Locate $\sqrt{3}$ on the number line.

(N/A) To locate $\sqrt{3}$ on the number line,follow these steps:
$1$. First,locate $\sqrt{2}$ on the number line by constructing a right-angled triangle with base $1$ unit and height $1$ unit. The hypotenuse will be $\sqrt{1^2 + 1^2} = \sqrt{2}$.
$2$. Now,construct a line segment $BD$ of unit length ($1$ unit) perpendicular to the hypotenuse $OB$ (where $OB = \sqrt{2}$).
$3$. Join $OD$. In the right-angled triangle $\triangle OBD$,by the Pythagoras theorem:
$OD^2 = OB^2 + BD^2$
$OD^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3$
$OD = \sqrt{3}$.
$4$. Using a compass,with centre $O$ and radius $OD$,draw an arc that intersects the number line at point $Q$. The point $Q$ represents $\sqrt{3}$ on the number line.
In the same way,you can locate $\sqrt{n}$ for any positive integer $n$,after $\sqrt{n-1}$ has been located.