115

CHAPTER -VI

VALUE OF n AND SOME MATHEMATICAL

CALCULATIONS IN VARIOUS TEXTS

6.1 PREAMBLE

The problem of squaring a circle or circling a square requires

knowledge about the area of the circle, the area of the square, and the

relation between the circumference and the diameter of the circle. The ratio

of the circumference to the diameter of the circle is known as JI.

At the time of Sulbasutras, Sulbakaras were aware of the necessity

to find the value of the ratio of the circumference and the diameter of a

circle which is now known as 71. Aryabatta(499 AD) gave the value of K as

71 =3.1416. At the beginning Indian mathematicians used better

approximations to the value of n.

The Greek letter n indicates the ratio between the circumference of

a circle and its diameter. The exact value of JT is not found in vedic

literature, but the value of K assumed at that time could be found out from

Sulbasutras. The periods of different Sulbasutras are 1) Baudhayana 800

BC 2) Manava 750 BC 3) Apastambha 600BC and Katyayana 200 BC^^^

R.P. Kulkarni, ^"-The value of n known lo Sulhasutrakaras'\ IJHS,Vol-13, 1978, p.32

116

There are evidences to show that the relation between

circumference and the diameter of a circle was known to the people of the

Indus civilization, the Rgvedic period and the people of the Brahmana m

period.

The bullock cart with wheels was known to the people of the Indus

cmWzaiion.^^^ Sphef\ca\ stone weights with f\a\. top and base made of

limestone and quartz were found in the excavation of the city of Mohenjo

Daro and the weights are accurate. This confirms our assumption that

these people might have known the different geometrical properties of a

circle, and the relation between volume and weight.

Chariots of different types were mentioned in the Rgveda^^^. The

number and variety of chariots that were manufactured increased by the

time of the Yajurveda.

At the time of construction, a metallic rim was fitted around the felly.

So it was very necessary to measure the perimeter very accurately

because a very small change in it causes oversize in the rim of the felly or

it is not possible to fit it around the felly. For different types of chariots, the

size of the wheels was also different. The chariot makers might have

known about the relationship between the diameter and perimeter of the

wheel.

" ' Ibid., p.33. ' " Ihid

117

There is a mention of three sacred fires in the Rgveda^^. The shape

of Garhapathya is circle of area one square purusa and the shape of

Ahavaniya is square of the same size^^^. The magnitude of n is necessary

to construct a square of an area equal to the area of a circle. This confirms

the fact that the idea of the value of n might have been developed at the

Brahmana period.

An approximate value of n is given in Baudhayana Sulbasutra as 3.

It is given that the diameter of the yupa is one pada and the circumference

of the pit in which the yupa is to be fixed is three padas^^^

According to Manava Sulbasutra, ^^ the square of two cubit square

is equivalent to a circle of radius one cubit and three angulas.

Area of square = Area of the circle with radius 1 cubit unit

i.e. 2* = 7t [1 + (1/8)] '

Therefore 7t =3.16049

Manava Sulba Sutra also gives the shapes and sizes of the three

sacred fires.^^° The shape of Ahavaniya is square and side length is

1Arafni= 24 Angula. The shape of G'arhapatya is circle and its radius is

13 +7/8 Angula. Daksinagni is semi circular in shape and its radius is IQV?

Angulas. Areas of these three are equal.

' * Rgveda (1.15-12,V. 11.2) (Ref. R.P Kulkami, The value O/K known to Sulbasutrakaras, IJHS,Vol-I3, 1978, p.33)

127

128

Datta B.B, The Sciences of the Sulba.a study in early Hindu Geometry(Ref. Ibid., p.34)

B.Sl.-\-\\3(Ref. Ibid, p.34) ^^^ M.Sl. 1-27, (Ref. . Ibid., p. 37)

''"' Mazumdar N.K,(I922) Mancn'a Sulha Sutram, Journal of dept of letters. Vol. Ill, pp.327-342. University of Culcutta, (Ref Ibid., p.37)

118

Area of Ahavanlya = Area of Garfiapatya => 7c (13 +7/8 y = 576 =>n = 2.99

Area of DaksinSgni = Area of Garfiapatya => n (19.5)^/2 = 576 => TI =3.029 0 m

6.2 APPROXIMATIONS OF K IN THE TRANSFORMATIONS OF CIRCLE

INTO A SQUARE OR SQUARE INTO A CIRCLE

In Katyayana Sulbasutram ^^\ a method is suggested to transform

a square into a circle of the same area. Draw half of the diagonal from the

centre towards a corner. By taking this as the radius draw a circle. By

adding one-third of the radius which lies outside the square with the half

width, draw another circle. This circle will have an area equal to that of the

square.

