2020-sided Polygon

Geometry Level 2

What is the perimeter of a regular polygon with 2020 2020 sides and an area of 365 square units? Write your answer to the nearest thousandths.


The answer is 67.725.

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3 solutions

Chew-Seong Cheong
Jul 24, 2020

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 365 π = 2 365 π 67.725 2\pi r = 2 \pi \cdot \sqrt{\frac {365}\pi} = 2 \cdot \sqrt{365\pi} \approx 67.725

The same answer up to three decimal places. So need no formula when the sides are many .

So, my thinking is right!!!! :)

Vinayak Srivastava - 10 months, 3 weeks ago

I don't know if what I did was right, but I assumed a 2020 2020- gon as a circle!

π r 2 = 365 r 10.778 \pi r^2 =365 \implies r \approx 10.778 2 π r 67.72 \implies 2\pi r \approx \boxed{67.72}

Kaizen Cyrus
Jul 23, 2020

The formula to get the area of a regular polygon is:

A = 1 4 n s 2 cot ( π n ) \large A = \frac{1}{4}ns^{2} \cot\left(\frac{π}{n}\right) n = the number of sides of the polygon. s = side length of the polygon. \small \begin{array}{lcl} n & = & \text{the number of sides of the polygon.} \\ s & = & \text{side length of the polygon.} \end{array}

365 = 1 4 ( 2020 ) ( s 2 ) cot ( π 2020 ) s = 2 365 tan ( π 2020 ) 2020 s 0.033527 \small \begin{array}{cccl} & 365 & = & \dfrac{1}{4}(2020)(s^{2}) \cot\left(\dfrac{π}{2020}\right) \\[1em] \implies & s & = & \dfrac{2 \sqrt{365} \sqrt{\tan \left( \dfrac{π}{2020} \right) }}{ \sqrt{2020}} \\[1em] & s & \approx & 0.033527 \end{array}

Multiplying the side length by 2020 2020 , we get the perimeter 67.725 \boxed{\approx 67.725} .

Refine how?

Pi Han Goh - 10 months, 3 weeks ago

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The array looks cramped. It doesn't look good.

Kaizen Cyrus - 10 months, 3 weeks ago

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Original: a = 1 2 = 3 6 \begin{array} { c c c } a & = & \frac12 \\ & = & \frac36 \end{array}

Modified 1: a = 1 2 = 3 6 \begin{array} { c c c } a & = & \frac12 \\[0.5em] & = & \frac36 \end{array}

Modified 2: a = 1 2 0 = 3 6 \begin{array} { c c c } a & = & \frac12 \\ \phantom{0} \\ & = & \frac36 \end{array}

Code:

Pi Han Goh - 10 months, 3 weeks ago

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@Pi Han Goh Thank you for this.

Kaizen Cyrus - 10 months, 3 weeks ago

@Pi Han Goh What does \phantom{0} do 0 \phantom{0}

Mahdi Raza - 10 months, 2 weeks ago

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@Mahdi Raza Tell the LaTeX system to put an invisible object that can't be seen.

Pi Han Goh - 10 months, 2 weeks ago

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@Pi Han Goh See this

Pi Han Goh - 10 months, 2 weeks ago

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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}}} \ ( \red{\text{This is a trial} \phantom{0 \text{ test}}} \ )

  • This is also a trial 0 test \red{\text{This is also a trial} \phantom{0} \text{ test}} \ ( (\red{\text{This is also a trial} \phantom{0} \text{ test}}\ )

Mahdi Raza - 10 months, 2 weeks ago

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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.

Section 4.3 and 5.4

Pi Han Goh - 10 months, 2 weeks ago

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@Pi Han Goh I'll check it out, looks interesting and well made

Mahdi Raza - 10 months, 2 weeks ago

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