Question 118400

given:

The length {{{L}}} of a rectangle is {{{3 cm}}} more than {{{2}}} times its width {{{W}}}.  

then {{{L = 2W + 3cm}}}

If the area {{{A}}} of the rectangle is {{{93 cm^2}}} 

to find: the dimensions of the rectangle to the nearest thousandth. 


since the area is {{{A = L*W}}}, we will have:

{{{93cm^2  = (2W + 3)*W}}}

{{{0 = 2W^2 + 3W - 93 cm^2}}}

{{{ 2W^2 + 3W - 93 cm^2 = 0}}}..


find only positive root ( the width cannot be negative)


{{{W[1,2]=(-3 +- sqrt (3^2 -4*2*(-93) )) / (2*2)}}}


{{{W[1,2]=(-3 +- sqrt (9 + 744 )) /4}}}


{{{W[1,2]=(-3 +- sqrt ( 753)) /4}}}


{{{W[1,2]=(-3 +- 27.44)/4}}}


{{{W[1]=(-3+ 27.44)/4}}}


{{{W[1]=( 24.44)/4}}}


{{{W[1]= 6.11cm}}}..then


{{{L = 2 * 6.11cm + 3cm}}}

{{{L = 12.22cm + 3cm}}}

{{{L = 15.22cm }}}

check:


{{{A = L*W}}}

{{{93cm^2= 15.22cm *6.11cm}}}

{{{93cm^2= 92.9942cm^2}}}.........we can round it

{{{93cm^2= 93cm^2}}}.........


{{{Width = 6.11cm}}}, {{{Length = 15.22cm }}}