We use the Pythagorean Theorem to find the first few hypotenuses:
[asy]
pair A=(0,0),B=(0,1),C=(0.4404,1.957), D=(1.2,2.9), E=(3.316,0);
draw(A--B--C--D--E--A);
draw(B--E--C);
draw(D--E);
draw(rightanglemark(E,A,B));
draw(rightanglemark(E,B,C));
draw(rightanglemark(E,C,D));
label("1",midpoint(A--B),W);
label("1",midpoint(C--B),NW);
label("1",midpoint(C--D),NW);
label("...",(2,2),E);
defaultpen(fontsize(8pt));
label("$\sqrt{11}$",midpoint(A--E),N);
label("$\sqrt{12}$",midpoint(B--E),N);
label("$\sqrt{13}$",midpoint(C--E),NE);
label("$\sqrt{14}$",midpoint(D--E),NE);
[/asy]
So, the first triangle has hypotenuse $\sqrt{12}$, the 2nd has hypotenuse $\sqrt{13}$, and so on. Thus, the 2014th will have hypotenuse $\sqrt{2014+11}=\sqrt{2025}=\boxed{45}$.