Class 8 · Mathematics · Chapter 2
📐 The Baudhāyana-Pythagoras Theorem
Complete NCERT notes — easy to understand, exam-ready, and student-friendly!
🏛️ Ancient Indian Maths
📐 Right Triangles
🔢 Pythagoras Theorem
✨ Baudhāyana Triples
∞ Fermat’s Last Theorem
📐 Right Triangles
🔢 Pythagoras Theorem
✨ Baudhāyana Triples
∞ Fermat’s Last Theorem
📋 Contents
- Introduction & Historical Background
- 2.1 Doubling a Square
- 2.2 Halving a Square
- 2.3 Hypotenuse of an Isosceles Right Triangle
- The Value of √2 (Decimal Representation)
- 2.4 Combining Two Different Squares
- The Baudhāyana-Pythagoras Theorem (Main Theorem)
- 2.5 Baudhāyana Triples (Pythagorean Triples)
- 2.6 Fermat’s Last Theorem
- 2.7 Applications of the Theorem
- Exam Questions & Answers
- Summary
Introduction & Historical Background
Who was Baudhāyana?
Baudhāyana was an ancient Indian mathematician who lived around 800 BCE. He wrote the famous Śulba-Sūtra (rules for the cord/rope), which contained geometric knowledge used for building fire altars. He was the first person in history to state what we now call the Pythagorean Theorem!
Baudhāyana was an ancient Indian mathematician who lived around 800 BCE. He wrote the famous Śulba-Sūtra (rules for the cord/rope), which contained geometric knowledge used for building fire altars. He was the first person in history to state what we now call the Pythagorean Theorem!
What about Pythagoras?
Pythagoras was a Greek philosopher-mathematician who lived around 500 BCE — that is about 300 years after Baudhāyana! He also studied and admired this theorem. Today the theorem is called the Baudhāyana-Pythagoras Theorem to honour both mathematicians.
Pythagoras was a Greek philosopher-mathematician who lived around 500 BCE — that is about 300 years after Baudhāyana! He also studied and admired this theorem. Today the theorem is called the Baudhāyana-Pythagoras Theorem to honour both mathematicians.
⭐ Key Terms to Know
- Right Triangle: A triangle with one angle equal to exactly 90°.
- Hypotenuse: The side opposite to the right angle. It is always the longest side.
- Perpendicular Sides (Legs): The two shorter sides that form the right angle.
- Square on a side: A square whose side length equals that side of the triangle.
2.1 Doubling a Square
Baudhāyana asks: “How can one construct a square having double the area of a given square?”
Common Mistake!
If you double the side length of a square, the area becomes 4 times, NOT 2 times! For example, a 2×2 square has area 4 — that’s 4× the area of a 1×1 square.
If you double the side length of a square, the area becomes 4 times, NOT 2 times! For example, a 2×2 square has area 4 — that’s 4× the area of a 1×1 square.
✅ Baudhāyana’s Correct Method (Verse 1.9)
“The diagonal of a square produces a square of double the area of the original square.”
If you draw a square on the diagonal of a given square, the new square has exactly double the area!
📖 Why Does This Work?
- The original square is made up of 2 small triangles.
- The new (tilted) square on the diagonal is made up of 4 small triangles.
- All 4 triangles are congruent (identical in shape and size).
- So the new square has exactly double the area of the original!
Doubling a Square — Paper Activity
Cut two identical squares. Divide Square 1 into 4 triangles (pieces 1–4) and Square 2 into 4 triangles (pieces 5–8). Place pieces 5, 6, 7, 8 around Square 1 to form a larger square with double the area!
Cut two identical squares. Divide Square 1 into 4 triangles (pieces 1–4) and Square 2 into 4 triangles (pieces 5–8). Place pieces 5, 6, 7, 8 around Square 1 to form a larger square with double the area!
📐 Sequence of Doubling
Square 1 → 2 small triangles
Square 2 → 4 small triangles (double area of Square 1)
Square 3 → 8 small triangles (double area of Square 2)
Each square has DOUBLE the area of the previous one!
Square 2 → 4 small triangles (double area of Square 1)
Square 3 → 8 small triangles (double area of Square 2)
Each square has DOUBLE the area of the previous one!
