Theory:

Let us recall the concept of the square number we learned so far.
If '\(a\)' is a natural number, then 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 a \(= b\). That is, \(b² = a\) and \(b > 0\).
Let us learn how to find x for any given positive real number '\(x\)' geometrically.
The general procedure to represent x geometrically:
 
Step 1: If the given positive number is '\(x\)' we have to mark '\(B\)' so that \(AB = x\) units.
 
7.png
 
Step 2: From point '\(B\)', mark a distance of \(1\) unit and label the point as '\(C\)'. That is \(BC =1\) unit.
 
4.png
 
Step 3: Now find the midpoint of \(AC\) and mark the midpoint as '\(O\)'.
 
6.png
 
Step 4: Draw a semicircle with centre '\(O\)' and radius \(OC\).
 
9.png
 
Step 5: Draw a line perpendicular to \(AC\) which passes through point '\(B\)' and intersect the semicircle at '\(D\)' such that \(∠OBD = 90°\).
 
5.PNG
 
Step 6: Now measure the length of \(BD\), which is nothing but x.
 
16.png
 
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 \(OB = OC - 1 = (x+1)/2 -1 = (x-1)/2\).
 
BD=OD2OB2BD=x+122x122BD=x2+1+2xx21+2x4BD=4x4BD=x
 
Thus, the length of \(BD\) gives the value of 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.
 
3.png
 
Step 2: From point '\(B\)', mark a distance of \(1\) unit and label the point as '\(C\)'. That is \(BC =1\) unit.
 
2.png
 
Step 3: Now find the midpoint of \(AC = 4.6+1 = 5.6\) units and mark the midpoint as '\(O\)'.
 
5.png
 
Step 4: Draw a semicircle with centre '\(O\)' and radius \(OC =5.6/2 = 2.8\) units.
 
11.png
 
Step 5: Draw a line perpendicular to \(AC\) which passes through point '\(B\)' and intersect the semicircle at '\(D\)' such that \(∠OBD = 90°\).
 
10.png
 
Step 6: Now measure the length of \(BD = 2.145\) which is nothing but x.
 
15.png
 
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.
 
BD=OD2OB2BD=2.821.82BD=7.843.24BD=4.6BD=2.145