Sulbasutra Geometry

For Prof. W. Casselman

By Susanna Kong

Math 309

April 2001


 The basis and inspiration for the whole of Indian mathematics is geometry. The beginnings of algebra can be traced to the constructional geometry of the Vedic priests preserved in the Sulbasutras, a manual of geometrical constructions from the 5th to the 8th centuries. Earlier remnants of geometrical knowledge of the Indus Valley Civilisation can be found in excavations at Harappa and Mohenjo-daro where there is evidence of circle-drawing instruments from as early as 2500 B.C.E. (Amma 1)

Early geometry of the Sulbasutras was based on religious needs with regards to the construction of altars such as agnicitis, vedis, mandapas etc. that are required for sacrificial ritual (Kulkarni 19, Amma 3).

The history of Indian geometry can be divided into three distinct periods:

§         pre-Aryan, such as excavated in Indus Valley.

§         Vedic or Sulbasutra

§         post-Christian.

Evidence of the pre-Aryan period includes well-planned towns and geometric designs including circles, squares and triangles. A link between this and the Vedic period can be found in the motif of a rectangle with the four sides curved inwards resembling a stretched hide; in the former period it can be seen as a decorative pattern while in the later period it is seen in the shape of the sacrificial altars or vedi. The vedi as well as the fireplaces or agni had such exact measurements and geometric shapes that they were codified and became the Sulbasutras. However, it is not known from how far back such knowledge originated as the sacrificial act is as old as the Vedas or older (Amma 5).

In the post-Christian era Aryabhata I (b. 475 C.E.) dominated Indian mathematics although what exactly his contributions were is somewhat unclear. Also important are Bhaskara I (c. 522 C.E.), Bramhagupta (628 C.E.) and Bhaskara II (b. 1114 C.E.). In this period, geometry took somewhat of a backseat to the pursuit of astronomy and algebra.

Sulbasutra Geometry

The Sulbasutras contain the geometry necessary for construction of the vedi and the agni for the obligatory and votive rites. These in turn are a part of the Kalpasutras , which are attached to the Vedas as one of the six V dangas or limbs of the Vedas. It is likely that the Sulbasutra sections were a part of the Srautasutras of the Yajur Veda, the Veda designed for the performance of sacrifices.

The three most primitive agni, Garhapatya, Aavaniya and Daksinagni, are older than the Rg Veda and the Mahavedi was likely known to the Indus Valley civilisation. It is generally recognised that the Pythagorean theorem was known in India at the latest in the 8th century B.C.E. (Amma 14-15).

The Sulbasutras deal with the correct construction of the vedi and agni including orientation, size, shape and areas and, as such, they are not meant as mathematical theorems or proofs. The geometry in the Sulbasutras can be categorised into that which expressly states theorems, constructions and implicit geometrical truths contained in constructions.

Theorem of the Square of the Diagonal or Pythagorean Theorem

Modern historical research has shown that what is known as Pythagoras’ Theorem was in fact known to the ancient Indians. Whereas Pythagoras (c. 540 B.C.E.) is given credit for this formula in modern western culture, the Vedic priests were using this principle to build their altars.

In the Black Yajur Veda (8th century B.C.E.) and the Satapath Brahmana, 36 units is the length of the east-west line or praci or line of symmetry or prsthya of the Mahavedi and 30 units as one of the north-south lines. The praci and half the side make the sides containing the right angle in the triangle with sides 36, 15, and 39 units. There is a possibility that this theorem was known even earlier by the construction of the three agni, Garhapatya, Ahavaniya and Daksina. Other rational right triangles as well as the irrational right triangle 1,1 2 and approximate right triangles are also mentioned in the Sulbasutras (Amma 17-18).

There are several Vedic proofs for this theorem which are all fairly simple (Tirthaji 349-351).

Proof I

The square AE = the square KG and the four congruent right-angled triangles all around it.

The areas are c2, (b-a)2 and 4( ½ ab) respectively.