We can approximate the value of TI from this.

K. SI, 3-13 (Ref. Katyayana Sulbasutra, Ed Khadilkar S.D, p.27)

19

Let ABCD be a square with side x.

OR =1/3 X (X/V2 - x/2 ) + x/2

= X/3V2 + x/3 = x/3 ( 1 + I/V2)

Area of the circle with radius OR = TU [X /3 (1+1/V2)f

=71 Xx^/9 (1+408/577)2

(byB.S.S. 1.61. V2= 577/408)

Area of circle = area of Square. So

71 xx2/9(1+408/577)2 =x2

So 71=3.0852

Another method is transforming a circle into a square. First divide the

diameter of the circle into 15 parts. Then construct a square with side

length 13 parts. ^2 w g can approximate the value of 71 from this.

If d is the diameter of the circle, then area = TI: x (d/2) ^= TC x (d^/4)

Side length of the square = (13/15) d

Area= (13/15)=^ x d^ =(169/225) x d^

' " K.Sl. 3-14 (Ref. Katyayana Sulbasutra, Ed. Khadilkar S.D, p.27) (MSI 10.3.2,13; Ap.SI 36-8,B.S 1-60) (Ref. R.P. Kulkami, The value of 7t Known to Sulbakaras, IJHS, Vol, 13,1978, p.34)

120

Area of Square =Area of Circle. So 71 c|2/4=169 d^/225.

So 71=3.004

Another method is described in Baudhayana Sulbasutra ^ ^ is given

as follows.

To transform a circle into a square, divide its diameter into 8 parts

and then divide one of these parts into 29 parts and subtract 28 of these

and also the sixth part of the preceding subdivision less the eighth part of

the last gives the side of the square of same area.

Side length x = 7d/8 + d / 8x29 + d/(8 x 29 x 6) - d/(8 x 29 x 6x8 )

= 7d/8 + d/(8 X 29) + 7d/(8 x 29 x 6 x 8)

= d/8x (9799/1392)

= 1.759877r

Area of square = Area of circle

x2=7ir2 => 71 = 3.0971

Dvarkanath Yajva, a great commentator of Sulbasutras tried to

improve the approximate value of n and he obtained the value of % as

71=3.157991.^^ Better values of TT as found by Manava is 71 = 3.1604 and

Dvarakanath Yajva is n=3.157991 and n =3.14159. Sulbasutras give the

value of 71 ranging from 3.09 to 3.16.

' " B.SJ. ]-59(Ref. R.P.KuikamiJhe value of w Known fo SuJbakaras, JJHS, Vol. 13, 1978, p.36).

' ^ Karthyayana Sulbasutra, Ch-3, p.67.

121

(In Baudhayana Sulbasutra^^^ there is a description of converting a

square into a chariot wheel. This starts with a very specific design using

bricks of a certain size, but ultimately the design is described in bricks of

entirely different shapes with circular arcs. First start with an area equal to

225 square bricks. Then continue by 64 more bricks so that one has a new

square equal to 17^ equal to 289 bricks. If we equate the square of side 15

units to a circle of diameter 17 we have 289 714 = 225. Therefore, TT =

900/289)

The construction of a circle of diameter 19 from a square of side 17

gives the value of n as 36l7t/4 = 289. Therefore, TI =1156/361. The above

value is very similar to the one value in Satapatha Brahmana which can be

expressed as 900/288.^^ The value of n as it appears in

Manavasulbasutra where the chariot wheel altar uses 344 bricks Instead of

the 289 of Baudhayana Sulbasulra is 1075/344. It is clear that the

representation of a square side 15 by a circle of diameter 17 is the best of

the three approximations of TI.

