The Baudhāyana–Pythagoras Theorem

Somewhere around 800 BCE, long before anyone in India had heard of Pythagoras, a set of building manuals was written down that we now call the Śulba-Sūtras. The word śulba means a cord or a measuring rope, and sūtra means a rule, so the name is honest: these are rope rules. The oldest one that has survived is Baudhāyana's Śulba-Sūtra, and this chapter carries his name.

What were the rules for? Building fire altars out of bricks. An altar had to have a fixed shape and, much more awkwardly, a fixed area. A builder might be told to make a square altar with the same area as a given rectangle, or to build a fresh altar with exactly double the area of the old one. You cannot do that by eye. You need geometry, and you need it to be reliable enough that a wall of bricks will actually close up at the end.

The builders also had almost no equipment. No protractor, no set square, no calculator. What they had was a cord and some pegs. With a cord you can do exactly three things: stretch it straight to mark a line, hold it at a fixed length, and swing it around a peg to sweep an arc. Nearly every construction in the Śulba-Sūtras is built out of those three moves, which is why the mathematics in them is so practical and so carefully worded.

The hardest everyday job was getting a corner truly square. A corner that is off by two degrees looks fine and ruins the altar. Baudhāyana's answer is the rule this chapter is about. Written out in today's language, it says that the area of the square drawn on the diagonal of a rectangle is equal to the areas of the squares drawn on its length and on its breadth, added together. In symbols:

diagonal² = length² + breadth²

Notice that he states it as a fact about areas, not as a formula about lengths. For an altar builder, area was the thing being ordered, so area was the natural language.

Here is how that statement turns into a practical trick on the ground. Suppose the altar is to be a rectangle 12 units long and 5 units wide. Before pegging anything, work out the diagonal:

12² + 5² = 144 + 25 = 169
√169 = 13

So cut a cord 13 units long. Peg out the 12-unit side, peg out the 5-unit side, and then swing the corner around until the 13-unit cord stretches exactly from one loose end to the other. The instant it fits, the corner is a perfect right angle. No angles were measured at all — only three lengths. That is the whole reason this rule mattered enough to be written down and memorised.

So why the double name, Baudhāyana–Pythagoras? Scholars date Baudhāyana's text to roughly 800 BCE, while Pythagoras lived in Greece about two centuries later. The rule was clearly known, stated and used in India first, so Indian textbooks now name both. What the Greek tradition added, and what Indian mathematicians such as Bhāskara did much later, was to write down a proof — an argument showing the rule must hold for every right triangle, not just for the handful of cord lengths anyone had tested. You will meet two of those proofs further on in this chapter.

One small point of language before we go on. A rectangle has two diagonals and they are equal, so it does not matter which one you talk about. A square is a rectangle whose length and breadth are the same, so Baudhāyana's rule covers it too — and the square case turns out to be the most interesting one of all.

The Śulba-Sūtras open with two constructions that are worth drawing yourself with a ruler, because the theorem is hiding inside both of them.

1. Doubling a square. You are given a square altar and told to build one with exactly twice the area. The Śulba answer is beautifully short: draw the diagonal of the square, and build the new square on that diagonal.

Why does it work? Take a square ABCD and draw the diagonal AC, which cuts it into two identical triangles, so each triangle is half the square. Now build the big square on AC, and draw its two diagonals. They cut the big square into four triangles that meet at its centre. Each of those four triangles has two equal sides meeting at a right angle, and each of those sides turns out to be exactly as long as a side of the original square. So each of the four is a copy of half the original square. Four halves make two wholes: the big square is exactly double.

Square of side 7 cm      area = 49 cm²
diagonal² = 7² + 7² = 49 + 49 = 98
Square on the diagonal   area = 98 cm² = 2 × 49 cm²

Look hard at the middle line. It says diagonal² = side² + side². That is the theorem, for the special case of a right triangle whose two short sides are equal. The doubling rule and the Baudhāyana–Pythagoras theorem are the same fact wearing different clothes.

