Question 187808
We basically have this triangle set up:



{{{drawing(500,500,-0.5,2,-0.5,3.2,
line(0,0,0,3),
line(0,3,2,0),
line(2,0,0,0),
locate(-0.2,1.5,x),
locate(1,-0.2,x+1),
locate(1,2,2x-1)
)}}}


Remember, the Pythagorean Theorem is {{{a^2+b^2=c^2}}} where "a" and "b" are the legs of a triangle and "c" is the hypotenuse.


Since the legs are {{{x}}} and {{{x+1}}} this means that {{{a=x}}} and {{{b=x+1}}}


   

Also, since the hypotenuse is {{{2x-1}}}, this means that {{{c=2x-1}}}.




{{{a^2+b^2=c^2}}} Start with the Pythagorean theorem.



{{{x^2+(x+1)^2=(2x-1)^2}}} Plug in {{{a=x+1}}}, {{{b=x+1}}}, and {{{c=2x-1}}} 



{{{x^2+x^2+2x+1=4x^2-4x+1}}} FOIL



{{{2x^2+2x+1=4x^2-4x+1}}} Combine like terms.



{{{2x^2+2x+1-4x^2+4x-1=0}}} Get all terms to the left side.



{{{-2x^2+6x=0}}} Combine like terms.



{{{-2x(x-3)=0}}} Factor the left side.



{{{-2x=0}}} or {{{x-3=0}}} Set each factor equal to zero



{{{x=0}}} or {{{x=3}}} Solve for "x" in each case



Ignoring the trivial solution {{{x=0}}}, the only answer is {{{x=3}}}



So the leg "x" is 3 units long


The leg {{{x+1}}} is 3+1=4 units long


and the hypotenuse {{{2x-1}}} is 2(3)-1=6-1=5 units long


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Answer:



So the legs are 3 and 4 units long while the hypotenuse is 5 units long.