So c2 = a2 – 2ab + b2 + 4( ½ ab) = a2 + b2

Proof I


Proof II


CD = AB = m; DE = BC = n

So, ABC and CDE are congruent and ACE is a right-angled isosceles.

The trapezium ABDE = ABC + CDE + ACE

So ½ mn + ½ h2 + ½ mn = ½ (m + n) = ½ m2 + mn + ½ n2

So, ½ h2 = ½ m2 + ½ n2

So, h2 = m2 + n2

Proof II


Proof III

AE = BF = CG = DH = m and EB = FC = GD = HA = n

The square AC = the square EG + the four congruent right-angled triangles around it.

So, h2 + 4 (½ mn) = (m + n)2 = m2 + 2mn + n2

So, h2 = m2 + n2

Proof III


Proof IV

BD is perpendicular to AC.

So, triangles ABC, ABD and BCD are similar.

So, AB2/AC2 = ADB/ABC and BC2/AC2 = BCD/ABC

So, (AB2 + BC2)/AC2 = (ADB + BCD)/ABC = ABC/ABC = 1

So, AB2 + BC2 = AC2

Proof IV


Proof V

Using coordinates, the distance between point A at (a,0) and point B at (0,b) is:

BA = [(a – 0)2 + (0 – b)2] = (a2 + b2).

Proof V


Square Root of 2

“The oldest value of 2 is obtained from one of the cuniform tablets from old Babylonian times (1600 B.C.) now in the Yale Babylonian collection. It shows a square with two diagonals and values in sexagesimal system from which 2 in that system worked out as 2 = 1; 24, 51, 10. The sexagesimal equivalent of the value of 2 as given in the Sulbasutra is 1; 24, 51, 10, 37…” (Kulkarni 99-106).

        Three derivations of the value of 2 as interpreted from the Sulbasutrakaras are given below:


1.                  Muller (1930)

ABED is the square with the length of side equal to one unit.

Another square ACFG is to be drawn such that its area is twice the area of the square ABED, i.e. the magnitude of AC = 2

The value of x is obtained.

The area of square ABED = area of rectangle BCHE + GJED + square JFHE.

So, 12 = 2(x) + x2

Say x = ½, then Ό + 1 > 12

Say x = , then 1/9 + = 7/9 < 12

So, 2 = 1 + + x1

So 2(4/3x1) + x12 = (2)2 – (4/3)2 = 2/9

2(4/3x1) < 2/9

so, x1 < 1/3.4

so, 2 = 1 + 1/3 + 1/3.4 = 17/12

now, 2(17/12)x2 – x22 = (17/12)2 – (2)2 = 289/144 – 2 = 1/144

2(17/12) > 1/144

so, x2 = (1/144)(12/17)(½) = (1/3)( Ό )(1/3)( Ό )

so, 2 = 1 + + 1/3.4 - (⅓)( Ό )(⅓)( Ό )


In decimal form this final calculation equals 1.40972222222 whereas today’s value of 2 is 1.41421356237, a difference of .00449134015.

I Muller


2.                  Dutta B. B. (1932)

Take two squares whose sides are of unit length.

Divide the second square into three equal stripes I, II an III.

Sub-divide the last strip into three small squares III1, III2, III3 of sides rd each. Then on placing I, II and III1 about the first square S in the position I’, II’ and III1’ a new square is formed. Now divide each of the portions III2 and III3 into four equal stripes.

Placing four and four of them about the square just formed, on its east and south side, say, and introducing a small square at the south-east corner, a large square will be formed, each side of which is equal to 1 + + 1/(3)(4).

Now this square is larger than the two original squares by an amount [1/(3)(4)]2, the areas of small square introduced at the corner.

In order to get the equivalent area remove two thin stripes say of width x from either side of the square.

2 x [1 + + 1/(3)(4) ] – x2 = [1/(3)(4)]2

Neglecting x2 as too small 2x(17/12) = [1/(3)(4)]2

So, x = [1/(3)(4)]2 (12/17) ( ½ ) = 1/(3)(4)(34)

Thus we have 2 = 1 + + 1/(3)(4) – 1/(3)(4)(34) nearly.


