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)
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).
history of Indian geometry can be divided into three distinct periods:
such as excavated in Indus Valley.
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).
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.
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.
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).
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.
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.
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
are several Vedic proofs for this theorem which are all fairly simple (Tirthaji
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
CD = AB = m; DE = BC = n
So, ABC and CDE are congruent and ACE is a
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
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
BD is perpendicular to AC.
So, triangles ABC, ABD and BCD are similar.
So, AB2/AC2 = ADB/ABC and BC2/AC2
So, (AB2 + BC2)/AC2
= (ADB + BCD)/ABC = ABC/ABC = 1
So, AB2 + BC2 = AC2
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).
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
derivations of the value of √2
as interpreted from the Sulbasutrakaras
are given below:
ABED is the square with the length of side equal to
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 = ⅓,
then 1/9 + ⅔
= 7/9 < 12
= 1 + ⅓ +
+ x12 = (√2)2
so, x1 < 1/3.4
= 1 + 1/3
x22 = (17/12)2
2 = 1/144
so, x2 = (1/144)(12/17)(½)
= 1 + ⅓
In decimal form this final calculation equals
1.40972222222 whereas todays value of √2
is 1.41421356237, a difference of .00449134015.
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 ⅓rd
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
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 + ⅓
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 + ⅓
] x2 = [1/(3)(4)]2
Neglecting x2 as too small 2x(17/12)
So, x = [1/(3)(4)]2
( ½ ) = 1/(3)(4)(34)
Thus we have √2
= 1 + ⅓
In decimal form this equals 1.41421568627 whereas
todays value of √2
is 1.41421356237, a difference of .0000021239.
ABC is a circle such that AB = BC = 1 and AC = √2,
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
So, AD = √2
In triangle ODB, BC2 = OD2 +
OB2 = (1 1/√2)2
So, BD2 = ½ (2 2√2
+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)
So, AG = (AD2 BD2 + AB2)/2AB
- )2 (2-√2)
DG2 = AD2 AG2
1) [(2 sqrt2)/2]2
= 1 ½ - √2
So, DG = √(3/2
= 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 2√2)/√2
Triangle AHJ and ADG are similar.
So, AH/AD = HJ/DG or HJ = (AH/AD)DG
So, HJ = (5√2
and HJ = HK
Drop KL perpendicular to AJ
Triangles AKL and AHJ are similar
AK = AC CD DH HK = (17√2
So, KL = (338 239√2)/(68
AC = CD + DH + HK =KL
= 1 + 1 + 1/√2
+ (338 239√2)/(68
= 1 + 1/3 + 1/12 1/(12)(34)
This last method yields the same result as the last
method 1.41421568627 whereas todays value of √2
is 1.41421356237, a difference of .0000021239.
Following are a selection of constructions given in
the Sulbasutras (Amma
two perpendicular diameters in a circle in Baudhayanas recipe for drawing a
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.
method is well known in later Indian mathematics as the fish method due to
construction of a square with a given side results in a beautiful geometrical
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.
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.
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
= 0.5690355 and the area of the circle (using todays value of π) is
One method is given by:
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.
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).
Area of pentagon = ABCDE = 5 (area of
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).
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.
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.
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.
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.
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).
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).
T. A. Sarasvati. Geometry in Ancient and Medieval India. Motilal
Banarsidass. Delhi: 1979.
R. P. Geometry According to Sulba Sutra. Vaidika Samsodhana Mandala. Pune:
Swami Sri Barati Krsna. Vedic Mathmatics: Sixteen simple mathematical
formulae from the Vedas. Motilal Banarsidass Publishers. Delhi: 1989.