Suryamg Master

Chapter 2: The Baudhāyana-Pythagoras Theorem

Simulation Topics

Simulation 1: Doubling a Square

Learning Objective: Explore how constructing a square on the diagonal of an original square yields exactly double its area.

Original Area = 25 sq cm | Dotted Square Area (on Diagonal) = 50 sq cm (Double!)

Simulation 2: Right Triangle Hypotenuse Calculator

Learning Objective: Verify Baudhāyana's Theorem (a2 + b2 = c2) interactively for any right-angled triangle.

a2 + b2 = 9 + 16 = 25 → Hypotenuse (c) = 5.00 cm

Simulation 3: Scaling Baudhāyana Triples

Learning Objective: Explore primitive Baudhāyana triples and their scaled versions (ka, kb, kc).

Scaled Triple: (6, 8, 10) → 62 + 82 = 36 + 64 = 100 = 102

Chapter Sections

2.1 & 2.2 Doubling and Halving a Square

In Baudhāyana's Śulba-Sūtra (c. 800 BCE), Baudhāyana considers constructing a square with double the area of a given square. Simply doubling each side length results in 4 times the area (2 × 2 = 4).

Baudhāyana's Rule (Verse 1.9) The diagonal of a square produces a square of double the area of the original square. The original square is made of 2 small triangles, while the new square constructed on its diagonal is made of 4 small triangles.

Halving a square is accomplished by drawing a tilted smaller square inside by connecting the midpoints of the sides, or folding paper inward along crease lines passing through side midpoints.

2.3 Hypotenuse of an Isosceles Right Triangle & Square Root of 2

The side opposite to the right angle is the hypotenuse. For an isosceles right triangle with side 1, the square constructed on its hypotenuse has an area of 2 sq. units. Thus, c2 = 2, giving c = √2.

Decimal Representation and Irrationality of √2 - Bounds: 1 < √2 < 2, refined to 1.414 < √2 < 1.415.
- Non-terminating decimal expansion: √2 = 1.41421356...
- Euclid's Proof: √2 cannot be expressed as a fraction m/n because prime factorization of squares requires even prime counts, leading to a contradiction in 2n2 = m2.

Complete Figure It Out Solutions (Page 39)

Figure It Out: Section 2.1 & 2.2

1. Cut two identical squares and arrange pieces to double area:
Step 1: Cut Square 1 into 4 triangles along diagonals (1, 2, 3, 4).
Step 2: Cut Identical Square 2 similarly into 4 triangles (5, 6, 7, 8).
Step 3: Place pieces 5, 6, 7, and 8 around Square 1 to form a larger square on the diagonal with double the area.

2. Length of two equal sides of an isosceles right triangle given, find hypotenuse and bounds:
(i) Side 3: c = √(32 + 32) = √18 ≈ 4.24 (between 4.2 and 4.3).
(ii) Side 4: c = √(42 + 42) = √32 ≈ 5.65 (between 5.6 and 5.7).
(iii) Side 6: c = √(62 + 62) = √72 ≈ 8.48 (between 8.4 and 8.5).
(iv) Side 8: c = √(82 + 82) = √128 ≈ 11.31 (between 11.3 and 11.4).
(v) Side 9: c = √(92 + 92) = √162 ≈ 12.72 (between 12.7 and 12.8).

3. The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths?
Step 1: c2 = 2a2 → 102 = 2a2 → 100 = 2a2 → a2 = 50.
Step 2: a = √50 ≈ 7.07 units.

2.4 Combining Different Squares & Baudhāyana's Theorem

Baudhāyana's Śulba-Sūtra (Verse 1.12) states: The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides.

Baudhāyana's Theorem on Right-angled Triangles If a right-angled triangle has sidelengths a, b, and c, where c is the length of the hypotenuse, then a2 + b2 = c2.

2.5 Baudhāyana Triples & Odd Numbers

Integer triples (a, b, c) satisfying a2 + b2 = c2 are called Baudhāyana triples. Examples include (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Scaling and Generating Triples - Scaled version (ka, kb, kc) is also a triple.
- Primitive triple has no common factor greater than 1.
- Odd number sum method: (n-1)2 + (2n-1) = n2 generates triples.

Complete Figure It Out Solutions (Page 47-48)

Figure It Out: Section 2.4 & 2.5 (Page 47-48)

1. Shorter sides 5 cm and 12 cm, find hypotenuse:
Step 1: c2 = 52 + 122 = 25 + 144 = 169.
Step 2: c = √169 = 13 cm.

2. Short side 8 cm and hypotenuse 17 cm, find third side:
Step 1: 82 + b2 = 172 → 64 + b2 = 289.
Step 2: b2 = 289 - 64 = 225 → b = 15 cm.

