Spiral of Theodore

We note that the hypotenuse of the $n$ th right triangle of Theodorus spiral is $h_n = \sqrt{n+1}$ . And $h_n = b_{n+1}$ the base of the $(n+1)$ th right triangle. Therefore the area of the $n$ th right triangle is $A_n=\dfrac {\sqrt n}2$ . And for $A_n = 1$ , we have $\dfrac {\sqrt n}2 = 1 \implies n = \boxed 4$ .

The triangle labelled $n$ in the diagram has shorter sides $1,\sqrt{n}$ , so its area is $\frac{\sqrt{n}}{2}$ . Hence the triangle we want is number $\boxed4$ .

Incidentally, it's nice to note that the six angles around the common point of the triangles in the diagram sum to just over $180^\circ$ (the angle is in fact $183.1^\circ$ , less than a $2\%$ error.)