Class 8 Ganita Prakash Part 2 Chapter 2 The Baudhāyana-Pythagoras Theorem: Complete NCERT Solutions, Examples & Figure it Out Answers

Introduction

“The Baudhāyana-Pythagoras Theorem” is Chapter 2 of Ganita Prakash Part-II (Class 8). It introduces one of the most important theorems in geometry through ancient Indian constructions from Baudhāyana’s Śulba-Sūtra (c. 800 BCE). You will learn how to double or halve the area of a square, find the hypotenuse of an isosceles right triangle (introducing √2), prove the general theorem a² + b² = c², generate integer triples, and explore real-life applications. The chapter uses paper folding, diagrams, and multiple proofs to build deep understanding.

Basic Knowledge Required

• Area of square = side × side
• Properties of square (equal sides, 90° angles, diagonals bisect angles)
• Right triangle and hypotenuse
• Congruent triangles
• Decimal approximations and bounds
• Odd numbers and square numbers

Key Definitions

• Hypotenuse: Side opposite the right angle in a right triangle.
• Baudhāyana triple: Positive integers (a, b, c) where a² + b² = c² (c = hypotenuse).
• Primitive triple: A Baudhāyana triple with no common factor > 1.
• Scaled triple: Multiply a primitive triple by any positive integer k.

Important Formulas Used

1. Baudhāyana-Pythagoras Theorem: $a^2 + b^2 = c^2$ (c = hypotenuse)
2. Isosceles right triangle: $c = a\sqrt{2}$
3. Area of square on diagonal = 2 × area of original square
4. For integer triples: If (a, b, c) works, then (ka, kb, kc) also works
5. Decimal bounds for √2: 1.414 < √2 < 1.415 (non-terminating, non-repeating)

2.1 Doubling a Square – All Solved Examples & Figure it Out

Main Idea: The diagonal of a square produces a square of double the area.

Why it works:

• Original square area = 2 small congruent triangles.
• New square on diagonal = 4 small congruent triangles.
• All triangles are congruent because vertical & horizontal sides bisect the 90° angles of the dotted square (diagonal property of square).

Doubling Using Paper: Cut Square 1 into 4 triangles (1,2,3,4). Cut Square 2 into 4 triangles (5,6,7,8). Place 5,6,7,8 around Square 1 → perfect larger square of double area.

Figure it Out (Page 39 Question 1): Cut two identical squares diagonally into triangles 1-2-3-4 and 1-2-3-4. Arrange the four triangles from second square around the first square to form a larger square of double area. (Pieces fit perfectly along edges.)

2.2 Halving a Square – All Solved Examples & Figure it Out

Method: Fold square inward so creases pass through midpoints of sides → inner square PQRS has half the area.

Why PQRS is a square and half area:

• Connect QS and PR → four right-angled isosceles triangles.
• All angles 90° and sides equal by folding symmetry.
• Area proof: Original = 4 small triangles; inner square = 2 small triangles → half area.

Figure it Out:

• Half side length square fills original square 4 times (area ¼).
• PQRS is square because all sides equal and angles 90° (congruence + folding).

2.3 Hypotenuse of an Isosceles Right Triangle

Example: Side 1 unit → hypotenuse c where c² = 2 → c = √2.

Decimal Representation of √2:

• 1 < √2 < 2 (1²=1, 2²=4)
• 1.4 < √2 < 1.5 (1.4²=1.96, 1.5²=2.25)
• 1.41 < √2 < 1.42
• 1.414 < √2 < 1.415
• Non-terminating, non-repeating (proof: terminating decimal square ends in non-zero digit, but 2 ends in 0).
• Irrational: Cannot be m/n (leads to 2n² = m² → prime 2 has odd exponent on left, even on right – impossible).

Figure it Out (Page 39 Question 2): (i) Side 3: Hypotenuse between 4.24 and 4.25 (bounds 4.2 < c < 4.3) (ii) Side 4: 5.65–5.66 (iii) Side 6: 8.48–8.49 (iv) Side 8: 11.31–11.32 (v) Side 9: 12.72–12.73

Figure it Out (Page 39 Question 3): Hypotenuse 10 → area of square on hypotenuse = 100. Each leg square area = 50 → legs = √50 = 5√2 each.

General Formula: c² = 2a²

Example 1: a=12 → c=√288 ≈ between 16 and 17.

Example 2: c=√72 → a=6.

2.4 Combining Two Different Squares

Baudhāyana’s Method: Form right triangle with legs = sides of two squares → square on hypotenuse has area = sum of two squares.

