The Baudhāyana–Pythagoras Theorem

CHAPTER 2: THE BAUDHĀYANA-PYTHAGORAS THEOREM

COMPREHENSIVE CLASS 8 NOTES

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2.1 DOUBLING A SQUARE

Historical Context and the Problem

**Baudhāyana** was an ancient Indian mathematician who lived around 800 BCE and documented his mathematical knowledge in a text called the **Śulba-Sūtra** (Verse 1.9). The Śulba-Sūtras were used as guides for constructing altars and temples in ancient India, requiring precise geometric methods.

**The Core Problem:** Given a square, how can we construct a new square that has exactly double the area of the original square?

Why Doubling the Side Length Doesn't Work

A common mistake is to assume that doubling the side length gives double the area.

**Example:** If the original square has side length 1 unit:

Original area = 1 × 1 = **1 square unit**

If we double the side length to 2 units: New area = 2 × 2 = **4 square units**

This gives us **4 times the area, not double!**

**Why this happens:** When you double a linear measurement (like side length), the area (a 2-dimensional measurement) increases by the square of that factor. So doubling the side length multiplies the area by 2² = 4.

Baudhāyana's Elegant Solution

**The Solution:** The diagonal of a square produces a square of **double the area** of the original square.

**Method:**

1. Take a square PQRS with side length 1 unit

2. Draw a square on its diagonal

3. This new square (shown with dotted lines) has exactly double the area

Why the Diagonal Square Has Double the Area

To understand why this works, we can draw horizontal and vertical lines (east-west and north-south lines) through the original square's sides and vertices.

**Key Insight:** The vertical and horizontal lines passing through the sides of the original square actually pass through the vertices of the new tilted square. This happens because:

The diagonals of a square bisect each corner angles (45° angles)

The sides of the original square, being perpendicular to each other, bisect the angles of the tilted square

**Visual Breakdown:**

The **original square** is divided into **2 small congruent triangles** along its diagonal

The **new tilted square** is divided into **4 small congruent triangles** (each identical to those in the original square)

Since all 4 triangles in the new square are congruent to the 2 triangles in the original, and the new square contains twice as many triangles: **Area of new square = 2 × Area of original square**

Sequence of Doubling Squares

Following the same method repeatedly creates a sequence:

Square 1: area = 1 square unit (made of 2 congruent triangles)

Square 2: area = 2 square units (made of 4 congruent triangles)

Square 3: area = 4 square units (made of 8 congruent triangles)

And so on...

**Real-Life Application:** In ancient India, when designing temple courtyards, architects needed to double or triple areas while maintaining geometric precision. Using Baudhāyana's method ensured exact proportions rather than approximations.

Paper-Cutting Verification

You can verify this by:

1. Cutting two identical squares from paper

2. Cutting the second square into 4 pieces (labeled 5, 6, 7, 8) based on the diagonal and perpendicular lines

3. Arranging these 4 pieces around the first square to form a larger square

4. The resulting square has double the area of either original square

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2.2 HALVING A SQUARE

The Reverse Problem

Now we consider the inverse problem: **Given a square, construct a square whose area is half that of the original square.**

This is the reverse of the doubling process from Section 2.1.

Why Halving the Side Length Doesn't Work

**Common Misconception:** If we halve the side length, does the area become half?

**Answer: No!**

**Example:** Original square has side = 2 units

Original area = 2 × 2 = 4 square units

Square with half the side length = 1 unit

Area of this smaller square = 1 × 1 = 1 square unit

This is 1/4 of the original area, not 1/2!

**Why:** When you halve a linear dimension, the area decreases by the square of that factor: (1/2)² = 1/4

**To fill the original square:** You would need **4 such smaller squares**, not 2.

Baudhāyana's Method for Halving

**The Method:**

1. Start with a square (let's call it ABCD)

2. Fold the paper inward so that creases pass through the **midpoints of all four sides**

3. The inner square formed (PQRS) has **exactly half the area** of the original square

Why This Method Works: The Mathematical Explanation

Let the original square have side length 2a (for convenience).