' " B.Sl, 16.6-11 (Ref. Subash. C. Kak,"rhree Old Indian Values of TU" UHS, 32(4) ,1997 , p.3IO) '^'' ^.^{Ref. Subash. C. Kak,"rhree Old Indian Values OIK" UHS, 32(4),I997.)

122

There ^ ^ is a method of drawing a semicircle, the area of which is equal to

the area of the given square.

D

B

R S

Draw a circle of radius r such that the area of this circle is equal to

the area of the square. Draw another square PQRS circumscribing this

circle. Then, draw another semicircle with OP as radius. Then the area of

this semicircle is equal to the area of the square ABCD. We can deduce the

value of n used in this transformation.

' " M.Sl. (10.1.1.8) (Ref. R.P Kulkami, "The value of Ji known to Sulbasutrakaras", IJHS, Vol-13, 1978,p.37)

Madhyako?j pramaneha mandalam parilekhayet | Atiriktatribhagena sarvam tu sahaniandlam | Caturasre"aksnay3"rajjurmadhyatah saiinipiatayet | Parilekhya tadardhenardhamandalameva tat |

123

Let X be the side length of ABCD. The measure of the radius r is

r = x/2+(1/3)[(x/V2)-(x/2)] = x/3+(x/3V2) = (1/6)I2x+V2x]^^

The length of side of PQRS = 2r

= (2x+V2x)/3

Radius of the semicircle = V2.r

= V2 [(2x+V2x)/6]

Area of semicircle = Area of ABCD => TT. [V2(2X+V2X)/6]^ (1/2) = x

o 7i/2[x(V2+1)/3]=^ = x

7t = 2.9/(V2+1)=^ = 3.088

That is in the transformation of the given square into a semicircle the

value of K assumed in Mahava Sulbasutra is 3.088.

Another method of drawing a circle of area equal to the area of a

square is given in Mahava Sulbasutra^^^. This says that "One shall divide a

square into 9 parts, the segments of the circumscribed circle into 3 parts

(By lengthening these lines), he shall remove the fifth part from the height

of such a lengthened line measured from its middle, draw a circle with this

as the radius". Then the area of the circle is equal to the area of the square.

' ^ M.Sl. 10.1.1.8, Baudhayana Sulbasutra - 1.58 M.Si. 10.3.2. IO(Ref. R.P Kulkami, "The value of TT known to Sulbasutrakaras", UHS, Vol-13, 1978, p.34)

' " M.Sl. (IO.I.8)(Ref. R.P Kulkami, "The value of TT known to Sulbasutrakaras",IJHS,Vol-l3, 1978, p.38) Caturasram navadha kuryaddhanuh kotyastridhatrfdha i Utsedhat parTcamam lumpet purisench tavasamam |

124

R

p

r

0

Q

M

B

Let ABCD be a square of side x. Let O be its centre. Draw a circumscribed

circle. P and Q are points on the side AB such that AP = PQ = QB = x/3.

Let M be the foot of the J_r from O on MR.

Radius of the circle is r = MR - (1/5) MR

Let the side be x, O be the center.

Therefore, AB = x

OA = X/V2 = OS = OR

OM = PQ/2 = x/6

MR = V(0R2-0M^)

MR = OS- TS = OT = V(OR^ - TR= )

= V[(x/V2) 2 - (x/6)']

=^[{xV2) - (x^/36)]

= V[(x^/2)(1-(1/18))]

=V( 17x736)

=(Vl7/6)x

The area of the circle is equal to the Area of the square

^ Tir = x => 71 [MR - (1/5)MR] ^ = x

^ n [(V17/6) X - (1/5) (V17/6) x] ^ = x=

o Tt[(Vl7/6)x]^ [1-(1/5)]^ = x

125

o 71 (17/36) (4/5)^=1

o 71= 25x9/(17x4) = 3.308

The^'"' period of Baudhayana is 800 BC and the period of Katyayana

is 200 BC. The values of TT obtained by different constructions are given

below 141

Sulbasutra

1) B.SI. 1-113

2) M.SI. 1.27

3) M.SI. (Mazumdar)

