Question

Locate $\sqrt{5}$ on the number line.

Answer 1

Here $5=2^2+1^2$
So, draw a right angled ∆OAB, in which OA = 2 units and AB=1 unit and $\angle OAB={90}^{\circ }$
By using Pythagoras theorem, we get
$OB=\sqrt{O{A}^2+A{B}^2}=\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt{5}$
Taking $OB=\sqrt{5}$ as radius and point O as centre, draw an arc which meets the number line at point P on the positive side of it.
Hence, it is clear that point P represents $\sqrt{5}$ on the number line.