Lecture 7: Worksheet 2: – Number System

 

📘 Lecture 7: Worksheet 2: – Number System

Class: 9 CBSE
Chapter: Number System
Topic: Mixed Problems
Worksheet Title: Consolidation and Skill Practice
Total Marks: 40
Time: 1 Hour

🔹 Section A: Fill in the Blanks (1 mark each)

1. $\sqrt{50} = \_\_\_$

2. $(a^3)^2 = \_\_\_$

3. $\sqrt{2} \times \sqrt{8} = \_\_\_$

4. A number that cannot be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, is called ___.

5. $\left(\frac{1}{4}\right)^{-1/2} = \_\_\_$

🔹 Section B: Match the Columns (2 marks each)

Match the expression in Column A with its simplified value in Column B.

Column A	Column B
a) $(5 - \sqrt{3})(5 + \sqrt{3})$	A. $8$
b) $\sqrt{12} \div \sqrt{3}$	B. $22$
c) $\frac{1}{\sqrt{7} - 2}$	C. $\frac{\sqrt{7} + 2}{3}$
d) $2^3 \cdot 2^{-2}$	D. $5^2 - 3$

🔹 Section C: Solve the Following (3 marks each)

6. Simplify:
   $(\sqrt{5} + 2)^2$

7. Rationalise the denominator and simplify:
   $\frac{2}{\sqrt{3} + \sqrt{2}}$

8. If $a = 2^{1/3}$ and $b = 2^{2/3}$, then evaluate $a \cdot b$

9. Classify the following as Rational or Irrational:
   a) $\sqrt{121}$
   b) $\frac{3\sqrt{2}}{7}$
   c) $\pi - 3$

🔹 Section D: Application-Based (4 marks each)

10. A rope is $\sqrt{75}$ metres long. Another rope is $\sqrt{27}$ metres long.
    a) Find the total length in simplest form.
    b) Classify the result as rational or irrational.

11. Simplify using exponent laws:
    a) $5^{1/2} \cdot 5^{3/2}$
    b) $\left( \frac{4}{9} \right)^{-3/2}$

🔹 Section E: Conceptual & Reasoning (5 marks each)

12. If a number is written as $2 + \sqrt{5}$, show that its conjugate is $2 - \sqrt{5}$.
    Now multiply the two and determine the result.
    What conclusion can you draw about the product of irrational conjugates?

13. A student claims:

“$\frac{22}{7}$ is the value of $\pi$, so it must be a rational number.”
Do you agree with this claim? Justify your answer and explain the difference between approximation and actual value of irrational numbers.

📌 Challenge Task (5 bonus marks)

Construct a proof that $\sqrt{3}$ is irrational using contradiction method (proof by assumption).

✅ Student Instructions:

• Read each question carefully.

• Use pencil and scale for diagrams.

• Try all sections for full understanding.

• Marks are mentioned next to each question.

📘 Lecture 7 Worksheet 2: – Number System: Solutions

🔹 Section A: Fill in the Blanks

1. $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$

2. $(a^3)^2 = a^{3 \cdot 2} = a^6$

3. $\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$

4. Irrational number

5. $\left(\frac{1}{4}\right)^{-1/2} = (4)^{1/2} = \sqrt{4} = 2$

🔹 Section B: Match the Columns

Column A	Column B
a) $(5 - \sqrt{3})(5 + \sqrt{3})$	B. $22$
b) $\sqrt{12} \div \sqrt{3}$	A. $2$
c) $\frac{1}{\sqrt{7} - 2}$	C. $\frac{\sqrt{7} + 2}{3}$
d) $2^3 \cdot 2^{-2}$	D. $2^{3 - 2} = 2^1 = 2$

So matching answers:
a–B, b–A, c–C, d–D

🔹 Section C: Solve the Following

6. $(\sqrt{5} + 2)^2 = (\sqrt{5})^2 + 2 \cdot \sqrt{5} \cdot 2 + 2^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}$

7. $\frac{2}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{2(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}$
   Denominator: $3 - 2 = 1$
   Answer: $2(\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}$

8. $a = 2^{1/3}, b = 2^{2/3} \Rightarrow a \cdot b = 2^{1/3 + 2/3} = 2^1 = 2$

a) $\sqrt{121} = 11$ – Rational
b) $\frac{3\sqrt{2}}{7}$ – Irrational
c) $\pi - 3$ – Irrational (since π is irrational)

