Class 8 Mathematics Chapter 9: The Baudhāyana-Pythagoras Theorem

A question from 800 BCE

In his Sulba-Sutra, Baudhayana asks how to build a square with twice the area of a given square. The question sounds like it should have an easy answer, and the easy answer is wrong.

The obvious guess fails

The first guess is to double the length of every side. But a square with twice the side has 2 times 2, which is 4 times the area. Doubling the side quadruples the area, so that square is far too large.

Baudhayana’s answer

His verse gives the construction in one line. A square raised on the diagonal covers twice the ground of the one it came from. So the side wanted is not twice the old side but the old diagonal.

Why that works

Draw the original square and the tilted square standing on its diagonal. Then draw horizontal and vertical lines through the middle, what Baudhayana calls east-west and north-south lines. The original square breaks into two small triangles and the tilted one into four.

Four against two

All the small triangles are congruent, so counting them is enough. The original is made of two of them and the new one of four, which is exactly double. No measuring and no arithmetic were needed.

Doing it again

Nothing stops the step being repeated. Build on the diagonal of the new square and you get a third, made of eight small triangles. So the areas run 2, then 4, then 8, each double the one before.

Running it backwards

Halving a square is the same construction in reverse. Draw a tilted square inside the original, with its corners at the midpoints of the four sides. That inner square has exactly half the area.

Why a half and not a quarter

Notice what halving is NOT. A square with half the side has a quarter of the area, since four of them fill the original. The midpoint square is a different square altogether, and folding the paper inwards along the creases shows it.

A triangle with no known side

Now take an isosceles right triangle whose two equal sides are 1 unit each. How long is its hypotenuse? Nothing measured so far gives it, but the doubling rule does.

Using the area instead

Two such triangles make a unit square, of area 1. The square built on the hypotenuse of one of them is the square on that unit square’s diagonal, so its area is 2. The length is still unknown, but its square is known.

Naming the length

Call the hypotenuse c. A square of side c has area c times c, which is c squared, and that area is 2. So c squared is 2, and the hypotenuse is the number whose square is 2, written as root 2.

The general formula

The same argument works for any equal sides. If the equal sides are a and the hypotenuse is c, the square on c has twice the area of the square on a, so c squared is 2a squared. That one line answers every such triangle.

Using it both ways

For equal sides of 12, c squared is 2 times 144, which is 288. Since 16 squared is 256 and 17 squared is 289, the hypotenuse lies between 16 and 17. And for a hypotenuse of root 72, a squared is 36, so each equal side is exactly 6.

How big is root 2

Trapping it is easy. One squared is 1 and 2 squared is 4, so root 2 lies between 1 and 2. We call 1 a lower bound and 2 an upper bound, and closer bounds come from squaring better guesses.

Closing in

Since 1.4 squared is 1.96 and 1.5 squared is 2.25, root 2 lies between 1.4 and 1.5. Then 1.41 squared is 1.9881 and 1.42 squared is 2.0164, so it lies between those. Carrying on, 1.414 squared is 1.999396 and 1.415 squared is 2.002225.

Why the decimal never stops

Suppose some decimal that STOPS had square exactly 2. Its last digit after the point would not be zero, and the square of such a decimal also ends in a non-zero digit after the point. But 2 is 2.000, with nothing but zeros, so no terminating decimal will do.

Why it is no fraction either

Suppose root 2 were m over n. Squaring gives 2n squared equal to m squared. In the prime factorisation of a square, every prime appears an even number of times. So on the left the prime 2 would appear an odd number of times and on the right an even number, which is impossible.

Scenario:

A mason is marking out a square shrine floor on a square plot of side 4 metres. The plan asks for a floor of twice that area. His first attempt marks a square of side 8 metres and he is surprised at how much ground it covers. His apprentice then suggests a square of side 2 metres for a later platform of half the area. They also need the diagonal of the original 4 metre square, and the apprentice writes it down as 4 metres times 2, which is 8.

Analysis:

Two of the three figures come from treating area as though it followed the side directly, and the third confuses a length with its square.

Title:

Trap root 2 by squaring

Safety:

Nothing here is risky, but write every square down before moving on.

What You Notice:

The bounds close in without ever meeting. No terminating decimal squares to exactly 2, because such a square always ends in a non-zero digit after the point.

Baudhayana set this down in his Sulba-Sutra around 800 BCE, in a single line: a square raised on the diagonal covers twice the ground.

The proof that root 2 is no fraction was given by Euclid in his Elements, around 300 BCE, and it still runs in four lines.

Diagonal doubles, midpoints halve — Build on the diagonal and the area doubles. Join the midpoints and it halves. One step out, one step in.

Two equal squares, already solved

Combining two copies of one square into a larger square is settled. The side of the larger square is the diagonal of either small one, and its area is the two areas together.

