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Represent $\sqrt{3.5},\sqrt{9.4}$ and $\sqrt{10.5}$ on the real number line.
Given:
Given numbers are $\sqrt{3.5},\sqrt{9.4}$ and $\sqrt{10.5}$
To do:
We have to represent $\sqrt{3.5},\sqrt{9.4}$ and $\sqrt{10.5}$ on the number line.
Solution:
1. Draw a line segment $AB=3.5$ units.
2. Produce $B$ till point $C$, such that $BC=1$ unit.
3. Find the mid-point of $\mathrm{AC}$, let it be $\mathrm{O}$.
4. Taking $O$ as the centre, draw a semi-circle, passing through $A$ and $C$.
5. Draw a line passing through $B$ perpendicular to $OB$, and cutting semicircle at $D$.
6. Consider $B$ as the centre and $BD$ as radius draw an arc cutting $OC$ produced at $E$.
In right angled triangle $\mathrm{OBD}$,
By Pythagoras theorem,
$\mathrm{BD}^{2}=\mathrm{OD}^{2}-\mathrm{OB}^{2}$
$=O C^{2}-(O C-B C)^{2}$ [Since $\mathrm{OD}=\mathrm{OC})$]
$\mathrm{BD}^{2}=2 \mathrm{OC} \times \mathrm{BC}-(\mathrm{BC})^{2}$
$=2 \times 2.25 \times 1-1$
$=3.5$
$\Rightarrow \mathrm{BD}=\sqrt{3.5}$
1. Draw a line segment $AB=9.4$ units.
Follow steps 2 to 6 as mentioned above.
$\mathrm{BD}^{2}=2 \mathrm{OC} \times \mathrm{BC}-(\mathrm{BC})^{2}$
$=2 \times 5.2 \times 1-1$
$=9.4$
$\Rightarrow \mathrm{BD}=\sqrt{9.4}$
1. Draw a line segment $AB=10.5$ units.
Follow steps 2 to 6 as mentioned above.
$\mathrm{BD}^{2}=2 \mathrm{OC} \times \mathrm{BC}-(\mathrm{BC})^{2}$
$=2 \times 5.75 \times 1-1$
$=10.5$
$\Rightarrow \mathrm{BD}=\sqrt{10.5}$