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The Baudhāyana–Pythagoras Theorem

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Chapter Notes

Overview
This chapter introduces the Baudhāyana-Pythagoras Theorem, which states the relationship between the sides of a right-angled triangle. It begins with ancient Indian methods for doubling and halving squares using diagonals, then generalizes to the theorem a² + b² = c². The big idea is that the square on the hypotenuse equals the sum of the squares on the other two sides, a fundamental principle in geometry used to find unknown side lengths and to generate integer triples.
Key Concepts (11)
Diagonal of a square
A line segment joining two opposite vertices of a square.
e.g. In a square of side 1 unit, the diagonal has length √2 units.
Doubling a square
Constructing a square whose area is twice that of a given square by building a square on the diagonal of the original square.
e.g. A square of side 2 cm has area 4 cm²; its diagonal square has area 8 cm².
Halving a square
Constructing a square whose area is half that of a given square by drawing a tilted inner square connecting midpoints of sides.
e.g. A square of side 4 cm has area 16 cm²; the inner square has area 8 cm².
Hypotenuse
In a right-angled triangle, the side opposite the right angle; it is the longest side.
e.g. In a triangle with sides 3 cm and 4 cm, the hypotenuse is 5 cm.
Isosceles right triangle
A right-angled triangle in which the two legs (shorter sides) are equal in length.
e.g. A triangle with legs of 1 unit each has a hypotenuse of √2 units.
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.
e.g. For a triangle with a=3, b=4, we have 3²+4²=9+16=25=c², so c=5.
Baudhāyana triple (Pythagorean triple)
A set of three positive integers (a, b, c) that satisfy a² + b² = c², forming the sides of a right-angled triangle.
e.g. (3, 4, 5), (5, 12, 13), (8, 15, 17).
Primitive Baudhāyana triple
A Baudhāyana triple in which the three numbers have no common factor greater than 1.
e.g. (3, 4, 5) is primitive; (6, 8, 10) is not because 2 is a common factor.
Scaled version of a triple
Multiplying each number in a Baudhāyana triple by the same positive integer k to get another triple (ka, kb, kc).
e.g. From (3,4,5), multiplying by 2 gives (6,8,10).
Fermat's Last Theorem
The statement that the equation xⁿ + yⁿ = zⁿ has no solution in positive integers for n > 2.
e.g. There are no positive integers x, y, z such that x³ + y³ = z³.
√2 (square root of 2)
The number which when multiplied by itself gives 2; it is non-terminating and non-repeating, approximately 1.41421356...
e.g. The diagonal of a unit square has length √2 ≈ 1.414 units.
Formulas & Equations (5)
c² = 2a²
Reads as: c squared equals 2 a squared
Variables: a = length of each equal side of an isosceles right triangle; c = length of the hypotenuse
Use when: Finding the hypotenuse when the equal side is known, or finding the equal side when the hypotenuse is known, for an isosceles right triangle.
⚠ Forgetting to take the square root at the end; students often write c = 2a instead of c = a√2.
a² + b² = c²
Reads as: a squared plus b squared equals c squared
Variables: a and b = lengths of the two shorter sides (legs) of a right triangle; c = length of the hypotenuse
Use when: Finding the third side of any right-angled triangle when two sides are known.
⚠ Adding a and b before squaring, e.g., writing (a+b)² = c² instead of a² + b² = c².
c = √(a² + b²)
Reads as: c equals the square root of a squared plus b squared
Variables: a and b = legs; c = hypotenuse
Use when: Directly calculating the hypotenuse when both legs are known.
⚠ Forgetting to take the square root after adding the squares.
a = √(c² - b²)
Reads as: a equals the square root of c squared minus b squared
Variables: c = hypotenuse; b = one leg; a = the other leg
Use when: Finding a missing leg when the hypotenuse and one leg are known.
⚠ Subtracting in the wrong order, e.g., b² - c² instead of c² - b².
1 + 3 + 5 + ... + (2n - 1) = n²
Reads as: the sum of the first n odd numbers equals n squared
Variables: n = number of terms; (2n-1) = the nth odd number
Use when: Generating Baudhāyana triples by expressing an odd square as the sum of consecutive odd numbers.
⚠ Using the wrong value of n when the odd number is given; e.g., for odd number 9, n=5, not 9.
How It Works
Doubling a square using the diagonal
  1. Take a given square of side length a.
  2. Draw its diagonal.
  3. Construct a new square using this diagonal as one side.
  4. The new square has area 2a², which is double the area of the original square.
💡 The new square is made of 4 congruent small triangles, while the original square is made of 2 such triangles.
Halving a square by folding
  1. Take a square paper.
  2. Fold the paper so that the crease lines pass through the midpoints of the sides.
  3. The inner quadrilateral PQRS formed by the creases is a square.
  4. Its area is exactly half the area of the original square.
💡 The inner square's side is the hypotenuse of an isosceles right triangle whose legs are half the side of the original square.
Finding the hypotenuse of an isosceles right triangle
  1. Let the equal sides be of length a.
  2. Construct a square on the hypotenuse; its area is c².
