Question 173453
The perimeter of a rectangle is given by {{{P=2l+2w}}}, so we know that {{{2l+2w=20}}}.


Solving for {{{l}}}:


{{{2l+2w=20}}}


{{{2l=20-2w}}}


{{{l=10-w}}}


The measure of a diagonal of a rectangle in terms of the length and width is given by the Pythagorean Theorem:


{{{d=sqrt(l^2+w^2)}}} where {{{d}}} is the measure of the diagonal.


But we know that:  {{{sqrt(l^2+w^2)=2sqrt(13)}}}


Square both sides:


{{{l^2+w^2=4*13=52}}}


Now substitute the expression for {{{l}}} in terms of {{{w}}} developed earlier:


{{{(10-w)^2+w^2=4*13=52}}}


Expand, collect terms, and put into standard form for a quadratic:


{{{100-20w+w^2+w^2=52}}}


{{{2w^2-20w+48=0}}}


Divide through by {{{2}}}


{{{w^2-10w+24=0}}}


Leaving you with a factorable quadratic to solve.  Hint:  {{{-4*-6=24}}} and {{{-4-6=-10}}}


Since you are solving for the width, pick the smaller of the two roots as your width and calculate the length from {{{l=10-w}}} (which you will find is the other root of your quadratic)