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Class 8 · Ganita Prakash · Chapter 2
The Baudhāyana-Pythagoras Theorem
India's greatest geometric discovery — from 800 BCE to today 🏆
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1 · Who is Baudhāyana?
The Indian mathematician who discovered it first
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Baudhāyana's Śulba-Sūtra (~800 BCE)
Baudhāyana was an ancient Indian mathematician who wrote the Śulba-Sūtra — a manual for constructing fire altars with precise geometric measurements. In this text, he stated what we now call the "Pythagorean Theorem" — about 300 years before Pythagoras!
~800 BCE
Baudhāyana writes the Śulba-Sūtra and states the theorem in India.
~500 BCE
Pythagoras of Greece studies and popularises the same theorem.
~300 BCE
Euclid proves √2 is irrational in his book Elements.
17th century
Fermat proposes his famous Last Theorem (inspired by Baudhāyana triples).
1994 CE
Andrew Wiles finally proves Fermat's Last Theorem — 357 years later!
🦉
Why "Baudhāyana-Pythagoras Theorem"?
The theorem is called by both names to honour the Indian mathematician Baudhāyana who stated it first, and the Greek mathematician Pythagoras who is more widely known for it.
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2 · Doubling a Square
Baudhāyana's clever method using the diagonal
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The Problem
How do you construct a square with exactly double the area of a given square?
Doubling the side length gives 4× the area, NOT 2× !
Doubling the side length gives 4× the area, NOT 2× !
@edugrown
⭐ Baudhāyana's Śulba-Sūtra (Verse 1.9)
"The diagonal of a square produces a square of double the area of the original square."
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Why does this work?
The original square is made up of 2 small triangles.
The new square (on the diagonal) is made up of 4 small triangles.
All 4 triangles are congruent → New area = 2 × original area ✅
The new square (on the diagonal) is made up of 4 small triangles.
All 4 triangles are congruent → New area = 2 × original area ✅
If original square has side a → diagonal = c
New square's side = c | New area = c² = 2a²
New square's side = c | New area = c² = 2a²
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Sequence of Doubling
Square 1: 2 small triangles → area = 1 unit
Square 2 (diagonal of 1): 4 small triangles → area = 2 units
Square 3 (diagonal of 2): 8 small triangles → area = 4 units
Each new square doubles the area of the previous one!
Square 2 (diagonal of 1): 4 small triangles → area = 2 units
Square 3 (diagonal of 2): 8 small triangles → area = 4 units
Each new square doubles the area of the previous one!
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3 · Halving a Square
The reverse of doubling — use midpoints
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Reverse the construction
Draw a tilted square inside the original square by connecting the midpoints of the four sides. This inner square has exactly half the area of the outer square.
@edugrown
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How to halve using paper folding
Fold the square paper inward so that the crease lines pass through the midpoints of the sides. The square PQRS formed this way has exactly half the area of the original square.
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Common Mistake!
A square with half the side length does NOT have half the area — it has only 1/4 the area! (Because area = side²; halving the side gives 1/4 the area.)
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4 · Hypotenuse of an Isosceles Right Triangle
Finding the mystery side using area
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Key Terms
Right Triangle — has one 90° angle
Hypotenuse — side opposite to the right angle (always the longest side)
Isosceles Right Triangle — both shorter sides are equal (legs = a, a)
Hypotenuse — side opposite to the right angle (always the longest side)
Isosceles Right Triangle — both shorter sides are equal (legs = a, a)
@edugrown
📐 Isosceles Right Triangle Rule
If the two equal sides (legs) each have length a, then the hypotenuse c satisfies:
c² = 2a² so c = a√2
Square on diagonal = 2 × Area of original square
c² = 2a²
c² = 2a²
∞
5 · √2 — An Irrational Number
It goes on forever — never terminates, never repeats
√2 = 1.41421356237… (goes on forever, never repeats!)
