The Baudhāyana–Pythagoras Theorem — Class 8 Mathematics (Ganita Prakash Part 2)
"The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides." — Baudhāyana, Śulba-Sūtra, Verse 1.12 (c. 800 BCE)
What the book actually covers (2026-27) This chapter is Chapter 2 of Ganita Prakash Part 2, pages 33–54. It is not a list of formulas to memorise. It builds the theorem the way Baudhāyana did — from a practical altar-building question — through seven sections: 2.1 Doubling a square · 2.2 Halving a square · 2.3 The hypotenuse of an isosceles right triangle, and the number √2 · 2.4 Combining two different squares, which is where the theorem itself appears · 2.5 Right triangles with integer sides — Baudhāyana triples, primitive and scaled · 2.6 Fermat's Last Theorem · 2.7 Applications, including a problem from Bhāskarāchārya's Līlāvatī. Two things the book deliberately does not do: it never proves the converse (that a² + b² = c² forces a right angle — that comes in Class 9), and it never uses similar triangles. Material of that kind has been moved to the appendix at the end of this page.
1. Where the chapter starts: doubling a square
Baudhāyana's Śulba-Sūtra was a manual for building fire altars, and altars sometimes had to be built to exactly twice the area of an existing one. So the opening question is a builder's question:
How do you construct a square with double the area of a given square?
The obvious first guess — double each side — is wrong, and wrong by a lot. Doubling the side gives a square of 4 times the area, not twice, because both factors in length × length have been doubled.
Baudhāyana's answer (Verse 1.9) is one line:
The diagonal of a square produces a square of double the area of the original square.
Why the diagonal works
Draw the original square, then build a square on its diagonal. Now draw the horizontal and vertical lines through the corner of the original square that sits at the centre of the new one. Those two lines turn out to be exactly the diagonals of the new square, and they cut it into four triangles. The original square's own diagonal cuts it into two triangles of the very same shape.
All these small triangles are right-angled, with two equal legs and the right angle between them, so they are congruent by SAS and have equal areas. Then the count settles everything:
- original square = 2 small triangles
- square on the diagonal = 4 small triangles
- therefore the new square has double the area
Repeat the construction and you get a sequence of squares made of 2, 4, 8, 16 … triangles, each double the last.
With paper
Cut two identical squares, each along one diagonal. That gives four congruent right isosceles triangles. Set their right-angle corners together at one point — the four right angles fill 360°, so they close up exactly — and the four hypotenuses form the outside of a square with double the area of either original.
2. Halving a square
The reverse question is just as practical, and it has the same trap.
Will a square of half the sidelength have half the area? No. A square of side L/2 has area L²/4 — a quarter, not a half. Four such squares fill the original. Halving a length halves the area twice over.
To genuinely halve the area, draw a tilted square inside the original, with its corners at the midpoints of the four sides.
The paper fold, and why it works
Fold the square inward so that the creases pass through the midpoints of the sides. What is left is a square PQRS. Two things need proving.
PQRS is a square. The four corner triangles folded away are congruent by SAS — each has two legs of length L/2 with a right angle between them — so their hypotenuses PQ, QR, RS, SP are all equal. Each corner triangle is right-angled and isosceles, so its base angles are 45°. At the point P three angles lie along a straight line, giving
∠SPQ = 180° − 45° − 45° = 90°
Four equal sides and four right angles make PQRS a square.
Its area is exactly half. Each corner triangle has area ½ × (L/2) × (L/2) = L²/8, and four of them total L²/2. So PQRS has area L² − L²/2 = L²/2.
Notice this is the doubling picture read backwards: the outer square's side is the inner square's diagonal.
3. The isosceles right triangle, and the number √2
Take a square of side 1. Its diagonal cuts it into two isosceles right triangles. How long is the hypotenuse?
We do not know the length, but we know the area of the square built on it — the doubling result gives it as 2. If c is the hypotenuse, that square has area c × c = c², so
c² = 2, and therefore c = √2.
