Question 962654


if the shorter leg {{{a}}} of a right triangle is {{{x+1}}}, 
the longer leg {{{b}}}is {{{x+2}}}, and the hypotenuse {{{c}}} is {{{x+3}}},then 

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

{{{(x+3)^2=(x+1)^2+(x+2)^2}}}

{{{x^2+6x+9=x^2+2x+1+x^2+4x+4}}}

{{{x^2+6x+9=2x^2+6x+5}}}

{{{0=2x^2+6x+5-x^2-6x-9}}}

{{{0=x^2-4}}}

{{{x^2=4}}}

{{{x=2}}}....we need only positive solution

now find the lengths of the sides of the triangle:

 {{{x+1}}}=>{{{2+1}}}=>{{{3}}}
 {{{x+2}}}=>{{{2+2}}}=>{{{4}}}
 {{{x+3}}}=>{{{2+3}}}=>{{{5}}}