Western biographers credit Archimedes of Syracuse (287-212 B.C) with the analytical evaluation of the factor Pi associated with circle, within a close range of 3^{1}/_{7} to 3^{10}/_{71}. However in the process of establishing this, they also recognize him as the first person to realize that the same factor is associated with both perimeter and area of the circle. He is also said to be the first person to propose and prove that “Area of the circle = ½ x Perimeter x Radius”. Is it really so? Let us look at another person, one Baudhayana, from ancient India (800BC) who also worked on the circles earlier to Archimedes, by a few centuries.

It was known to Baudhayana, or even to people earlier to him, that the perimeter of the circle depends only on its radius or diameter and that it is actually proportional to the radius or diameter. Though it has not been stated explicitly, it is clear from various sutras that they were well aware that for similar figures, the ratio of the areas equals the square of the ratio of the lengths of the corresponding sides. It was also known that the area of the circle depends only on its radius or diameter and that it is actually proportional to the square of the radius. That is, for a circle, it was known that: – Area = Ka x r^{2}, and Perimeter P = Kp x (2r). But do they know that, Ka = Kp = Pi ?.

During 800 BC, this Indian high priest, Baudhayana, has formulated, in his Sulvasutra (I-48), the so-called Pythagoras theorem, centuries before Pythagoras (572 BC). In another sutra (I-51) he has given a general rule for finding the square root of any number, both geometrically and arithmetically. In his Sutra (I-61) he found the value of √2 to a great accuracy and has given the procedure for the same. This Indian mathematician could construct a circle almost equal in area to a square and vice versa. He has described such procedures in his sutras (I-58 and I-59). All these were achieved in 800BC!

As Baudhayana was designing a religious altar for performing the Hindu rites, he constructed a square within a square as below:

He observed the inner square is exactly half of the bigger square in area. This led him to formulate, in his Sutra I-48, the so called Pythagoras theorem, which was reinvented by Pythagoras a few centuries later. Please refer to my earlier blog on “Baudhayana’s (Pythagoras) Theorem”.

Subsequently Baudhayana wanted to evolve procedures for constructing circular altars. He constructed two circles circumscribing the two squares shown above. Now, just as the areas of the squares, he realized that the inner circle should be exactly half of the bigger circle in area. Yes, he knows that the area of the circle is proportional to the square of its radius and the above construction proves the same. By the same logic, just as the perimeters of the two squares, the perimeter of outer circle should also be √2 times the perimeter of the inner circle. This proves the known fact, that the perimeter of the circle is proportional to its radius. Now it is known beyond any doubt that for a circle,

Area = Ka x r^{2}, and Perimeter P = Kp x (2r).

But that is not enough. Baudhayana wants to construct circular altars of specific areas. He needs to know the values of Ka and Kp. Baudhayana and his ilk were more interested in the area of the circle than its perimeter.

At this point, Baudhayana would make an important observation. Areas and perimeters of many regular polygons, including the squares above, can be related to each other just as the case of circles. The perimeters and areas of some simple regular polygons are listed below (‘r’ is the distance from the centre of the polygon to its sides):

Equi. Triangle– Perimeter = K3(2r) & Area = K3(r^{2}); with K3 = 3√3

Square- Perimeter = K4(2r) & Area = K4(r^{2}); with K4 = 4

Hexagon- Perimeter = K6(2r) & Area = K6(r^{2}); with K6 = 2√3

Octagon- Perimeter = K8(2r) & Area = K8 (r^{2}); with K8 = 8(√2-1)

It may also be noticed that the values of the constants, Ki’s, are gradually reducing from about 5 to about 3.3, by the time we reach the octagon. Another fascinating feature with these polygons is: all their areas are ‘½ r’ times their perimeters. Anybody would have been tempted to conclude from above that for circles also, Kp= Ka= K0. This will automatically make Area of the circle = ½ r x (Perimeter). However Baudhayana wanted to prove this.

Let us now consider an N-gon, a regular polygon of N-sides. Let ‘r’ be the distance of the sides from the centre. Let each side be equal to ‘s’. The area of the triangle one side makes along with the centre is (½)sr. Hence the total area of this N-gon is ½Nsr.