🔹 Section D: Application-Based

a) $\sqrt{75} + \sqrt{27} = \sqrt{25 \cdot 3} + \sqrt{9 \cdot 3} = 5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3}$
b) Since $\sqrt{3}$ is irrational, $8\sqrt{3}$ is also irrational.

a) $5^{1/2} \cdot 5^{3/2} = 5^{1/2 + 3/2} = 5^2 = 25$
b) $\left( \frac{4}{9} \right)^{-3/2} = \left( \frac{9}{4} \right)^{3/2} = \left( \sqrt{9}/\sqrt{4} \right)^3 = (3/2)^3 = 27/8$

🔹 Section E: Conceptual & Reasoning

Given number = $2 + \sqrt{5}$, its conjugate is $2 - \sqrt{5}$
Product:
$(2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1$
Conclusion: The product of conjugates $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$, always gives a rational number.

No, we do not agree.

• $\pi \approx \frac{22}{7}$, but this is only an approximation.

• Actual value of $\pi$ is non-terminating, non-repeating = irrational

• Rational approximations are used for calculations, but the true nature of $\pi$ remains irrational.

📌 Challenge Task: Proof $\sqrt{3}$ is irrational

Assume $\sqrt{3}$ is rational.
Then $\sqrt{3} = \frac{p}{q}$, where p, q are integers with no common factor and $q \neq 0$.
Squaring both sides:
$3 = \frac{p^2}{q^2} \Rightarrow p^2 = 3q^2$
So p² is divisible by 3 → p is divisible by 3 → p = 3k
Then $p^2 = 9k^2 = 3q^2 \Rightarrow q^2 = 3k^2 \Rightarrow q$ is also divisible by 3.

So both p and q are divisible by 3, contradicting that they have no common factor.
Hence, $\sqrt{3}$ is irrational.

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Solved Worksheet 1: Number System

✅ Solved Worksheet: Number System – Mixed Problems

Class: 9 CBSE
Total Marks: 40 | Time: 60 Minutes

🔹 Section A: Multiple Choice Questions (1 mark each)

1. $(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) = 7 - 5 = \boxed{2}$ → Answer: A

2. Rational number: $\boxed{\frac{3}{4}}$ → Answer: C

3. $(2\sqrt{3})^2 = 4 \times 3 = \boxed{12}$ → Answer: A

4. $16^{3/4} = (2^4)^{3/4} = 2^3 = \boxed{8}$ → Answer: B

5. $\left( \frac{1}{5^2} \right)^{-1} = 5^2 = \boxed{25}$ → Answer: B

🔹 Section B: Very Short Answer (2 marks each)

6. Rationalise:

$$\frac{1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{3 - 1} = \boxed{\frac{\sqrt{3} - 1}{2}}$$

7. $3 - \sqrt{11}$: Irrational, because $\sqrt{11}$ is irrational and difference with rational number remains irrational. → \boxed{\text{Irrational}}

8. $(5 + \sqrt{2})^2 = 25 + 2 \times 5 \times \sqrt{2} + 2 = \boxed{27 + 10\sqrt{2}}$

9. Geometrical Representation of $\sqrt{2}$:
   Draw a right-angled triangle with both legs of 1 unit. The hypotenuse is $\sqrt{2}$.
   Using compass, draw an arc from origin with radius $\sqrt{2}$. The intersection point on number line is $\boxed{\sqrt{2}}$.

a) $27^{1/3} = \boxed{3}$
b) $81^{3/4} = (3^4)^{3/4} = 3^3 = \boxed{27}$

🔹 Section C: Short Answer (3 marks each)

$$\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \frac{(\sqrt{5} + \sqrt{2})^2}{5 - 2} = \frac{5 + 2 + 2\sqrt{10}}{3} = \boxed{\frac{7 + 2\sqrt{10}}{3}}$$

a) $(3^2 \cdot 3^3) \div 3^4 = 3^{2+3-4} = 3^1 = \boxed{3}$

b) $(2^3)^2 \cdot 2^{-4} = 2^{6} \cdot 2^{-4} = 2^{2} = \boxed{4}$

a) $\sqrt{49} = 7$ → \boxed{\text{Rational}}
b) $\frac{2}{\sqrt{7}}$ → Denominator irrational → \boxed{\text{Irrational}}
c) $\sqrt{3} + \sqrt{7}$ → sum of irrationals not simplifiable → \boxed{\text{Irrational}}

