Question 81780
Since we know the diagonal is 3 feet we can use Pythagoreans theorem


{{{a^2+b^2=c^2}}}


and replace {{{c}}} with 3 and {{{a^2+b^2}}} with {{{x^2+x^2}}} like this:


{{{x^2+x^2=3^2}}}


{{{2x^2=9}}} Square 3


{{{2x^2=9}}} Combine like terms


{{{x^2=9/2}}} Divide both sides by 2


{{{x=sqrt(9/2)}}} Take the square root of both sides


*[Tex \LARGE \pm\frac{3}{sqrt{2}}] Reduce

Since a negative length doesn't make sense our only answer is 


{{{x=3/sqrt(2)}}}


Check:

Plug in the leg {{{3/sqrt(2)}}} into Pythagoreans theorem


{{{(3/sqrt(2))^2+(3/sqrt(2))^2=3^2}}}


{{{9/2+9/2=3^2}}} Square each individual term


{{{18/2=3^2}}} Add


{{{9=3^2}}} Reduce


{{{sqrt(9)=sqrt(3^2)}}} Take the square root of both sides


{{{3=3}}} works. This verifies our answer.