals as sides.


The formula would be d = 3 * s


Since d = s * (s-3) / 2, then this formula becomes:


s * (s-3)/2 = 3 * s


Multiply both sides of this equation by 2 to get:


s * (s-3) = 6 * s


Divide both sides of this equation by s to get:


s-3 = 6


Add 3 to both sides of this equation to get:


s = 9


This suggests that a polygon with 9 sides will have 3 times the number of diagonals as sides.


The formula for the number of diagonals is:


d = s * (s-3) / 2


Let s = 9 and this formula becomes:


d = 9 * 6 / 2 which becomes d = 54 / 2 which becomes d = 27


since 27 = 3 * 9, then this is your answer.


The polygon with 9 sides will have 3 times the number of diagonals as sides.