4) M.SI. 10.1.1.8

5) M.SI. 10.1.8

6) K.SI. 3-13

7) K.SI. 3-14

8) B.SI. 1-60

9) B.SI. 1-59

10)Dvarkanath Yajva

11)Aryabhatiyam2-10

12)B.SI. 2-10

13)Leelavathy(p-277)

14)K.SI. 3-13

15)B.SI. 16-6-11

Value of 71

3

3.16049

2.99,3.029

3.088

3.308

3.0852

3.004

3.004

3.0971

3.157991

3.1416

3.14159

3.1415926535

3.088

3.114

The concept of n is closely related to the concept of the circle. Early

Indian mathematics was developed in accordance with practical need.

R.P Kulkami, 'The value of K known to Sulhasutrakaras\\mSyo\-n, 1978, p.40) Ibid..

126

6.3 VALUE OF 7C IN LILAVATY AND IN YUKTI DEEPIKA

71 is given as a series in Leelavaty.^^^

The perimeter of a circle is obtained by multiplying the diameter by 4 and

dividing this by 1,3,5,... and giving + and - sign alternatively.

i.e.,2 7rr = 4d/1-4d/3 + 4d/5-4d/7 +

= 4d(1 - 1 / 3 + 1/5-1/7+ )

i.e.,7i = 1-1/3 +1 /5 -1 /7 +

This series is now known as Euler Series. But Madhavan (1320 - 1425)

defined this about 200 years before the time of Euler.

Anothei series for TT is found in YuktiDeepika^*^. This is as follows.

The perimeter is given as a series in terms of the diameter. The first term is

obtained by multiplying the diameter of a circle vi -::.The 2 '^ term is

obtained by dividing the 1®' term by 3 next term is obtained by dividing this

term by 3 and so on. These terms are multiplied respectively by 1,3,5 etc.

and adding the term in the odd places and subtracting the terms in the

even places, we get the perimeter of the circle.

i.e., 27rr =Vl2d/1 - Vl2d/(3x3) + Vl2d/(5x32) - Vl2d/(7x3^) +

Therefore, n=Vl2 x (1 -1/3x3+ 1/5x3^-1/7x3^+ )

''*' C.Krishanan Namboodiri, BharathiyaSas{racintaJ99S, Arshaprikasam Prasidheekarana Samithi, Kozhikode, p.255

'" YuktiDeepika -2. 212-214 (Ref. C.Krishanan Namboodiri, Bharathiya SaslracinlaJ99&, Arshaprikasam Prasidheekarana Samithi, Kozhikode, p.257)

127

More accurate value of TI is given in Leelavaty.^^ This says that the

perimeter of a circle of diameter 9x10^^ is 2827433388233. From this we

can find the value of TT as n= perimeter/diameter= 2827433388233/(9x10)^''

=3.1415926535

The approximate circumference of a circle of diameter 20000 is 62832. '*^

The history of Indian mathematics generally begins with the

geometry of Sulbasutras. Thus geometry is related to the construction of

altars described in Satapatha Brahmana. The transformation of a square

altar into a circular one is a part of the construction of altars. Satapatha

Brahmana 7.1.1.118 - 31 describes the construction of a circular

Garhapathya altar using different types of bricks. Garhapathya altar is

circular in shape and Ahavaniya is square in shape and are of equal area.

According to Sulbasutras^'^, their area is equal to one vyama. One

Purusa= one Vyama = Five Aratnis =120 Arigulas. The construction of

Garhapatya altar is using square and oblong bricks. The square bricks are

of 24x24 square arigulas. The oblong bricks are of 48x24 square ahgulas.

