Question 1125597

let the length be {{{L}}}, and the width {{{W}}}

if the width of a rectangle is {{{9}}} less than {{{twice}}} its length,we have

{{{W=2L-9}}}....eq.1


if the area of the rectangle is {{{A=101cm^2}}}, we have

{{{L*W=101cm^2}}}....eq.2

substitute {{{W}}} from eq.1


{{{L(2L-9)=101cm^2}}}...solve for {{{L}}}


{{{2L^2-9L=101cm^2}}}

{{{2L^2-9L-101cm^2=0}}}


use quadratic formula


{{{L = (-(-9) +- sqrt( (-9)^2-4*2*(-101) ))/(2*2) }}}

{{{L = (9 +- sqrt(81+808 ))/4 }}}


{{{L = (9 +- sqrt(889 ))/4 }}}

since length, we need only positive solution

{{{L = (9 + sqrt(889 ))/4 }}}

{{{L = (9 + 29.8161)/4 }}}

{{{L = 38.8161/4 }}}

{{{L = 9.704025 }}}


now find {{{W}}}

{{{W=2*9.704025-9}}}....eq.1

{{{W=19.40805-9}}}

{{{W=10.40805}}}


check the area:

{{{L*W=101cm^2}}}
{{{10.40805*9.704025=101cm^2}}}
{{{100.99997740125=101cm^2}}}...round it
{{{101=101cm^2}}}