So, for N-gon -> Perimeter = Ns & Area = ½Nsr; with Kn = Ns/2r

Here again, Area = (^{r}/_{2}) x Perimeter. In all the above regular polygons, a circle of radius r can be inscribed. Now let us assume that, for this circle, constants Kp and Ka are different. The areas and perimeters of the above polygons are steadily reducing but still remaining more than those of this circle. (i.e.) Kn(2r) > Kp(2r) and Kn(r^{2}) > Ka(r^{2}). Hence any Kn will have to be greater than both K1 and K2. In the N-gon last considered, Ns being the perimeter, it reduces constantly as number of sides N increases. Kn = Ns/2r will also reduce gradually and finally will converge to a finite value, now known as Pi. Baudhayana realized that there is an N-gon with such a perimeter, whose Kn is just more than Kp by any arbitrarily small amount ‘∂’. Similarly Baudhayana concluded that there is an N-gon of such an area, whose Kn is just more than Ka by any small amount ‘ε’. However both these Kn’s are greater than Kp and Ka. (i.e.) Kp + ∂ > Ka, and, Ka + ε > Kp, for any small amounts of ∂’s and ε’s. The above is possible only when Kp=Ka.

Thus Baudhayana concluded, Kp = Ka = Ko, which is now known as Pi. This automatically makes, Area of the circle = ½ r x (Perimeter).

So we may conclude that, the above facts about the circles were already known to Baudhayana and other ancient Indian mathematicians even before Archimedes. Credit is surely due to Archimedes for narrowing down the value of Pi between 3^{1}/_{7 } and 3^{10}/_{71}. However, Baudhayana on his own has narrowed down the value of Pi to be between 40/12 and 40/13 (i.e. 3^{1}/3 and 3^{1}/_{13}). It is the value 40/13 he has used in his Sulva Sutra I-58, (for Circling the Square or to find a circle equal in area to a square), as will be demonstrated later. Before going to I-58, let us see how he derived the above values for Pi.

Ancient mathematicians could have already found the value of Pi with limited accuracy, by actual measurements of diameter and perimeter of the circle by using ropes as per the practices existing in those days. It could have given them, at best, a value between 3.11 and 3.17 (i.e. Pi +/- 1%). So, the attempts continued to analytically find the value of Pi.

Even before Baudhayana, the upper limit for the value of Pi was fixed as 4 by considering a circle inscribed in a square. The lower limit was fixed as 3 by considering a regular hexagon along with its circum-circle. But after Baudhayana’s (Pythagoras) theorem, these limits could be narrowed down to be between 3 and 2√3 by considering both circum-circle and in-circle of a hexagon as below.

The circum-radius is ‘a’ and the in-radius is √3/2(a). The perimeter of the hexagon is larger than that of in-circle and less than that of circum-circle. i.e. π(2xa) > 6a and π(2ax√3/2) < 6a. Hence, 2√3 > π > 3. There is an indirect reference to the value of 3 in an earlier sutra of Baudhayana for constructing altars: “The pits for the sacrificial posts are 1 pada in diameter, 3 padas in circumference.” This gives an approximate value of 3 for Pi. But they knew it is more than 3, as 3 pada is already known as the perimeter for a hexagon inscribed in the above circle. Hence 3 is just the lower limit for value of Pi.

Baudhayana went one step further by considering a regular octagon along with its circum-circle and in-circle. Just as Archimedes found the bounds for value of Pi, by considering 96-gon circum-scribing and in-scribing a given circle, Baudhayana found the bounds, by considering circum-circle and in-circle of an octagon. (yes, a complimentary procedure to what was done by Archimedes, five centuries later). Baudhayana used for √2, the value of 17/12 (normally used in those days) to arrive at these simple fractions, as below. He first considered a square with 12 units as half side. Hence half diagonal will be 17 units considering 17/12 as √2. Referring to the diagram below, the octagon inscribed within this square will have its side as 10.