2. Halving a square. Now the opposite job: build a square with exactly half the area of a given one. Drawing a single diagonal is no help — it does halve the area, but it leaves you a triangle, not a square. The Śulba method is to join the midpoints of the four sides. The tilted square you get inside has exactly half the area of the one you started with.

The picture proves it on its own. Joining the midpoints slices four small triangles off the corners, and each corner triangle has two sides of half the original length meeting at a right angle. Now look at the tilted square in the middle: draw its two diagonals and it splits into four triangles, and each of those also has two sides of half the original length meeting at a right angle. So the four pieces you cut off the corners are exactly the four pieces the inner square is made of. Half is thrown away and half is kept.

Square of side 12 cm     area = 144 cm²
inner square side = √(6² + 6²) = √72 ≈ 8.49 cm
inner square area = 72 cm² = half of 144 cm²

Both constructions hand you a fact worth remembering on its own: for any square, area = diagonal² ÷ 2. If a square window pane has a diagonal of 10 cm, its area is 100 ÷ 2 = 50 cm², and you never needed to find the side at all. Check it the long way if you like: side² + side² = 10² = 100, so 2 × side² = 100 and side² = 50 — and side² is the area.

Finish with a warning that catches out a lot of people. Doubling the area is not the same as doubling the side. A square of side 7 cm has area 49 cm². A square of side 14 cm has area 196 cm², which is four times as much, not twice. To double the area you must multiply the side by the diagonal factor, which is about 1.414, giving a side of about 9.9 cm. That factor is the subject of the next section, and it is the number the Śulba authors worked hardest on.

Take the simplest square there is, with side 1 unit. Its diagonal d satisfies

d² = 1² + 1² = 2,   so   d = √2

and now the trouble starts, because √2 is not a tidy number. It is not 1.4, because 1.4² = 1.96, which is too small. It is not 1.5, because 1.5² = 2.25, which is too big. Its decimal begins 1.41421356... and it never stops and never settles into a repeating block, so no fraction is exactly equal to it. The best anyone can do, then or now, is to get close and know how close.

Baudhāyana's Śulba-Sūtra gives a value for it, and the value is startlingly good. His recipe, put into plain words, is this: take the side, add a third of it, then add a fourth of that third, and finally take away a thirty-fourth of that fourth. Written as numbers:

√2 ≈ 1 + 1/3 + 1/(3 × 4) − 1/(3 × 4 × 34)
   = 1 + 1/3 + 1/12 − 1/408

Work it out one step at a time, the way you would in class. First 1 + 1/3 = 4/3. Next add 1/12: rewrite 4/3 as 16/12, so the running total is 17/12, which is 1.41666... Finally subtract 1/408: rewrite 17/12 as 578/408, and taking away 1/408 leaves 577/408.

577 ÷ 408 = 1.4142157   (to 7 decimal places)
√2        = 1.4142136   (to 7 decimal places)

He is out by about 0.0000021, that is about two millionths, and he is out on the high side. Peg out the diagonal of a square whose side is 10 m and that error grows only to about two hundredths of a millimetre — far finer than anything a person could actually mark on the ground, then or now.

There is a neat way to see how good the fraction is without doing any long division. Just square it:

577² = 332929
408² = 166464
2 × 166464 = 332928

So 577² is exactly 1 more than twice 408². In other words (577/408)² = 2 + 1/166464, a hair above 2 — which is precisely what an extremely good approximation to √2 should look like. Squaring your answer to see how near 2 it lands is a check you can run on any approximation you meet.

What you will actually use. For school work, remember √2 ≈ 1.414 and this rule:

diagonal of a square = side × √2 ≈ side × 1.414

A square floor tile of side 30 cm has a diagonal of about 30 × 1.414 = 42.42 cm. A square park of side 50 m has a diagonal of about 50 × 1.414 = 70.7 m. That second one quietly explains why everybody cuts across parks instead of walking round: two sides come to 50 + 50 = 100 m, while the diagonal is only about 70.7 m, so you save roughly 29 m on every crossing.