The harder question

What if the two squares have DIFFERENT sizes? There is no shared diagonal to build on, so the earlier trick gives nothing. Baudhayana gives a rule for this case too, in his Sulba-Sutra.

Baudhayana’s rule

His verse says that a square raised on the diagonal covers as much ground as the two squares raised on the sides put together. In other words, stand the two side lengths as the legs of a right triangle.

What that means in practice

Make a right-angled triangle whose two perpendicular sides are the sides of the two squares. Then the square whose side is the hypotenuse of that triangle has an area equal to the two original areas added together.

Checking it against the easy case

The rule should agree with what we already know. If the two squares are the same size, the right triangle is isosceles, and the square on its hypotenuse has twice the area of either. That is exactly the earlier result.

Baudhayana’s second verse

Another verse explains why the rule works. To join squares of unlike size, cut from the larger one a rectangle as wide as the smaller square’s side. The diagonal of that rectangle is the side wanted.

Following the instruction

Join the two squares along a side, with sides a and b, where b is the larger. Inside the larger square, mark off a strip as wide as a. Drawing the diagonal of that strip gives a right triangle whose two legs measure a and b.

Four copies, not one

Now draw three more triangles just like it, each on the hypotenuse of the last, so that four of them surround a four-sided figure. The claim is that this figure is a square of exactly the area wanted.

Why the four sides are equal

All four triangles are congruent, since each was drawn from the same two lengths a and b. So every side of the new four-sided figure is a hypotenuse of one of them, which means all four sides have the same length.

Why every corner is square

At each corner, two angles of the triangles meet the angle of the figure. Those two angles are the acute angles of a right triangle, so they come to x and 90 minus x, which is 90 in all. The remaining angle is therefore a right angle.

Equal sides and right angles

A four-sided figure with four equal sides and four right angles is a square. So the figure built on the hypotenuse really is a square, and its side is the hypotenuse of the triangle with legs a and b.

Why its area is the sum

Label the pieces. The square on the hypotenuse is made of the pieces T, U and V. The same three areas can be rearranged as V, W and X, which are exactly the two original squares. So the area on the hypotenuse is the two areas together.

The theorem itself

Put that in symbols. Let a right triangle have shorter sides a and b and hypotenuse c. Then the two small squares of areas a squared and b squared add up to the large square of area c squared. So a squared plus b squared equals c squared.

Checking with a ruler

Draw a right triangle with shorter sides 3 cm and 4 cm and measure the hypotenuse. It reads about 5 cm. The theorem predicts it exactly: 9 plus 16 is 25, and 25 is 5 squared, so the hypotenuse is 5 cm.

Running it the other way

The theorem also recovers a shorter side. If one leg is 8 and the hypotenuse is 17, then 64 plus b squared is 289, so b squared is 225 and b is 15. Subtracting rather than adding is the whole difference.

When the answer is not whole

Most triangles give untidy answers, and that is no objection. Legs of 5 and 7 give c squared equal to 74, and since 8.6 squared is 73.96 and 8.7 squared is 75.69, the hypotenuse lies between 8.6 and 8.7.

Who it is named after

Baudhayana was the first in recorded history to state this result in full generality and in essentially modern form. Pythagoras, who also admired and studied it, lived around 500 BCE, a couple of hundred years later. Hence the joint name.

Scenario:

A carpenter is laying two square boards, of sides 9 and 12 units. He wants one square board of the same total area. His apprentice suggests a board of side 21 units, adding the two sides. They also have a right-angled brace. Its shorter arm is 16 units and its long arm 20, and they want the third arm. Finally the apprentice measures a triangular offcut with legs of 5 and 7. He writes its longest side down as 74.

Analysis:

One figure adds lengths where squares were needed, one needs the subtraction, and one reports a square as a length.

Title:

Test the theorem with a ruler

Safety:

Use the set square carefully and keep the ruler flat on the table.

What You Notice:

The first three come out at exactly 5, 10 and 13. The fourth does not land on a tidy number at all, which is normal: its square is 74, so the side lies between 8.6 and 8.7.

Baudhayana was the first person in recorded history to state this theorem in full generality and in essentially its modern form, around 800 BCE.

Pythagoras lived around 500 BCE, a couple of hundred years after Baudhayana, which is why the double name is now often used.

Legs square, add, root — Square both legs, add them, and the answer is the square of the hypotenuse. Going back, subtract instead of adding.

Triangles with whole-number sides

In his Sulba-Sutra, Baudhayana lists triples of whole numbers that work as the sides of a right triangle. They include 3, 4 and 5; 5, 12 and 13; 8, 15 and 17; 7, 24 and 25; 12, 35 and 37; and 15, 36 and 39.

What they are called

Because he listed them, such triples are called Baudhayana triples. They also go by Baudhayana-Pythagoras triples, Pythagorean triples, and simply right-angled triangle triples. Each satisfies a squared plus b squared equal to c squared.

