Question
Locate $\sqrt{10}$ on the number line.
✓
Answer
Draw a number line as shown.
On the number line, take point $O$ corresponding to zero.
Now take point $A$ on number line such that $OA = 3$ units.
Draw perpendicular $AZ$ at $A$ on the number line and cut-off arc $AB = 1$ unit.
By Pythagoras Theorem, $OB^2 = OA^2 + AB^2 = 3^2 + 1^2= 9 + 1 = 10 \Rightarrow\text{OB}=\sqrt{10}$
Taking $O$ as centre and $\text{OB}=\sqrt{10}$ as radius draw an arc cutting real line at $C$.
Clearly, $\text{OC}=\text{OB}=\sqrt{10}$
Hence, $C$ represents $\sqrt{10}$ on the number line.
On the number line, take point $O$ corresponding to zero.
Now take point $A$ on number line such that $OA = 3$ units.
Draw perpendicular $AZ$ at $A$ on the number line and cut-off arc $AB = 1$ unit.
By Pythagoras Theorem, $OB^2 = OA^2 + AB^2 = 3^2 + 1^2= 9 + 1 = 10 \Rightarrow\text{OB}=\sqrt{10}$
Taking $O$ as centre and $\text{OB}=\sqrt{10}$ as radius draw an arc cutting real line at $C$.
Clearly, $\text{OC}=\text{OB}=\sqrt{10}$
Hence, $C$ represents $\sqrt{10}$ on the number line.
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