Question 119171
I presume x to be the short leg, so your 2x - 1 representation would be correct for the longer leg, but the hypotenuse is then x + 1 rather than 1 - x.


Using Pythagoras:
{{{(x+1)^2=x^2+(2x-1)^2}}}


Expand the binomials, collect terms, and solve
{{{x^2+2x+1=x^2+4x^2-4x+1}}}
{{{4x^2-6x=0}}}
{{{x(4x-6)=0}}}


So {{{x=0}}} or {{{x=3/2}}}.  But we can exclude the {{{x=0}}} root because that would mean that all three sides of the triangle were of length 0 -- a very uninteresting triangle indeed.  So, the short leg is {{{3/2}}}cm, the hypotenuse is {{{3/2+1=5/2}}}cm and the long leg is {{{2(3/2)-1=3-1=2}}}cm.


Check:


{{{sqrt((3/2)^2+(4/2)^2))}}} =
{{{sqrt((9/4)+(16/4))}}} =
{{{sqrt(25)/sqrt(4)=5/2}}}, Check!


Hope that helps.
John