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

In a square-root spiral, start with a right triangle having both legs of length 1. Each new triangle is right-angled, with one leg being the previous hypotenuse and the other leg of length 1.

Triangle 1: legs 1 and

$$\Rightarrow \text{hypotenuse} = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Triangle 2: legs $\sqrt{2}$ and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{2+1} = \sqrt{3}$$

Triangle 3: legs $\sqrt{3}$ and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{3+1} = \sqrt{4} = 2$$

Triangle 4: legs 2 and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{4+1} = \sqrt{5}$$

Triangle 5: legs $\sqrt{5}$ and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{6}$$

Triangle 6: legs $\sqrt{6}$ and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{7}$$

Triangle 7: legs $\sqrt{7}$ and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{8} = 2\sqrt{2}$$

Triangle 8: legs $\sqrt{8}$ and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{9} = 3$$

Triangle 9: legs 3 and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{10}$$

Triangle 10: legs $\sqrt{10}$ and 1

$$\Rightarrow \text{hypotenuse} = \sqrt{11}$$

Concept Applied

In a square-root spiral, the length of each new hypotenuse is found using the Baudhāyana-Pythagoras theorem, giving $\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}$, and so on.