Question
Locate $\sqrt{10}$ on the number line.
✓
Answer
Here, $10 = 3^2 + 1 $
So, draw a right angled $\triangle\text{OAB}$
in which $OA = 3 $units and $AB = 1$ unit and $\angle\text{OAB}=90^\circ$
By using Pythagoras theorem.
we get $\text{OB}=\sqrt{\text{OA}^2+\text{AB}^2}$
$=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}$
Taking $\text{OB}=\sqrt{10}$ as radius and point $O$ as centre, draw an arc which meets the number line at point $P$ on the positive side of it. The point $P$ represents $\sqrt{10}$ on the number line.
So, draw a right angled $\triangle\text{OAB}$
in which $OA = 3 $units and $AB = 1$ unit and $\angle\text{OAB}=90^\circ$
By using Pythagoras theorem.
we get $\text{OB}=\sqrt{\text{OA}^2+\text{AB}^2}$
$=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}$
Taking $\text{OB}=\sqrt{10}$ as radius and point $O$ as centre, draw an arc which meets the number line at point $P$ on the positive side of it. The point $P$ represents $\sqrt{10}$ on the number line.
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