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 $O A=1$ unit.
Draw perpendicular $A Z$ at $A$ on the number line and cut-off arc $A B=1$ unit.By Pythagoras Theorem,
$O B^2=O A^2+A B^2$
$=1^2+1^2=1+1=2$
$\Rightarrow OB=\sqrt{2}$
Taking O as centre and $OB =\sqrt{2}$ as radius draw an arc cutting real line at C .
Clearly, $OC = OB =\sqrt{2}$ 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,
$OE^2=OC^2+CE^2$
$=(\sqrt{2})^2+1^2=2+1=3$
$\Rightarrow OE=\sqrt{3}$
Taking O as centre and $OE =\sqrt{3}$ as radius draw an arc cutting real line at D .
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 $O A=1$ unit.
Draw perpendicular $A Z$ at $A$ on the number line and cut-off arc $A B=1$ unit.By Pythagoras Theorem,
$O B^2=O A^2+A B^2$
$=1^2+1^2=1+1=2$
$\Rightarrow OB=\sqrt{2}$
Taking O as centre and $OB =\sqrt{2}$ as radius draw an arc cutting real line at C .
Clearly, $OC = OB =\sqrt{2}$ 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,
$OE^2=OC^2+CE^2$
$=(\sqrt{2})^2+1^2=2+1=3$
$\Rightarrow OE=\sqrt{3}$
Taking O as centre and $OE =\sqrt{3}$ as radius draw an arc cutting real line at D .
Hence, D represents $\sqrt{3}$ on the number line.
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