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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
5 months, 2 weeks ago

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It should be \(\dfrac14 N s^2 \cot \dfrac \pi N \). Pi Han Goh · 5 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 · 5 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 · 5 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 · 5 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 · 5 months, 2 weeks ago

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@Siddhartha Srivastava Thats what I wrote. Sorry about the handwritring😁 Andrés Dextre · 5 months, 1 week ago

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@Andrés Dextre Haha no problem! Pi Han Goh · 5 months, 1 week ago

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