Question
Locate $\sqrt{5}$ on the number line.
✓
Answer
Here, $5=2^2+1^2 9$So, draw a right angled $\triangle\text{OAB}$ in which $OA = 2$ units and $AB = 1$ unit and $\angle\text{OAB}=90^\circ$ By using Pythagoras theorem, we get $\text{OB}=\sqrt{\text{OA}^2+\text{AB}^2}$ $=\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt{5}$
Taking $\text{OB}=\sqrt{5}$ as radius and point $O$ as centre, draw an arc which meets the number line at point $P$ on the positive side of it. Hence, it is clear that point P represents $\sqrt{5}$ on the number line.
Taking $\text{OB}=\sqrt{5}$ as radius and point $O$ as centre, draw an arc which meets the number line at point $P$ on the positive side of it. Hence, it is clear that point P represents $\sqrt{5}$ on the number line.
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