### Theory:

Let us recall the concept of the square number we learned so far.
If '$$a$$' is a natural number, then $\sqrt{a}$ $$= b$$. That is, $$b² = a$$ and $$b > 0$$.
We are now going to extend the same definition to positive real numbers.
If $$a > 0$$ be a real number, then $\sqrt{a}$ $$= b$$. That is, $$b² = a$$ and $$b > 0$$.
Let us learn how to find $\sqrt{x}$ for any given positive real number '$$x$$' geometrically.
The general procedure to represent $\sqrt{x}$ geometrically:

Step 1: If the given positive number is '$$x$$' we have to mark '$$B$$' so that $$AB = x$$ units.

Step 2: From point '$$B$$', mark a distance of $$1$$ unit and label the point as '$$C$$'. That is $$BC =1$$ unit.

Step 3: Now find the midpoint of $$AC$$ and mark the midpoint as '$$O$$'.

Step 4: Draw a semicircle with centre '$$O$$' and radius $$OC$$.

Step 5: Draw a line perpendicular to $$AC$$ which passes through point '$$B$$' and intersect the semicircle at '$$D$$' such that $$∠OBD = 90°$$.

Step 6: Now measure the length of $$BD$$, which is nothing but $\sqrt{x}$.

Let us verify the value by applying Pythagoras theorem.

Consider the right triangle $$OBD$$.

$$OD²=OB²+BD²$$

We know that $$OD = OC$$(radius).

Here $$AC = AB +BC = x+1$$. So $$OC = (x+1)/2 = OD$$ .

Then $$OB = OC - 1 = (x+1)/2 -1 = (x-1)/2$$.

$\begin{array}{l}\mathit{BD}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{{\mathit{OD}}^{2}-{\mathit{OB}}^{2}}\\ \\ \mathit{BD}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{{\left(\frac{x+1}{2}\right)}^{2}-{\left(\frac{x-1}{2}\right)}^{2}}\\ \\ \mathit{BD}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{\frac{{x}^{2}+1+2x-{x}^{2}-1+2x}{4}}\\ \\ \mathit{BD}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{\frac{4x}{4}}\\ \\ \mathit{BD}\phantom{\rule{0.147em}{0ex}}=\sqrt{x}\end{array}$

Thus, the length of $$BD$$ gives the value of $\sqrt{x}$.
Example:
Now we will try to find square root for the number $$4.6$$

Step 1: If the given positive number is '$$x$$' we have to mark '$$B$$' so that $$AB = 4.6$$ units.

Step 2: From point '$$B$$', mark a distance of $$1$$ unit and label the point as '$$C$$'. That is $$BC =1$$ unit.

Step 3: Now find the midpoint of $$AC = 4.6+1 = 5.6$$ units and mark the midpoint as '$$O$$'.

Step 4: Draw a semicircle with centre '$$O$$' and radius $$OC =5.6/2 = 2.8$$ units.

Step 5: Draw a line perpendicular to $$AC$$ which passes through point '$$B$$' and intersect the semicircle at '$$D$$' such that $$∠OBD = 90°$$.

Step 6: Now measure the length of $$BD = 2.145$$ which is nothing but $\sqrt{x}$.

Let us verify the value by applying Pythagoras theorem.

Consider the right triangle $$OBD$$.

$$OD²=OB²+BD²$$

We know that $$OD = OC = 2.8$$ (radius).

Now $$OB = OC - BC = 2.8-1 = 1.8$$ units.

$\begin{array}{l}\mathit{BD}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{{\mathit{OD}}^{2}-{\mathit{OB}}^{2}}\\ \\ \mathit{BD}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{{\left(2.8\right)}^{2}-{\left(1.8\right)}^{2}}\\ \\ \mathit{BD}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{7.84-3.24}\\ \\ \mathit{BD}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{4.6}\\ \\ \mathit{BD}\phantom{\rule{0.147em}{0ex}}=2.145\end{array}$