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A regular hexagon inscribed in a circle has an area of $$54*3^\frac{1}{3} \text{sq.in}$$ Prove that the circumference of a circle is $$25\pi$$

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Hint: The area of a regular $n $-gon inscribed in a circle of radius $R$ is: $$ A_n=\frac n2 R^2\sin\frac{2\pi}n. $$ Can you take it from here?

Warning: the answer will appear to be different from $25\pi$ (in), since the latter value is wrong.

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