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,
OB2 = OA2 + AB2
= 12 + 12 = 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.
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,
OB2 = OA2 + AB2
= 12 + 12 = 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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