🔹 Section D: Long Answer (4 marks each)

$$(2 + \sqrt{3})^2 - (2 - \sqrt{3})^2 = [4 + 4\sqrt{3} + 3] - [4 - 4\sqrt{3} + 3] = 7 + 4\sqrt{3} - (7 - 4\sqrt{3}) = \boxed{8\sqrt{3}}$$

$$\frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}} \\ = \frac{3(\sqrt{5} + \sqrt{2}) + 2(\sqrt{5} - \sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2} \\ = \frac{3\sqrt{5} + 3\sqrt{2} + 2\sqrt{5} - 2\sqrt{2}}{5 - 2} \\ = \frac{5\sqrt{5} + \sqrt{2}}{3} = \boxed{\frac{5\sqrt{5} + \sqrt{2}}{3}}$$

16. Prove $\sqrt{2}$ is irrational – contradiction method:

Assume $\sqrt{2} = \frac{p}{q}$, where $p, q$ are integers, $\gcd(p, q) = 1$

Squaring:
$2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2$

So, $p^2$ even ⇒ $p$ even ⇒ $p = 2k$
Then: $p^2 = 4k^2 = 2q^2 ⇒ q^2 = 2k^2$ ⇒ $q$ also even

⇒ $p$ and $q$ both even ⇒ Contradiction to assumption that they are coprime.

∴ $\boxed{\sqrt{2} \text{ is irrational}}$

a)

$$\left( \frac{8}{27} \right)^{2/3} = \frac{8^{2/3}}{27^{2/3}} = \frac{(2^3)^{2/3}}{(3^3)^{2/3}} = \frac{2^2}{3^2} = \frac{4}{9} \Rightarrow \boxed{\frac{4}{9}}$$

b)

$$8^{2/3} \div 2^{4/3} = (2^3)^{2/3} \div 2^{4/3} = 2^2 \div 2^{4/3} = 2^{2 - 4/3} = 2^{2/3} = \boxed{2^{2/3}}$$

🔹 Section E: Challenge Question (5 marks)

Rohit's statement:
“π is defined as the ratio of circumference to diameter ⇒ it must be rational” is incorrect.

Explanation:

• Though $\pi = \frac{C}{d}$, both circumference and diameter are real numbers.

• The ratio is not always rational.

• π has a non-terminating, non-repeating decimal expansion, and cannot be expressed as a fraction ⇒ Irrational.

Approximation Check:

Radius $r = 7 \text{ cm}$
Circumference $C = 2\pi r \approx 2 \times \frac{22}{7} \times 7 = 44 \text{ cm}$

True value using $\pi \approx 3.1416$:
$C = 2 \times 3.1416 \times 7 = 43.9824 \approx 44 \text{ cm}$

Hence, $\frac{22}{7}$ is a very good rational approximation of π, but π itself remains irrational.

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Worksheet: Number System – Mixed Problems

📄 Worksheet 1: Number System – Mixed Problems

Class: 9 CBSE
Chapter: Number System
Topic: Simplification, Rational/Irrational, Rationalisation, Exponents
Total Marks: 40
Time: 60 Minutes

🔹 Section A: Multiple Choice Questions (1 mark each)

Choose the correct option and write the answer.

1. $(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5})$ equals:

   • A. 2

   • B. $\sqrt{2}$

   • C. 12

   • D. $\sqrt{35}$

2. Which of the following is a rational number?

   • A. $\sqrt{2}$

   • B. $\frac{1}{\sqrt{3}}$

   • C. $\frac{3}{4}$

   • D. $\pi$

3. $(2\sqrt{3})^2 =$

   • A. 12

   • B. 6

   • C. $4\sqrt{3}$

   • D. 9

4. $16^{3/4}$ is equal to:

   • A. 4

   • B. 8

   • C. 16

   • D. 2

5. $\left( \frac{1}{5^2} \right)^{-1} =$

   • A. 5

   • B. 25

   • C. $\frac{1}{25}$

   • D. $-25$

🔹 Section B: Very Short Answer (2 marks each)

Answer in one step wherever possible.

6. Rationalise: $\frac{1}{\sqrt{3} + 1}$

7. Classify as rational or irrational: $3 - \sqrt{11}$

8. Simplify: $(5 + \sqrt{2})^2$

9. Represent $\sqrt{2}$ geometrically on a number line. (Sketch or describe method)

10. Find:
    a) $27^{1/3}$
    b) $81^{3/4}$

🔹 Section C: Short Answer Questions (3 marks each)

Show working where required.

11. Simplify and express in simplest form:
    $\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

12. Evaluate:
    a) $(3^2 \cdot 3^3) \div 3^4$
    b) $(2^3)^2 \cdot 2^{-4}$

13. Without actual calculation, state whether the following are rational or irrational. Justify:

    • a) $\sqrt{49}$

    • b) $\frac{2}{\sqrt{7}}$

    • c) $\sqrt{3} + \sqrt{7}$

🔹 Section D: Long Answer Questions (4 marks each)

Explain each step clearly.

