Locate $\sqrt{13}$ on the number line.
Locate $\sqrt{13}$ on the number line.
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 $OB$ (see $Fig.$).
By Pythagoras theorem,
$OB =\sqrt{13}$
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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