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2. How to represent square roots of real numbers geometrically

Let us recall the concept of the square number we learned so far.

$a$ 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.

$a > 0$ 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

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

$\begin{array}{l} BD = \sqrt{{OD}^{2} - {OB}^{2}} \\ BD = \sqrt{{(\frac{x + 1}{2})}^{2} - {(\frac{x - 1}{2})}^{2}} \\ BD = \sqrt{\frac{x^{2} + 1 + 2 x - x^{2} - 1 + 2 x}{4}} \\ BD = \sqrt{\frac{4 x}{4}} \\ BD = \sqrt{x} \end{array}$

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

$\begin{array}{l} BD = \sqrt{{OD}^{2} - {OB}^{2}} \\ BD = \sqrt{{(2.8)}^{2} - {(1.8)}^{2}} \\ BD = \sqrt{7.84 - 3.24} \\ BD = \sqrt{4.6} \\ BD = 2.145 \end{array}$

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2. How to represent square roots of real numbers geometrically

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