14. Simplify the expression:

$$(2 + \sqrt{3})^2 - (2 - \sqrt{3})^2$$

15. Rationalise and simplify:

$$\frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}}$$

16. Prove that $\sqrt{2}$ is irrational.
    (Use contradiction method)

17. Evaluate:

$$\left( \frac{8}{27} \right)^{2/3}$$

Then use exponent laws to show:

$$8^{2/3} \div 2^{4/3}$$

🔹 Section E: Challenge Question (5 marks)

This tests higher-order thinking.

18. Rohit says that since π is defined as the ratio of circumference to diameter (C/d), it must be a rational number. Do you agree? Justify with reasoning. Then calculate an approximate value of π using a circle of radius 7 cm (Use $\pi \approx \frac{22}{7}$) and check the accuracy of this approximation.

✅ Instructions for Students:

• Solve each section step by step.

• For geometry-related questions, draw figures neatly.

• Highlight final answers.

• Use separate sheets if necessary.

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Worksheet: Number System – Mixed Problems

📄 Worksheet: Number System – Mixed Problems

Class: 9 CBSE

Chapter: Number System

Topic: Simplification, Rational/Irrational, Rationalisation, Exponents

Total Marks: 40

Time: 60 Minutes

🔹 Section A: Multiple Choice Questions (1 mark each)

Choose the correct option and write the answer.

1. \( (\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) \) equals:

• 2
• \( \sqrt{2} \)
• 12
• \( \sqrt{35} \)

2. Which of the following is a rational number?

• \( \sqrt{2} \)
• \( \frac{1}{\sqrt{3}} \)
• \( \frac{3}{4} \)
• \( \pi \)

3. \( (2\sqrt{3})^2 = \)

• 12
• 6
• \( 4\sqrt{3} \)
• 9

4. \( 16^{3/4} \) is equal to:

• 4
• 8
• 16
• 2

5. \( \left( \frac{1}{5^2} \right)^{-1} = \)

• 5
• 25
• \( \frac{1}{25} \)
• -25

🔹 Section B: Very Short Answer (2 marks each)

Answer in one step wherever possible.

6. Rationalise: \( \frac{1}{\sqrt{3} + 1} \)
7. Classify as rational or irrational: \( 3 - \sqrt{11} \)
8. Simplify: \( (5 + \sqrt{2})^2 \)
9. Represent \( \sqrt{2} \) geometrically on a number line. (Sketch or describe method)
10. Find:
    • a) \( 27^{1/3} \)
    • b) \( 81^{3/4} \)

🔹 Section C: Short Answer Questions (3 marks each)

11. Simplify and express in simplest form: \( \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \)
12. Evaluate:
    • a) \( (3^2 \cdot 3^3) \div 3^4 \)
    • b) \( (2^3)^2 \cdot 2^{-4} \)
13. Without actual calculation, state whether the following are rational or irrational. Justify:
    • a) \( \sqrt{49} \)
    • b) \( \frac{2}{\sqrt{7}} \)
    • c) \( \sqrt{3} + \sqrt{7} \)

🔹 Section D: Long Answer Questions (4 marks each)

14. Simplify the expression:
    \( (2 + \sqrt{3})^2 - (2 - \sqrt{3})^2 \)
15. Rationalise and simplify:
    \( \frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}} \)
16. Prove that \( \sqrt{2} \) is irrational. (Use contradiction method)
17. Evaluate:
    • \( \left( \frac{8}{27} \right)^{2/3} \)
    • Then use exponent laws to show: \( 8^{2/3} \div 2^{4/3} \)

🔹 Section E: Challenge Question (5 marks)

18. Rohit says that since \( \pi \) is defined as the ratio of circumference to diameter (C/d), it must be a rational number. Do you agree? Justify with reasoning. Then calculate an approximate value of \( \pi \) using a circle of radius 7 cm (Use \( \pi \approx \frac{22}{7} \)) and check the accuracy of this approximation.

✅ Instructions for Students:

• Solve each section step by step.
• For geometry-related questions, draw figures neatly.
• Highlight final answers.
• Use separate sheets if necessary.

Lecture 6: Simplification, Rational/Irrational Numbers, and Exponents

 This lecture includes simplification of expressions with square roots, classification of numbers, rationalisation of denominators, and laws of exponents.

