SOLUTION: What polygon has 3 times as many diagonals as sides?

Question 328662: What polygon has 3 times as many diagonals as sides?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.

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