General formula
For an isosceles right triangle with equal sides a:
c² = 2a², so c = a√2
Read the other way, a² = c²/2. This one formula handles both directions — finding the hypotenuse from the sides, and finding the sides from the hypotenuse.
What kind of number is √2?
The chapter now does something unusual for Class 8: it proves two negative facts.
√2 has no terminating decimal. Trapping it between squares gives 1 < √2 < 2, then 1.4 < √2 < 1.5, then 1.41 < √2 < 1.42, then 1.414 < √2 < 1.415 — and the process never ends. The reason is that a terminating decimal has a non-zero last digit, so its square also has a non-zero last digit after the decimal point, and can therefore never be exactly 2.000….
√2 is not a fraction. Suppose √2 = m/n for counting numbers m and n. Squaring gives 2n² = m². Now count the prime 2 on each side. In any square number every prime appears an even number of times — so 2 appears an even number of times in both n² and m². But the left-hand side is 2 × n², carrying one extra 2, an odd number of them. That is impossible, so no such fraction exists. This is Euclid's proof, from about 300 BCE.
So √2 = 1.41421356… is a genuinely new kind of number — the first irrational number most students meet. Class 9 picks the idea up properly.
Trapping a surd between one-decimal bounds
This is a technique the exercises use constantly. To bound √72, square one-decimal candidates until one falls below and the next rises above:
8.4² = 70.56 < 72 and 8.5² = 72.25 > 72, so 8.4 < √72 < 8.5
| Equal side a | c² = 2a² | Hypotenuse | Bounds |
|---|---|---|---|
| 3 | 18 | √18 = 3√2 | 4.2 < c < 4.3 |
| 4 | 32 | √32 = 4√2 | 5.6 < c < 5.7 |
| 6 | 72 | √72 = 6√2 | 8.4 < c < 8.5 |
| 8 | 128 | √128 = 8√2 | 11.3 < c < 11.4 |
| 9 | 162 | √162 = 9√2 | 12.7 < c < 12.8 |
Every entry is the side multiplied by √2 ≈ 1.414. The multiplying finds the answer; the squaring proves the bounds.
4. Combining two different squares — the theorem itself
Two squares of the same size can be combined into one square, whose side is the diagonal of either. What about two squares of different sizes? Baudhāyana's Verse 1.12 answers it:
The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides.
In practice: build a right-angled triangle whose two perpendicular sides are the sides of the two squares. The square on its hypotenuse has area equal to the sum of the two.
Why it works
Follow Verse 2.1. Join the two squares, mark a rectangle inside the larger one using a side of the smaller, and draw its diagonal — this creates a right triangle with legs a and b. Draw three more copies of that triangle around the hypotenuse.
The four-sided figure they enclose is a square. Its sides are the four hypotenuses, all equal because the triangles are congruent. Its angles are right angles because at each vertex the acute angles x and 90° − x from two neighbouring triangles sit on a straight line with the vertex angle:
vertex angle = 180° − x − (90° − x) = 90°
The x cancels, so the result holds whatever the triangle's shape. Rearranging the pieces then shows this square has area equal to the two original squares combined.
The theorem
Baudhāyana's Theorem on right-angled triangles. If a right-angled triangle has sidelengths a, b and c, where c is the length of the hypotenuse, then a² + b² = c²
Baudhāyana was the first person in recorded history to state this in full generality and in essentially modern form. Pythagoras (c. 500 BCE), who lived a couple of centuries later, also studied and admired it, and Western tradition attached his name. The joined name Baudhāyana-Pythagoras Theorem is used so that everyone knows which theorem is meant.
Sanity check on the same-size case: put b = a and the formula gives c² = 2a², which is the doubling result from §2.1. The general rule contains the special one.
Using it in the two directions
The whole skill is reading the figure correctly.
| What is given | What to do | Worked example |
|---|---|---|
| both legs | add the squares | legs 5 and 12 → c² = 25 + 144 = 169, c = 13 |
| one leg + hypotenuse | subtract the squares | leg 8, hyp 17 → b² = 289 − 64 = 225, b = 15 |
The side that faces the right angle is the hypotenuse and always sits alone on the right of a² + b² = c². Getting this backwards is the single commonest error in the chapter — and it announces itself, because the "hypotenuse" comes out shorter than a leg.
