Class 8 (Part 2) — Chapter 2: The Baudhāyana-Pythagoras Theorem

2.1 Doubling a Square

Baudhāyana’s Śulba-Sūtra (c. 800 BCE) poses the fundamental question: How can one construct a square having double the area of a given square?
Simply doubling the side length of a square does not double its area; it quadruples it (Area = $2 \times 2 = 4$ times the original area).
Baudhāyana's Rule (Verse 1.9): "The diagonal of a square produces a square of double the area of the original square."
Geometric Proof: An original square of side $1$ can be divided into $2$ congruent right isosceles triangles. A square constructed on its diagonal is composed of $4$ such congruent triangles, thus having exactly double the area.

2.2 Halving a Square

To construct a square with half the area of a given square, the process of doubling is reversed.
By connecting the midpoints of the sides of a square, a tilted inner square (PQRS) is formed.
The area of this inner square is exactly half of the outer square because the outer square is composed of 8 congruent triangles, while the inner square contains 4 of them.

2.3 Hypotenuse of an Isosceles Right Triangle

In a right-angled triangle, the side opposite to the right angle is the hypotenuse.
For an isosceles right triangle with equal sides of $1$ unit:
The square constructed on the hypotenuse has an area equal to $2$ square units.
If $c$ is the length of the hypotenuse, then:
c² = 2
c = √2 units.

Bounds of √2:
- $1 < \sqrt{2} < 2$ (since $1^2 = 1$ and $2^2 = 4$)
- $1.4 < \sqrt{2} < 1.5$ (since $1.4^2 = 1.96$ and $1.5^2 = 2.25$)
- $1.41 < \sqrt{2} < 1.42$ (since $1.41^2 = 1.9881$ and $1.42^2 = 2.0164$)
- $1.414 < \sqrt{2} < 1.415$ (since $1.414^2 = 1.999396$ and $1.415^2 = 2.002225$)
Irrationality: $\sqrt{2}$ cannot be expressed as a terminating decimal or a fraction $m/n$ where $m$ and $n$ are counting numbers. It is a non-terminating, non-recurring decimal: $\sqrt{2} \approx 1.41421356...$

2.4 Combining Two Different Squares

Śulba-Sūtra (Verse 1.12): "The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides."
To combine two squares of different areas (sides $a$ and $b$), construct a right-angled triangle with perpendicular sides $a$ and $b$. The square on the hypotenuse $c$ will have an area equal to the sum of the areas of the two original squares:
a² + b² = c²
This is the fundamental statement of the Baudhāyana-Pythagoras Theorem.

2.5 Right-Triangles Having Integer Sidelengths

Integer triples $(a, b, c)$ that satisfy $a^2 + b^2 = c^2$ are called Baudhāyana triples (or Pythagorean triples).
Common examples: $(3, 4, 5)$, $(5, 12, 13)$, $(8, 15, 17)$, $(7, 24, 25)$, $(12, 35, 37)$, $(15, 36, 39)$.
Scaling Property: If $(a, b, c)$ is a triple, then $(ka, kb, kc)$ is also a triple for any positive integer $k$.
Primitive Triple: A triple with no common factor greater than $1$ (e.g., $(3, 4, 5)$ is primitive, while $(6, 8, 10)$ is a scaled version).
Generation Method: Triples can be generated using the sum of consecutive odd numbers:
1 + 3 + 5 + ... + (2n - 1) = n²

2.6 A Long-Standing Open Problem

Inspired by Baudhāyana triples, Pierre de Fermat conjectured that the equation xⁿ + yⁿ = zⁿ has no positive integer solutions for n > 2.
This famous conjecture, known as Fermat's Last Theorem, remained unsolved for over 300 years until it was proven by Andrew Wiles in 1994.

2.7 Applications of the Theorem

The Lotus Problem: A lotus stem in a lake is $1$ unit above the water. When swayed by the wind, it touches the water surface $3$ units away from its original vertical position. Find the depth of the lake.
Solution: Let $x$ be the depth of the lake. The total length of the stem is $x + 1$. Applying the theorem:
3² + x² = (x + 1)²
9 + x² = x² + 2x + 1
9 = 2x + 1
x = 4 units.
The depth of the lake is $4$ units.

Mathematics: Class 8 (Part 2) — Chapter 2: The Baudhāyana-Pythagoras Theorem