  3. Note that this square has twice the area of the square on side a, so c² = 2a².
  4. Take square root: c = a√2.
💡 For a=1, c=√2 ≈ 1.414; the decimal never terminates.
Combining two different squares (Baudhāyana's method)
  1. Take two squares of side lengths a and b (a < b).
  2. Join them side by side.
  3. Mark a rectangle of sides a and b on the larger square.
  4. Draw the diagonal of this rectangle; this diagonal becomes the side of a new square.
  5. The new square's area equals a² + b², the sum of the areas of the two original squares.
💡 This works because the diagonal forms a right triangle with legs a and b, and by the theorem, the square on the diagonal has area a² + b².
Generating Baudhāyana triples using odd numbers
  1. Recall that the sum of the first n odd numbers equals n².
  2. If the nth odd number (2n-1) is itself a perfect square, say m², then we have (n-1)² + m² = n².
  3. This gives a triple (n-1, m, n).
  4. Example: 9 is the 5th odd number and is a square (3²). So (4, 3, 5) is a triple.
💡 This method always gives triples where the hypotenuse is one more than one leg.
Solving Bhāskarāchārya's lotus problem
  1. Let x be the depth of the water (stem underwater).
  2. Total stem length = x + 1 (since 1 unit is above water).
  3. When the tip touches water 3 units away, a right triangle is formed: legs 3 and x, hypotenuse x+1.
  4. Apply theorem: 3² + x² = (x+1)².
  5. Solve: 9 + x² = x² + 2x + 1 → 9 = 2x + 1 → x = 4 units.
💡 The key assumption is that the stem is initially perpendicular to the water surface.
Exam Traps — Don't Make These Mistakes
⚠ Confusing the hypotenuse with a leg
✗ In a right triangle with sides 3 and 4, they write 3² + 4² = 5², but then label 5 as a leg.
→ The hypotenuse is always the longest side and opposite the right angle; 5 is the longest, so it must be the hypotenuse.
✓ Clearly identify c as the side opposite the right angle. For sides 3 and 4, the hypotenuse is 5.
⚠ Forgetting to take the square root
✗ For a=3, b=4, they write c² = 25, and then stop, writing c = 25.
→ c² = 25 means c = √25 = 5, not 25. The variable c is the side length, not its square.
✓ Always take the square root: c = √(a² + b²). For 3 and 4, c = √25 = 5.
⚠ Misapplying the formula for isosceles right triangles
✗ For an isosceles right triangle with legs 5, they write c² = 5² + 5² = 50, then write c = 50.
→ Again, they forget to take the square root. c = √50 = 5√2, not 50.
✓ c = √(a² + a²) = √(2a²) = a√2. For a=5, c = 5√2.
⚠ Incorrectly scaling triples
✗ They claim (3,4,5) scaled by 2 gives (6,8,10) but then say (6,8,10) is primitive.
→ A primitive triple has no common factor >1. (6,8,10) has a common factor 2, so it is not primitive.
✓ Check for common factors. (6,8,10) is a scaled version, not primitive. Primitive triples are like (3,4,5), (5,12,13).
⚠ Using the wrong side in the lotus problem
✗ They set up the equation as x² + (x+1)² = 3².
→ The hypotenuse is the total stem length (x+1), not the distance 3. The distance 3 is a leg.
✓ The right triangle has legs 3 and x, and hypotenuse (x+1). So 3² + x² = (x+1)².
Compare & Contrast
Method of construction
Doubling a square
Halving a square
Doubling uses the diagonal of the original square as the side of the new square; halving uses the midpoints of the sides to create an inner tilted square.
Common factor among the three numbers
Primitive Baudhāyana triple
Non-primitive (scaled) Baudhāyana triple
A primitive triple has no common factor greater than 1 (e.g., 3,4,5); a non-primitive triple is a multiple of a primitive triple (e.g., 6,8,10 is 2×3,4,5).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total 12 marks
Most asked: Applying a² + b² = c² to find missing sides of right triangles · Identifying and generating Baudhāyana triples · Finding the diagonal of a square using √2 · Solving word problems (like the lotus problem)
Don't skip: Word problems involving the theorem (e.g., lotus problem, ladder against a wall); Finding missing sides and stating whether a triple is primitive or scaled
One-Line Revision (read aloud before the exam)
  1. The diagonal of a square produces a square of double the area.
  2. Halving a square: the inner square formed by joining midpoints has half the area.
  3. In a right triangle, the side opposite the right angle is called the hypotenuse.
  4. Baudhāyana-Pythagoras Theorem: a² + b² = c², where c is the hypotenuse.
  5. For an isosceles right triangle with equal sides a, the hypotenuse c = a√2.
  6. √2 is approximately 1.414 and cannot be written as a terminating decimal or a simple fraction.
  7. A Baudhāyana triple is a set of three positive integers satisfying a² + b² = c².
  8. Examples of Baudhāyana triples: (3,4,5), (5,12,13), (8,15,17), (7,24,25).
  9. A primitive triple has no common factor greater than 1; (3,4,5) is primitive.
  10. Multiplying a triple by a positive integer k gives another triple (ka, kb, kc).
  11. Fermat's Last Theorem states xⁿ + yⁿ = zⁿ has no positive integer solutions for n > 2.
  12. The lotus problem: depth of water = 4 units when stem is 1 unit above water and tip touches 3 units away.
  13. The sum of the first n odd numbers equals n².
  14. If the nth odd number is a perfect square, it generates a Baudhāyana triple.