Two cautions about rounding. First, 1.414² = 1.999396, which is a shade under 2, so 1.414 is very slightly small; if a question wants more accuracy, use 1.41421. Second, keep the √2 in your working and turn it into a decimal only at the very last step, because rounding early lets small errors pile up through a long calculation. And if a question asks for an exact answer, leaving it as 30√2 cm is a complete answer, not an unfinished one.

The statement. In a right-angled triangle, the side opposite the right angle is called the hypotenuse, and it is always the longest of the three sides. Call it c, and call the two sides that meet at the right angle a and b. Then

a² + b² = c²

Said in the older area language: if you draw a square on each of the three sides, the square on the hypotenuse has exactly the same area as the other two squares put together. For a 3, 4, 5 triangle the three squares have areas 9, 16 and 25, and 9 + 16 = 25.

        B
        |\
        | \
      a |  \ c     c is the hypotenuse:
        |   \      the side opposite
        |____\     the right angle at A
        A  b  C

Two things are worth noticing before any proof. First, the right angle has to be there. The rule says nothing at all about a triangle with angles of 80, 60 and 40 degrees. Second, c is the longest side, so if you ever work out a hypotenuse that comes out shorter than one of the other two sides, stop, because you have made a slip somewhere.

Proof 1: the square of side (a + b). Draw a big square whose side is a + b. In each corner place a copy of your right triangle, each one turned a quarter turn from the last, so that along every side of the big square you see a length a followed by a length b. Four triangles go in, and a tilted hole is left in the middle.

That hole is a square. Every one of its four sides is a hypotenuse of one of the triangles, so all four sides are c. Its angles are right angles too: at each point where two triangles meet along an edge of the big square, one triangle contributes one of its acute angles and the neighbour contributes the other. The two acute angles of a right triangle add up to 90 degrees, because all three angles add to 180 and one of them is already 90. So the angle left over for the hole is 180 − 90 = 90 degrees.

Now count the area of the big square in two different ways and set the answers equal.

Way 1:  area = (a + b)² = a² + 2ab + b²

Way 2:  area = four triangles + the hole
             = 4 × (½ × a × b) + c²
             = 2ab + c²

So      a² + 2ab + b² = 2ab + c²
Take 2ab off both sides:
        a² + b² = c²

Test the picture with real numbers, a = 3 and b = 4. The big square has side 7, so its area is 49. The four triangles come to 4 × ½ × 3 × 4 = 24. That leaves 49 − 24 = 25 for the hole, and √25 = 5. The hypotenuse is 5, exactly as it should be.

Proof 2: the pinwheel inside the square on the hypotenuse. This is the dissection Bhāskara is famous for. Start with the square on the hypotenuse, so its side is c, and fit four copies of the same triangle inside it arranged like a pinwheel. A small square is left in the middle, and its side is b − a, the difference between the two shorter sides.

c² = 4 × (½ × a × b) + (b − a)²
   = 2ab + (b² − 2ab + a²)
   = a² + b²

The middle step uses the identity (b − a)² = b² − 2ab + a², which you have already met when multiplying brackets. Check the whole thing with numbers. With a = 3, b = 4 and c = 5: the big square is 25, the four triangles come to 24, and the middle square has side 4 − 3 = 1 and area 1, and 24 + 1 = 25. With a = 6, b = 8 and c = 10: the big square is 100, the four triangles come to 4 × ½ × 6 × 8 = 96, the middle square has side 2 and area 4, and 96 + 4 = 100.

Why bother with proofs at all, when you could just measure a few triangles? Because measuring only ever tests the triangles you happened to draw. These two arguments never mention a particular number: they work for any a and b at all, so they settle the question for every right triangle that could ever exist, including ones far too big or too small to draw.

