Question 329105
Use the Pythagorean theorem,
{{{(3n)^2+(3(n+1))^2=(3(n+2))^2}}}
{{{9n^2+9(n^2+2n+1)=9(n^2+4n+4)}}}
{{{2n^2+2n+1=n^2+4n+4}}}
{{{n^2-2n-3=0}}}
{{{(n-3)(n+1)=0}}}
Only the positive solution makes sense in this problem,
{{{n-3=0}}}
{{{n=3}}}
The hypotenuse is {{{3(n+2)=15}}}.