So the area of the square in the figure below = 4/5 x 4/5 square purusas. m

Therefore, the diameter of the circle = V [(4/5) * + (4/5) * ]

= V32/5ptymsa

Therefore, radius =2V2 /5

Therefore, area= 7t r = 8 7i/25

'"^ Leelavathy, (Ref. C.Krishanan Naniboodiri, Bharathiya Sastracinta, 1998, Arshaprikasam Prasidheekarana Samithi, Kozhikode, p.259)

"" Aryahhaliya (ganitapada 10) [Scientific Heritage of India, Mathematics, Edited by KG Paulose, published by Govt. Sanskrit College committee Ihripunithura, p. 30]

'•"•Zf.SV 7.4-5 (Ref Subash.C.Kak," Three Old Indian Values of TT" I JHS,32 ( 4 ) , I 9 9 7 )

128

But the area is taken to be 1 square purusa. So 8 7i/25 = 1 or TT

=25/8=3.125

W

6.4 SOME NON INDIAN VALUES FOR 71

6.4.1 Egyptian Values^*^

Some mathematical ideas are obtained from the Egyptian hierographics

(sacred carvings) on tombs and monuments. The main sources of

information are the Egyptian Papyri. One of the problems of Rhind Papyrus

says that the area of the circle with radius 9/2 is equal to the area of a

square with side length 8.

From this,

n (9/2)' = 8'

=> 81 71 = 256

71 = 256/81 =3.16

'• ^ Ramakrishnan, ''The concept of Pi through the ages", M.Phil Dissertation, Dept of Mathematics, CUSAT

129

6.4.2 Chinese Values^**

The Chinese found various values for 71. Chou Pei Suan Ching, the oldest

mathennatical classic, gives the value of 71 as 3. Some other approximations

are VIO, 92/29, 142/45,3.14159 and soon.

6.4.3 Greek Values^^^

The first scientific attempt to compute the value of n is done by Archimedes

of Syracuse. According to him it is 3 + 10/71 < n < 3 + 1/7. Marcus Vitruvius

Pollio, the author of 'De architecture' approximated the value of n used by

Claudius Rolemy of Alexandria and it is 377/120.

6.4.4 Gregory Series and approximations of n^^

The Scotch mathematician James Gregory obtained an infinite series for

tan'^x.

ie, tan-'x = x-(x^/3)+( x^/5)-(x^/7)+

But this was discovered by Madhava about 300 years ago.

When x=1, the series becomes

TT/4 = 1-(1/3)+(1/5H1/7)+ which is known as Leibnitz Series.

From this we can approximate the value of 71.

The result 7id = C = V(12d=^)-[V(12d^)]/3.3 + [V(12d2)]/3 .5 +

""' /hid, ''" Ibid..

"'Ibid.,

130

given in 'Tantrasaiigraha' of Nilakanta Somayaji about 200 years ago is

obtained by putting x = 1/V3 in the Gregory series.

6.5 IRRATIONALITY OF 71 151

Aryabhata I has given the asanna (approximate) value for n as

62832/20,000. From the term asanna, the irrationality of n is implied.

Sangamagrama of Madhava says that "Multiply the square of the diameter

by 12 and extract the square root of the product. This is the first term.

Divide the first term by 3 to obtain the second, the second by 3 to obtain

the third and so on. Divide the terms, in order by the odd numbers 1,3,5.

And so on. Add the odd order term to and subtract the even order terms

*rom the proceeding. The result will give the circumference'.

ie, C = TT D = (Vl2D^) - (Vl2D=^)/3.3 + (Vl2D^)/5.32 - (Vl2D'^)/7.3^ +

Therefore TI =Vl2[1-(1/3.3)+(1/5.32)-<1/7.3=')+ ]

This is an infinite series. The irrationality of TI is recognized from this.

6.6 SOME OTHER INDIAN VALUES

6.6.1 The Value of n in the Mahabharata^^^

The approximate value of n used in the Mahabharata is 3. There are

some verses about TI in the sixth pan/a called Bhisma pan/a of

Mahabharata. This says that the diameter of Rahu is 12000 yojanas and

perimeter is 36000 yojanas.

'" Ibid., ' " Ibid.,

131

Perimeter Diameter

Moon 33000 11000

Sun 30000 10000

6.6.2 The Value of n In the Puranas^^

Brahmanda pumna states that:

nava yojana sahasro vistaro Bhaskarasya tu

vistarat trigunascasya parinahastu mandala".

i.e., 9000 yqjanas is the perimeter of the Sun and three times the

diameter in the circumference in the peripheral circle. Matsya, Aditya and

Vayu Puranas state that the circumference of the circle Is three times its

diameter.

