Question 919246
w for width, L for length; and d for diagonal.
2w+2L=32 and d=8+w.


The diagonal serves as hypotenuse and w and L serve as legs of a right-triangle.
w^2+L^2=d^2.


Simplify the perimeter equation and substitute for d into the other two equations:
{{{system(w+L=16,w^2+L^2=(w+8)^2)}}};
Still needs a bit of simplification, and then just solve the system.



{{{L=16-w}}};
Substitute...
{{{w^2+(16-w)^2=w^2+16w+64}}}
{{{(16-w)^2=16w+64}}}
{{{16^2-32w+w^2=16w+64}}}
{{{w^2-32w+16^2-16w-64=0}}}
{{{w^2-48w+16^2-64=0}}}
{{{highlight_green(w^2-48w+192=0)}}}


{{{w=(48+- sqrt(48^2-4*192))/2}}}
{{{w=(48+- 8*sqrt(6))/2}}}
{{{highlight(w=24+- 4*sqrt(6))}}}, which are actually the two dimensions.