Question 82817
Let x = shortest side (a leg of the right triangle)
x+3 = longer side (a leg of the right triangle)
{{{x*sqrt(3)}}} = longest side (hypotenuse of right triangle)


{{{x^2 +(x+3)^2 = (x*sqrt(3))^2}}}
{{{x^2 + x^2+6x + 9 = x^2 *3 }}}
{{{2x^2 + 6x + 9 = 3x^2 }}}


Set equal to zero, moving everything to the right side of the equation:

{{{2x^2  + 6x + 9-2x^2-6x -9 = 3x^2 -2x^2 -6x - 9}}}

{{{0=x^2-6x-9}}}


This does NOT factor (does this have an ugly answer??), so you'll have to complete the square or use the quadratic formula.  I will complete the square:
{{{x^2-6x-9=0}}}
{{{x^2 -6x + ____ = 9 + ____}}}
{{{x^2 -6x + 9 = 9 + 9}}}
{{{(x-3)^2= 18}}}
{{{x-3 = 0+-sqrt(18) }}}
{{{x =3 +- 3*sqrt(2) }}}


Reject the negative answer, since a side of a triangle cannot be negative:
{{{x =3 + 3*sqrt(2) }}} meters (shortest side)
{{{x + 3 =6 + 3*sqrt(2) }}} meters (second leg of the right triangle)
{{{x*sqrt(3) =(sqrt(3))*(3 + 3*sqrt(2)) }}} meters (hypotenuse)
{{{x*sqrt(3) =(3 *sqrt(3) + 3*sqrt(6)) }}} meters (hypotenuse)


R^2 at SCC