2.2 Halving a Square
Now we reverse the process: given a square, construct a square with half the area.
✅ Method
Draw a smaller, tilted (rotated) square inside the larger square, connecting the midpoints of the four sides. This inner square has exactly half the area of the outer square.
Don’t confuse this!
A square with half the side length does NOT have half the area. It has one-quarter (¼) the area. Four such squares fit inside the original — not two!
A square with half the side length does NOT have half the area. It has one-quarter (¼) the area. Four such squares fit inside the original — not two!
Paper Activity — Halving
Fold the square inward so the fold lines (crease lines) pass through the midpoints of all four sides. The square PQRS formed inside is the required square with half the area.
Fold the square inward so the fold lines (crease lines) pass through the midpoints of all four sides. The square PQRS formed inside is the required square with half the area.
📖 Why is PQRS a Square?
- Connect QS and PR (the diagonals of PQRS).
- Use triangle congruence to show all angles of PQRS are 90°.
- All sides of PQRS are equal (they’re all diagonals of identical smaller squares).
- Since all sides are equal and all angles are 90°, PQRS is a square!
2.3 Hypotenuse of an Isosceles Right Triangle
Remember!
An isosceles right triangle has two equal sides (called legs) and one right angle (90°). The side opposite the right angle is the hypotenuse.
An isosceles right triangle has two equal sides (called legs) and one right angle (90°). The side opposite the right angle is the hypotenuse.
🔍 Finding the Hypotenuse When Legs = 1 unit
A square of side 1 unit is made up of two isosceles right triangles. The square on its diagonal has area = 2 × 1 = 2 sq. units.
If c = hypotenuse, then:
Area of square on hypotenuse = c × c = c²
So, c² = 2 → c = √2
So, c² = 2 → c = √2
📐 General Formula for Isosceles Right Triangle
Let a = length of the two equal sides, and c = hypotenuse.
c² = 2a² or equivalently c = a√2
🔸 Example 1
Equal sides = 12 units
c = √(2 × 12²) = √288
Since 16² = 256 and 17² = 289
∴ c is between 16 and 17
Equal sides = 12 units
c = √(2 × 12²) = √288
Since 16² = 256 and 17² = 289
∴ c is between 16 and 17
🔸 Example 2
Hypotenuse = √72
c² = 2a² → 72 = 2a²
a² = 36 → a = √36
∴ Each equal side = 6 units
Hypotenuse = √72
c² = 2a² → 72 = 2a²
a² = 36 → a = √36
∴ Each equal side = 6 units
The Value of √2 — Decimal Representation
📊 Finding Bounds for √2
1² = 1 < 2 < 4 = 2²
So: 1 < √2 < 2Going further:
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√2 = 1.41421356… (non-terminating decimal)
So: 1 < √2 < 2Going further:
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√2 = 1.41421356… (non-terminating decimal)
Important facts about √2:
√2 cannot be expressed as a terminating decimal. It also cannot be expressed as a fraction m/n (where m and n are counting numbers). This was proved by Euclid around 300 BCE using prime factorization.
√2 cannot be expressed as a terminating decimal. It also cannot be expressed as a fraction m/n (where m and n are counting numbers). This was proved by Euclid around 300 BCE using prime factorization.
🧠 Why √2 ≠ m/n (Proof Idea)
- If √2 = m/n, then 2 = m²/n², so 2n² = m².
- In any square number’s prime factorization, every prime appears an even number of times.
- On the left side (2n²), the prime 2 appears an odd number of times; on the right side (m²), it appears an even number of times.
- This is a contradiction — so √2 cannot be a fraction!
2.4 Combining Two Different Squares
Problem: Given two squares of different sizes, how do you make one larger square whose area equals the sum of both?
Baudhāyana’s Solution (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.”
“The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides.”
📐 Method — Step by Step
- Let the two squares have side lengths a and b (where b > a).
- Join the two squares side by side.
- Mark a rectangle of dimensions a × b inside the larger square.
- Draw the diagonal of this rectangle — this diagonal has length c (the hypotenuse of a right triangle with legs a and b).
- Construct a square on this diagonal. Its area = a² + b²!
- Draw 3 more such right triangles around the hypotenuse to get a 4-sided figure.
- This 4-sided figure is a square with side length = c.
