Question 49205
Let the length of the rectangle be L and its width be W.
The length, L, is 1 cm longer than the width, W, so L = W+1
The sides of the rectangle can be written in terms of the width: L=(W+1) and W=(W).
The diagonal of the rectangle forms the hypotenuse (c) of a right triangle whose sides are (W+1) and (W).

Using the Pythagorean theorem: {{{c^2 = a^2+b^2}}}, you can find W:

{{{4^2 = (W+1)^2 + W^2}}} Simplify.
{{{16 = W^2+2W+1 + W^2}}} Collect like-terms.
{{{2W^2+2W+1 = 16}}} Subtract 16 from both sides of the equation.
{{{2W^2+2W-15 = 0}}} Use the quadratic formula to solve: {{{W=(-b+-sqrt(b^2-4ac))/2a}}}

{{{W = (-2+-sqrt(2^2-4(2)(-15)))/2(2)}}}
{{{W = (-2+-sqrt(4+120))/4}}}
{{{W = (-2+-sqrt(124))/4}}}
{{{W = (-2+-2sqrt(31))/4}}}
{{{W = -1/2-(1/2)sqrt(31)}}} Discard this solution as the width must be a positive number.
{{{W = -1/2 + (1/2)sqrt(31)}}}

For an approximate answer:
{{{W = -1/2+(1/2)(5.57)}}}
{{{W = 2.28}}}cm
{{{L = W+1}}}
{{{L = 3.28}}}cm