Question 683894


Let the width of the walkway be {{{xft}}}

Then, {{{(30 + 2x)(40 + 2x) = 1800}}}

Or {{{2(15 + x)*2(20 + x) = 1800}}}

{{{4(15 + x)(20 + x) = 1800}}}

{{{cross(4)(15 + x)(20 + x) = cross(1800)450}}}

{{{(15 + x)(20 + x) = 450}}}

{{{x^2 + 35x + 300 = 450}}}

{{{x^2 + 35x + 300-450=0}}}

{{{x^2 + 35x - 150 = 0}}}......use quadratic formula


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x = (-35 +- sqrt( 35^2-4*1*(-150) ))/(2*1) }}}


{{{x = (-35 +- sqrt( 1225+600 ))/2 }}}


{{{x = (-35 +- sqrt( 1825 ))/2 }}}


{{{x = (-35 +- 42.72)/2 }}}....find only positive root since width could be only positive number

{{{x = (-35 + 42.72)/2 }}}

{{{x = 7.72/2 }}}

{{{highlight(x = 3.86ft) }}}


so, your answer is c. {{{3.860 ft}}}