四十角形
S = 10 a 2 cot  π 40 = 10 ( 1 + 5 + 5 + 2 5 + ( 1 + 5 + 5 + 2 5 ) 2 + 1 ) a 2 = 10 ( 1 + 5 + 5 + 2 5 + ( 1 + 5 ) 2 + ( 2 1 ) ( 1 + 5 ) 5 + 2 5 + ( 5 + 2 5 ) 2 + 1 ) a 2 = 10 ( 1 + 5 + 5 + 2 5 + ( 6 + ( 2 1 ) 5 ) + ( 2 1 ) ( 1 + 5 ) 5 + 2 5 + ( 5 + 2 5 ) + 1 ) a 2 = 10 ( 1 + 5 + 5 + 2 5 + ( 11 + 4 5 + ( 2 1 ) ( 1 + 5 ) 5 + 2 5 ) + 1 ) a 2 = 10 ( 1 + 5 + 5 + 2 5 + 12 + 4 5 + ( 2 1 ) ( 1 + 5 ) 5 + 2 5 ) a 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}}}