Constructing multiples of a square
Once you can combine any two squares, you can build a square of any whole multiple of a given area (Verse 1.10):
- Triple: right triangle with legs s and s√2 (the diagonal), since s² + 2s² = 3s²
- Five times: right triangle with legs s and 2s, since s² + 4s² = 5s²
- Difference: put the larger length on the hypotenuse. A right triangle with hypotenuse 7 and leg 5 has third side √24, and 24 = 49 − 25.
5. Baudhāyana triples
In Verse 1.13 Baudhāyana lists integer triples that are the sides of a right triangle:
(3, 4, 5) · (5, 12, 13) · (8, 15, 17) · (7, 24, 25) · (12, 35, 37) · (15, 36, 39)
Any triple (a, b, c) of positive integers with a² + b² = c² is called a Baudhāyana triple (also Pythagorean triple).
Scaling
If (a, b, c) is a triple then so is (ka, kb, kc), because
(ka)² + (kb)² = k²(a² + b²) = k²c² = (kc)²
So (3, 4, 5) generates (6, 8, 10), (9, 12, 15), (12, 16, 20), … without end — there are infinitely many triples. The rule runs backwards too: dividing a triple by any common factor gives another triple.
Primitive triples
A triple with no common factor greater than 1 is primitive. (3, 4, 5) is primitive; (9, 12, 15) is not.
Careful here — the book's printed list is not complete, and the exercise knows it. Page 48 says the triples with all numbers at most 20 "contain" (3,4,5), (6,8,10), (9,12,15) and (12,16,20). Those are all multiples of (3,4,5). But two more qualify — (5, 12, 13) and (8, 15, 17) — and neither is a multiple of (3,4,5). The complete list has six entries: (3,4,5), (5,12,13), (6,8,10), (8,15,17), (9,12,15), (12,16,20). Of these, exactly three are primitive: (3,4,5), (5,12,13) and (8,15,17). Finding the two the book leaves out is the point of the exercise — it is what shows that not every triple is a scaled (3,4,5).
Every triple is either primitive or a scaled primitive. Divide by the largest common factor: what is left has no common factor, so it is primitive, and the original is that primitive multiplied back up.
Generating new primitives — the odd-square method
The sum of the first n odd numbers is n². Equivalently
(n − 1)² + (2n − 1) = n²
If the nth odd number 2n − 1 happens to be a perfect square, this becomes a sum of two squares equal to a square — a triple.
Take any odd number m. Then m² is odd, so it is the nth odd number for n = (m² + 1)/2, and (m, n − 1, n) is a triple:
| m | m² | n = (m²+1)/2 | Triple | Check |
|---|---|---|---|---|
| 3 | 9 | 5 | (3, 4, 5) | 9 + 16 = 25 ✓ |
| 5 | 25 | 13 | (5, 12, 13) | 25 + 144 = 169 ✓ |
| 7 | 49 | 25 | (7, 24, 25) | 49 + 576 = 625 ✓ |
| 9 | 81 | 41 | (9, 40, 41) | 81 + 1600 = 1681 ✓ |
| 11 | 121 | 61 | (11, 60, 61) | 121 + 3600 = 3721 ✓ |
| 13 | 169 | 85 | (13, 84, 85) | 169 + 7056 = 7225 ✓ |
| 15 | 225 | 113 | (15, 112, 113) | 225 + 12544 = 12769 ✓ |
Every triple this produces is primitive. The two larger members differ by 1, so any common divisor would have to divide n − (n − 1) = 1, forcing it to be 1.
But the method is not complete. It only ever gives triples with the hypotenuse one more than a leg. So it can never produce (8, 15, 17) — differences 2 and 9 — nor (20, 21, 29), (12, 35, 37), (28, 45, 53) or (33, 56, 65). A complete generator exists (see the appendix), but it belongs to a later class.