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Finding bounds of √2 step by step
1² = 1 < 2 < 4 = 2² → 1 < √2 < 2
1.4² = 1.96 < 2 < 2.25 = 1.5² → 1.4 < √2 < 1.5
1.41² = 1.9881 < 2 < 2.0164 = 1.42² → 1.41 < √2 < 1.42
1.414² = 1.999396 < 2 < 2.002225 = 1.415² → 1.414 < √2 < 1.415
1.4² = 1.96 < 2 < 2.25 = 1.5² → 1.4 < √2 < 1.5
1.41² = 1.9881 < 2 < 2.0164 = 1.42² → 1.41 < √2 < 1.42
1.414² = 1.999396 < 2 < 2.002225 = 1.415² → 1.414 < √2 < 1.415
Not a fraction
√2 ≠ m/n for any integers m, n (proved by Euclid ~300 BCE)
Non-terminating
The decimal expansion never ends or repeats — it's irrational!
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Why can't √2 be a fraction?
If √2 = m/n, then 2 = m²/n², so 2n² = m².
In prime factorization of any perfect square, every prime occurs an even number of times.
But 2 appears an odd number of times on the left — contradiction!
∴ √2 cannot be expressed as any fraction. (Euclid's proof, ~300 BCE)
In prime factorization of any perfect square, every prime occurs an even number of times.
But 2 appears an odd number of times on the left — contradiction!
∴ √2 cannot be expressed as any fraction. (Euclid's proof, ~300 BCE)
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√2 is IRRATIONAL
It cannot be expressed as a terminating decimal OR as a fraction p/q. Numbers like this are called irrational numbers. They exist on the number line but cannot be written exactly as a fraction!
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6 · The Main Theorem: a² + b² = c²
The most famous theorem in all of geometry
🏆 Baudhāyana-Pythagoras Theorem
If a right-angled triangle has sidelengths a, b, and c, where c is the hypotenuse, then:
a² + b² = c²
@edugrown
🎯
Classic Example — The 3-4-5 Triangle
a = 3 cm, b = 4 cm → find c
a² + b² = c²
9 + 16 = c²
25 = c² → c = 5 cm ✅
a² + b² = c²
9 + 16 = c²
25 = c² → c = 5 cm ✅
| What we know | What to find | Formula to use |
|---|---|---|
| Both legs a and b | Hypotenuse c | c = √(a² + b²) |
| One leg a, hypotenuse c | Other leg b | b = √(c² − a²) |
| Equal legs (a = b) | Hypotenuse c | c = a√2 |
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The hypotenuse is ALWAYS the longest side!
c² = a² + b² means c² > a² and c² > b², so c > a and c > b always. The hypotenuse is opposite the largest angle (90°).
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7 · Baudhāyana (Pythagorean) Triples
Integer sets that satisfy a² + b² = c²
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What is a Baudhāyana Triple?
A set of three positive integers (a, b, c) such that a² + b² = c².
Also called: Pythagorean triples / right-angled triangle triples.
Also called: Pythagorean triples / right-angled triangle triples.
✅ Common Baudhāyana Triples
✓ (3, 4, 5)
✓ (5, 12, 13)
✓ (8, 15, 17)
✓ (7, 24, 25)
✓ (6, 8, 10)
✓ (9, 12, 15)
✓ (12, 16, 20)
✓ (12, 35, 37)
✓ (15, 36, 39)
| Type | Definition | Example |
|---|---|---|
| Primitive Triple | No common factor > 1 among a, b, c | (3, 4, 5) ✅ primitive |
| Scaled Triple | k × (a, b, c) for any positive integer k | (6, 8, 10) = 2×(3,4,5) |
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Big Fact: Infinitely many Baudhāyana triples exist!
If (a, b, c) is a triple, then (ka, kb, kc) is also a triple for any positive integer k.
Proof: (ka)² + (kb)² = k²a² + k²b² = k²(a²+b²) = k²c² = (kc)² ✅
Proof: (ka)² + (kb)² = k²a² + k²b² = k²(a²+b²) = k²c² = (kc)² ✅
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8 · Generating Baudhāyana Triples
A clever method using odd square numbers
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Key Identity
Sum of first n odd numbers = n²
1 + 3 + 5 + … + (2n−1) = n²
So: (n−1)² + (2n−1) = n²
1 + 3 + 5 + … + (2n−1) = n²
So: (n−1)² + (2n−1) = n²
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The Method
If the nth odd number (2n−1) is itself a perfect square, then we get a Baudhāyana triple: (n−1, √(2n−1), n)
| Odd Square | n value | Triple Generated | Check |
|---|---|---|---|
| 9 = 3² | n = 5 (since 9 = 2×5−1) | (4, 3, 5) | 9 + 16 = 25 ✅ |
| 25 = 5² | n = 13 (since 25 = 2×13−1) | (12, 5, 13) | 144 + 25 = 169 ✅ |
| 49 = 7² | n = 25 (since 49 = 2×25−1) | (24, 7, 25) | 576 + 49 = 625 ✅ |
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Pattern in generated triples
In this method, one of the smaller sides is always exactly one less than the hypotenuse. These triples are always primitive.