In decimal form this equals 1.41421568627 whereas today’s value of 2 is 1.41421356237, a difference of .0000021239.

II Dutta



ABC is a circle such that AB = BC = 1 and AC = 2, and ΠABC = 90

OC = OB = 1/2

Taking C as centre and CB as radius draw an arc which cuts AC at D.

CD = CB = 1

OD = 1 – 1/2 = 2-1/2

So, AD = 2 –1

In triangle ODB, BC2 = OD2 + OB2 = (1 – 1/2)2 + (1/2)2 = [(2-1)/2]2 + (1/2)2.

So, BD2 = ½ (2 – 22 +1 ) + ½ = 2 – 2

So, BD = (2-2)

Drop DG perpendicular to AB

DG2 = AD2 – AG2 = BD2 – BG2, BG = AB – AG

So, AD2 – AG2 = BC2 – (AB – AG)2 = BD2 – AB2 + 2(AB)(AG) – AG2

So, AG = (AD2 – BD2 + AB2)/2AB = [(2 - )2 – (2-2) + 12]/(2)(1) = (2 – 2)/2

DG2 = AD2 – AG2 = (2 – 1) – [(2 – sqrt2)/2]2 = 1 ½ - 2

So, DG = (3/2 – 2) = 1 – 1/2

Taking D as centre and DG as radius, draw an arc, which cuts AC at H.

So DH = DG

Drop HJ perpendicular to AG

AH = AC – CD – DH = 2 – 1 – (1 – 1/2) = (3 – 22)/2

Triangle AHJ and ADG are similar.

So, AH/AD = HJ/DG or HJ = (AH/AD)DG

So, HJ = (52 – 7)/2(2-1) and HJ = HK

Drop KL perpendicular to AJ

Triangles AKL and AHJ are similar

AK = AC – CD – DH – HK = (172 – 24)/2(2 – 1)

So, KL = (338 – 2392)/(68 – 482)

AC = CD + DH + HK =KL

So, 2 = 1 + 1 + 1/2 + (52 – 7)/2(2-1) + (338 – 2392)/(68 – 482)

So, 2 = 1 + 1/3 + 1/12 – 1/(12)(34)


This last method yields the same result as the last method 1.41421568627 whereas today’s value of 2 is 1.41421356237, a difference of .0000021239.



Sulbasutra Constructions

Following are a selection of constructions given in the Sulbasutras (Amma 23-46).


i)  Drawing two perpendicular diameters in a circle in Baudhayana’s recipe for drawing a square.

        Drawing a line one fixes a pin at its middle. Slipping the end ties on to this pin, one draws a circle with the mark (the middle of the cord) and fixes pins at the ends of the diameter. With the end-tie on the eastern pin one draws a circle with the whole cord. Similarly at the western pin. The second diameter should be stretched through the points where these (circles) intersect.

        This method is well known in later Indian mathematics as the ‘fish’ method due to its shape.

'Fish' Method


ii)  The construction of a square with a given side results in a beautiful geometrical pattern.

         Wishing to construct a square one should make ties at both ends of a string as long as the desired side and make a mark at its middle. One should draw a line and fix a pin at its middle. Fixing the ties on this pin one should draw a circle by the middle mark (of the cord) and at the ends of the diameter (formed by the praci) one should fix pins. Fixing one tie on the eastern pin one should draw a circle with the other tie. Similarly round the western pin. Through the points where they meet the second diameter should be drawn and pins should be fixed at this end. With the ties on the eastern pin a circle is to be drawn with the middle mark. Similarly round the southern, western and northern (pins). Their outer points of intersection form the square.

         This method is found in the Baudhayana Sulbasutra.

Construction of a Square


Transformation of Figures

 Different shapes were prescribed for the fire-altars depending on the benefit sought: the falcon or syenacit for attaining heaven, the isosceles triangle or praugacit for destroying enemies etc. However, the shapes had to have the same area of 7 ½ square purusas and therefore the Sulbasutras described different methods of changing the shapes of figures while retaining the same areas.


