Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.

In the square root spiral, each right triangle is formed by taking the hypotenuse of the previous triangle as one leg and adding a perpendicular of length 1 unit as the other leg.

Triangle 1 : Both legs = 1 unit.

By the Pythagoras theorem :

$\Rightarrow h$1^2 = 1^2 + 1^2 = 2 \\[1em] \Rightarrow h1 = \sqrt{2}.

Triangle 2 : Legs = $\sqrt{2}$ and 1.

$\Rightarrow h$2^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3 \\[1em] \Rightarrow h2 = \sqrt{3}.

Triangle 3 : Legs = $\sqrt{3}$ and 1.

$\Rightarrow h$3^2 = (\sqrt{3})^2 + 1^2 = 3 + 1 = 4 \\[1em] \Rightarrow h3 = \sqrt{4} = 2.

Triangle 4 : Legs = 2 and 1.

$\Rightarrow h$4^2 = 2^2 + 1^2 = 4 + 1 = 5 \\[1em] \Rightarrow h4 = \sqrt{5}.

Triangle 5 : Legs = $\sqrt{5}$ and 1.

$\Rightarrow h$5^2 = (\sqrt{5})^2 + 1^2 = 5 + 1 = 6 \\[1em] \Rightarrow h5 = \sqrt{6}.

Triangle 6 : Legs = $\sqrt{6}$ and 1.

$\Rightarrow h$6^2 = (\sqrt{6})^2 + 1^2 = 6 + 1 = 7 \\[1em] \Rightarrow h6 = \sqrt{7}.

Triangle 7 : Legs = $\sqrt{7}$ and 1.

$\Rightarrow h$7^2 = (\sqrt{7})^2 + 1^2 = 7 + 1 = 8 \\[1em] \Rightarrow h7 = \sqrt{8}.

Triangle 8 : Legs = $\sqrt{8}$ and 1.

$\Rightarrow h$8^2 = (\sqrt{8})^2 + 1^2 = 8 + 1 = 9 \\[1em] \Rightarrow h8 = \sqrt{9} = 3.

Triangle 9 : Legs = 3 and 1.

$\Rightarrow h$9^2 = 3^2 + 1^2 = 9 + 1 = 10 \\[1em] \Rightarrow h9 = \sqrt{10}.

Triangle 10 : Legs = $\sqrt{10}$ and 1.

$\Rightarrow h${10}^2 = (\sqrt{10})^2 + 1^2 = 10 + 1 = 11 \\[1em] \Rightarrow h{10} = \sqrt{11}.

In general, the n^th hypotenuse is $\sqrt{n + 1}$.

Hence, the lengths of the hypotenuses of the right triangles in the square root spiral are $\sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, 3, \sqrt{10}$ and $\sqrt{11}$.