Question 119202
The length is 1 cm longer than the width, so if the width is w, the length is w +1.  The diagonal of the rectangle forms a right triangle with the length and the width, so Pythagoras says:


{{{w^2+(w+1)^2=4^2}}}


Expand the binomial and collect terms:


{{{w^2+w^2+2w+1=16}}}


{{{2w^2+2w+1-16=0}}}


{{{2w^2+2w-15=0}}}


Use the quadratic formula:


{{{w = (-2 +- sqrt( 4-4*2*(-15) ))/(4) }}} 


{{{w=(-2+-sqrt(124))/4}}}


{{{w=(-1+-sqrt(31))/2}}}


{{{w=(-1-sqrt(31))/2<0}}} so this won't do as an answer because a rectangle can't have a negative width.  So, {{{w=(-1+sqrt(31))/2}}} is the width.


{{{l=w+1=(-1+sqrt(31))/2+1=((-1+sqrt(31))/2)+(2/2)=(1+sqrt(31))/2}}} is then the length.