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.

$\mathrm{BD}^{2}=2 \mathrm{OC} \times \mathrm{BC}-(\mathrm{BC})^{2}$

$=2 \times 5.75 \times 1-1$

$=10.5$

$\Rightarrow \mathrm{BD}=\sqrt{10.5}$

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Updated on: 10-Oct-2022

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