3. Construct square of triple/five times area:
Use right triangles where perpendicular sides combine areas using Baudhāyana's theorem.

4. Find missing sidelengths:
(i) a = 5, b = 7 → c = √(25 + 49) = √74.
(ii) a = 8, b = 12 → c = √(64 + 144) = √208.
(iii) a = 9, c = 15 → 81 + b2 = 225 → b2 = 144 → b = 12.
(iv) a = 7, b = 12 → c = √(49 + 144) = √193.
(v) a = 1.5, b = 3.5 → c = √(2.25 + 12.25) = √14.5.

Complete Figure It Out Solutions (Page 50)

Figure It Out: Section 2.5 (Page 50)

1. Find 5 more Baudhāyana triples using the odd number sum method:
- 3rd odd square (49th odd / n=25): 242 + 72 = 252 (7, 24, 25).
- 4th odd square (81st odd / n=41): 402 + 92 = 412 (9, 40, 41).
- Others include (11, 60, 61), (13, 84, 85), (15, 112, 113).

2. Does this method yield non-primitive Baudhāyana triples?
No, when n is odd, this algebraic construction produces primitive Pythagorean triples.

3. Are there primitive triples that cannot be obtained through this method?
Yes, e.g., (5, 12, 13) cannot be generated directly because 12 is not one less than a square in that form.

2.6 & 2.7 Fermat's Last Theorem & Further Applications

Fermat stated that xn + yn = zn has no natural number solutions for n > 2. Proven by Andrew Wiles in 1994 (Fermat's Last Theorem).

Applications (e.g., Bhāskarāchārya's Līlāvatī) Lotus stem problem: 32 + x2 = (x+1)2 → 9 + x2 = x2 + 2x + 1 → 2x = 8 → x = 4 units depth.

Complete Figure It Out Solutions (Page 52-53)

Figure It Out: Section 2.7 (Page 52-53)

1. Find the diagonal of a square with sidelength 5 cm:
c = √(52 + 52) = √50 ≈ 7.07 cm.

2. Find missing sidelengths in given right triangles:
- Triangle with sides 7 and 9: Hypotenuse = √(49 + 81) = √130.
- Triangle with hypotenuse 41 and side 40: Third side = √(412 - 402) = √(1681 - 1600) = √81 = 9.
- Triangle with side 10 and hypotenuse √200: Third side = √(200 - 100) = 10.
- Triangle with side 10 and other side √150: Hypotenuse = √(100 + 150) = √250.
- Triangle with side 27 and hypotenuse 45: Third side = √(452 - 272) = √(2025 - 729) = √1296 = 36.

3. Find sidelength of a rhombus whose diagonals are of length 24 units and 70 units:
Half-diagonals are 12 and 35. Sidelength c = √(122 + 352) = √(144 + 1225) = √1369 = 37 units.

4. Is the hypotenuse the longest side of a right triangle? Justify:
Yes, because it is opposite to the right angle (the largest angle in a right triangle).

5. True or False - Every Baudhāyana triple is either primitive or a scaled version of a primitive triple:
True.

6. Give 5 examples of rectangles whose sidelengths and diagonals are all integers:
(3, 4, 5), (6, 8, 10), (5, 12, 13), (8, 15, 17), (7, 24, 25).

7. Construct a square whose area is equal to the difference of squares of sidelengths 5 and 7 units:
Side length = √(72 - 52) = √(49 - 25) = √24 units.

8. Grid squares (area 2, 3, 4, 5):
(a) Area 2: formed by connecting opposite corners of 1x1 grid squares (diagonal = √2, area = 2).
(b) Area 3: derived from hypotenuse of triangle with sides 1 and √2.
(c) Area 4: 2x2 grid square.
(d) Area 5: formed by triangle with sides 1 and 2 (12 + 22 = 5).
Possible integer-valued areas: Sum of two squares (e.g., 1, 4, 5, 8, 9, 10, 13, etc.).

9. Find the area of an equilateral triangle with sidelength 6 units:
Step 1: Altitude bisects base into 3 and 3.
Step 2: Height h = √(62 - 32) = √(36 - 9) = √27 = 3√3.
Step 3: Area = (1/2) * base * height = (1/2) * 6 * 3√3 = 9√3 sq units.