The Baudhayana-Pythagoras Theorem

Chapter at a Glance

This chapter details the geometric and algebraic principles of the Baudhayana-Pythagoras Theorem, tracing its historical roots in ancient Indian altar-building texts (the Śulba-Sūtras) and early Greek mathematics. Through hands-on paper folding, geometric dissection, and algebraic notation, it explores how to double and halve squares, combine squares of different sizes, and calculate hypotenuse lengths in right-angled triangles. Additionally, the chapter covers the properties of irrational numbers like $\sqrt{2}$, details the properties of Baudhayana triples, mentions Fermat’s Last Theorem, and applies the theorem to solve classic word problems.

Key Definitions & Terminology

• Hypotenuse: The longest side of a right-angled triangle, situated directly opposite the right ($90^\circ$) angle.
• Baudhayana-Pythagoras Theorem: The geometric theorem stating that in any right-angled triangle, the area of the square constructed on the hypotenuse is equal to the sum of the areas of the squares constructed on the other two sides.
• Baudhayana (Pythagorean) Triple: A set of three positive integers $(a, b, c)$ satisfying the equation $a^2 + b^2 = c^2$.
• Primitive Triple: A Baudhayana triple whose terms have no common factors greater than $1$ (e.g., $(3, 4, 5)$).
• Scaled Triple: A triple formed by multiplying each term of a primitive triple by a positive integer $k$ (e.g., $(6, 8, 10)$).
• Irrational Number: A number (like $\sqrt{2}$) that cannot be expressed as a simple fraction $\frac{m}{n}$ of two counting integers or as a terminating decimal.
• Lower/Upper Bound: Values that restrict the range of an unknown quantity from below and above (e.g., $1.4 < \sqrt{2} < 1.5$).
• Dissection: The process of cutting a geometric figure into pieces and rearranging them to form a different shape of equal area.

Formulas, Rules & Properties

• The Baudhayana-Pythagoras Theorem Formula:
• For a right triangle with legs of length $a$ and $b$, and hypotenuse of length $c$:
  $$a^2 + b^2 = c^2 \quad \rightarrow \quad c = \sqrt{a^2 + b^2}$$
• Isosceles Right Triangle Relationship:
• In a right-angled triangle where the two legs are equal ($a = b$):
  $$c^2 = 2a^2 \quad \rightarrow \quad c = a\sqrt{2}$$
• Baudhayana Triple Scaling Property:
• If $(a, b, c)$ is a Baudhayana triple, then for any positive integer $k$, the triple $(ka, kb, kc)$ is also a Baudhayana triple:
  $$(ka)^2 + (kb)^2 = k^2(a^2 + b^2) = k^2 c^2 = (kc)^2$$
• Triple Generation Formula:
• For any positive integer $n \ge 2$, if the $n$-th odd number $(2n - 1)$ is a perfect square $y^2$, it generates a triple:
  $$(n - 1)^2 + y^2 = n^2$$
  where the generated triple is $(n - 1, y, n)$.
• Fermat's Last Theorem:
• The equation $x^n + y^n = z^n$ has no non-zero integer solutions for $n > 2$.

Core Concepts & Topics

• Doubling and Halving Squares (Śulba-Sūtras):
• Doubling: Constructing a square on the diagonal of an existing square doubles the area. An original square of side $1$ has area $1$, while a square built on its diagonal of length $\sqrt{2}$ has area $2$. This is shown by dividing the shapes into congruent triangles: the original square contains $2$ triangles, while the diagonal square contains $4$.
• Halving: Connecting the midpoints of the sides of a square creates a tilted inner square whose area is exactly half that of the outer square.
• Irrationality of $\sqrt{2}$:
• In a square of side $1$, the diagonal length is $\sqrt{2}$.
• Euclid's prime factorization proof: If $\sqrt{2} = \frac{m}{n}$, then $2n^2 = m^2$. In a perfect square's prime factorization, every prime factor occurs an even number of times. The left side has an odd number of $2$s, and the right side has an even number, which is a contradiction. Thus, $\sqrt{2}$ cannot be expressed as a fraction.
• Historical Context of the Theorem:
• Baudhayana (c. 800 BCE) wrote the first known statement of this theorem in the Śulba-Sūtras: "The diagonal of a rectangle produces both areas which the flank and the width produce separately."
• Pythagoras (c. 500 BCE) studied and proved the theorem in Greece, living a couple hundred years after Baudhayana.
• Rhombus Diagonal Property:
• Since the diagonals of a rhombus bisect each other perpendicularly, they divide the rhombus into four congruent right-angled triangles, enabling the sidelength to be calculated from the half-diagonal values.