**Step 1 - Identify the Vertices of the Halved Square:**

The four vertices P, Q, R, S are located at the midpoints of the four sides of the original square

Each midpoint divides a side into two segments of length a

**Step 2 - Prove PQRS is a Square:**

Using the right triangles formed in the corners:

Each corner of the original square is cut off by a right isosceles triangle

For example, at corner A: the triangle has two perpendicular sides of length a

By symmetry, all four corner triangles are congruent right isosceles triangles

**Using Triangle Congruence (Method of proof):**

1. Consider the four corner triangles formed (at each vertex of the original square)

2. Each corner triangle has legs of length a (half the side)

3. The hypotenuses of these triangles form the sides of PQRS

4. All four corner triangles are **congruent by SAS (Side-Angle-Side):**

Two perpendicular sides of length a

Right angle between them

5. Therefore, all hypotenuses are equal, making PQRS a rhombus

6. Each angle of PQRS is a right angle (can be proven by analyzing angles in the corner triangles)

7. Thus, PQRS is a **square**

**Step 3 - Calculate the Area of PQRS:**

Area of original square = (2a)² = 4a²

Area of each corner triangle = (1/2) × a × a = a²/2

Total area of 4 corner triangles = 4 × (a²/2) = 2a²

Area of PQRS = 4a² - 2a² = **2a²**

This is exactly **half** of the original square's area (4a²)

Paper-Folding Verification

**Practical Method:**

1. Take a square piece of paper

2. Mark the midpoint of each side

3. Fold each corner inward so that the fold line passes through the two adjacent midpoints

4. The inner square PQRS has half the area of the original square

**Indian Context Example:** A cloth merchant in Delhi needs to cut a square piece of cloth into two equal square pieces for two customers. Using this folding method, they can fold the original square along the midpoints and cut along the creases to get two squares of equal area.

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2.3 HYPOTENUSE OF AN ISOSCELES RIGHT TRIANGLE

Understanding Right Triangles

In a **right triangle:**

**Right angle:** The 90° angle

**Hypotenuse:** The side opposite the right angle (the longest side)

**Other two sides:** Called the perpendicular sides or legs

In an **isosceles right triangle:**

The right angle is between the two equal sides

The two legs have equal length (let's call it a)

The hypotenuse is the side opposite the right angle

Finding the Hypotenuse of a Unit Isosceles Right Triangle

**Given:** Isosceles right triangle with both equal sides = 1 unit

**Goal:** Find the length of the hypotenuse

Key Geometric Relationship

**Important Discovery:**

A square with side 1 unit is made up of exactly **two identical isosceles right triangles** when cut along its diagonal

The diagonal of this square becomes the hypotenuse of each triangle

Therefore:

Area of square PEAR (with side 1 unit) = 1 × 1 = 1 square unit

Area of square REST (built on the diagonal) = 2 × 1 = 2 square units (from Section 2.1)

Calculating the Hypotenuse Length

If c is the length of the hypotenuse ER, and we know the area of the square built on it:

Area of square REST = c × c = c²

Since Area of REST = 2 square units:

**c² = 2**

Therefore: **c = √2 units**

What is √2 (Square Root of 2)?

**Definition:** √2 is the positive number that when multiplied by itself equals 2. It's written as √2 and read as "square root of 2" or sometimes "root 2."

#### Finding the Decimal Representation of √2

**Method: Successive Approximation (Bounding Method)**

This technique finds increasingly narrow bounds on the value of √2.

**Step 1: Is √2 less than or greater than 1?**

1² = 1

2² = 4

Since 1 < 2 < 4, we have: **1 < √2 < 2**

**Step 2: Is √2 less than or greater than 2?**

2² = 4 > 2

So √2 < 2 ✓

Therefore: **1 < √2 < 2**

**Step 3: Narrow the bounds using decimals (one decimal place):**

Testing numbers between 1 and 2:

1.1² = 1.21 < 2

1.2² = 1.44 < 2

1.3² = 1.69 < 2

1.4² = 1.96 < 2

1.5² = 2.25 > 2

Therefore: **1.4 < √2 < 1.5**

**Step 4: Narrow to two decimal places:**

Testing between 1.4 and 1.5:

1.41² = 1.9881 < 2

1.42² = 2.0164 > 2

Therefore: **1.41 < √2 < 1.42**

**Step 5: Narrow to three decimal places:**

Testing between 1.41 and 1.42:

