Question
Represent $\sqrt{4.5}$ on the number line.
✓
Answer
Consider, $AB =4.5$ units.
Extend $AB$ upto point $C$ such that $BC =1$ unit.
$\therefore AC=4.5+1=5.5 \text { units. }$
Now mark $O$ as the midpoint of $AC .$
With $O$ as centre and radius $OC$ draw a semicircle.
Draw perpendicular $BD$ on $AC$ which intersect the semicircle at $D .$
This length $BD =\sqrt{4.5}$ units.
To show $BD$ on the number line, consider line $\text{ABC}$ as number line with point $B$ as zero.
Therefore, $BC=1$ unit.
With $B$ as centre and radius $B D$ draw an arc which intersects number line $\text{ABC}$ at $E.$
So, point $E$ represents $\sqrt{4.5}$
$AB =4.5 \text { units }$
$BC =1 \text { unit }$
$\text{BD=BE}=\sqrt{4.5} \text { units }$
Extend $AB$ upto point $C$ such that $BC =1$ unit.
$\therefore AC=4.5+1=5.5 \text { units. }$
Now mark $O$ as the midpoint of $AC .$
With $O$ as centre and radius $OC$ draw a semicircle.
Draw perpendicular $BD$ on $AC$ which intersect the semicircle at $D .$
This length $BD =\sqrt{4.5}$ units.
To show $BD$ on the number line, consider line $\text{ABC}$ as number line with point $B$ as zero.
Therefore, $BC=1$ unit.
With $B$ as centre and radius $B D$ draw an arc which intersects number line $\text{ABC}$ at $E.$
So, point $E$ represents $\sqrt{4.5}$
$AB =4.5 \text { units }$
$BC =1 \text { unit }$
$\text{BD=BE}=\sqrt{4.5} \text { units }$
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