Question 78137
Let h=height, w=width

So we have

{{{w=2+h}}}


Since the diagonal is the hypotenuse, we can say

{{{w^2+h^2=(sqrt(6))^2}}}


{{{(2+h)^2+h^2=(sqrt(6))^2}}} Substitute 2+h into w


{{{4+4h+h^2+h^2=(sqrt(6))^2}}} foil the parenthesis


{{{4+4h+2h^2=6}}} Combine like terms and reduce {{{(sqrt(6))^2}}} to 6


{{{4+4h+2h^2-6=0}}} Subtract 6 from both sides


{{{2h^2+4h-2=0}}} Combine like terms

Now use the quadratic formula to find the value of L:

*[invoke quadratic "h", 2, 4, -2 ]

Ignoring the negative answer we get:


{{{h=0.414}}}

So use this to find w

{{{w=2+0.414=2.414}}}

So the height is 0.414 m and the width is 2.414 m approximately