Question 939411

the area of a trapezoid is:

{{{A=((a+b)/2)*h }}}

given:

{{{A=360m^2}}}
{{{h=20m}}}
{{{a:b=4:5}}}=> {{{5a=4b}}}=> {{{a=4b/5}}}


{{{360m^2=((4b/5+b)/2)*20m }}}....solve for {{{b}}}


{{{360m^cross(2)=((4b/5+5b/5)/cross(2))*cross(20)10cross(m) }}}


{{{360m=(9b/5)*10 }}}


{{{360m*5=90b }}}


{{{1800m/90=b }}}


{{{highlight(b=20m )}}}

now find {{{a}}}


 {{{a=4b/5}}}

 {{{a=(4*20m)/5}}}

{{{a=(4*cross(20)4m)/cross(5)}}}

{{{a=4*4m)}}}

{{{highlight(a=16m)}}} =>the lengths of the shorter bases