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9 · Applications of the Theorem
Solving real-world geometry problems
🪷 Classic Problem from Bhāskarāchārya's Līlāvatī
"A lotus peeps 1 unit above the water. Swayed by breeze, its tip touches water 3 units away. Find depth of the lake."
Solution:
Let depth = x → total stem length = x + 1
Right triangle: legs = 3 and x, hypotenuse = x + 1
3² + x² = (x+1)²
9 + x² = x² + 2x + 1
9 = 2x + 1 → x = 4 units 🎉
Let depth = x → total stem length = x + 1
Right triangle: legs = 3 and x, hypotenuse = x + 1
3² + x² = (x+1)²
9 + x² = x² + 2x + 1
9 = 2x + 1 → x = 4 units 🎉
@edugrown
📌 Other Applications
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Diagonal of a square (side s): d = s√2
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Diagonal of a rectangle (sides a, b): d = √(a² + b²)
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Side of a rhombus (diagonals d1, d2): side = √((d1/2)² + (d2/2)²)
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Height of equilateral triangle (side a): h = (√3/2)a
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10 · Fermat's Last Theorem
357 years to prove — the greatest math mystery
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Fermat's Question (17th century)
We know a² + b² = c² has infinitely many integer solutions (Baudhāyana triples).
Fermat asked: what about a³ + b³ = c³? Or a⁴ + b⁴ = c⁴? etc.
Fermat asked: what about a³ + b³ = c³? Or a⁴ + b⁴ = c⁴? etc.
🔐 Fermat's Last Theorem
The equation aⁿ + bⁿ = cⁿ has NO solution in positive integers when n > 2.
aⁿ + bⁿ = cⁿ has no solution for n > 2
~1637 CE — Fermat's Claim
Fermat wrote in a book's margin: "I have a marvellous proof… but the margin is too small to contain it!" 😄
1637–1994 — 357 Years of Attempts
Greatest mathematicians tried and failed to prove it. No one could find Fermat's proof either.
1963 — A 10-year-old boy
Andrew Wiles reads about Fermat's Last Theorem and resolves to prove it someday.
1994 — PROVED! 🎉
Andrew Wiles finally proves Fermat's Last Theorem after 7 years of secret work. The mystery is solved!
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Lesson: Never give up on hard problems!
Andrew Wiles spent 7 years working in secret on this problem. The study of Baudhāyana triples (from 800 BCE!) ultimately led to one of the greatest mathematical achievements of the 20th century.
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11 · Quick Summary
Everything at a glance
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The diagonal of a square = side × √2. Square on diagonal has double the area of original.
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Tilted square connecting midpoints of sides has half the area of the original square.
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Isosceles right triangle: c² = 2a² → c = a√2 (legs = a, hypotenuse = c).
✓
√2 = 1.41421… is irrational — not a fraction, not a terminating decimal.
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Baudhāyana-Pythagoras Theorem: In a right triangle, a² + b² = c² (c = hypotenuse).
✓
Baudhāyana triples: integer sets (a,b,c) with a²+b²=c². Examples: (3,4,5), (5,12,13), (8,15,17).
✓
If (a,b,c) is a triple → (ka,kb,kc) is also a triple for any integer k. Infinitely many exist!
✓
Fermat's Last Theorem: aⁿ+bⁿ=cⁿ has no integer solution for n>2. Proved by Andrew Wiles (1994).
📐
Baudhāyana-Pythagoras Theorem
a² + b² = c²
Stated in India ~800 BCE · One of the most important theorems in all of mathematics
c = √(a²+b²)
c² = 2a² (isosceles)
(3,4,5) triple
@edugrown