The discovery of Gregory Series is an important factor for calculating

the value of TI.

6.7 SOME IMPORTANT RESULTS GIVEN IN YUKTIBHASA

Mathematics and astronomy flourished in Kerala at a higher level.

Yuktibhcisa based on Tantra Sangraha contains proofs for many

mathematical rules. The mathematicians of Kerala achieved these results

about 300 years before when the Europeans rediscovered them.

The concern for accuracy in the results of computations is reflected in the

calculations.

' " /hid..

132

It is believed that Tantrasamgraham is a revised edition of

"Drugganitapaddhati" oi \/a6assery Parameswaran Namboodiri. The author

of Tantrasamgraham is Kelallur Somayajippad, and was written around AD

1500. Yuktibhasa is based on this text and was written around 1639. Its «

author is Brahmadattan Namboodiri.^^ It is believed that he belongs to the

"Parangode lllam" of Alathur village. The nnethod of illustration in this book

is very appreciable. It describes about the mathematical part of Jyotisastra.

Chapters 6 and 7 describe the theorem of hypotenuse, properties of

triangles, construction of a circle from a square and so on. All these are

applicable in Vastusastra especially in the construction of the roof and in

the construction of Yajnakundas. Yuktibhasa describes many properties

which we have seen now in modern mathematics like the theorem of

hypotenuse (Pythagoras theorem). From these we can assume that there

were many scholars among the ancient Indians.

The fact that "The product of the height of a segment of a circle and

its complementary segment is equal to the square of the half of the

chord^^^" is given in Yuktibhasa.

\ ^ c

w

^

B

' Yuktibhasa, Ed. A.R Akhileswara Iyer, Mangalodayam Ltd. Publishers, Fhrissur

IbuL, ch-7 p. 273.

133

We can prove this using modern theory.

Consider two complementary arcs QAP and QBP. Let O be the centre of

the circle and C be the midpoint of PQ.

Clearly AC is perpendicular to the chord PQ and bisects it, Since AC

passes through O.

Consider AAPB, a triangle inscribed within a semicircle. Therefore

<APB=90°

Therefore AB2=AP^ + PB^ (1)

AAPC is a right angled triangle. Therefore

AP^ =AC^ + CP^ (2)

Substituting in 8quation(1), we have AB^ =AC^ + CP^ + PB^

(AC+CB) 2 = AC ' + CP= +( PC^+CB^)

(From APCB, PB^ = PC^+CB^)

Therefore AC= +BC= +2 AC.BC=AC^+2PC2+BC=^

Therefore AC.BC=PC^

That is the product of the height of a segment of a circle and its

complementary segment is equal to the square of half of the chord. Some

important results in Yuktibhasa are given below.

The common chord of two intersecting circle is perpendicular to the line of

centres. ^^

"" Ibid..

134

The surface area of a sphere is given by the product of circumference and

Its diameter ^ ^ and its volume is given by 1/6 ^ of the product of its

Diameter ^^ These are evident since

Surface area = 4 jir^ = (2 TT r)x 2r

=Circumference x Diameter

Volume = 4/37ur3 = 1/6 x 2 TC r x (2r) =

=1/6 X Circumference x (Diameter)

In modern mathematics, we can prove this using integral calculus.

The area of a circle is given by the product of semi circumference

and the semi diameter. Some formulae and their proofs are given in

Yuktibhasa. Yuktibhasa is based on Tantrasamgraham by Kelallur

Nilakantha Somayajipad. This book gives illustration and proof of the

*

operations and subjects described in Tantrasamgraham. The author unifies

the proofs of subjects of Yuktibhasa which were scattered in different

families and places and transferred heretically. The style of description is

very appreciable. Starting from the basic point and illustrating different

branches based on this point and re-examining these after illustration is a

very good method.

We can see the trigonometrical results,

Sin^A - Sin^ B = Sin(A+B) x Sin(A-B)

' " Ibid.. " " /hid.

135

This also states that the area of a cyclic quadrilateral is the product of

the three diagonals divided by twice the circum diameter^^®.

SinA. Sin B = Sin^ (A+B / 2) - Sin^ (A-B / 2) with the help of the cyclic

quadrilateral. ^^ Let ABCD be the quadrilateral. DE || to BC AE is the third

diagonal.

