SOLUTION: Let x represent the side length of a square. Find a regular polygon with side length x whose perimeter is twice the perimeter of the square. Find a regular polygon with side length
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Question 987678: Let x represent the side length of a square. Find a regular polygon with side length x whose perimeter is twice the perimeter of the square. Find a regular polygon with side length x whose perimeter is three times the length of the square.
I just can't figure out how to calculate that when I don't know what type of polygon and how many sides it has?? Thanks for any help
Found 2 solutions by solver91311, macston:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
If the side lengths are the same, the only way for a polygon to have a perimeter twice as large as the square is for the polygon to have twice as many sides as the square. There are 4 sides on a square, so the polygon with twice the perimeter must have how many sides?
John
My calculator said it, I believe it, that settles it
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
x=length of side; n=number of sides
perimeter=(number of sides)(length of side)
Perimeter=nx
perimeter(square)=4x
Regular polygon with perimeter 2 times square:
perimeter=2 times perimeter of square=2(4x)=8x
Perimeter=nx
8x=nx
8=n Number of sides=8
ANSWER 1: The polygon is an octagon.
.
length of square=x
perimeter=3x
perimeter=nx
3x=nx
n=3 Number of sides=3
ANSWER 2: The polygon is a triangle.
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