Locate $\sqrt{5}, \sqrt{10}$ and $\sqrt{17}$ on the number line.

Presentation of $\sqrt{5}$ on number line:

We write $5$ as the sum of the square of two natural numbers:

$5=1+4=1^{2}+2^{2}$

On the number line, take $O A=2$ units.

Draw $BA =1$ unit, perpendicular to OA. Join $OB$.

By Pythagoras theorem, $OB =\sqrt{5}$

Using a compass with centre $O$ and radius $O B$, draw an arc which intersects the number line at the point $C$. Then, $C$ corresponds to $\sqrt{5}$.

Presentation of $\sqrt{10}$ on the number line:

We write 10 as the sun of the square of two natural numbers:

$10=1+9=1^{2}+3^{2}$

On the number line, taken $O A=3$ units.

Draw $BA = 1 unit,$ perpendicular to $OA,$ Join $OB.$

By Pythagoras theorem, $OB =\sqrt{10}$

Using a compass with centre $O$ and radius $O B$, draw an arc which intersects the number line at the point $C$. Then, $C$ corresponds to $\sqrt{10}$.

Presentation of $\sqrt{17}$ on the number line:

We write $17$ as the sum of the square of two natural numbers:

$17=1+16=1^{2}+4^{2}$

On the number line, take $O A=4$ units.

Draw $BA =1$ units, perpendicular to $OA$. Join $OB$.

By Pythagoras theorem, $OB =\sqrt{17}$

Using a compass with centre $O$ and radius $O B$, draw an arc which intersects the number line at the point $C$. Then, $C$ corresponds to $\sqrt{17}$.