Locate 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,

$\mathrm{OA}=\sqrt{\mathrm{OB}^{2}+\mathrm{AB}^{2}}$

$=\sqrt{3^{2}+1^{2}}$

$=\sqrt{9+1}$

 

$=\sqrt{10}$