basically, it is when legnth is [tex] \frac{1+ \sqrt{5} }{2} [/tex] the width
legnth=12
legnth= [tex] \frac{1+ \sqrt{5} }{2} [/tex] times width
12= [tex] \frac{1+ \sqrt{5} }{2} [/tex] times width
multiply both sides by 2
24= [tex] 1+ \sqrt{5} [/tex] times width
subtract 1
23=√5 times width
divide both sides by √5
23/(√5)=width
multiply by (√5)/(√5) to rationalize denomenator
[tex] \frac{23 \sqrt{5} }{5} [/tex]=width
aprox 10.28 so 10.3 ft
w=[tex] \frac{23 \sqrt{5} }{5} [/tex] ft or aprox 10.3ft