A Baudhāyana triple — you will also see it called a Pythagorean triple — is a set of three whole numbers a, b and c, each bigger than 0, with a² + b² = c². They matter because whole numbers are what you can actually knot into a rope and count off on the ground. The Śulba-Sūtras list cords in exactly this way, and here is that list with every check written out:

 a     b      c        check
 3     4      5        9   +   16   =   25
 5    12     13       25   +  144   =  169
 8    15     17       64   +  225   =  289
 7    24     25       49   +  576   =  625
12    35     37      144   + 1225   = 1369
15    36     39      225   + 1296   = 1521

Each right-hand answer is a perfect square: 25 = 5², 169 = 13², 289 = 17², 625 = 25², 1369 = 37² and 1521 = 39². Work a couple of them out yourself rather than taking my word for it, because that is exactly what a question will ask you to do.

Multiples come free. If (a, b, c) is a triple, then so is (ka, kb, kc) for any whole number k bigger than 0. The reason is one line of algebra:

(ka)² + (kb)² = k²a² + k²b² = k²(a² + b²) = k²c² = (kc)²

So from 3, 4, 5 you instantly get 6, 8, 10 and 9, 12, 15 and 12, 16, 20 and 30, 40, 50, without checking any of them. This is why a mason on site can use 60 cm, 80 cm and 100 cm: it is simply 3, 4, 5 with everything multiplied by 20, and it fits a real wall better than a 5 cm triangle would.

Now look back at the Śulba list with that in mind. The last row, 15, 36, 39, is just 5, 12, 13 with everything multiplied by 3. It is a perfectly genuine triple, but it is not a new one.

Primitive triples. A triple is called primitive when its three numbers share no common factor bigger than 1. So 3, 4, 5 is primitive while 6, 8, 10 is not; 5, 12, 13 is primitive while 15, 36, 39 is not; 8, 15, 17 and 7, 24, 25 are both primitive. To test a triple, find the HCF of the three numbers: if the HCF is 1 it is primitive, and if it is anything more, divide right through by it to find the primitive triple underneath. Every triple is either primitive or a primitive one scaled up, so the primitive ones are the real family and everything else is a copy at a different size.

Recipe one, from odd numbers. Take any odd number n of 3 or more and work out these two:

 n     n²    (n² − 1)/2   (n² + 1)/2     triple
 3      9         4            5        3, 4, 5
 5     25        12           13        5, 12, 13
 7     49        24           25        7, 24, 25
 9     81        40           41        9, 40, 41
11    121        60           61        11, 60, 61

Since n is odd, n² is odd too, so n² − 1 and n² + 1 are both even and halving them always gives whole numbers. It works for every odd n from 3 upwards, and here is the reason, using the difference of two squares x² − y² = (x + y)(x − y):

((n² + 1)/2)² − ((n² − 1)/2)²
  = [(n² + 1)² − (n² − 1)²] ÷ 4
  = [(n² + 1 + n² − 1) × (n² + 1 − n² + 1)] ÷ 4
  = [2n² × 2] ÷ 4
  = n²

Rearranged, that says n² + ((n² − 1)/2)² = ((n² + 1)/2)², which is exactly the triple. Check it once with n = 11: 121 + 3600 = 3721, and 61² = 3721.

Recipe two, from even numbers. Take n even and 4 or more, and use (n/2)² − 1 and (n/2)² + 1:

 n    n/2   (n/2)² − 1   (n/2)² + 1    triple
 4     2         3            5       3, 4, 5
 6     3         8           10       6, 8, 10
 8     4        15           17       8, 15, 17
12     6        35           37       12, 35, 37
16     8        63           65       16, 63, 65

Between them, these two recipes produce five of the six cords in the Śulba list. The one they miss is 15, 36, 39, and the reason is easy to see. Every triple the odd recipe builds contains two numbers exactly 1 apart, and every triple the even recipe builds contains two numbers exactly 2 apart. In 15, 36, 39 the gaps are 36 − 15 = 21, 39 − 36 = 3 and 39 − 15 = 24, so no two of them are that close. That cord is only 5, 12, 13 tripled, and the odd recipe does hand you 5, 12, 13.

