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Area of a regular polygon

Could someone tell me if this formula is ok? \(N\) is the number of sides and \(s\) is the length of the side

Note by Andrés Dextre
10 months, 2 weeks ago

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It should be \(\dfrac14 N s^2 \cot \dfrac \pi N \). Pi Han Goh · 10 months, 2 weeks ago

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@Pi Han Goh They're both the same

\( \tan(\frac{90^{\circ}(n-2)}{n}) = \tan( 90^{\circ} - \frac{180^{\circ}}{n}) = \cot(\frac{180^{\circ}}{n}) \) Siddhartha Srivastava · 10 months, 2 weeks ago

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@Siddhartha Srivastava OP wrote \(A = \dfrac19 ns^2 \tan \left( \dfrac{\text{gorn}-2}n \right ) \). Pi Han Goh · 10 months, 2 weeks ago

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@Pi Han Goh I'll let OP clarify, but from what I see, the "9" is more likely a "4", "go" is "90" and the "r" is a bracket.

So what I thought he wrote is \( A = \frac{1}{4} ns^2 \tan( \frac{90(n-2)}{n} ) \) Siddhartha Srivastava · 10 months, 2 weeks ago

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@Siddhartha Srivastava If you think he wrote is true, then his expression is indeed correct. Mine is just simpler/standard. Pi Han Goh · 10 months, 2 weeks ago

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@Siddhartha Srivastava Thats what I wrote. Sorry about the handwritring😁 Andrés Dextre · 10 months ago

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@Andrés Dextre Haha no problem! Pi Han Goh · 10 months ago

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