MathsCBSE Class 8 · The Baudhayana-Pythagoras Theorem

Baudhayana-Pythagoras Theorem, Triples and Applications for Class 8

Updated 17 October 2026 · Fact-checked

In a right triangle with perpendicular sides a and b and hypotenuse c, a squared + b squared = c squared. Integer triples such as (3, 4, 5), (5, 12, 13), (8, 15, 17) and (7, 24, 25) satisfy it; if (a, b, c) is a triple so is (ka, kb, kc). Exams ask for missing sides, rhombus and rectangle sides from diagonals, ladder and pole problems, and testing or generating triples.

Notice the two smaller squares, 9 and 16, together have the same area as the square on the hypotenuse, 25.

Illustration drawn for QuestionBuddy.

Understand Baudhayana-Pythagoras Theorem, Triples and Applications

Combining two different squares. In his *Sulba-sutra* (verse 1.12 as cited in the chapter) Baudhayana says that the square on the diagonal of a rectangle has the area of the squares on the two sides together. Form a right triangle whose perpendicular sides are $a$ and $b$ and place three more congruent copies around a square on the hypotenuse $c$ . The big figure is a square (its sides are all $c$ and its angles are right angles) with area made of the area of $a^{2}$ and $b^{2}$ pieces, which gives
$ $a^{2} + b^{2} = c^{2} .$ $
This is the Baudhayana-Pythagoras theorem: *in a right-angled triangle, the sum of the squares of the two shorter sides equals the square of the hypotenuse.* The chapter notes that Baudhayana stated it in this general and essentially modern form and that Pythagoras (c. 500 BCE) lived a couple of hundred years later. The hypotenuse is always the longest side, since $c^{2} = a^{2} + b^{2}$ is larger than $a^{2}$ and $b^{2}$ . With $a = 3$ and $b = 4$ we get $c = 5$ exactly.

Finding a missing side. If $c$ is unknown, $c = a^{2} + b^{2}$ ; if a leg is unknown, $a = c^{2} - b^{2}$ . For legs 5 and 12, $c^{2} = 25 + 144 = 169$ and $c = 13$ .

Baudhayana triples. Integers $(a, b, c)$ with $a^{2} + b^{2} = c^{2}$ are Baudhayana triples (also Baudhayana-Pythagoras triples). Baudhayana's verse 1.13, as quoted in the chapter, lists $(3, 4, 5)$ , $(5, 12, 13)$ , $(8, 15, 17)$ , $(7, 24, 25)$ , $(12, 35, 37)$ and $(15, 36, 39)$ . The triples with all numbers up to 20 include $(3, 4, 5)$ , $(6, 8, 10)$ , $(9, 12, 15)$ , $(12, 16, 20)$ , $(5, 12, 13)$ , $(8, 15, 17)$ .

Scaling. If $(a, b, c)$ is a triple then $(k a, k b, k c)$ is also one for every positive integer $k$ , because $(k a)^{2} + (k b)^{2} = k^{2} (a^{2} + b^{2}) = k^{2} c^{2} = (k c)^{2}$ . So there are infinitely many triples. For example $(300, 400, 500) = 100 \times (3, 4, 5)$ .
Primitive triples have no common factor greater than 1: $(3, 4, 5)$ and $(5, 12, 13)$ are primitive, $(9, 12, 15)$ is not (it is $3 \times (3, 4, 5)$ ). Finding all primitive triples finds them all by scaling.

Generating triples from odd squares. The sum of the first $n$ odd numbers is $n^{2}$ , so $(n - 1)^{2} + (2 n - 1) = n^{2}$ . If the odd number $2 n - 1$ is itself a square, we get a triple. For $2 n - 1 = 9$ , $n = 5$ : $4^{2} + 3^{2} = 5^{2}$ . For $2 n - 1 = 25$ , $n = 13$ : $1 2^{2} + 5^{2} = 1 3^{2}$ . This method gives triples where one leg is 1 less than the hypotenuse, so it does not give every primitive triple (for example $(8, 15, 17)$ has $17 - 15 = 2$ ).

Applications. The theorem is used whenever perpendicular lengths occur.

*Bhaskara II's lotus problem (Lilavati):* a lotus stem rises 1 unit above the water; swayed by the breeze, its tip touches the water 3 units from the original position. If the depth is $x$ , the stem has length $x + 1$ and $x^{2} + 3^{2} = (x + 1)^{2}$ , so $2 x = 8$ and the depth is 4 units.
*Ladders and poles:* the wall, the ground and the ladder form a right triangle.
*Rhombus:* its diagonals are perpendicular bisectors, so the side is the hypotenuse of a right triangle with legs equal to half the diagonals; a rhombus with diagonals 24 and 70 has side $1 2^{2} + 3 5^{2} = 37$ .
*Rectangle and square diagonals.*

Fermat's Last Theorem was inspired by triples: it says $x^{n} + y^{n} = z^{n}$ has no solution in natural numbers for $n > 2$ ; the chapter says that more than 300 years of attempts to prove it failed.

