SOLUTION: For this exercise, consider a circle of radius 1, and corresponding inscribed and circumscribed polygons with the number of sides n = 3, 4, 5, 6, and 8. For each n = 3, 4, 5, 6 &

Algebra ->  Polygons -> SOLUTION: For this exercise, consider a circle of radius 1, and corresponding inscribed and circumscribed polygons with the number of sides n = 3, 4, 5, 6, and 8. For each n = 3, 4, 5, 6 &      Log On


   

Question 739054: For this exercise, consider a circle of radius 1, and corresponding inscribed and circumscribed polygons with the number of sides n = 3, 4, 5, 6, and 8.
For each n = 3, 4, 5, 6 & 8, what are the perimeters of the inscribed and circumscribed polygons with n sides?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
NOTE:
I do not know what is expected of you because I do not know if this is homework for fifth grade, for college, or somewhere in between. I assume that using trigonometric functions is acceptable. I suspect that the intention is showing (by examples) how the perimeter approaches the circumference of a circle as the number of sides increases. (Note how 2pi=about6.28 is always in between the perimeter for the inscribed and circumscribed polygons, an how both approach 2pi=about6.28 as the number of sides increases)

THE SOLUTION:
We can only solve the problem if those are regular (symmetrical) polygons, because otherwise the perimeter would change with the shape.
Connecting the vertices to the center, we can split a polygon with n sides into n congruent isosceles triangles. If we draw the altitudes, we split each isosceles triangle into two congruent right triangles, for a total of 2n right triangles.
O is the center of the circle and the polygon; AB is the side of the polygon
The angles AOP and POB measure pi%2Fn (or 180%5Eo%2Fn if you prefer degrees rather than radians)

FOR AN INSCRIBED POLYGON:
A and B are points on the circle and OA and OB are radii with length
OA=OB=1
The length of the side of the polygon is
AB=AP%2BPB=2%2APB=2%2A%28OB%2Asin%28pi%2Fn%29%29=2%2A1%2Asin%28pi%2Fn%29=2sin%28pi%2Fn%29
and the perimeter is
highlight%28P%28n%29=2n%2Asin%28pi%2Fn%29%29
P%283%29=6sin%28pi%2F3%29=6sin%2860%5Eo%29=6%28sqrt%283%29%2F2%29=3sqrt%283%29=about5.20
P%284%29=8sin%28pi%2F4%29=8sin%2845%5Eo%29=8%28sqrt%282%29%2F2%29=4sqrt%282%29=about5.66
P%285%29=10sin%28pi%2F5%29=10sin%2836%5Eo%29=about5.88
P%286%29=12sin%28pi%2F6%29=12sin%2830%5Eo%29=12%281%2F2%29=6
P%288%29=16sin%28pi%2F8%29=16sin%2822.5%5Eo%29=about6.12

FOR A CIRCUMSCRIBED POLYGON:
P is on the circle, OP is a radius with lenght OP=1
The length of the side of the polygon is
AB=AP%2BPB=2%2APB=2%2A%28OP%2Atan%28pi%2Fn%29%29=2%2A1%2Atan%28pi%2Fn%29=2tan%28pi%2Fn%29
and the perimeter is
highlight%28P%28n%29=2n%2Atan%28pi%2Fn%29%29
P%283%29=6tan%28pi%2F3%29=6tan%2860%5Eo%29=6sqrt%283%29=6sqrt%283%29=about10.39
P%284%29=8tan%28pi%2F4%29=8tan%2845%5Eo%29=8%2A1=8
P%285%29=10tan%28pi%2F5%29=10tan%2836%5Eo%29=about7.27
P%286%29=12tan%28pi%2F6%29=12tan%2830%5Eo%29=12%28sqrt%283%29%2F3%29=4sqrt%283%29=about6.93
P%288%29=16tan%28pi%2F8%29=16tan%2822.5%5Eo%29=about6.63