Locate $\sqrt{13}$ on the number line.

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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 $\sqrt{13}$

1099-s36

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