The doubling of squares, the isosceles right triangle and the nature of root 2 are on the previous topic page of this chapter.

Baudhayana-Pythagoras Theorem, Triples and Applications: formulas to remember

Baudhayana-Pythagoras theorem: $a^{2} + b^{2} = c^{2}$; c is the hypotenuse, a and b the perpendicular sides.
Missing leg: $a = c^{2} - b^{2}$; Subtract, never add, when the hypotenuse is given.
Scaled triple: $(a, b, c) \to (k a, k b, k c)$; k is a positive integer.
Odd-square method: $(n - 1)^{2} + (2 n - 1) = n^{2}$; Needs 2n - 1 to be a perfect square.
Rhombus side from diagonals: $s^{2} = (\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2}$; The diagonals meet at right angles at the centre.

How to solve Baudhayana-Pythagoras Theorem, Triples and Applications questions

A routine for right-triangle word problems.

1Draw the figure and mark the right angle; label the hypotenuse (opposite the right angle).
2Write a squared + b squared = c squared with the hypotenuse on the right.
3Substitute the known values and solve for the unknown square.
4Take the square root; if it is not exact, give bounds with one decimal.
5For problems with an unknown length, name it x and write all sides in terms of x (as in the lotus problem).
6To test a triple, check that the largest number squared equals the sum of the other two squares.
7Check that the hypotenuse is the longest side.

Baudhayana-Pythagoras Theorem, Triples and Applications in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Ladder against a wall

Suppose a 25 m long ladder leans against a wall and its top touches the wall at a height of 24 m.

How far is the foot of the ladder from the wall? If the top slides down 4 m, how far does the foot move out?

Try it first, then show the solutionSolution

Base $^{2} = 2 5^{2} - 2 4^{2} = 625 - 576 = 49$ , so the base is 7 m.
After the slip the top is at 20 m: base $^{2} = 625 - 400 = 225$ , base $= 15$ m.
The foot moves out by $15 - 7 = 8$ m.

Answer: 7 m at first; the foot moves out 8 m.

Memory hook: Ladder, wall and ground form a right triangle with the ladder as the hypotenuse.

Broken pole

Imagine a 12 m pole that snaps at some height; the top part bends over and touches the ground 6 m from the foot of the pole.

At what height above the ground did the pole break?

Try it first, then show the solutionSolution

Let the break be at height $x$ . The bent part has length $12 - x$ and is the hypotenuse.
$x^{2} + 6^{2} = (12 - x)^{2}$ , so $x^{2} + 36 = 144 - 24 x + x^{2}$ , giving $24 x = 108$ and $x = 4.5$ m.
Check: $4. 5^{2} + 6^{2} = 20.25 + 36 = 56.25 = 7. 5^{2}$ , and $12 - 4.5 = 7.5$ .

Answer: 4.5 m

Memory hook: Name the unknown, express every side, and let the theorem make the equation.

Tricks to remember Baudhayana-Pythagoras Theorem, Triples and Applications

Short memory aids. Say each one aloud once, then use it on a question the same day.

Hypotenuse sits alone
In a squared + b squared = c squared, the hypotenuse is on its own on the right, the two legs on the left.
Use it when: Avoiding adding when subtracting is needed.
Learn the five families
(3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), and their multiples cover most questions.
Use it when: Spotting exact answers quickly.
Scale by k
Multiply every side of a triple by the same integer to get another triple.
Use it when: Generating and testing triples.
Name the unknown
In worded problems let x be the unknown length and write every other side using x.
Use it when: Lotus and pole problems.

Baudhayana-Pythagoras Theorem, Triples and Applications: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Hypotenuse from legs · 1 mark · MCQBoard level

The legs of a right triangle are 9 cm and 40 cm. The hypotenuse is: (a) 31 cm (b) 41 cm (c) 49 cm (d) 360 cm

Try it first, then show the solutionSolution

$c^{2} = 81 + 1600 = 1681 = 4 1^{2}$ .

Answer: (b) 41 cm

Example 2: Missing leg · 2 marks · Short answerBoard level

One side of a rectangle is 7 cm and its diagonal is 25 cm. Find the other side and the area of the rectangle.

