# 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
2 years, 4 months ago

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It should be $$\dfrac14 N s^2 \cot \dfrac \pi N$$.

- 2 years, 4 months ago

They're both the same

$$\tan(\frac{90^{\circ}(n-2)}{n}) = \tan( 90^{\circ} - \frac{180^{\circ}}{n}) = \cot(\frac{180^{\circ}}{n})$$

- 2 years, 4 months ago

OP wrote $$A = \dfrac19 ns^2 \tan \left( \dfrac{\text{gorn}-2}n \right )$$.

- 2 years, 4 months ago

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 $$A = \frac{1}{4} ns^2 \tan( \frac{90(n-2)}{n} )$$

- 2 years, 4 months ago

If you think he wrote is true, then his expression is indeed correct. Mine is just simpler/standard.

- 2 years, 4 months ago

Thats what I wrote. Sorry about the handwritring😁

- 2 years, 3 months ago

Haha no problem!

- 2 years, 3 months ago