Question 39247
Start with the Pythagorean theorem.
The diagonal (D) of a rectangle is the hypotenuse of a right triangle whose legs are the rectangle's length (L) and width (W). So we can write:

{{{D^2 = L^2 + W^2}}}

But we are also given that the length (L) is 2 cm longer than its width (W), so we can write: L = W+2  Substituting this into the above equation, we have:

{{{D^2 = (W+2)^2 + W^2}}} and we know that D = 10 cm, so...

{{{10^2 = (W+2)^2 + W^2}}} Simplify and solve for W.
{{{100 = W^2+4W+4+W^2}}}
{{{100 = 2W^2+4W+4}}} Subtract 100 from both sides of the equation.
{{{2W^2+4W-96 = 0}}} Divide both sides by 2 to simplify.
{{{W^2+2W-48 = 0}}} Factor the quadratic equation.
{{{(W-6)(W+8) = 0}}} Apply the zero products principle.
{{{W-6 = 0}}} and/or {{{W+8 = 0}}}
If {{{W-6 = 0}}} then {{{W = 6}}}
If {{{W+8 = 0}}} then {{{W = -8}}} Discard this solution as the width must be a positive number.

The width is 6 cm.
The length is width + 2 cm = 6 + 2 = 8 cm.