6. Fermat's Last Theorem
Infinitely many squares are the sum of two squares. Fermat, in the 17th century, asked whether the same could ever happen for higher powers:
xⁿ + yⁿ = zⁿ, with x, y, z natural numbers and n > 2 — does any solution exist?
In the margin of a book he wrote that none does, and added: "I have found a truly marvellous proof of this statement, but the margin is too small to contain it." No one ever found that proof.
Three hundred years of failed attempts followed. In 1963 a ten-year-old boy named Andrew Wiles read about the problem, and resolved to solve it. He finally proved it in 1994.
The chapter includes this to show something worth carrying: an ordinary-looking pattern in a Class 8 textbook can open directly onto one of the hardest problems in the history of mathematics.
7. Applications
Bhāskarāchārya's lotus problem (Līlāvatī)
"In a lake surrounded by chakra and krauñcha birds, there is a lotus flower peeping out of the water, with the tip of its stem 1 unit above the water. On being swayed by a gentle breeze, the tip touches the water 3 units away from its original position. Quickly tell the depth of the lake."
The stem length is never given, so the problem looks under-determined. It is not.
Let x be the depth, which is the stem below water. The whole stem is x + 1. When swayed, the stem does not stretch, so it is still x + 1 long and now forms the hypotenuse of a right triangle with legs 3 and x. Assuming the upright stem was perpendicular to the water:
3² + x² = (x + 1)² 9 + x² = x² + 2x + 1 9 = 2x + 1 x = 4
The depth is 4 units. The x² terms cancel, which is exactly why the missing stem length never mattered — squaring puts the unknown on both sides and it disappears.
Reading a figure before calculating
The six triangles in the closing exercise are worth studying, because each tests whether you found the right-angle mark first:
| Given | Which rule | Answer |
|---|---|---|
| legs 7 and 9 | add | √130 ≈ 11.40 (11.4 < c < 11.5) |
| legs 4 and 10 | add | √116 = 2√29 ≈ 10.77 |
| leg 40, hyp 41 | subtract | 9 exactly |
| leg 10, hyp √200 | subtract | 10 exactly (isosceles) |
| legs 10 and √150 | add | √250 = 5√10 ≈ 15.81 |
| leg 27, hyp 45 | subtract | 36 exactly |
Two shortcuts show up here. For 41² − 40², the identity c² − a² = (c + a)(c − a) gives 81 × 1 = 81 in one step. And 27 and 45 are 9 × 3 and 9 × 5, so the answer is 9 × 4 = 36 with no arithmetic at all. Note also that a side labelled √200 makes the work easier, not harder — squaring it just gives 200.
Shapes that need a right angle created first
The theorem needs a right angle to exist before it can be used. In several standard figures you have to make one.
- Rhombus — the diagonals bisect each other at right angles. Diagonals 24 and 70 give a right triangle with legs 12 and 35 (halved!), so the side is √1369 = 37. Using 24 and 70 directly is a classic error and gives 74.
- Equilateral triangle — drop an altitude. It bisects the base (RHS congruence), so a triangle of side 6 has height √(36 − 9) = √27 = 3√3 ≈ 5.196 and area ½ × 6 × 3√3 = 9√3 ≈ 15.59. In general: height a√3/2, area a²√3/4.
- Square or rectangle — the diagonal is already a hypotenuse. Square of side 5: diagonal √50 = 5√2 ≈ 7.07.
Squares on a grid
If a square has all four corners on grid dots, one edge moves p across and q up, so
area = p² + q²
That single fact answers a surprisingly deep question. Areas 2 (p=q=1), 4 (p=2, q=0) and 5 (p=1, q=2) are all drawable. Area 3 is impossible — the only squares up to 3 are 0 and 1, and no combination gives 3.
Achievable areas: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, … Impossible: 3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, 27, 28, 30, …
8. Common mistakes
- Putting the hypotenuse on the wrong side of the equation. Find the right-angle mark first. If your answer exceeds the given hypotenuse, you added when you should have subtracted.
- Thinking half the side gives half the area. It gives a quarter; four such squares fill the original.
- Stopping at c². c² = 1296 means c = 36. The theorem is about squares of sides; the question asks for a side.