The in-radius of this octagon will be 12 units and circum-radius will be 13 units. (‘5,12,13’ , the Pythagoras triple, as known even in Baudhayana’s times defines this octagon!). The perimeter of the octagon is 80 units. As this octagon is sandwiched between the circum-circles and in-circle, (ref Figure above), we may compare their perimeters as below:

Pi(2×13) > 80 > Pi(2×12)

**Hence, **

**Baudhayana’s values of Pi are given by: 40/12 > Pi > 40/13.**

Accuracy of Pi was always sought to be improved throughout the history, even after much closer estimates by Archimedes. A later mathematician Manava (650 ~ 300 BC) has stated in his sutra:

*Viskambhah pa**n**cabhaagasca viskambhastrigunasca yah.*

*sa mandalapariksepo na vaalamatiricyate **|| *

(Manava Sulvasutra 10.3.2.13)

*(a fifth of the diameter plus three times the diameter, is*

*the circumference of **the circle, not a hair-breadth remains.)*

It gives a value of 3^{1}/_{5} as the upper limit for value of Pi, a better value than Baudhayana’s 3^{1}/_{3, }and approaching 3^{1}/_{7 }of a later day Archimedes: (as per Manava’s statement “not a bit remains” after 3.2 times the diameter).

Baudhayana cleverly used his values of Pi for finding the area of a circle and for drawing a circle of a given area. He found the value of 40/13 to be closer to the eventual value of Pi and has used the same to derive his Sulvasutra I-58 for “Circling the Square”. Baudhayana would have derived this formula as below:

Let ‘a’ be the distance of the sides of the square from its centre. i.e. each side of the square is ‘2a’. To find a circle of radius ‘r’ with an area equivalent to this square, one may write

πr^{2} = 4a^{2}; (i.e.), r = (2/√π)a

Baudhayana used 40/13 as value of Pi, and 17/12 as value of √2.

From the picture above we see, a < r < √2a.

Hence he assumed, r = a + (1/x)(√2a-a) = a [1 + (1/x)(√2 – 1)]

For the above square, πr^{2} = 4a^{2}

r^{2} = 4a^{2}/π = 4a^{2} *(13/40) = a^{2} * 130/100

Hence, r = a * (√130)/10

By Baudhayana’s method (I-51) for finding the square root of any number,

√130 = √(12^{2} – 14) = 12 – (14/24) = 10 + 17/12

So, r = a(1 + 17/120) = a [1 + (1/3)(51/120)]

≈ a[1 + (1/3)(5/12)] = a[1 + (1/3)(√2 – 1)]

Thus, r = [a + (1/3)(√2a – a)]

So Baudhayana formulates his sulvasūtra I-58 as below:

*caturaśraṃ maṇḍalaṃ cikīrṣann*

*akṣṇayārdhaṃ madhyātprācīmabhyāpātayet* |

*yadatiśiṣyate tasya saha *

*tṛtīyena maṇḍalaṃ parilikhet*

* *

*caturaśraṃ = Square, Mandalam = Circle, *

*akṣṇayārdhaṃ = Half Diagonal, *

*madhyātprācīm = From centre towards east, *

*abhyāpātayet = laiddown, **yadatiśiṣyate = portion in excess, *

*tasya saha tṛtīyena = using only a third of this, *

*parilikhet – Draw around*

*To make a square into a circle, draw half its diagonal from the centre towards the East; then describe a circle using only a third of the portion which is in excess.*

i.e. Using the above formula your are able to draw a circle of given area (=4a^{2}), where a is the measure of half the side of the above square. The radius of this circle is given as:

r = [a+1/3(√2a – a)] = [1+1/3(√2 – 1)] a

Let us see with value of Pi as 40/13, how the areas of square and circle compare.

Area of Square = 4a^{2}

Area of circle = 40/13 x [1 + (1/3)(√2-1)]a^{2}

= (40/13) x (41/36)^{2} a^{2 }= (67240/16848) a^{2} = 3.9909 a^{2}

Remarkably close.