The recipes do not produce every triple that exists either. For instance 20, 21, 29 is a perfectly good primitive triple, since 400 + 441 = 841 = 29², and neither recipe ever reaches it.

One last fact that saves real time in a test: a triple can never be made of three odd numbers. If a and b are both odd, then a² and b² are both odd, so a² + b² is even, which makes c² even and c even as well. So a set like 3, 5, 7 or 9, 11, 15 can be rejected on sight, without squaring anything at all.

Almost every question on this theorem is one of two jobs, and the first thing to do is decide which one you are looking at.

Job A, the hypotenuse is missing. You know the two sides that meet at the right angle, so you add: c = √(a² + b²).

Job B, a shorter side is missing. You know the hypotenuse and one shorter side, so you subtract: a = √(c² − b²).

Getting these two the wrong way round is an easy mistake to make. There is a simple guard against it. The hypotenuse is the longest side, so an answer for a hypotenuse must come out bigger than either of the other two, and an answer for a shorter side must come out smaller than the hypotenuse. If your answer breaks that, you have added where you should have subtracted.

A reliable order of work:

1. Draw the triangle and mark the right angle.
2. Label the side opposite the right angle as the hypotenuse.
3. Write a² + b² = c² with the numbers put in.
4. Square each number, then add or subtract.
5. Take the square root.
6. Write the unit.

Example 1, finding the hypotenuse. The two shorter sides are 9 cm and 40 cm.

c² = 9² + 40² = 81 + 1600 = 1681
c  = √1681 = 41 cm

Example 2, finding a shorter side. The hypotenuse is 26 cm and one shorter side is 24 cm.

a² = 26² − 24² = 676 − 576 = 100
a  = √100 = 10 cm

Example 3, the ladder. A ladder 13 m long leans against a wall with its foot 5 m from the wall. How high does it reach? The wall is vertical and the ground is level, so the corner between them is the right angle and the ladder is the hypotenuse.

height² = 13² − 5² = 169 − 25 = 144
height  = √144 = 12 m

Now suppose the foot slips out to 12 m from the wall. The ladder is still 13 m long, so the new height is √(169 − 144) = √25 = 5 m, and the top has slid down 12 − 5 = 7 m. The numbers 5 and 12 have simply swapped places, which is a good sign you have done it right.

Example 4, the shortest distance. Ravi walks 8 km due east and then 6 km due north. East and north are at right angles, so his straight-line distance from the start is √(8² + 6²) = √(64 + 36) = √100 = 10 km, even though his feet covered 14 km. The gap between 14 and 10 is what a shortcut is worth.

Example 5, the diagonal of a rectangle. A rectangle is 24 cm long and 7 cm wide. A diagonal cuts it into two right triangles whose shorter sides are the length and the breadth, so the diagonal is √(576 + 49) = √625 = 25 cm.

Example 6, two poles. Two poles, 13 m and 8 m tall, stand upright on level ground 12 m apart, and a wire runs from the top of one to the top of the other. Draw a horizontal line from the shorter top across to the taller pole. That gives a right triangle whose shorter sides are 12 m across and 13 − 8 = 5 m up.

wire² = 12² + 5² = 144 + 25 = 169
wire  = √169 = 13 m

Example 7, working out a cost. A rectangular garden is 15 m by 20 m, and a straight path is to be laid along its diagonal at ₹180 per metre. First the length: √(225 + 400) = √625 = 25 m. Then the money: 25 × 180 = ₹4,500. Notice the order — never multiply by the rate until the length is finished and the square root has been taken.

