What is the perimeter of a regular polygon with 2 0 2 0 sides and an area of 365 square units? Write your answer to the nearest thousandths.
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So, my thinking is right!!!! :)
I don't know if what I did was right, but I assumed a 2 0 2 0 − gon as a circle!
π r 2 = 3 6 5 ⟹ r ≈ 1 0 . 7 7 8 ⟹ 2 π r ≈ 6 7 . 7 2
The formula to get the area of a regular polygon is:
A = 4 1 n s 2 cot ( n π ) n s = = the number of sides of the polygon. side length of the polygon.
⟹ 3 6 5 s s = = ≈ 4 1 ( 2 0 2 0 ) ( s 2 ) cot ( 2 0 2 0 π ) 2 0 2 0 2 3 6 5 tan ( 2 0 2 0 π ) 0 . 0 3 3 5 2 7
Multiplying the side length by 2 0 2 0 , we get the perimeter ≈ 6 7 . 7 2 5 .
Refine how?
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The array looks cramped. It doesn't look good.
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Original: a = = 2 1 6 3
Modified 1: a = = 2 1 6 3
Modified 2: a 0 = = 2 1 6 3
Code:
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@Pi Han Goh – Thank you for this.
@Pi Han Goh – What does \phantom{0} do 0
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@Mahdi Raza – Tell the LaTeX system to put an invisible object that can't be seen.
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@Pi Han Goh – Alright. Any colour as well? This is a trial 0 test \ ( \red{\text{This is a trial} \phantom{0 \text{ test}}} \ )
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@Mahdi Raza – What you did is basically correct. You should see Pall Marton's note about coloring too, I got to learn something new from his note too.
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@Pi Han Goh – I'll check it out, looks interesting and well made
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Since there are so many sides, we can assume that the regular polygon is a circle. Then the perimeter is the circumference or
2 π r = 2 π ⋅ π 3 6 5 = 2 ⋅ 3 6 5 π ≈ 6 7 . 7 2 5
The same answer up to three decimal places. So need no formula when the sides are many .