Do these formulas for area and perimeter work?
Given the radius (r) and the amount of sides (s) a polygon has would area = (s*r^2*sin(360/s))/2 and would the perimeter = 2*r*s*sin(360/s)?
Your question needs more information before we can do anything. First, radius doesn’t really apply to polygons. You may be referring to a regular polygon inscribed in a circle or a regular polygon circumscribed in a circle.
Either way, the sin(360/n) factors look illogical. Whatever the question really calls for, half angles would be involved, so sin(180/n) is probably needed.
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Looking at your question again a few hours later, I think you are asking about regular polygons inscribed in a circle. If you break the polygon into s isosceles triangles with vertices at the center and bases corresponding to the sides of the polygon, you can figure out the area and perimeter.
I get area = s * (r * sin (180/s) * (r * cos(180/s) =
s * r² * sin (180/s) * cos (180/s).
However, from here you can use the formula sin(2x) = 2 * sin(x) * cos(x) to simplify it further. Let x = 180/s and you get sin(360/s) = 2 * sin(180/s) * cos(180/s), so sin(180/s) * cos(180/s) = sin(360/s) / 2. Substitute that into the above formula, and you get area =
s * r² * sin(360/s) / 2 — lo and behold, there’s your formula!
The perimeter looks a little different, though. For that I get:
2 * r * s * sin(180/s)
In other words, I get 180/s, not 360/s for the variable in the sine function — no double angle substitution this time.
Your question needs more information before we can do anything. First, radius doesn’t really apply to polygons. You may be referring to a regular polygon inscribed in a circle or a regular polygon circumscribed in a circle.
Either way, the sin(360/n) factors look illogical. Whatever the question really calls for, half angles would be involved, so sin(180/n) is probably needed.
* * * * *
Looking at your question again a few hours later, I think you are asking about regular polygons inscribed in a circle. If you break the polygon into s isosceles triangles with vertices at the center and bases corresponding to the sides of the polygon, you can figure out the area and perimeter.
I get area = s * (r * sin (180/s) * (r * cos(180/s) =
s * r² * sin (180/s) * cos (180/s).
However, from here you can use the formula sin(2x) = 2 * sin(x) * cos(x) to simplify it further. Let x = 180/s and you get sin(360/s) = 2 * sin(180/s) * cos(180/s), so sin(180/s) * cos(180/s) = sin(360/s) / 2. Substitute that into the above formula, and you get area =
s * r² * sin(360/s) / 2 — lo and behold, there’s your formula!
The perimeter looks a little different, though. For that I get:
2 * r * s * sin(180/s)
In other words, I get 180/s, not 360/s for the variable in the sine function — no double angle substitution this time.
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