- Expecting a whole-number answer. Triples are the exception. √74 and √193 are correct final answers — give one-decimal bounds if asked, but do not round.
- Calling a triple primitive after checking only the legs. Primitive means no factor above 1 divides all three.
- Forgetting to halve a rhombus's diagonals. 24 and 70 become 12 and 35.
- Using the theorem where there is no right angle. Draw a diagonal or an altitude first.
9. Chapter summary
- a² + b² = c², where c faces the right angle. Only for right-angled triangles.
- The square on a diagonal has double the area; the square on half the side has a quarter of the area.
- Isosceles right triangle: c² = 2a², so c = a√2.
- √2 = 1.41421356… has no terminating decimal and is not a fraction — both are proved in the chapter.
- Baudhāyana triples: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (12,35,37), (15,36,39) and all their multiples.
- Primitive = no common factor above 1. Every triple is primitive or a scaled primitive.
- Odd-square method: odd m gives n = (m² + 1)/2 and the primitive triple (m, n − 1, n) — always primitive, never complete.
- Fermat's Last Theorem: xⁿ + yⁿ = zⁿ has no positive-integer solution for n > 2. Proved by Andrew Wiles, 1994.
- Grid squares have area p² + q², so area 3 is impossible.
Appendix — beyond the current syllabus
Everything below is true and useful, but it is not part of Chapter 2 of Ganita Prakash Part 2. Read it for interest, or when you reach the class where it belongs — but do not present it as this chapter's content in a Class 8 exam.
The converse (Class 9)
If a² + b² = c² for a triangle, then the triangle is right-angled, with the right angle facing c.
This chapter proves only the forward direction. The converse is what lets you test a triangle for a right angle — sides 9, 12, 15 satisfy 81 + 144 = 225, so the triangle is right-angled. It is genuinely true, and it is proved in Class 9. Use it if a question asks you to check a triangle, but know that this chapter has not established it.
The similar-triangles proof (Class 10)
Drop a perpendicular from the right angle to the hypotenuse. The two smaller triangles created are each similar to the original, and the resulting side ratios give the theorem directly. This is a beautiful proof, but similarity is not introduced until Class 10, so the Class 8 book uses the dissection argument instead.
Euclid's complete generator for primitive triples
For coprime integers m > n of opposite parity,
(m² − n², 2mn, m² + n²)
is a primitive triple, and every primitive triple arises this way exactly once.
The chapter's odd-square method is the special case m = n + 1. Check it: m=2, n=1 gives (3, 4, 5); m=3, n=2 gives (5, 12, 13); m=4, n=3 gives (7, 24, 25); m=5, n=4 gives (9, 40, 41) — exactly the list the odd-square method produces, and the reason every one of them has a hypotenuse one more than a leg, since c − b = m² + n² − 2mn = (m − n)² = 1.
The triples the chapter's method misses are the ones where m and n differ by more than 1. Take m = 4, n = 1: (16 − 1, 2×4×1, 16 + 1) = (15, 8, 17). Or m = 5, n = 2: (25 − 4, 2×5×2, 25 + 4) = (21, 20, 29).
When there is no right angle (Class 11)
For a general triangle the theorem becomes the cosine rule:
c² = a² + b² − 2ab·cos C
When C = 90°, cos C = 0 and the correction term vanishes, giving back a² + b² = c². So Baudhāyana's theorem is the right-angled case of a more general law.
Where the theorem goes next
- Class 9 Number Systems — √2 is the entry point to irrational numbers.
- Class 9 Coordinate Geometry — the distance formula √((x₂−x₁)² + (y₂−y₁)²) is this theorem with coordinates.
- Class 10 Trigonometry — sin²θ + cos²θ = 1 is this theorem on a triangle with hypotenuse 1.
- Class 11 Three-Dimensional Geometry — extends to √(x² + y² + z²) by applying the theorem twice.
- Fermat's two-square theorem — a whole number is a sum of two squares exactly when every prime factor of the form 4k + 3 appears an even number of times. This is what decides which grid-square areas are possible.