A pattern in the small ones

List the triples whose numbers are all at most 20 and something stands out. Four of them are 3, 4, 5; then 6, 8, 10; then 9, 12, 15; then 12, 16, 20. Every one is the first triple with each member multiplied by the same number.

Turning that into a claim

So here is a guess: 3k, 4k and 5k is a triple for every counting number k. Checking is easy. The square of 3k is 9 times k squared and the square of 4k is 16 times k squared, and together they make 25 times k squared, which is the square of 5k.

Proving it in general

Nothing in that check used the numbers 3, 4 and 5. Take any triple a, b and c. Then the square of ka is k squared times a squared, and similarly for kb, so their sum is k squared times the quantity a squared plus b squared.

Finishing the proof

But a squared plus b squared is c squared, since a, b and c was a triple. So the sum is k squared times c squared, which is the square of kc. Hence ka, kb and kc is a triple too, and we call it a scaled version.

An endless supply

That settles a question worth asking. Since every counting number k gives a new triple from an old one, there are infinitely many Baudhayana triples, and a short argument proved it.

Primitive triples

Scaled triples carry no new information, so the interesting ones are those that are not scaled copies. A triple whose three numbers share no common factor above 1 is called primitive. So 3, 4 and 5 is primitive while 9, 12 and 15 is not.

Why primitives are the real question

Among triples with numbers at most 20, the primitive ones are just 3, 4, 5; then 5, 12, 13; then 8, 15, 17. Every other triple in that range is one of these scaled. So finding all the primitives would find them all.

A way to grow new ones

Here is an old fact worth reusing. Adding the odd numbers from 1 up to the nth gives exactly n squared. So 1 is 1 squared, 1 plus 3 is 2 squared, 1 plus 3 plus 5 is 3 squared, and so on.

Writing that as an equation

The nth odd number is 2n minus 1. The first n odd numbers total n squared, and the first n minus 1 of them total n minus 1, all squared. So the equation reads: n minus 1, squared, plus 2n minus 1, equals n squared.

Spotting the opportunity

Look at that equation as a sum of two things making a square. If the odd number 2n minus 1 happens to be a SQUARE as well, then all three terms are squares, and a triple has appeared.

Growing one

Take 9, which is an odd square. Since 9 is 2 times 5 minus 1, it is the fifth odd number, so n is 5. The equation gives 4 squared plus 9 equal to 5 squared, which is the triple 3, 4 and 5. Taking 25 instead, the thirteenth odd number, gives 12 squared plus 25 equal to 13 squared, so the triple 5, 12 and 13.

What the method always produces

Notice what these have in common: one leg is exactly one less than the hypotenuse. That means the method only ever produces primitive triples, and also that it misses some. The primitive triple 8, 15 and 17 can never appear, since 15 and 17 differ by 2.

Fermat’s question

There are infinitely many squares that are a sum of two squares. Fermat, in the seventeenth century, asked whether any CUBE is a sum of two cubes, or any fourth power the sum of two fourth powers, and upwards from there.

Three hundred years of silence

He wrote in a book margin that no such number exists, adding that he had a marvellous proof which the margin was too small to hold. No one ever found that proof, and more than three centuries of attempts failed. Andrew Wiles finally proved it in 1994.

A lotus in a lake

Bhaskaracharya sets this problem in his Lilavati. A lotus stands 1 unit above the water. A breeze pushes its tip down until it touches the water 3 units from where it was. How deep is the lake?

Solving it

Let x be the stem under water, which is the depth wanted, so the whole stem is x plus 1. The stem swings about its root, giving a right triangle with sides 3, x and x plus 1. So 9 plus x squared equals x squared plus 2x plus 1, which leaves 9 equal to 2x plus 1, so x is 4.

Scenario:

A surveyor has three tasks and one theorem. She needs the depth of a village pond. A lotus there stands 1 unit above the water, and a breeze tips it until the tip meets the water 3 units away. She also has a rhombus-shaped plot whose diagonals measure 24 and 70 units, and wants the side. Her assistant claims that 15, 36 and 39 cannot be a triple because it is not in Baudhayana’s list of primitive ones.

Analysis:

One problem looks short of information, one needs a property of diagonals, and one confuses being a triple with being primitive.

Title:

Find the height of an equilateral triangle

Safety:

Keep the compass point away from your fingers and work on a flat table.

What You Notice:

The altitude lands exactly at the midpoint, so the right triangle has legs of 3 and the height, with hypotenuse 6. That gives a height whose square is 27, about 5.2, and an area of about 15.59 square centimetres.

Andrew Wiles read about Fermat’s Last Theorem at the age of ten in 1963 and resolved to prove it. He finally did so in 1994, after more than three centuries of failed attempts.

Bhaskaracharya set the lotus problem in his Lilavati. It looks as though it has too little information to solve, and it has exactly enough.

Scale it, or grow it — Multiply a known triple to get another, or find an odd square among the odd numbers to grow a brand new one.