Lecture 5: Geometrical Representation of Square Roots

📘 Lecture 5: Geometrical Representation of Square Roots & Properties of Square Roots

🔹 Topic 1: What does √a mean?

For any real number $a > 0$,
$\sqrt{a} = b \text{ means } b^2 = a \text{ and } b > 0$

This definition works for both:

• Natural numbers (e.g., √4 = 2, because 2² = 4)

• Real numbers (e.g., √3.5)

🔹 Topic 2: How to Represent √x Geometrically?

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

📐 Example: Constructing √3.5 geometrically

Refer to Fig. 1.11

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. Draw a perpendicular line from point B to the semicircle. Let it intersect at point D.

6. Then, BD = √3.5

✅ This gives a geometric way to represent square root of 3.5.

📐 General Case: Represent √x geometrically

Refer to Fig. 1.12

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 = √x

🧠 Proof using Pythagoras Theorem

In △OBD, ∠OBD = 90°, so it’s a right triangle.

Let’s calculate:

• Radius of semicircle =
  $OC = \frac{x + 1}{2}$

• OB =
  $x - \frac{x + 1}{2} = \frac{x - 1}{2}$

Using Pythagoras:

$$BD^2 = OD^2 - OB^2 = \left( \frac{x+1}{2} \right)^2 - \left( \frac{x-1}{2} \right)^2 = \frac{(x+1)^2 - (x-1)^2}{4} = \frac{4x}{4} = x$$

Thus,

$$BD = \sqrt{x}$$

📘 Topic 3: Properties of Square Roots

Let $a, b > 0$ be real numbers.

1. $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$

2. $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

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 $\sqrt{8} \cdot \sqrt{2} = \sqrt{16}$

📌 Key Takeaway

We can visualize square roots of real numbers on a number line using semicircles and perpendicular lines, and use algebraic identities to simplify square root expressions.