Question 818996
{{{15^2+(sqrt(145))^2=hypotnuse^2}}}
For convenience, h = hypotenuse.


{{{15^2+(sqrt(29*5))^2=h^2}}}
{{{h=sqrt(15^2+29*5)}}}
{{{h=sqrt(3*5*3*5+5*29)}}}
{{{h=sqrt(225+145)=sqrt(370)}}}
Recognize the divisibility by 10 and 2 and 5, but that will not help in any simplifcation.  


{{{highlight(h=sqrt(370))}}} or {{{h=sqrt(2*5*37)}}}