1.414² = 1.999396 < 2

1.415² = 2.002225 > 2

Therefore: **1.414 < √2 < 1.415**

Important Property: √2 Is Not a Terminating Decimal

**Question:** Can √2 be written as a terminating decimal like 1.414?

**Answer: No**

**Proof (by contradiction):**

1. Suppose √2 could be written as a terminating decimal ending with a non-zero digit

2. For example: √2 = 1.414...4 (ending in 4)

3. If we square this number: (1.414...4)² would also have to end in a non-zero digit after the decimal point

4. But we know √2 squared equals exactly 2, which is 2.000... (has infinitely many zeros after the decimal point)

5. This is a contradiction!

6. Therefore, √2 cannot be expressed as a terminating decimal

√2 Cannot Be Written as a Fraction (Proof by Euclid)

**Question:** Can √2 be expressed as a fraction m/n where m and n are positive integers?

**Answer: No** (This remarkable proof was given by **Euclid** around 300 BCE)

**Proof by Contradiction:**

Assume √2 can be written as a fraction:

√2 = m/n (where m and n are positive integers)

Squaring both sides:

2 = m²/n²

**2n² = m²** ... (equation 1)

**Key Fact About Prime Factorization:**

In the prime factorization of any perfect square (like m² or n²), every prime number appears an **even number of times**.

**Analyzing Equation 1 (2n² = m²):**

On the **left side (2n²):**

The prime 2 appears at least once (from the factor 2)

The prime 2 appears an even number of times in n² (from prime factorization of perfect square)

Together: 2 appears an **odd number of times** on the left side

On the **right side (m²):**

Every prime, including 2, appears an **even number of times** (since m² is a perfect square)

**Contradiction:**

Left side: 2 appears an odd number of times

Right side: 2 appears an even number of times

This is impossible! Therefore, our assumption was wrong.

**Conclusion:** √2 cannot be expressed as a fraction m/n.

Non-Terminating, Non-Repeating Decimal

Since √2 is:

NOT a terminating decimal (like 1.5 or 2.75)

NOT a repeating decimal (like 0.333... or 0.142857142857...)

NOT a fraction

It must be an **irrational number** with a **non-terminating, non-repeating decimal expansion**.

**Decimal Value (to 8 places):**

**√2 = 1.41421356...**

This decimal goes on forever without repeating any pattern.

Real-Life Application

**Carpentry and Construction:** A carpenter in Mumbai needs to find the diagonal of a square floor tile with side 1 meter. Using this method, they know the diagonal is √2 ≈ 1.414 meters, helping them plan the layout of larger floor designs.

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2.4 COMBINING TWO DIFFERENT SIZED SQUARES

The Problem

We've learned how to:

1. **Double a square** (combine two identical squares)

2. **Halve a square** (reverse process)

Now: **How do we combine two squares of different sizes to create a new square whose area equals the sum of the two original areas?**

Visual representation:

Square with area A + Square with area B = Square with area (A + B)

Baudhāyana's Remarkable Solution

From Baudhāyana's Śulba-Sūtra (Verse 1.12), he provides an elegant method:

**"The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides."**

**Method in Simple Terms:**

1. Take two squares with side lengths a and b (where a < b)

2. Construct a right-angled triangle with:

One perpendicular side of length a

Other perpendicular side of length b

3. The hypotenuse of this right triangle becomes the side of a new square

4. This new square has area equal to (area of first square) + (area of second square)

Verification Using Same-Sized Squares

**Does this work when a = b?**

When both squares are identical:

Right triangle has both perpendicular sides = a

This forms an **isosceles right triangle**

Hypotenuse = √(a² + a²) = √(2a²) = a√2

Square on hypotenuse has area = (a√2)² = 2a²

Sum of two identical squares = a² + a² = 2a²

**Yes! It agrees perfectly with our earlier doubling method.**

The Geometric Proof: Baudhāyana's Construction

**Historical Note:** Baudhāyana provided a second verse (Verse 2.1) giving detailed instructions on why his method works.

**Step-by-Step Construction:**

**Step 1: Join the Two Squares**

Place two squares with sides a and b adjacent to each other. Imagine square with side a is inside or overlapping with square with side b.

