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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