Why is it a square?
Triangles T, U, X, and W are all congruent (equal). So all sides of the new figure are equal. The angles at each corner = x° + (90° − x°) = 90°. Equal sides + all right angles = it’s a square!
Triangles T, U, X, and W are all congruent (equal). So all sides of the new figure are equal. The angles at each corner = x° + (90° − x°) = 90°. Equal sides + all right angles = it’s a square!
Area of new square = Area of square on side a + Area of square on side b
c² = a² + b²
The Baudhāyana-Pythagoras Theorem
Baudhāyana-Pythagoras Theorem:
If a right-angled triangle has side lengths a, b, and c,
where c is the length of the hypotenuse, then:
a² + b² = c²
In simple words:
In a right-angled triangle, the square on the hypotenuse = the sum of the squares on the other two sides.
In a right-angled triangle, the square on the hypotenuse = the sum of the squares on the other two sides.
🔍 Classic Example: The 3-4-5 Triangle
Let a = 3 cm, b = 4 cm. Find hypotenuse c.a² + b² = c²
3² + 4² = c²
9 + 16 = c²
25 = c²
c = 5 cm ✓
3² + 4² = c²
9 + 16 = c²
25 = c²
c = 5 cm ✓
📚 More Practice Examples
Example: Find hypotenuse
a = 5 cm, b = 12 cm
c² = 5² + 12² = 25 + 144 = 169
c = 13 cm
a = 5 cm, b = 12 cm
c² = 5² + 12² = 25 + 144 = 169
c = 13 cm
Example: Find a side
a = 8 cm, c = 17 cm
b² = 17² − 8² = 289 − 64 = 225
b = 15 cm
a = 8 cm, c = 17 cm
b² = 17² − 8² = 289 − 64 = 225
b = 15 cm
Key Rule:
The theorem works ONLY for right-angled triangles. Always identify the hypotenuse (opposite the right angle) correctly before applying the formula!
The theorem works ONLY for right-angled triangles. Always identify the hypotenuse (opposite the right angle) correctly before applying the formula!
2.5 Baudhāyana Triples (Pythagorean Triples)
A Baudhāyana triple is a set of three positive integers (a, b, c) that satisfy a² + b² = c². They can be the sides of a right-angled triangle.
📋 Famous Baudhāyana Triples (from Śulba-Sūtra)
| (a, b, c) | Check: a² + b² = c² | Type |
|---|---|---|
| (3, 4, 5) | 9 + 16 = 25 ✓ | Primitive |
| (5, 12, 13) | 25 + 144 = 169 ✓ | Primitive |
| (8, 15, 17) | 64 + 225 = 289 ✓ | Primitive |
| (7, 24, 25) | 49 + 576 = 625 ✓ | Primitive |
| (6, 8, 10) | 36 + 64 = 100 ✓ | Scaled (×2 of 3,4,5) |
| (9, 12, 15) | 81 + 144 = 225 ✓ | Scaled (×3 of 3,4,5) |
| (12, 16, 20) | 144 + 256 = 400 ✓ | Scaled (×4 of 3,4,5) |
🔑 Two Key Definitions
🔹 Primitive Triple
A triple where a, b, c have no common factor greater than 1.
Examples: (3, 4, 5), (5, 12, 13), (8, 15, 17)
A triple where a, b, c have no common factor greater than 1.
Examples: (3, 4, 5), (5, 12, 13), (8, 15, 17)
🔸 Scaled Triple
Obtained by multiplying a primitive triple by a positive integer k.
(3k, 4k, 5k) for any k — e.g., (6, 8, 10) when k=2
Obtained by multiplying a primitive triple by a positive integer k.
(3k, 4k, 5k) for any k — e.g., (6, 8, 10) when k=2
📖 Important Theorem on Scaling
If (a, b, c) is a Baudhāyana triple, then (ka, kb, kc) is also a Baudhāyana triple for any positive integer k.
📖 Proof:
Given: a² + b² = c²
Show: (ka)² + (kb)² = (kc)²(ka)² + (kb)² = k²a² + k²b²
= k²(a² + b²)
= k²c²
= (kc)² ✓ Proved!
