Question:
Locate $\sqrt{10}$ on the number line
Solution:
To represent $\sqrt{10}$ on the number line, follow the following steps of construction:
(i) Mark points 0 and 3 as $O$ and $B$, respectively.
(ii) At point $A$, draw $A B \perp O A$ such that $A B=1$ units.
(iii) Join OA.
(iv) With $O$ as centre and radius $O A$, draw an arc intersecting the number line at point $P$.
Thus, point P represents $\sqrt{10}$ on the number line.
Justification:
In right $\triangle O A B$,
Using Pythagoras theorem,
$\mathrm{OA}=\sqrt{\mathrm{OB}^{2}+\mathrm{AB}^{2}}$
$=\sqrt{3^{2}+1^{2}}$
$=\sqrt{9+1}$
$=\sqrt{10}$