Worked Examples

• Finding Bounds for Isosceles Hypotenuse (Page 39 Q2):
• Problem: Equal sides of an isosceles right triangle are $3\text{ cm}$. Find the hypotenuse bounds.
• Solution:
  $$c^2 = 2a^2 = 2(3^2) = 18 \rightarrow c = \sqrt{18} = 3\sqrt{2}$$
  • Since $1.414 < \sqrt{2} < 1.415 \rightarrow 3 \times 1.414 < c < 3 \times 1.415 \rightarrow 4.242 < c < 4.245$.
  • Bounds: $4.2\text{ cm} < c < 4.3\text{ cm}$.
• Bhāskarāchārya's Lotus Stem Problem (Page 52):
• Problem: "In a lake, a lotus stem peeps out $1\text{ unit}$ above the water. When swayed by the wind, the tip touches the water surface $3\text{ units}$ away from its original vertical line. Find the lake's depth."
• Solution: Let the depth of the lake be $x$. The total length of the stem is $x + 1$.
  • The vertical position, the surface displacement, and the tilted stem form a right-angled triangle.
  • Applying the theorem:
    $$x^2 + 3^2 = (x + 1)^2 \rightarrow x^2 + 9 = x^2 + 2x + 1$$
    $$9 = 2x + 1 \rightarrow 2x = 8 \rightarrow x = 4\text{ units}$$
  • Answer: The depth of the lake is $4\text{ units}$.
• Generating Triples using Odd Squares (Page 50):
• Problem: Generate the triple associated with the odd square $25$.
• Solution:
  • $25$ is the $13$-th odd number ($2n - 1 = 25 \rightarrow 2n = 26 \rightarrow n = 13$).
  • Using $(n-1)^2 + (2n-1) = n^2$:
    $$(13 - 1)^2 + 25 = 13^2 \rightarrow 12^2 + 5^2 = 13^2$$
  • Triple: $(5, 12, 13)$.
• Missing Sidelengths (Figure it Out Q2, Page 52):
• Problem (iii): If $a = 9$ and $c = 15$, find the leg $b$.
  • Solution:
    $$b^2 = c^2 - a^2 = 15^2 - 9^2 = 225 - 81 = 144 \rightarrow b = \sqrt{144} = 12$$
• Problem (iv): If $a = 7$ and $b = 12$, find the hypotenuse $c$.
  • Solution:
    $$c^2 = a^2 + b^2 = 7^2 + 12^2 = 49 + 144 = 193 \rightarrow c = \sqrt{193} \approx 13.89$$
• Sidelength of a Rhombus (Figure it Out Q3, Page 53):
• Problem: Find the sidelength of a rhombus whose diagonals are $24\text{ units}$ and $70\text{ units}$.
• Solution: The diagonals bisect each other perpendicularly, forming four right triangles with legs equal to the half-diagonals: $\frac{24}{2} = 12$ and $\frac{70}{2} = 35$.
  • Sidelength $s = \sqrt{12^2 + 35^2} = \sqrt{144 + 1225} = \sqrt{1369} = 37\text{ units}$.
• Area of an Equilateral Triangle (Page 54 Q9):
• Problem: Find the area of an equilateral triangle with sidelength $6\text{ units}$.
• Solution:
  • The altitude splits the base of $6$ into two segments of $3$.
  • Height $h = \sqrt{6^2 - 3^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}$.
  • $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}\text{ sq. units}$.

Practical Activities & Experiments

• Baudhayana Paper Dissection: Cut two square pieces of paper with sidelengths $3\text{ cm}$ and $4\text{ cm}$. Align them side-by-side. Make cuts along the diagonal of the resulting rectangle to split them into three pieces. Rearrange the pieces to form a single larger square of side $5\text{ cm}$, demonstrating that $3^2 + 4^2 = 5^2$.
• Lotus Stem Basin Model: Fill a shallow basin with water. Place a straight stick in the center so it protrudes slightly above the water. Tip the stick until the top touches the water surface, and measure the depth and displacement to verify Bhāskara's lotus stem formula.