Show: (ka)² + (kb)² = (kc)²(ka)² + (kb)² = k²a² + k²b²
= k²(a² + b²)
= k²c²
= (kc)² ✓ Proved!
🌟 Generating Triples Using Odd Squares
Using the identity: (n-1)² + (2n-1) = n²
If the nth odd number (2n−1) is a perfect square, we get a Baudhāyana triple!
Odd square 9 → 9 = 2×5−1, so n=5
(5-1)² + 9 = 5² → 4² + 3² = 5² → (3, 4, 5) ✓Odd square 25 → 25 = 2×13−1, so n=13
(13-1)² + 25 = 13² → 12² + 5² = 13² → (5, 12, 13) ✓
(5-1)² + 9 = 5² → 4² + 3² = 5² → (3, 4, 5) ✓Odd square 25 → 25 = 2×13−1, so n=13
(13-1)² + 25 = 13² → 12² + 5² = 13² → (5, 12, 13) ✓
2.6 Fermat’s Last Theorem
The study of Baudhāyana triples inspired the great French mathematician Pierre de Fermat (17th century) to ask a bigger question.
❓ Fermat’s Question
We know there are infinitely many solutions to x² + y² = z² (Baudhāyana triples). What about higher powers?
Fermat’s Last Theorem:
The equation xⁿ + yⁿ = zⁿ has no solution in positive integers x, y, z when n > 2.
Fermat’s Famous Note
In the margin of a book, Fermat wrote: “I have found a truly marvellous proof of this statement, but the margin is too small to contain it.” No one ever found his proof!
In the margin of a book, Fermat wrote: “I have found a truly marvellous proof of this statement, but the margin is too small to contain it.” No one ever found his proof!
🧒
Andrew Wiles — A Childhood Dream!
In 1963, a 10-year-old boy named Andrew Wiles read about Fermat’s Last Theorem and resolved to prove it one day. After more than 300 years of failed attempts by mathematicians worldwide, Wiles finally proved it in 1994! His proof was over 100 pages long.
In 1963, a 10-year-old boy named Andrew Wiles read about Fermat’s Last Theorem and resolved to prove it one day. After more than 300 years of failed attempts by mathematicians worldwide, Wiles finally proved it in 1994! His proof was over 100 pages long.
2.7 Applications of the Theorem
🌸 Classic Problem: The Lotus Flower (from Bhāskarāchārya’s Līlāvatī)
Problem:
A lotus sticks 1 unit above the water. When pushed by a breeze, its tip touches the water 3 units away. Find the depth of the lake.
A lotus sticks 1 unit above the water. When pushed by a breeze, its tip touches the water 3 units away. Find the depth of the lake.
Solution:
Let x = depth of lake (length of stem inside water). Total stem length = x + 1.
This forms a right triangle with legs 3 and x, and hypotenuse x + 1.
3² + x² = (x + 1)²
9 + x² = x² + 2x + 1
9 = 2x + 1 (subtracting x² from both sides)
2x = 8
x = 4 units
9 + x² = x² + 2x + 1
9 = 2x + 1 (subtracting x² from both sides)
2x = 8
x = 4 units
📐 Other Key Applications
🔷 Diagonal of a Square
Square with side a.
Diagonal d = √(a² + a²) = a√2
Side 5 cm → diagonal = 5√2 ≈ 7.07 cm
Square with side a.
Diagonal d = √(a² + a²) = a√2
Side 5 cm → diagonal = 5√2 ≈ 7.07 cm
🔷 Rhombus Side Length
Diagonals of rhombus bisect at 90°. Each half-diagonal forms a right triangle.
Example: Diagonals 24 & 70 → side = √(12² + 35²) = √(144+1225) = √1369 = 37
Diagonals of rhombus bisect at 90°. Each half-diagonal forms a right triangle.
Example: Diagonals 24 & 70 → side = √(12² + 35²) = √(144+1225) = √1369 = 37
Equilateral Triangle Area
For an equilateral triangle with side a, draw the altitude. It bisects the base, forming a right triangle with legs (a/2) and h (height). Use the theorem: h² + (a/2)² = a² → h = (√3/2)a. Area = (√3/4)a².
For an equilateral triangle with side a, draw the altitude. It bisects the base, forming a right triangle with legs (a/2) and h (height). Use the theorem: h² + (a/2)² = a² → h = (√3/2)a. Area = (√3/4)a².