📘 Lecture 6: Simplification, Rational/Irrational Numbers, and Exponents

🔹 Topic 1: Simplification Involving Square Roots

Let’s begin with examples that show how to simplify algebraic expressions containing square roots.

📍 Example 15: Simplify

1. $(5 + \sqrt{7})(2 + \sqrt{5})$
   Use distributive law (FOIL method):

   $$= 5 \cdot 2 + 5 \cdot \sqrt{5} + \sqrt{7} \cdot 2 + \sqrt{7} \cdot \sqrt{5} = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}$$

2. $(5 + \sqrt{5})(5 - \sqrt{5})$
   Identity: $(a + b)(a - b) = a^2 - b^2$

   $$= 25 - 5 = 20$$

3. $(\sqrt{3} + \sqrt{7})^2$
   Identity: $(a + b)^2 = a^2 + 2ab + b^2$

   $$= 3 + 2\sqrt{21} + 7 = 10 + 2\sqrt{21}$$

4. $(\sqrt{11} - \sqrt{7})(\sqrt{11} + \sqrt{7})$
   Identity: $(a - b)(a + b) = a^2 - b^2$

   $$= 11 - 7 = 4$$

🔹 Topic 2: Classifying Numbers as Rational or Irrational

A number is irrational if it cannot be expressed as a fraction $\frac{p}{q}$, where p and q are integers and $q \ne 0$.

📍 Classify the following:

1. $2 - \sqrt{5}$ → Irrational
   (Rational - Irrational = Irrational)

2. $(3 + \sqrt{23}) - \sqrt{23} = 3$ → Rational

3. $\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}$ → Rational

4. $\frac{1}{\sqrt{2}}$ → Irrational

5. $2\pi$ → Irrational

🔹 Topic 3: Simplifying Square Root Expressions

Use standard identities and combine like terms.

Examples:

1. $(3 + \sqrt{3})(2 + \sqrt{2})$

2. $(3 + \sqrt{3})(3 - \sqrt{3})$

3. $(\sqrt{5} + \sqrt{2})^2$

4. $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$

👉 Use formulas like:

• $(a + b)^2 = a^2 + 2ab + b^2$

• $(a + b)(a - b) = a^2 - b^2$

🔹 Topic 4: Think and Discuss

📌 Why is π irrational, even though it's defined as a ratio (C/d)?

• π is defined as the ratio of circumference to diameter, but it's not a ratio of two integers.

• So while it appears rational in form, it is not expressible exactly as a fraction.

• Its decimal form is non-terminating and non-repeating, making it irrational.

🔹 Topic 5: Number Line Representation

📍 Represent $\sqrt{3}$ on the number line using geometry (similar to the method in Lecture 5).

1. Mark 0 and 1 on the number line.

2. Construct segment 1.5 units from 0, form a right triangle.

3. Use Pythagoras’ Theorem to locate $\sqrt{3}$.

🔹 Topic 6: Rationalising the Denominator

To rationalise means to eliminate the square root from the denominator.

Examples:

1. $\frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

2. $\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}$

3. $\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{3}$

4. $\frac{1}{\sqrt{7} - 2} = \text{Multiply numerator and denominator by } \sqrt{7} + 2$

🔹 Topic 7: Laws of Exponents

Let $a \ne 0$, and m, n are integers.

1. $a^m \cdot a^n = a^{m+n}$

2. $(a^m)^n = a^{mn}$

3. $\frac{a^m}{a^n} = a^{m-n}$

4. $a^m \cdot b^m = (ab)^m$

🔹 Exercise Practice (from Ex 1.5)

1. Find:

• $64^{\frac{2}{3}} = (4^3)^{2/3} = 4^2 = 16$

• $32^{\frac{3}{5}} = (2^5)^{3/5} = 2^3 = 8$

2. Simplify powers using laws of exponents:

• $2^3 \cdot 2^5 = 2^8$

• $\left( \frac{1}{3^7} \right)^{-1} = 3^7$

• $\frac{11^{\frac{1}{2}} \cdot 11^{\frac{1}{4}}}{11^{\frac{3}{4}}} = 11^{(1/2 + 1/4 - 3/4)} = 11^0 = 1$

📝 Homework

1. Rationalise: $\frac{1}{\sqrt{3} + \sqrt{2}}$

2. Classify: $\sqrt{49} + \pi$

3. Simplify: $(\sqrt{3} + 1)^2$

4. Prove: $\sqrt{2}$ is irrational.

Lecture 5: Geometrical Representation of Square Roots

  Edunes Online Education

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?

>

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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