四十角形

正四十角形においては、中心角と外角は9°で、内角は171°となる。一辺の長さが a の正四十角形の面積 S は

${\displaystyle S={\frac {40}{4}}a^{2}\cot {\frac {\pi }{40}}\simeq 127.06205a^{2}}$

${\displaystyle {\begin{aligned}S=10a^{2}\cot {\frac {\pi }{40}}=&10\left(1+{\sqrt {5}}+{\sqrt {5+2{\sqrt {5}}}}+{\sqrt {\left(1+{\sqrt {5}}+{\sqrt {5+2{\sqrt {5}}}}\right)^{2}+1}}\right)a^{2}\\=&10\left(1+{\sqrt {5}}+{\sqrt {5+2{\sqrt {5}}}}+{\sqrt {\left(1+{\sqrt {5}}\right)^{2}+{\binom {2}{1}}\left(1+{\sqrt {5}}\right){\sqrt {5+2{\sqrt {5}}}}+\left({\sqrt {5+2{\sqrt {5}}}}\right)^{2}+1}}\right)a^{2}\\=&10\left(1+{\sqrt {5}}+{\sqrt {5+2{\sqrt {5}}}}+{\sqrt {\left(6+{\binom {2}{1}}{\sqrt {5}}\right)^{}+{\binom {2}{1}}\left(1+{\sqrt {5}}\right){\sqrt {5+2{\sqrt {5}}}}+\left(5+2{\sqrt {5}}\right)^{}+1}}\right)a^{2}\\=&10\left(1+{\sqrt {5}}+{\sqrt {5+2{\sqrt {5}}}}+{\sqrt {\left(11+4{\sqrt {5}}+{\binom {2}{1}}\left(1+{\sqrt {5}}\right){\sqrt {5+2{\sqrt {5}}}}\right)+1}}\right)a^{2}\\=&10\left(1+{\sqrt {5}}+{\sqrt {5+2{\sqrt {5}}}}+{\sqrt {12+4{\sqrt {5}}+{\binom {2}{1}}\left(1+{\sqrt {5}}\right){\sqrt {5+2{\sqrt {5}}}}}}\right)a^{2}\end{aligned}}}$

${\displaystyle \cos(2\pi /40)}$を有理数と平方根で表すことが可能である。

${\displaystyle \cos {\frac {2\pi }{40}}=\cos {\frac {\pi }{20}}=\cos 9^{\circ }={\frac {1}{2}}{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}$

• 五角形
• 十角形
• 二十角形

• Weisstein, Eric W. "Tetracontagon". mathworld.wolfram.com (英語).