Important Exam Questions & Answers
Q1. A right-angled triangle has shorter sides 5 cm and 12 cm. Find the hypotenuse.
Ans: c² = 5² + 12² = 25 + 144 = 169 → c = 13 cm
Q2. A right triangle has one short side 8 cm and hypotenuse 17 cm. Find the third side.
Ans: b² = 17² − 8² = 289 − 64 = 225 → b = 15 cm
Q3. Find the diagonal of a square with side 5 cm.
Ans: d² = 5² + 5² = 50 → d = √50 = 5√2 ≈ 7.07 cm
Q4. Is (5, 12, 13) a Baudhāyana triple? Is it primitive?
Ans: 5² + 12² = 25 + 144 = 169 = 13² ✓. GCD(5,12,13) = 1, so YES it is primitive!
Q5. Find the sidelength of a rhombus with diagonals 24 and 70 units.
Ans: Half-diagonals = 12 and 35. Side = √(12² + 35²) = √(144+1225) = √1369 = 37 units
Q6. Is the hypotenuse always the longest side? Justify.
Ans: YES. Since a² + b² = c², we have c² > a² and c² > b², so c > a and c > b. The hypotenuse is always the longest side of a right triangle.
Q7. The hypotenuse of an isosceles right triangle is 10. Find the other two sides.
Ans: c² = 2a² → 100 = 2a² → a² = 50 → a = √50 = 5√2 ≈ 7.07 units
Q8. Prove that (3k, 4k, 5k) is a Baudhāyana triple for any positive integer k.
Ans: (3k)² + (4k)² = 9k² + 16k² = 25k² = (5k)². Since a² + b² = c², it is a Baudhāyana triple. ✓
Q9. Find the missing side: a = 9, c = 15.
Ans: b² = 15² − 9² = 225 − 81 = 144 → b = 12
Q10. Who proved Fermat’s Last Theorem and when?
Ans: Andrew Wiles proved Fermat’s Last Theorem in 1994, after over 300 years of failed attempts by mathematicians.
Chapter Summary
🔲 Doubling a Square
Build a square on the diagonal of the original — it has double the area. (Baudhāyana, Verse 1.9)
Build a square on the diagonal of the original — it has double the area. (Baudhāyana, Verse 1.9)
✂️ Halving a Square
Connect midpoints of sides to get an inner square with half the area.
Connect midpoints of sides to get an inner square with half the area.
⭐ The Main Theorem
In a right triangle with sides a, b (legs) and c (hypotenuse): a² + b² = c²
In a right triangle with sides a, b (legs) and c (hypotenuse): a² + b² = c²
📐 Isosceles Right Triangle
c = a√2. The number √2 ≈ 1.41421… is non-terminating and cannot be a fraction.
c = a√2. The number √2 ≈ 1.41421… is non-terminating and cannot be a fraction.
🔢 Baudhāyana Triples
Integer triples (a, b, c) with a²+b²=c². Primitive: no common factor > 1. Scaled: (ka, kb, kc).
Integer triples (a, b, c) with a²+b²=c². Primitive: no common factor > 1. Scaled: (ka, kb, kc).
🔭 Fermat’s Last Theorem
xⁿ + yⁿ = zⁿ has no solution in positive integers for n > 2. Proved by Andrew Wiles in 1994.
xⁿ + yⁿ = zⁿ has no solution in positive integers for n > 2. Proved by Andrew Wiles in 1994.
Top 5 Things to Remember for Exams
1. a² + b² = c² only for RIGHT-ANGLED triangles. 2. Hypotenuse is always the LONGEST side. 3. Diagonal of a square with side a = a√2. 4. (3,4,5), (5,12,13), (8,15,17) are key primitive Baudhāyana triples. 5. √2 ≈ 1.414 is irrational — it’s non-terminating and non-repeating.
1. a² + b² = c² only for RIGHT-ANGLED triangles. 2. Hypotenuse is always the LONGEST side. 3. Diagonal of a square with side a = a√2. 4. (3,4,5), (5,12,13), (8,15,17) are key primitive Baudhāyana triples. 5. √2 ≈ 1.414 is irrational — it’s non-terminating and non-repeating.
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