MCQs — 10 Questions with Answers

Q1. If a square has side length 1 unit, what is the area of the square constructed on its diagonal?

  • A. 1 square unit
  • B. 2 square units ✓
  • C. √2 square units
  • D. 4 square units

Answer: B — Baudhāyana showed that the square on the diagonal has exactly double the area of the original square, so 2 × 1 = 2 square units.

Q2. What happens to the area when you double the side length of a square?

  • A. Area becomes 2 times
  • B. Area becomes 4 times ✓
  • C. Area becomes √2 times
  • D. Area remains the same

Answer: B — If original side is a, area is a². If new side is 2a, new area is (2a)² = 4a², which is 4 times the original area.

Q3. For an isosceles right triangle with equal sides of 6 units, which formula gives the hypotenuse c?

  • A. c = 2 × 6
  • B. c² = 6 + 6
  • C. c² = 2 × 6² ✓
  • D. c = 6 + 6

Answer: C — The relationship for isosceles right triangles is c² = 2a², where a is the length of the equal sides.

Q4. Between which two consecutive whole numbers does √2 lie?

  • A. 0 and 1
  • B. 1 and 2 ✓
  • C. 2 and 3
  • D. √2 and 2

Answer: B — Since 1² = 1 < 2 and 2² = 4 > 2, we have 1 < √2 < 2.

Q5. A farmer in Maharashtra wants to double his square plot area. He originally has a square plot with side 20 meters. According to Baudhāyana's method, what should be the side of his new square plot?