**Step 2: Mark the Rectangular Portion**

Use the side of the smaller square (side a) to mark a rectangular portion of the larger square

This creates a rectangle with dimensions a × b

Draw the diagonal of this rectangle

This diagonal is the hypotenuse of a right triangle with:

Perpendicular side 1: length a

Perpendicular side 2: length b

Hypotenuse: length c (to be determined)

**Step 3: Create Four Congruent Triangles**

The crucial insight: Create three more congruent copies of the right triangle we just made. Arrange all four triangles around a central figure to form a square.

**The four triangles are labeled T, U, V, X** (arranged around the perimeter) with a central square labeled W.

**Step 4: The Central Configuration**

When arranged correctly:

The four right triangles T, U, V, X are all **congruent** (identical in shape and size)

Each has perpendicular sides of length a and b

The hypotenuse of each triangle has length c

Why the Central Figure Is a Square

**Claim:** The central figure TUVXW is actually a square with all sides of length c.

**Proof of This Claim:**

**1. All sides have equal length:**

Each side of the outer boundary is the hypotenuse of one of the congruent triangles

All four hypotenuses equal c

Therefore, the outer figure is a **rhombus**

**2. All angles are right angles:**

Consider where two triangles meet

At each vertex of the central figure, two angles from different right triangles come together

If a right triangle has legs of length a and b, the two acute angles are complementary (sum to 90°)

When two triangles share a vertex: angle from first triangle + angle from second triangle = 90°

These angles plus the angle from the central square at that point must equal 360°

This forces each angle of the outer boundary to be 90°

**3. Conclusion:**

A rhombus with all right angles is a **square**

This square has side length c (the hypotenuse length)

Area Analysis

**The key insight:**

The area of the large square built on the hypotenuse equals:

Area of 4 congruent right triangles + Area of central square

= 4 × (1/2 × a × b) + (c² - 2ab)

= 2ab + c² - 2ab

= c²

But also:

The original two squares can be rearranged into the same space

**Area equation:**

**a² + b² = c²**

Paper-Cutting Verification

**Practical Demonstration:**

1. **Cut two different-sized squares** from paper with sides a and b

2. **Make two strategic cuts:**

Cut along a line that creates the rectangle of dimensions a by b

Cut along the diagonal of this rectangle

3. **Rearrange the three pieces:**

You get exactly three pieces: the two original squares transformed, and they can be rearranged

These three pieces form a larger square

4. **Verification using right triangle:**

Make a right triangle with sides a and b

Draw a square on the hypotenuse (side c)

Use the three pieces from cutting the original squares to cover this square perfectly

**This physical demonstration proves the relationship: a² + b² = c²**

Real-Life Application

**Land Measurement in India:** A farmer in Punjab has two square fields:

Field 1: side 300 meters (area = 90,000 m²)

Field 2: side 400 meters (area = 160,000 m²)

Using Baudhāyana's method, they could arrange these into a single square field with area 250,000 m², which has side length 500 meters (the hypotenuse of the 300-400 right triangle).

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2.5 THE BAUDHĀYANA-PYTHAGORAS THEOREM

The Universal Theorem

**Baudhāyana's Theorem on Right-Angled Triangles:**

For any right-angled triangle with:

Perpendicular sides of length a and b

Hypotenuse of length c

The relationship is:

**a² + b² = c²**

This is the most important theorem in this chapter.

Historical Significance

**Baudhāyana** stated this theorem in its most general form around 800 BCE in his Śulba-Sūtra.

**Later Development:**

**Pythagoras** (c. 500 BCE), the Greek philosopher-mathematician, also studied and promoted this theorem about 300 years after Baudhāyana

This is why the theorem is often called the **Pythagorean Theorem** in Western mathematics

**Inclusive Naming:**

Modern mathematicians use the term **Baudhāyana-Pythagoras Theorem** to:

Acknowledge both mathematicians

Credit the earlier discoverer (Baudhāyana)

Recognize the wider adoption through Pythagoras

Application: The 3-4-5 Triangle

**Example Setup:** Draw a right-angled triangle with:

One perpendicular side = 3 cm

Other perpendicular side = 4 cm

Hypotenuse = ? (to be determined)

**Using Baudhāyana's Theorem:**

Let c = length of hypotenuse

a² + b² = c²

3² + 4² = c²

9 + 16 = c²

**25 = c²**

**c = 5 cm**

**Verification by Measurement:** If you draw this triangle accurately and measure the hypotenuse, it will measure exactly 5 cm!