Try it first, then show the solutionSolution

The diagonal is the hypotenuse of a right triangle with the two sides as legs: $b^{2} = 2 5^{2} - 7^{2} = 625 - 49 = 576$ , so $b = 24$ cm. (1 mark)
Area $= 7 \times 24 = 168$ square cm. (1 mark)

Answer: Other side 24 cm; area 168 square cm

Example 3: Rhombus from its diagonals · 3 marks · Short answerHarder board level

The diagonals of a rhombus are 48 cm and 20 cm. Find its side and perimeter.

Try it first, then show the solutionSolution

The diagonals of a rhombus bisect each other at right angles, giving right triangles with legs 24 cm and 10 cm. (1 mark)
Side $^{2} = 2 4^{2} + 1 0^{2} = 576 + 100 = 676$ , so the side is 26 cm. (1 mark)
Perimeter $= 4 \times 26 = 104$ cm. (1 mark)

Answer: Side 26 cm; perimeter 104 cm

Example 4: Testing triples · 3 marks · Short answerHarder board level

Check whether (20, 21, 29) and (24, 45, 51) are Baudhayana triples. Which is primitive? If the second is a scaled version, of which triple?

Try it first, then show the solutionSolution

$2 0^{2} + 2 1^{2} = 400 + 441 = 841 = 2 9^{2}$ : a triple. $2 4^{2} + 4 5^{2} = 576 + 2025 = 2601 = 5 1^{2}$ : also a triple. (1 mark)
(20, 21, 29) has no common factor above 1, so it is primitive; (24, 45, 51) has the common factor 3. (1 mark)
Dividing by 3 gives (8, 15, 17), so (24, 45, 51) is $3 \times (8, 15, 17)$ . (1 mark)

Answer: Both are triples; (20, 21, 29) is primitive; (24, 45, 51) is 3 times (8, 15, 17).

Example 5: A broken pole · 4 marks · Long answerHarder board level

A 12 m pole breaks at a point and its top touches the ground 6 m from the foot, with the break still attached. Find the height of the break.

Try it first, then show the solutionSolution

Let the break be at height $x$ m; the bent part is $(12 - x)$ m. (1 mark)
The pole, ground and bent part form a right triangle with hypotenuse $12 - x$ : $x^{2} + 6^{2} = (12 - x)^{2}$ . (1 mark)
Expand: $x^{2} + 36 = 144 - 24 x + x^{2}$ , so $24 x = 108$ . (1 mark)
$x = 4.5$ m. Check: $4. 5^{2} + 6^{2} = 56.25 = 7. 5^{2}$ . (1 mark)

Answer: The pole broke at 4.5 m above the ground.

Where students lose marks in Baudhayana-Pythagoras Theorem, Triples and Applications

The slips examiners see most often, and what to write instead.

Mistake: Adding the squares when the hypotenuse is known
When the hypotenuse is given, subtract the other leg's square from it.
Wrong working
"c = 25, a = 7: b^2 = 625 + 49."
Right working
$b^{2} = 625 - 49 = 576$ , $b = 24$ .
Do this: Write the theorem first and substitute into it.
Usually costs: The answer.
Mistake: Treating the longest side as a leg
The hypotenuse is opposite the right angle and is the longest side.
Wrong working
"Sides 5, 12, 13: 5^2 + 13^2 = 12^2."
Right working
$5^{2} + 1 2^{2} = 1 3^{2}$ with 13 as the hypotenuse.
Do this: Put the largest number on the right.
Usually costs: 2 marks.
Mistake: Leaving the answer as a square
The theorem gives c squared; take the square root.
Wrong working
"The hypotenuse is 169."
Right working
$c^{2} = 169$ , so $c = 13$ .
Do this: Underline "length" in the question and write the root as the last step.
Usually costs: 1 mark.
Mistake: Using full diagonals in the rhombus triangle
The right triangle at the centre has legs equal to half the diagonals.
Wrong working
"Side $^{2} = 4 8^{2} + 2 0^{2}$ ."
Right working
Side $^{2} = 2 4^{2} + 1 0^{2} = 676$ , side $= 26$ .
Do this: Mark the centre and halve each diagonal.
Usually costs: The answer.

Exam tips

Always sketch and label the right angle; name the hypotenuse before writing the equation.
Memorise a handful of triples and their multiples to recognise exact answers, then verify by squaring.
In worded problems introduce a variable, write every side, form the equation and check by substitution.
For non-exact answers give bounds between two decimals and show the squares.
Quote the theorem by name when you use it.

