Question 435486

given:

The length of a rectangle: {{{L}}} is {{{3cm}}} more than the width {{{W}}}. 

The area is {{{A=70 cm^2}}}. 

Find:

the dimentions of the rectangle

{{{A=L*W}}}

{{{70cm^2=L*W}}}.........{{{L=W+3cm}}}..substitute in {{{70 cm^2=L*W}}}

{{{70cm^2=(W+3cm)*W}}}

{{{70cm^2=W^2+W*3cm}}}

{{{W^2+3cm*W-70 cm^2=0}}}..solve quadratic for {{{W}}} and use only positive root because {{{W}}} could not be negative

{{{W = (-(3cm) +- sqrt( (3cm)^2-4*1*(-70 cm^2) ))/(2*1) }}} 

{{{W = (-3cm +- sqrt( 9cm^2+280cm^2))/2 }}}

{{{W = (-3cm +- sqrt( 289cm^2))/2 }}}

{{{W = (-3cm +- 17cm)/2 }}}


{{{W = (-3cm + 17cm) ))/2 }}}

{{{W = 14cm/2 }}}

{{{W = 7cm }}}......now find {{{L}}}

{{{L=7cm+3cm}}}

{{{L= 10cm}}}