Example 8, a square from its diagonal. A square handkerchief has a diagonal of 10 cm. Using area = diagonal² ÷ 2 from earlier, its area is 100 ÷ 2 = 50 cm², and its side is √50 = 5√2 ≈ 7.07 cm. You could also get there the long way: side² + side² = 100, so 2 × side² = 100 and side² = 50.

A closing note on units. If a question mixes metres and centimetres, convert everything to one unit before you square anything, and keep that unit all the way to the answer. Squaring a mixture is the fastest way to a wrong answer that still looks tidy.

The converse. Everything so far started with a right angle and worked out a length. The rule also runs the other way, and that is what makes it useful on a building site: if the three sides of a triangle satisfy a² + b² = c², where c is the longest side, then the triangle is right-angled, and the right angle is the one opposite c.

Test 12, 16, 20. The longest side is 20, so square that on its own and the other two together.

12² + 16² = 144 + 256 = 400
20²       = 400
Equal, so the triangle is right-angled.

Test 5, 12, 14. The longest side is 14.

5² + 12² = 25 + 144 = 169
14²      = 196
Not equal, so it is not right-angled.

Always square the longest side on its own. If you carelessly test 5² + 14² against 12², you will reject a triangle that may well have been right-angled, and you will never spot the error.

The knotted rope. This is the converse in the hands of a mason, and it is thousands of years old. Take a rope with 12 equal parts marked off by 13 knots, counting both ends. Peg it out into a triangle with sides of 3, 4 and 5 parts. Since 3 + 4 + 5 = 12, the rope closes exactly with nothing left over, and because 9 + 16 = 25, the corner between the 3 and the 4 is a true right angle. On a real site people scale the same triple up: measure 60 cm along one wall from the corner, 80 cm along the other, and if the distance between the two marks is exactly 100 cm the corner is square. If it comes to 101 cm the corner is slightly open, more than a right angle; if it comes to 99 cm it is slightly closed, less than a right angle.

A bonus test. Comparing c² with a² + b² tells you more than just yes or no:

c² equal to a² + b²    the biggest angle is exactly 90 degrees
c² more than a² + b²   the biggest angle is more than 90 degrees
c² less than a² + b²   every angle is less than 90 degrees

So 5, 12, 14 has an obtuse angle, because 196 is more than 169. And 6, 7, 8 has none, because 64 is less than 36 + 49 = 85. Here is a sharp example worth remembering: 10, 24, 26 is right-angled, because 100 + 576 = 676 = 26². But 10, 24, 25 is not, because 625 is less than 676. Being close is not the same as being right, which is exactly why builders measure instead of eyeballing.

Slips to avoid.

• Picking the wrong hypotenuse. The hypotenuse is the side opposite the right angle, not simply the side drawn along the bottom of your diagram. Mark the right angle first, every time, and then read off the side facing it.
• Adding sides instead of squares. c is not a + b. In a 3, 4, 5 triangle, 3 + 4 = 7, which is nowhere near 5. Square first, then add, then take the root.
• Forgetting the square root. Writing c = 625 when you meant c = √625 = 25 throws away a correct calculation at the last step. If your final number looks enormous, you have probably left it squared.
• Mixing units. A question might give 1.2 m and 50 cm. Convert both first: 120 cm and 50 cm give √(14400 + 2500) = √16900 = 130 cm, which is 1.3 m.
• Rounding too early. Keep √2 or √5 in the working and turn it into a decimal only in the last line.
• Using the rule where there is no right angle. A triangle with sides 6, 7 and 10 has no right angle at all, since 36 + 49 = 85 while 100 is bigger, so the rule a² + b² = c² does not apply to it.

Two quick checks are worth running on every answer you write. First, the hypotenuse must be longer than either of the other two sides. Second, it must be shorter than the two of them added together — in a 3, 4, 5 triangle, 5 sits comfortably between 4 and 7. If an answer breaks either check, go back and hunt for the slip before you move on to the next question.