we are to assume that the righmost side is also 1
so
use pythagorean theorem repeatedly
k, so
a²+b²=c²
we know that one of the legs will always be 1 so
replace b or a (I'll pick b) with 1
a²+1²=c²
a²+1=c²
now, we are going to be given the 2 sides and we need the hyptonuse
so
if we sqrt both sides
√(a²+1)=c
where a is the unknown leg
see attchment
first hypotonuse is √2
2nd is √((√2)²+1)=√(2+1)=√3
3rd is , hmm, there is a pattern
the nth hyptonuse will be √(n+1)
so there are 7 triangles so the 7th hyptonuse is √(7+1)=√8=2√2
x=2√2