To convert a square into a circle

The Sulbasutras give an approximate construction, which results in a value of π of 3.088. Whether or not it was known at the time that this method indeed results in an approximation is not known.

            If a is the side of the square and r the radius of the circle then r = a(2 + Φ2)/6.

If a = 1, then the area of the square is 1, r = (2 + Φ2)/6 = 0.5690355 and the area of the circle (using today’s value of π) is 1.0178256.

Convert a Square to a Circle


To convert a square into a rectangle

 One method is given by:

         Wishing to transform a square into a rectangle one should cut diagonally in the middle, divide one part again and place the two halves to the north and east of the other part. If the figure is a quadrilateral one should place together as it fits. This is the distribution.

Convert a Square to a Rectangle


Rectilinear Figures and the Circle

 Knowledge of determining the areas and measuring of rectilinear figures it seems is derived from a functional need in terms of land or ownership thereof. In Rg Veda there are many instances where the division of land by Asuras and Gods are mentioned. Likewise, the three fires or agni, Garhapatya, Ahavaniya and Daksina were developed from pre-Vedic times and their shapes, circular, square and semicircular, are of equal area (Kulkarni 68).


The area of a regular pentagon, hexagon etc. inscribed in a circle

 Area of pentagon = ABCDE = 5 (area of triangle AOB).

OF2 = OB2 – BF2 = OB2 – Ό AB2

The construction of a pentagon inscribed by a circle was known.

Area of pentagon ABCDE = 5 ( ½ OF AB)


By following this procedure the area of any sided figures inscribed in a circle can be found (Kulkarni 70).

Area of a Regular Pentagon


Area of a citi

 The area of a normal citi is 7 ½ square purusa which is equal to 108 000 square angula (7 ½ times 1202). The area of a citi is divided into 15 equal parts or 7200 square angula.

The cross line of the Smasana cita towards east is 3 purusa in length and the one on the western side is 2 purusa and praci (east-west) is 6 purusa long. The area of this trapezium is ½ (3 + 2) 6 = 15 square purusa or 108 000 square angula as all citi.

Likewise, the volume of all citi is expected to be equal. As other citi are the same height, this problem does not arise but for the Smasana cita its height on the eastern side is greater than the western side.

The volume of a standard citi is 108 000 times 32 = 3 456 000 cubic angula. The Smasana cita height is increased by one fifth = 32 + 32/5 = 38.4 angula and then divided by three = 38.4/3 = 12.8 angula. Twice this height is divided by 4 = 6.4 angula. Four layers of these bricks are laid. The height of theses 4 layers of bricks = 6.4 times 4 = 25.6 angula. The fifth layer of bricks is placed with bricks of thickness 1/3 times 38.4 = 12.8 angula. The total height of the citi will be 25.6 + 12.8 = 38.4 angula.

The thickness of bricks of the top layer (fifth layer) is decreased by drawing a diagonal plane through it so that the height of citi on one side is 38.4 angula and 25.6 angula on the other side. Thus the volume of the Smasana citi = (108 000)(25.6) + (108 000)(12.8)( ½ ) = (108 000)(32) = 3 456 000 cubic angula which is equal to the standard volume of citi (Kulkarni 71-71).

Smasana citi


Note on Measurements

 The length of measure angula occurs frequently in the three Vedas and seems to have been in use from ancient times. The Sulbasutras give the relations of different units to each other and in particular the angula for example the purusa is said to be 120 angula. However there is a dissenting measurement of the purusa, which is 125 angula or the height of a man with upraised hands when standing on toes (Kulkarni 30-31).



Works Cited

 Amma, T. A. Sarasvati. Geometry in Ancient and Medieval India. Motilal Banarsidass. Delhi: 1979.

 Kulkarni, R. P. Geometry According to Sulba Sutra. Vaidika Samsodhana Mandala. Pune: 1983.

 Tirthaji, Swami Sri Barati Krsna. Vedic Mathmatics: Sixteen simple mathematical formulae from the Vedas. Motilal Banarsidass Publishers. Delhi: 1989.