Question 145447
let x =width
Also x+2 = length (2meters longer)
Then diagonal, which served as the hypotenuse = 10 meters. Marked as "y".
by Pyth. Theorem, 
{{{y^2 = (x+2)^2 + x^2}}} ----------- eqn 1
{{{10^2 = x^2 + 4x + 4 + x^2}}}
{{{100 = 2x^2 + 4x +4}}}
{{{2x^2 + 4x -96=0}}}
{{{x^2 + 2x -48 = 0}}}
{{{(x+8)(x-6)=0}}}
{{{x= -8}}}, not to use being negative (-)
{{{x= 6}}}, perfect!
So, the {{{width = x = 6 meters}}}
And the {{{Length = 6+2 = 8 meters}}}
In doubt? Go back eqn 1,
{{{10^2 = (6+2)^2 + 6^2}}}
{{{100= 64 + 36}}}
{{{100=100}}}
</pre><font size = 4 = "indigo"><b>Thank you,
Jojo</pre>