Try to extend this method for constructing line segments of lengths $\sqrt{3}$ and $\sqrt{5}$ using a ruler and a compass. Generalise this method to construct a line segment of any length of the form $\sqrt{n}$, where $n$ is a positive integer.

Construction of $\sqrt{3}$ :

Step 1 : Construct OB = $\sqrt{2}$ (as shown previously).

Step 2 : At B, draw a perpendicular BD of length 1 unit.

Step 3 : Join O to D. By the Pythagoras theorem :

⇒ OD² = OB² + BD²

⇒ OD² = ($\sqrt{2}$)² + 1²

⇒ OD² = 2 + 1 = 3

⇒ OD = $\sqrt{3}$.

Step 4 : With O as the centre and OD as the radius, draw an arc cutting the number line at point Q. Then OQ = $\sqrt{3}$.

Construction of $\sqrt{5}$ :

Step 1: On the number line, measure a horizontal segment OA = 2 units.

Step 2: At point A, draw a perpendicular line segment AC of length 1 unit.

Step 3: Join O to C. The length OC is:

⇒ OC² = OA² + AC²

⇒ OC² = (2)² + 1²

⇒ OC² = 4 + 1 = 5

⇒ OC = $\sqrt{5}$.

Step 4 : With O as the centre and OC as the radius, draw an arc cutting the number line at point R. Then OR = $\sqrt{5}$.

Generalisation to construct $\sqrt{n}$ :

Suppose we have a line segment of length $\sqrt{n - 1}$. Then at one end of this segment, draw a perpendicular of length 1 unit. The hypotenuse of the resulting right triangle will have length :

$\Rightarrow \sqrt{(\sqrt{n - 1})^2 + 1^2} \\[1em] \Rightarrow \sqrt{(n - 1) + 1} \\[1em] \Rightarrow \sqrt{n}.$

This method, when extended repeatedly starting from a unit segment, gives the famous "square root spiral".

Hence, by repeatedly applying the Pythagoras theorem with a unit perpendicular to a known segment of length $\sqrt{n - 1}$, we can construct a line segment of any length of the form $\sqrt{n}$ for any positive integer n.