Practice questions on Baudhayana-Pythagoras Theorem, Triples and Applications

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonBoard level
Assertion (A): (6, 8, 10) is a Baudhayana triple.
Reason (R): (6, 8, 10) is 2 times the triple (3, 4, 5).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Check 36 + 64 and use scaling.
Answer
(a). $36 + 64 = 100 = 1 0^{2}$ , and scaling a triple by 2 gives another triple.
Q2 · 2 marks · Very short answerBoard level
Use the odd-square method to find a Baudhayana triple in which the odd square is 49.
Hint
2n - 1 = 49 gives n.
Answer
$2 n - 1 = 49$ gives $n = 25$ . Then $(n - 1)^{2} + 49 = n^{2}$ , i.e. $2 4^{2} + 7^{2} = 2 5^{2}$ : the triple $(7, 24, 25)$ ( $576 + 49 = 625$ ).
Q3 · 3 marks · Short answerHarder board level
An equilateral triangle has side 10 units. Find the height of the triangle, giving two decimals (one digit after the point) between which it lies, and bounds for the area.
Hint
The altitude bisects the base. Area = (1/2) x base x height.
Answer
The altitude bisects the base, giving a right triangle with legs $h$ and 5 and hypotenuse 10: $h^{2} = 100 - 25 = 75$ . Since $8. 6^{2} = 73.96 < 75 < 75.69 = 8. 7^{2}$ , $8.6 < h < 8.7$ . Area $= \frac{1}{2} \times 10 \times h = 5 h$ , between $43.0$ and $43.5$ square units.
Q4 · 4 marks · Long answerHOTS / competency
Prove that if (a, b, c) is a Baudhayana triple then so is (ka, kb, kc). List all scaled versions of (3, 4, 5) with every number at most 30, and find the smallest scaled version of (3, 4, 5) whose shortest side exceeds 100.
Hint
Use (ka)^2 + (kb)^2 = k^2(a^2 + b^2).
Answer
$(k a)^{2} + (k b)^{2} = k^{2} a^{2} + k^{2} b^{2} = k^{2} (a^{2} + b^{2}) = k^{2} c^{2} = (k c)^{2}$ . With every number at most 30 (so $5 k \leq 30$ , $k \leq 6$ ): $(3, 4, 5)$ , $(6, 8, 10)$ , $(9, 12, 15)$ , $(12, 16, 20)$ , $(15, 20, 25)$ , $(18, 24, 30)$ . The shortest side $3 k > 100$ needs $k \geq 34$ , giving $(102, 136, 170)$ ; check $10 2^{2} + 13 6^{2} = 10404 + 18496 = 28900 = 17 0^{2}$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case study: a 25 m ladder leans against a vertical wall and its top touches the wall at a height of 24 m.
(i) How far is the foot of the ladder from the wall? (1 mark)
(ii) The top slides down 4 m. How far from the wall is the foot now? (1 mark)
(iii) How far has the foot moved? (1 mark)
(iv) Name the Baudhayana triples involved. (1 mark)
Hint
The ladder is the hypotenuse in both positions.
Answer
(i) $2 5^{2} - 2 4^{2} = 49$ , so 7 m. (ii) The top is at 20 m: $2 5^{2} - 2 0^{2} = 625 - 400 = 225$ , so 15 m. (iii) $15 - 7 = 8$ m. (iv) $(7, 24, 25)$ and $(15, 20, 25)$ , the latter being $5 \times (3, 4, 5)$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the Baudhayana-Pythagoras theorem.Show answer
In a right triangle, a squared + b squared = c squared, where c is the hypotenuse.
2.What is a primitive Baudhayana triple?Show answer
A triple with no common factor greater than 1.
3.If (a, b, c) is a triple, what about (ka, kb, kc)?Show answer
It is also a triple for every positive integer k.
4.Name two triples from Baudhayana's list.Show answer
(3, 4, 5) and (5, 12, 13) (also (8, 15, 17), (7, 24, 25), (12, 35, 37), (15, 36, 39)).
5.What is the depth of the lotus lake in Bhaskara II's problem?Show answer
4 units.
6.Which side of a right triangle is longest?Show answer
The hypotenuse.

Baudhayana-Pythagoras Theorem, Triples and Applications: frequently asked questions

Why is it called the Baudhayana-Pythagoras theorem?

The chapter credits Baudhayana (c. 800 BCE) with stating it in its general form, long before Pythagoras (c. 500 BCE), who also studied it.

Do all triples come from the odd-square method?

No. That method gives triples where one leg is one less than the hypotenuse, so primitive triples such as (8, 15, 17) are not produced.

Can the three sides of a right triangle all be consecutive integers?

Only (3, 4, 5). It is the only right triangle with three consecutive whole-number sides.

How do I recognise a right triangle from its sides?

Check whether the square of the largest side equals the sum of the squares of the other two.