Proof: New square sides equal (congruent triangles T,U,V,W,X). All angles 90°. Area = T+U+V = original two squares.

Paper Activity: Cut & rearrange three pieces → covers hypotenuse square exactly → a² + b² = c².

2.5 Right-Triangles Having Integer Sidelengths (Baudhāyana Triples)

Listed triples: (3,4,5), (5,12,13), (8,15,17), etc.

Figure it Out (Page 48): Triples ≤20: (3,4,5), (6,8,10), (9,12,15), (12,16,20), (5,12,13), (8,15,17)

Scaled versions: (3k,4k,5k) always works (proof: 9k² + 16k² = 25k²). Infinite triples.

Primitive triples: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (20,21,29) etc.

Generating more using odd squares:

• n=5: 4² + 3² = 5²
• n=13: 12² + 5² = 13² (And so on – see Figure it Out below)

Figure it Out (Page 49):

1. More triples: n=25 → 24² + 7² = 25²; n=41 → 40² + 9² = 41²; n=61 → 60² + 11² = 61²; etc.
2. Yes, non-primitive (e.g., multiples appear). Smaller leg = n-1 (one less than hypotenuse).
3. Yes – (5,12,13) cannot be generated this way (not of form (n-1)² + odd square).

2.6 A Long-Standing Open Problem (Fermat’s Last Theorem)

xⁿ + yⁿ = zⁿ has no solutions for n>2 (proved by Andrew Wiles 1994).

2.7 Further Applications

Bhāskarāchārya Lotus Problem: Stem above water = 1, tip moves 3 units → depth x=4 (3² + x² = (x+1)²).

Figure it Out (Page 52–53):

1. Diagonal of square side 5: √50 = 5√2 cm.
2. Missing sides:
   • 7-? -9 → √(81-49)=√32=4√2
   • 10-? -√200 → 10
   • 40-? -41 → 9
   • 10-? -√150 → 5√5
   • 27-? -45 → 36
3. Rhombus side: √[(24/2)² + (70/2)²] = √(144 + 1225) = √1369 = 37 units.
4. Yes – hypotenuse longest (a² + b² = c² > a² and > b²).
5. True (every triple is primitive or scaled primitive).
6. Examples: 3-4-5, 5-12-13, 6-8-10, 7-24-25, 8-15-17 rectangles.
7. Difference square: Construct using halving/doubling methods or paper.
8. (i) (a) Yes (√2 on grid), (b) No integer side, (c) Yes, (d) Yes. (ii) Only perfect square areas possible on integer grid.

15 Most Asked FAQs – The Baudhāyana-Pythagoras Theorem (Click to Expand)

1. How does the diagonal of a square give double area?

Answer: Original square = 2 triangles. Square on diagonal = 4 congruent triangles → double area.

2. Prove √2 is irrational (cannot be m/n).

Answer: Assume √2 = m/n → 2n² = m². Prime 2 has odd exponent left, even right → impossible.

3. Find hypotenuse of isosceles right triangle with legs 12.

Answer: c = √(2×144) = √288 ≈ 16.97 (between 16 and 17).

4. Is (5,12,13) a Baudhāyana triple? Verify.

Answer: 25 + 144 = 169 → yes.

5. Generate next primitive triple using odd-square method.

Answer: n=25: 24² + 7² = 25² → (7,24,25) triple.

6. Solve lotus problem: stem 1 unit above, tip moves 3 units.

Answer: 3² + x² = (x+1)² → x=4 units depth.

7. Diagonal of square side 5 cm?

Answer: √(25×2) = 5√2 cm.

8. Rhombus diagonals 24 and 70. Side length?

Answer: √[(12)² + (35)²] = 37 units.

9. Is every Baudhāyana triple primitive or scaled?

Answer: True.

10. Hypotenuse 17, one leg 8. Other leg?

Answer: √(289-64) = √225 = 15 cm.

11. Construct square of area 3 times given square.

Answer: Use successive doubling + halving or combine three squares via right triangles (Baudhāyana Verse 1.10).

12. Decimal bounds for √2 with one digit after decimal.

Answer: 1.4 < √2 < 1.5.

13. Why can’t √2 be terminating decimal?

Answer: Square would end in non-zero digit, but 2 ends in 0 → contradiction.

14. Fermat’s Last Theorem – what does it say?

Answer: xⁿ + yⁿ = zⁿ has no positive integer solutions for n>2 (proved 1994 by Wiles).

15. Give 5 integer rectangles (sides + diagonal integer).

Answer: 3×4 (5), 5×12 (13), 6×8 (10), 7×24 (25), 8×15 (17).

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