**Why 3-4-5 Is Special:** This is the most famous right triangle because:

All three sides are whole numbers (integers)

It's the smallest Pythagorean triple

It's frequently used in construction and architecture

Common Right-Angled Triangle Relationships to Remember

The most commonly used Pythagorean triples (explained later) include:

(3, 4, 5): used frequently in construction

(5, 12, 13): common in Indian textile measurements

(8, 15, 17): used in architectural designs

(7, 24, 25): found in ancient Indian manuscripts

Using the Theorem to Find Unknown Sides

**Case 1: Finding the Hypotenuse (c is unknown)**

**Given:** Both perpendicular sides a and b

**Formula:** c = √(a² + b²)

**Example:** If a = 6 and b = 8

c² = 6² + 8² = 36 + 64 = 100

c = √100 = 10

**Case 2: Finding a Missing Perpendicular Side**

**Given:** Hypotenuse c and one perpendicular side a

**Formula:** b = √(c² - a²)

**Example:** If c = 13 and a = 5

b² = 13² - 5² = 169 - 25 = 144

b = √144 = 12

Common Mistakes to Avoid

**Mistake 1:** Confusing which side is the hypotenuse

The hypotenuse is ALWAYS the longest side

It's ALWAYS opposite the right angle

In the equation a² + b² = c², c is always the hypotenuse

**Mistake 2:** Forgetting to take the square root

The equation gives you c², not c

You must find √(a² + b²) to get the actual length

**Mistake 3:** Using this theorem for non-right triangles

This theorem ONLY works for right triangles

Don't apply it to acute or obtuse triangles

Real-Life Applications in India

**1. Construction and Carpentry:**

A carpenter in Chennai needs to check if a corner of a room is truly 90°. They measure:

3 meters along one wall

4 meters along the adjacent wall

The diagonal across the corner

Using the 3-4-5 triangle:

If the diagonal measures exactly 5 meters, the corner is a perfect right angle

If not, the corner needs adjustment

**2. Ladder Safety:**

A painter in Delhi leans a 13-meter ladder against a building:

The base of the ladder is 5 meters from the wall

The ladder reaches 12 meters up the wall (since 5² + 12² = 13²)

Using this 5-12-13 triangle ensures the ladder is at the correct angle (not too steep, not too shallow)

**3. Agricultural Land Measurement:**

A surveyor in Tamil Nadu uses the 3-4-5 triangle to mark out a square field:

Measure 300 meters along one direction

Measure 400 meters perpendicular

The diagonal is 500 meters (scaled version of 3-4-5)

This ensures the corner is exactly 90°

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2.5(a) BAUDHĀYANA TRIPLES: SPECIAL INTEGER SOLUTIONS

What Are Baudhāyana Triples?

A **Baudhāyana triple** (also called a **Pythagorean triple**) is a set of three positive integers (a, b, c) that satisfy the equation:

**a² + b² = c²**

These triples represent the side lengths of right-angled triangles where **all three sides are whole numbers**.

**Alternative Names:**

Baudhāyana-Pythagoras triples

Pythagorean triples

Right-angled triangle triples

Integer triples

List of Common Baudhāyana Triples (from Baudhāyana's Śulba-Sūtra)

Baudhāyana himself listed several examples in Verse 1.13:

1. **(3, 4, 5)** — The most famous triple

Check: 3² + 4² = 9 + 16 = 25 = 5² ✓

2. **(5, 12, 13)** — Used in textiles

Check: 5² + 12² = 25 + 144 = 169 = 13² ✓

3. **(8, 15, 17)** — Found in Indian architecture

Check: 8² + 15² = 64 + 225 = 289 = 17² ✓

4. **(7, 24, 25)** — From ancient manuscripts

Check: 7² + 24² = 49 + 576 = 625 = 25² ✓

5. **(12, 35, 37)**

Check: 12² + 35² = 144 + 1,225 = 1,369 = 37² ✓

6. **(15, 36, 39)**

Check: 15² + 36² = 225 + 1,296 = 1,521 = 39² ✓

Finding All Triples with Numbers ≤ 20

**Complete List:**

(3, 4, 5)

(6, 8, 10)

(9, 12, 15)

(12, 16, 20)

Notice: Each of these can be obtained by multiplying (3, 4, 5) by a whole number:

(3, 4, 5) × 1 = (3, 4, 5)

(3, 4, 5) × 2 = (6, 8, 10)

(3, 4, 5) × 3 = (9, 12, 15)

(3, 4, 5) × 4 = (12, 16, 20)

Scaled (Non-Primitive) Triples

If (a, b, c) is a Baudhāyana triple, then **(ka, kb, kc) is also a Baudhāyana triple** for any positive integer k.