  • A. 20√2 meters ✓
  • B. 40 meters
  • C. 20 meters
  • D. 10√2 meters

Answer: A — The diagonal of the original square (which is 20√2 meters) becomes the side of the new square with double the area.

Q6. When cutting two identical squares and rearranging pieces to create a double-area square, how many congruent triangles form the original square?

  • A. 1 triangle
  • B. 2 triangles ✓
  • C. 4 triangles
  • D. 8 triangles

Answer: B — The diagonal of a square divides it into exactly 2 congruent isosceles right triangles.

Q7. The hypotenuse of an isosceles right triangle is 10 units. What is the length of each equal side?

  • A. 5 units
  • B. 5√2 units
  • C. 10/√2 units ✓
  • D. 20 units

Answer: C — Using c² = 2a², we get 10² = 2a², so 100 = 2a², giving a² = 50, and a = √50 = 5√2 = 10/√2 units.

Q8. In a paper-folding method to create a square with half the area, the four midpoints of the original square's sides become vertices of a new square. Why is this new square exactly half the area?

  • A. Because the original square is divided into 4 equal parts
  • B. Because the new square's diagonal is half the original side
  • C. Because the 4 corner triangles created have total area equal to the inner square's area ✓
  • D. Because the side of the new square is half the original

Answer: C — The folding creates 4 congruent right triangles at the corners; when their combined area equals the inner square's area, the inner square is exactly half the original.

Q9. Which statement best explains why √2 cannot be written as a fraction m/n?

  • A. Because √2 is larger than 1
  • B. Because the prime 2 would appear an odd number of times on one side and even times on the other in the equation 2n² = m²
  • C. Because √2 is irrational and all fractions are rational
  • D. Both B and C are correct ✓

Answer: D — Both statements are true: C defines what irrationality means, and B provides the mathematical proof using prime factorization.

Q10. When Baudhāyana combines two different-sized squares with sides a and b (where b > a), the resulting large square has side length equal to what?

  • A. a + b
  • B. The hypotenuse of a right triangle with perpendicular sides a and b ✓
  • C. b - a
  • D. √(a² × b²)

Answer: B — Baudhāyana's Verse 1.12 states that the hypotenuse of a right triangle with perpendicular sides a and b is the side of a square whose area equals the sum of the areas of the two given squares.

Flashcards

What is the area of a square constructed on the diagonal of a unit square?

The area is 2 square units because the diagonal creates a square with double the original area.

If an isosceles right triangle has equal sides of length 5 units, what is c² (hypotenuse squared)?

c² = 2 × 5² = 50, so c² = 50.

Why does √2 have a non-terminating decimal representation?

Because no terminating decimal when squared gives exactly 2, as any terminating decimal squared would have a non-zero last digit.

What does Baudhāyana's Verse 1.12 state about two squares?

The area of the square on the hypotenuse equals the sum of the areas of the squares on the two perpendicular sides of a right triangle.

If you double the side of a square from 1 unit to 2 units, how many times does the area increase?

The area increases 4 times because (2)² = 4 while (1)² = 1.

What is the approximate decimal value of √2 to three decimal places?

√2 ≈ 1.414 (calculated from 1.414² = 1.999396, which is closest to 2).

In the halving square paper folding method, which points form the vertices of the inner square?

The midpoints of the four sides of the original square form the vertices (labeled P, Q, R, S) of the inner square with half the area.

What relationship exists between the hypotenuse c and equal side a of an isosceles right triangle?

c = √(2a²) = a√2, meaning the hypotenuse is √2 times the length of each equal side.

Can √2 be expressed as a fraction m/n where m and n are whole numbers?

No, because if 2n² = m², the prime 2 would occur an odd number of times on the left and even times on the right, which is impossible.

When combining two squares of different sizes using Baudhāyana's method, what forms the hypotenuse?

The side of the new square whose area equals the sum of the two original squares' areas forms a right triangle with the two original side lengths as perpendicular sides.

