Lecture 5: Geometrical Representation of Square Roots

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Geometrical Representation of Square Roots | Properties of Square Roots
Mathematics | Class 9 | CBSE & SEBA Board

Geometrical Representation of Square Roots
Properties of Square Roots

Edunes Online Education

BOARD: CBSE | CLASS: 9 | Mathematics | Square Roots
📌 Geometrical Representation of Square Roots

Topic 1 : what does \( \sqrt{a} \) mean?

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For any real number a > 0,
\( \sqrt{a} = b \) means b² = a and b > 0
This definition works for both:

1. Natural numbers (e.g., √4 = 2, because 2² = 4)
2. Real numbers {e.g., \( \sqrt{3.5} \)

Topic 2 : How to represent \( \sqrt{x} \) geometrically ?

We now explore how to construct \( \sqrt{x} \) geometrically when x is a positive real number.

Construct \( \sqrt{3.5} \) geometrically

Steps:

1. Draw a line segment AB = 3.5 units on a number line.
2. From point B . mark 1 unit to the right. Let that point be C.
3. Find midpoint of AC, mark it as O.
4. Draw a semicircle with center O and radius OC.
5. Then , draw BD = \( \sqrt{3.5} \)

This gives a geometric way to represent \( \sqrt{3.5} \)

General Case: Represent \( \sqrt{x} \) geometrically

Steps for any real number x > 0:

1. Mark AB = x units.
2. From B, mark 1 unit to the right, name the point as C.
3. Find the midpoint O of AC.
4. Draw a semicircle with diameter AC.
5. Draw a perpendicular from B to the semicircle and mark intersection point as D.
6. Then, BD = \( \sqrt{x} \)

Proof using Pythagoras Theorem

In △OBD, \( \angle{OBD} \) = 90°, so it's a right triangle.
Let's calculate:

1. Radius of semicicrcle, OC = \( \dfrac{x + 1}{2} \)


2. OB = \( x - \dfrac{x + 1}{2} \) = \( \dfrac{x - 1}{2} \)

Using Pythagoras:
\( {BD}^2 = {OD}^2 - {OB}^2 \)

\( = {(\dfrac{x +1}{2})}^2 - {(\dfrac{x - 1}{2})}^2 \)

\( = \dfrac{{(x + 1)}^2 - {(x - 1)}^2}{4} \)

\( = \dfrac{4x}{4} = x \)

\( BD = \sqrt{x} \)

Topic 3: Properties of Square Roots

Let a, b > 0 be real numbers

1. \( \sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \)
2. \( \sqrt{ \dfrac{a}{b}} = \dfrac{ \sqrt{a}}{ \sqrt{b}} \)
3. \( (\sqrt{a} + \sqrt{b})( \sqrt{a} - \sqrt{b} ) = a - b \)
4. \( ( a + \sqrt{b})( a - \sqrt{b} ) = a^2 - b \)
5. \( ( \sqrt{a} + \sqrt{b} )( \sqrt{c} + \sqrt{d} ) \) = \( \sqrt{ac} + \sqrt{ad} + \sqrt{bc} + \sqrt{bd} \)
6. \( {( \sqrt{a} + \sqrt{b} )}^2 \) = \( a + 2\sqrt{ab} + b \)

Exercise for practice

1. Geometrically construct \( \sqrt{2.5} \) using the method described
2. Prove that \( {( \sqrt{5} + \sqrt{3} )}^2 = 5 + 2\sqrt{15} + 3 \)
3. Use Pythagoras Theorem to verify that your geometric construction is correct
4. Show that \( \sqrt{8} \cdot \sqrt{2} = \sqrt{16} \)

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