Locate $\sqrt{13}$ on the number line.
(N/A) We write $13$ as the sum of the squares of two natural numbers:
$13 = 9 + 4 = 3^{2} + 2^{2}$
On the number line,take $OA = 3$ units.
Draw $BA = 2$ units,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{OA^{2} + AB^{2}} = \sqrt{3^{2} + 2^{2}} = \sqrt{9 + 4} = \sqrt{13}$.
Using a compass with centre $O$ and radius $OB$,draw an arc which intersects the number line at the point $C$. Then,$C$ corresponds to $\sqrt{13}$.
$13 = 9 + 4 = 3^{2} + 2^{2}$
On the number line,take $OA = 3$ units.
Draw $BA = 2$ units,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{OA^{2} + AB^{2}} = \sqrt{3^{2} + 2^{2}} = \sqrt{9 + 4} = \sqrt{13}$.
Using a compass with centre $O$ and radius $OB$,draw an arc which intersects the number line at the point $C$. Then,$C$ corresponds to $\sqrt{13}$.
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