As per this construction the value of Pi we obtain as per today’s value of √2 is, Pi = 3.088312

However in the reverse process of squaring the circle, he has gone for corrective fractions as will be demonstrated by Baudhayana’s sutra I-59, as given below:

*maṇḍalaṃ caturaśraṃ cikīrṣanviṣkambhamaṣṭau*

*bhāgānkṛtvā bhāgamekonatriṃśadhā *

*vibhajyāṣṭāviṃśatibhāgānuddharet* |

*bhāgasya ca ṣaṣṭhamaṣṭamabhāgonam*

*viṣkambham = Diameter; aṣṭau bhāgānkṛtvā = making eight parts; *

*bhāgaekona = take out one part,*

*triṃśadhā vibhajya = 29 parts of this part;*

* āṣṭāviṃśatibhāgānuddharet = of these remove 28 parts; *

*bhāgasya ca = from this part also,*

*ṣaṣṭhamaṣṭamabhāgonam = remove (1/6 minus 1/8 of 1/6).*

*If you wish to turn a circle into a square, divide the diameter into eight parts and one of these parts into twenty-nine parts: of these twenty-nine parts remove twenty-eight and moreover the sixth part (of the one part left) less the eighth part (of the sixth part). *

The above formula is to make a square of area, equal to a given circle. Baudhayana could have really inverted the earlier formula I-58 to obtain this. But this sutra appears quite complicated. It is so only because Baudhayana in this case used a better value for √2. As per his sutra I-61,

√2 = 1 + (1/3) + (1/3*4) – (1/3*4*34) = 577/408

We may be wondering how handicapped the ancient mathematicians were without the present day decimal point system. However ancient Indians were so facile with fractions they never needed the decimal point system. Even as late as 19^{th} century, Indians were using fractional multiplications tables of ½, ¼, ^{1}/_{8} , ^{1}/_{16} , ^{1}/_{32} and even ^{3}/_{16 } in their every day arithmetic calculations. Baudhayana uses this amazing fractional arithmetic to arrive at the above formula.

We know from I.58, r = a + (1/3)(√2a-a) = [1+1/3(√2 – 1)]a

So we may write, a = r/[1+(1/3)(√2-1)]

With √2 = 577/408,

we get, a = r/[1+(1/3)(169/408)]= (1224/1393)r

Now for fractional magic:

1224/1393 = (1224/1392)*(1392/1393) = (51/58)/(1393/1392)

= [1-(7/56)(56/58)]/[1 + (1/1392)]

= [1-(1/8)(28/29)]*[1- (1/1392)]

= [1-(28/8*29)]*[1 – (1/8*29*6)]

(assuming 28/29 ≈ 1, in the last term),

= 1 – [28/(8*29)] – [1/(8*29*6)] + [1/(8*29*8*6)], .

Thus Baudhayana obtains the final formula of I-59 as:

(With Side of square as ‘s’ and the diameter of the circle as ‘d’)

** s = [1- 28/(8*29) – 1/(8*29*6) + 1/(8*29*6*8)] x d**

i.e., s = 0.878682 * d

This increases the value of Pi marginally from 3.088312 to 3.088326.

**Conclusion:**

It was after studying the book “Journey Through Genius” by William Dunham, I got interested in the History of Mathematics. I read several books to know more on this subject. In order to create interest among our genext, I started to write a few blogs on this subject. The present one is on ancient Indian Mathematician Baudhayana (800BC) and his works on Circles. We seem to know him only through his Sulva-sutras. However by reading extensive material on him, I could make out a few narrations of his works. Of course this narration includes a few of my imaginations and intuitions. Ancient Indians in 800BC were well aware of the basic properties of the circle. Baudhayana’s sulvasutras I-58 and I-59 give ample proof of this. Baudhayana was also able to fix the value of Pi to be between 40/12 and 40/13 (i.e. between 3.33 and 3.08). Baudhayana’s name is still uttered during many Hindu rituals. Even my own family is linked to Baudhayana through his line of disciples as mentioned often by us in our prayers as, Apasthamba, Aangirasa, Baragaspathya and Bharatwaja.

**References:**

1. A history of Ancient Indian Mathematics – C N Srinivasaiengar, The World Press Private Ltd. Calcutta. (1967)

2. Journey Through Genius – William Dunham, Penguin Books 1990.

3. S.G. Dani, **Geometry in Sulvas_sutras**, in ‘*Studies in the history of Indian mathematics*’, Cult. Hist. Math. 5, Hindustan Book Agency, New Delhi, 2010.