Question 204198

Let the length of the rectangle be L
Since its width is 4 less than twice its length, then width = 2L - 4
Since its area = 183 cm^2, then we’’ll have: 


L(2L – 4) = 183
{{{2L^2-4L-183 = 0}}}


Using the quadratic equation:  {{{x=(-b+-sqrt(b^2-4*a*c))/(2*a)}}}, where:


a  =  2  ;  b  =  - 4  ;  c =  - 183, we have:


{{{L=(--4+-sqrt(-4^2-4*2*-183))/(2*2)}}}


{{{L=(4+-sqrt(16+1464))/(4)}}}


{{{L=(4+-sqrt(1480))/(4)}}}


{{{L=(4+38.47)/4}}}  or  {{{L=(4-38.47)/4}}}


{{{L=42.47/4}}}  or  {{{L=-34.47/4}}}


L  =  {{{highlight_green(10.62)}}}  or  {{{highlight_green(-8.62)}}}


Since we CANNOT have a negative measurement, we reject L = - 8.62


Therefore, the length (or base) of the rectangle is 10.62 cm, and the width (or height) = 2(10.62) – 4  =  21.24 – 4 = 17.24 cm.


To now find the diagonal, we use the Pythagorean formula: {{{a^2+b^2=c^2}}}, using the length and width as “a” & “b” and the diagonal as “c.”


{{{a^2+b^2=c^2}}} 


{{{10.62^2+17.24^2=c^2}}}


{{{112.78 + 297.22 = c^2}}}


{{{410=c^2}}}


{{{sqrt(410)=c}}}


c, or the diagonal  =  {{{highlight_green(20.25)}}} cm.