Question
Locate $\sqrt{8}$ 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 $OA = 2$ units.
Draw perpendicular $AZ$ at $A$ on the number line and cut-off arc $AB = 2$ units.
By Pythagoras Theorem,
$OB^2 = OA^2 + AB^2$
$= 2^2 + 2^2= 4 + 4 = 8$
$\Rightarrow\text{OB}=\sqrt{8}$
Taking $O$ as centre and $\text{OB}=\sqrt{8}$ as radius draw an arc cutting real line at $C.$
Clearly, $\text{OC}=\text{OB}=\sqrt{8}$
Hence, $C$ represents $\sqrt{8}$ on the number line.
On the number line, take point $O$ corresponding to zero.
Now take point $A$ on number line such that $OA = 2$ units.
Draw perpendicular $AZ$ at $A$ on the number line and cut-off arc $AB = 2$ units.
By Pythagoras Theorem,
$OB^2 = OA^2 + AB^2$
$= 2^2 + 2^2= 4 + 4 = 8$
$\Rightarrow\text{OB}=\sqrt{8}$
Taking $O$ as centre and $\text{OB}=\sqrt{8}$ as radius draw an arc cutting real line at $C.$
Clearly, $\text{OC}=\text{OB}=\sqrt{8}$
Hence, $C$ represents $\sqrt{8}$ on the number line.
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