Important Board Questions

State the relationship between the hypotenuse and equal sides of an isosceles right triangle. [1 mark]

Use the formula c² = 2a² where c is the hypotenuse and a is the length of each equal side; express c in terms of a.

Why does doubling a square's side length result in an area that is 4 times larger, not 2 times? Explain with an example. [2 marks]

Show that if original side is a (area = a²), new side is 2a (area = 4a²); use a numerical example like a = 3.

Explain why √2 cannot be expressed as a fraction m/n where m and n are whole numbers. Use the concept of prime factorization in your explanation. [3 marks]

Start by assuming √2 = m/n, square both sides to get 2n² = m², then analyze how many times the prime 2 appears on each side; show the contradiction.

Draw and label two identical squares of side lengths a and b. Using Baudhāyana's method described in Verses 1.12 and 2.1, show step-by-step how to construct a square whose area equals the sum of both squares. Explain why the 4-sided figure formed is a square and why its area equals a² + b². [5 marks]

Show how 4 congruent right triangles with sides a and b are arranged around a central square of side (b − a); prove all angles are 90° using angle properties; verify area by adding: 4 × (½ab) + (b − a)² = a² + b².

Assertion-Reason Questions for The Baudhāyana–Pythagoras Theorem

Q1. Assertion (A):

The diagonal of a square with side length 1 unit has length √2 units.

Reason (R):

The square constructed on the diagonal of a unit square has double the area, so if c is the diagonal length, then c² = 2 × 1 = 2, giving c = √2.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason directly explains why the assertion must be correct by using the doubling-area property from Baudhāyana's construction and the definition of square area.

Q2. Assertion (A):

√2 cannot be expressed as a fraction m/n where m and n are counting numbers.

Reason (R):

√2 is irrational because Euclid proved using prime factorization that if 2n² = m², the prime 2 appears an odd number of times on the left and an even number of times on the right, which is impossible.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains the assertion by showing the contradiction in the prime factorization argument presented in the material.

Q3. Assertion (A):

In paper folding to halve a square, the inscribed square PQRS formed by connecting the midpoints of the original square's sides has vertices at those midpoints.

Reason (R):

PQRS must be a square because the fold creates right angles, and the area is halved because the four corner triangles are congruent to each other and to PQRS.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—PQRS does have vertices at the midpoints. The reason is partially true (the vertices form a square with right angles) but does not correctly explain why the area is halved. The area is halved because the four corner triangles together equal the area of PQRS, not because they are congruent to PQRS itself.

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Case-Based Questions (CBQ) for The Baudhāyana–Pythagoras Theorem

Case Study:

Harini is tiling the floor of her living room using two different square tile patterns. Pattern A uses small square tiles of side 1 meter, and she wants to create a Pattern B using larger tiles such that each new tile has exactly double the area of Pattern A. Following Baudhāyana's construction method, she arranges the tiles by placing them along the diagonal of the original Pattern A square. She finds that she needs exactly 2 tiles from Pattern A to form one tile of Pattern B. The cost of Pattern A tiles is ₹200 per tile, and Pattern B tiles cost ₹350 per tile. Harini wants to tile her room completely using Pattern B.

Q1. What is the area of one tile in Pattern B?

  • A. 1 square meter
  • B. 2 square meters ✓
  • C. 4 square meters
  • D. √2 square meters

Answer: B

Q2. If Harini's room has an area of 20 square meters, how many Pattern B tiles does she need to cover the entire floor?

  • A. 5 tiles
  • B. 10 tiles ✓
  • C. 20 tiles
  • D. 40 tiles

Answer: B

Q3. What is the side length of one Pattern B tile?

  • A. √2 meters ✓
  • B. 2 meters
  • C. 1.414 meters
  • D. 2√2 meters

Answer: A

Q4. What is the total cost of Pattern B tiles needed to tile Harini's entire living room?

  • A. ₹3500 ✓
  • B. ₹4000
  • C. ₹3000
  • D. ₹7000

Answer: A

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