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Omni-MATH / Consider a regular n-gon with radius r. Let x be the side length of the n-gon. So, since the central angle is (2π)/(n) (see diagram below), use the Law of Cosines to find that …

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Consider a regular nn-gon with radius rr. Let xx be the side length of the nn-gon. So, since the central angle is 2πn\frac{2 \pi}{n} (see diagram below), use the Law of Cosines to find that $x^{2}=r^{2}+r^{2}-2 r * r \cos \frac{2 \pi}{n}$, so x2=2r2(1−cos2πn)x^{2}=2 r^{2}\left(1-\cos \frac{2 \pi}{n}\right). Thus, x=r21−cos2πnx=r \sqrt{2} \sqrt{1-\cos \frac{2 \pi}{n}}. So, the total perimeter of the nn-gon is nx=nr21−cos2πnn x=n r \sqrt{2} \sqrt{1-\cos \frac{2 \pi}{n}}. Now, if we take \lim _{n \rightarrow \infty}oftheperimeter,theresultwillbe of the perimeter, the result will be 2 \pi n,sincethe, since the n−gonapproachesacirle,so-gon approaches a cirle, so \lim _{n \rightarrow \infty} n r \sqrt{2} \sqrt{1-\cos \frac{2 \pi}{n}}=2 \pi r,andso, and so \lim _{n \rightarrow \infty} \boldsymbol{n} \boldsymbol{r} \sqrt{\mathbf{1 - \operatorname { c o s } \frac { 2 \pi } { n }}}=\pi r \sqrt{2}$.
Plain-text mathematical notation (without MathML)
Consider a regular n-gon with radius r. Let x be the side length of the n-gon. So, since the central angle is (2π)/(n) (see diagram below), use the Law of Cosines to find that $x^{2}=r^{2}+r^{2}-2 r * r \cos \frac{2 \pi}{n}$, so x²=2r²(1−cos(2π)/(n)). Thus, x=r√(2)√(1−cos(2π)/(n)). So, the total perimeter of the n-gon is nx=nr√(2)√(1−cos(2π)/(n)). Now, if we take \lim _{n \rightarrow \infty}oftheperimeter,theresultwillbe2 \pi n,sincethen−gonapproachesacirle,so\lim _{n \rightarrow \infty} n r \sqrt{2} \sqrt{1-\cos \frac{2 \pi}{n}}=2 \pi r,andso\lim _{n \rightarrow \infty} \boldsymbol{n} \boldsymbol{r} \sqrt{\mathbf{1 - \operatorname { c o s } \frac { 2 \pi } { n }}}=\pi r \sqrt{2}$.
Original LaTeX notation
Consider a regular $n$-gon with radius $r$. Let $x$ be the side length of the $n$-gon. So, since the central angle is $\frac{2 \pi}{n}$ (see diagram below), use the Law of Cosines to find that $x^{2}=r^{2}+r^{2}-2 r * r \cos \frac{2 \pi}{n}$, so $x^{2}=2 r^{2}\left(1-\cos \frac{2 \pi}{n}\right)$. Thus, $x=r \sqrt{2} \sqrt{1-\cos \frac{2 \pi}{n}}$. So, the total perimeter of the $n$-gon is $n x=n r \sqrt{2} \sqrt{1-\cos \frac{2 \pi}{n}}$. Now, if we take \lim _{n \rightarrow \infty}$ of the perimeter, the result will be $2 \pi n$, since the $n$-gon approaches a cirle, so $\lim _{n \rightarrow \infty} n r \sqrt{2} \sqrt{1-\cos \frac{2 \pi}{n}}=2 \pi r$, and so $\lim _{n \rightarrow \infty} \boldsymbol{n} \boldsymbol{r} \sqrt{\mathbf{1 - \operatorname { c o s } \frac { 2 \pi } { n }}}=\pi r \sqrt{2}$.

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task_id
e6dbf1f3-e607-5da5-8404-71fa774057eb
task_key
test--e6dbf1f3-e607-5da5-8404-71fa774057eb
task_revision_id
1
config / split
/ test

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