Question
Locate $\sqrt{3}$ 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 $OA = 1$ unit.
Draw perpendicular $AZ$ at $A$ on the number line and cut-off arc $AB = 1 $unit.
By Pythagoras Theorem, $OB^2 = OA^2 + AB^2 = 1^2 + 1^2= 1 + 1 = 2$
$\Rightarrow\text{OB}=\sqrt{2}$ Taking $O$ as centre and $\text{OB}=\sqrt{2}$ as radius draw an arc cutting real line at $C$.
Clearly, $\text{OC}=\text{OB}=\sqrt{2}$ Thus, $C$ represents $\sqrt{2}$ on the number line. .
Now, draw perpendicular $CY$ at $C$ on the number line and cut-off arc $CE = 1$ unit.
By Pythagoras Theorem, $\text{OE}^2=\text{OC}^2+\text{CE}^2$
$=\big(\sqrt{2}\big)^2+1^2=2+1=3$
$\Rightarrow\text{OE}=\sqrt{3}$ Taking $O$ as centre and $\text{OE}=\sqrt{3}$ as radius draw an arc cutting real line at $D$.
Clearly, $\text{OD}=\text{OE}=\sqrt{3}$
Hence, $D$ represents $\sqrt{3}$ on the number line.
Now take point $A$ on number line such that $OA = 1$ unit.
Draw perpendicular $AZ$ at $A$ on the number line and cut-off arc $AB = 1 $unit.
By Pythagoras Theorem, $OB^2 = OA^2 + AB^2 = 1^2 + 1^2= 1 + 1 = 2$
$\Rightarrow\text{OB}=\sqrt{2}$ Taking $O$ as centre and $\text{OB}=\sqrt{2}$ as radius draw an arc cutting real line at $C$.
Clearly, $\text{OC}=\text{OB}=\sqrt{2}$ Thus, $C$ represents $\sqrt{2}$ on the number line. .
Now, draw perpendicular $CY$ at $C$ on the number line and cut-off arc $CE = 1$ unit.
By Pythagoras Theorem, $\text{OE}^2=\text{OC}^2+\text{CE}^2$
$=\big(\sqrt{2}\big)^2+1^2=2+1=3$
$\Rightarrow\text{OE}=\sqrt{3}$ Taking $O$ as centre and $\text{OE}=\sqrt{3}$ as radius draw an arc cutting real line at $D$.
Clearly, $\text{OD}=\text{OE}=\sqrt{3}$
Hence, $D$ represents $\sqrt{3}$ on the number line.
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