Question 110343
the area of a rectangle is:

{{{A = L*W}}}

given:

{{{A = 61cm^2}}}

and the {{{length=L}}} of a rectangle is {{{4cm}}}{{{ more}}} that {{{2}}}{{{ times}}} its {{{width=W}}}, and we can write it this way:

{{{L = 2W + 4cm}}}............we will substitute this in formula for area
in order to find the dimensions of the rectangle

 {{{A = L*W}}}

{{{61cm^2 = (2W + 4cm)*W}}}

{{{61cm^2 = 2W^2 + 4Wcm}}}.........move {{{61cm^2}}} to the right

{{{0 = 2W^2 + 4Wcm - 61cm^2}}}

or

{{{ 2W^2 + 4Wcm - 61cm^2 = 0}}}.............now we can use quadratic formula to solve for {{{W}}}


{{{W[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{W[1,2]=(-4 +- sqrt (4^2 -4*2*(-61) )) / (2*2)}}}

{{{W[1,2]=(-4 +- sqrt (16 + 488 ))/4}}}

{{{W[1,2]=(-4 +- sqrt (504 ))/4}}}

{{{W[1,2]=(-4 +- (22.449))/4}}}

then:

{{{W[1]=(-4 + (22.449))/4}}}

{{{W[1]=(18.449)/4}}}

{{{W[1]= 4.612cm}}}


{{{W[2]=(-4 - (22.449))/4}}}..........we don't need negative value, lenght is positive value


so {{{W = 4.612cm}}}

now find {{{L}}}


{{{L = 2W + 4cm}}}

{{{L = 2(4.612)cm + 4cm}}}

{{{L =(9.224)cm + 4cm}}}

{{{L =13.224cm}}}