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Sagot :
[tex]a\ golden\ rectangle:\ \ \ \frac{a+b}{a}= \frac{a}{b} \\----------------\\a+b=16\ [ft]\ \ \ \ \Rightarrow \ \ \ b=16-a\\\\ \frac{16}{a} = \frac{a}{16-a} \ \ \ \ \Leftrightarrow\ \ \ 16(16-a)=a^2\\\\a^2+16a-256=0\\\\\ \ \ \Rightarrow\ \ \ \Delta=16^2-4\cdot(-256)=256\cdot5\ \ \ \Rightarrow\ \ \ \sqrt{\Delta} =16 \sqrt{5} \\\\a_1= \frac{-16-16 \sqrt{5} }{2} =-8-8 \sqrt{5} <0\\\\a_2= \frac{-16+16 \sqrt{5} }{2} =-8+8 \sqrt{5}=8( \sqrt{5} -1)>0\\\\[/tex]
[tex]a=8( \sqrt{5} -1)\ \ \Rightarrow\ \ b=16-8( \sqrt{5} -1)=16-8\sqrt{5} +8=8(3- \sqrt{5} )\\\\Ans.\ This\ golden\ ractangle\ has\ sides:\\.\ \ \ \ \ \ \ 8( \sqrt{5} -1)\ [ft]\ \ and\ \ 8(3- \sqrt{5} )\ [ft][/tex]
[tex]a=8( \sqrt{5} -1)\ \ \Rightarrow\ \ b=16-8( \sqrt{5} -1)=16-8\sqrt{5} +8=8(3- \sqrt{5} )\\\\Ans.\ This\ golden\ ractangle\ has\ sides:\\.\ \ \ \ \ \ \ 8( \sqrt{5} -1)\ [ft]\ \ and\ \ 8(3- \sqrt{5} )\ [ft][/tex]
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