**Mathematical Proof:**

**Given:** (a, b, c) is a Baudhāyana triple, so a² + b² = c²

**To Prove:** (ka)² + (kb)² = (kc)²

**Calculation:**

(ka)² + (kb)² = k²a² + k²b²

= k²(a² + b²)

= k²(c²) [since a² + b² = c²]

= (kc)²

**Therefore, (ka, kb, kc) is a Baudhāyana triple. ✓**

**Definition:** We call (ka, kb, kc) a **scaled version** of (a, b, c).

**Examples:**

(3, 4, 5) scaled by 2 → (6, 8, 10)

(3, 4, 5) scaled by 10 → (30, 40, 50)

(5, 12, 13) scaled by 3 → (15, 36, 39)

Primitive Baudhāyana Triples

A **primitive Baudhāyana triple** is a triple (a, b, c) where **a, b, and c have no common factor greater than 1**.

In other words, the three numbers are **relatively prime** (they share no common divisor except 1).

**Examples of Primitive Triples:**

(3, 4, 5) — primitive

Common factors? GCD(3, 4, 5) = 1 ✓

(5, 12, 13) — primitive

Common factors? GCD(5, 12, 13) = 1 ✓

**Examples of Non-Primitive Triples:**

(6, 8, 10) — not primitive

Common factors? All three are even. GCD(6, 8, 10) = 2

This is (3, 4, 5) × 2

(9, 12, 15) — not primitive

Common factors? All divisible by 3. GCD(9, 12, 15) = 3

This is (3, 4, 5) × 3

(15, 36, 39) — not primitive

Common factors? All divisible by 3. GCD(15, 36, 39) = 3

This is (5, 12, 13) × 3

Relationship Between Primitive and Non-Primitive Triples

**Key Principle:** Every non-primitive Baudhāyana triple is a scaled version of a primitive triple.

**Converting Non-Primitive to Primitive:**

If (a, b, c) is non-primitive with common factor f > 1, then:

(a/f, b/f, c/f) is a primitive Baudhāyana triple

**Example:**

Given (9, 12, 15) — common factor is 3

Primitive form: (9÷3, 12÷3, 15÷3) = (3, 4, 5)

Verify: 3² + 4² = 9 + 16 = 25 = 5² ✓

**Why This Matters:** To find all Baudhāyana triples, we need to:

1. Find all primitive triples

2. Then generate scaled versions of each primitive triple

This is infinitely efficient because there are infinitely many triples but only finitely many primitive ones in each "family."

Is There an Unending Sequence of Triples?

**Question:** Can we generate infinitely many Baudhāyana triples?

**Answer: Yes!**

**Proof:**

We know (3, 4, 5) is a primitive triple

For any positive integer k:

(3k, 4k, 5k) is also a Baudhāyana triple

Since k can be any positive integer (1, 2, 3, 4, ...), there are **infinitely many** triples

**Examples:**

k = 1: (3, 4, 5)

k = 2: (6, 8, 10)

k = 3: (9, 12, 15)

k = 4: (12, 16, 20)

...continuing forever

Generating Primitive Triples Using Odd Numbers

**The Method:** Mathematicians discovered that primitive Baudhāyana triples can be generated using **odd square numbers**.

#### Understanding the Algebraic Relationship

**Key Formula:**

For any positive integer n:

**1 + 3 + 5 + ... + (2n - 1) = n²**

This is the sum of the first n odd numbers.

**Examples:**

1 = 1²

1 + 3 = 4 = 2²

1 + 3 + 5 = 9 = 3²

1 + 3 + 5 + 7 = 16 = 4²

**Algebraic Derivation:**

The nth odd number is: **2n - 1**

Sum of first (n-1) odd numbers: **(n-1)²