Question
Locate $\sqrt{10}$ on the number line.
✓
Answer
We can write $10$ as
$10=9+1=3^2+1^2$
Draw $OA =3$ units, on the number line
Draw $BA =1$ unit, perpendicular to $OA .$
Join $OB$
Figure:
Now, by Pythagoras theorem,
$\ce{OB^2=AB^2 + OA^2}$
$OB^2=1^2+3^2=10$
$\Rightarrow OB=\sqrt{10}$
Taking $O$ as centre and $OB$ as a radius, draw an arc which intersects the number line at point $C.$
Clearly, $OC$ corresponds to $\sqrt{10}$ on the number line.
$10=9+1=3^2+1^2$
Draw $OA =3$ units, on the number line
Draw $BA =1$ unit, perpendicular to $OA .$
Join $OB$
Figure:
Now, by Pythagoras theorem,
$\ce{OB^2=AB^2 + OA^2}$
$OB^2=1^2+3^2=10$
$\Rightarrow OB=\sqrt{10}$
Taking $O$ as centre and $OB$ as a radius, draw an arc which intersects the number line at point $C.$
Clearly, $OC$ corresponds to $\sqrt{10}$ on the number line.
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