Question 118067
Use the pythagorean formula
{{{c^2 = a^2 + b^2}}}
where c is the diagonal, a is the width, b is the height
{{{(sqrt(6))^2 = (x + 2)^2 + x^2}}}
{{{6 = x^2 + 4x + 4 + x^2}}}
{{{2x^2 + 4x - 2 = 0}}}
{{{x^2 + 2x - 1 = 0}}}
{{{x^2 + 2x  = 1}}}
Complete the square. Take 1/2 of the coefficient of the {{{x}}} term
square it, and add it to both sides
{{{x^2 + 2x + 1 = 2}}}
{{{(x + 1)^2 = 2}}}
{{{x + 1 = 0 +- sqrt(2)}}}
{{{x = -1 +- sqrt(2)}}}
{{{x = -1 + sqrt(2)}}} The other solution is negative,
so it must be rejected
The width is {{{x + 2}}}, so
width = {{{2 - 1 + sqrt(2)}}}
width = {{{1 + sqrt(2)}}} answer
Height = {{{x}}}
height = {{{-1 + sqrt(2)}}} answer
check
{{{(sqrt(6))^2 = (x + 2)^2 + x^2}}}
{{{(sqrt(6))^2 = (-1 + sqrt(2) + 2)^2 + (-1 + sqrt(2))^2}}}
{{{6 = 1 + 2*sqrt(2) + 2 + 1 - 2*sqrt(2) + 2}}}
{{{6 = 6}}}
OK