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

(N/A) Presentation of $\sqrt{5}$ on the number line:
We write $5$ as the sum of the squares of two natural numbers:
$5 = 1 + 4 = 1^{2} + 2^{2}$
On the number line,take $OA = 2$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{5}$.
Presentation of $\sqrt{10}$ on the number line:
We write $10$ as the sum of the squares of two natural numbers:
$10 = 1 + 9 = 1^{2} + 3^{2}$
On the number line,take $OA = 3$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{10}$.
Presentation of $\sqrt{17}$ on the number line:
We write $17$ as the sum of the squares of two natural numbers:
$17 = 1 + 16 = 1^{2} + 4^{2}$
On the number line,take $OA = 4$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{4^{2} + 1^{2}} = \sqrt{17}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{17}$.