AM is x'to BC .-.I'to DE. DN x'to BC.

.-.Area of the quadrilateral = V^ BC x AL + V^ BC(DN)

= Va BC (AL + DN) = 72 BC (AL + LM)

= VaBCxAM

from triangle ADE, AM = (AD x AE)/2R, where 2R is the diameter of the

circum circle.

[For AABC a/Sin A = b / SinB = c / Sin C = 2R]

.-.Area of quadrilateral = V BC x (AD x AE)/2R

= 72 BC X AD X AE/Diameter

or Diameter = 14 BC(AD) AE/ Area of the quadrilateral.

Ibid.xh-l pp.228-238. Ibid.., ch-7 pp.224-227.

136

Yuktibhasa also says that the product of the sides of a triangle

divided by the circum diameter is the altitude to the base.^^V

6.7.1 Statement and proof of the formula (a+b)*= a* + 2ab + b*

We can see the statement and proof of the formula (a+b)^ = a^ + 2ab + b

in Yuktibhasa. It is said that to square a number, split this into two and

add twice the product of these numbers to the sum of the squares of

them.^^^

Proof:

There is given a geometric proof for this.

Consider a square of side x+y say D APQS. Let ABND be a square

of side 'x and MNRQ a square of side 'y'.

Area of QAPQS = Area of DABND + Area of D BPMN + Area of D MNRQ

+ Area of D RNDS.

Therefore, (x+y) * = x* + xy +y* + xy

=x' + y* + 2xy^^^

/hid.,ch-l pp. 231

"'" /hid, ch-],p.\9. "'' /bid,ch-],p.2\

I.U

D N

M

0

S R

There is a reference of another formula (a+b) * = (a-b) * + 4ab. i.e., If

four times the product of two numbers is added to the square of the

difference of these two, we will get the square of the sum of these

numbers. We can see a geometric proof for this in Yuktibhasa.

/ ^

s

A 8

5>' c

L Let OP=x and OQ=y.

AM=y, PM=y

Therefore, AB = x-y,

SM = x = QN

a

A/

M

138

Therefore, Area of D PQRS = Area of u ABCD + Area of D POBM +

Area of D OQNC + Area of U NRLD + Area of D LSMA

Therefore, (x+y)' = (x-y) * + xy + xy + xy + xy

= (x-y) ' + 4xy

6.7.2 Statement and proof of the Formula x*+y' - 2xy = (x-y)^.

Another property given in Yuktibhasa is that if we subtract twice the product

of two numbers from the sum of their squares, we obtain the square of their

difference.

x*+y 2 - 2xy = (x-y)^ ^^

The geometrical proof given in Yuktibhasa is given below.

p s •

Let PQ = x+y

PB = x = CQ

BQ = y = AP = CR = DS

Area of rABCD = Area of nPQRS - Area [AAPB - ABQC - ACDR - AASD]

=Area of nPQRS-Area of [ 72 r]PBNA+ Yz nBQCL + 72; iCTDR + 72 i AMDS]

/hid.

139

= (x+y) * - V2 [xy + xy + xy + xy]

= (x+y) * - 2xy

= X* + y*

from the above formula.

Area of D MNLT = Area of D ABCD - Area of AABN - Area of ABCL - Area

of ACTD - Area of ADAM

= Area of OABCD -Area of[ Ya D PBNA + 72 DBQCL + V^ nCTDR + VzU MDS]

i.e. (x-y) * = (x* + y*)-[ !/ xy + V xy + Vaxy +Vixy ]

= X* + y* - 2xy

6.7.3 Some trignometrical results in Yuktibhasa

Another property given in Yuktibhasa is that, the difference of the

squares of the two sides of a triangle is equal to the difference of the

squares of their projections on the base.^^

We can prove this using trigonometry. B

A

Let ABC be a triangle and let AB = x, BC = y. Let AP be the projection of

AB on the base AC and PC be the projection of BC on the base AC.

Clearly ABP and BCP are right angled triangles. Then

/hid, ch-\,pJ5.

140

BP2 = AB2-AP^ fromAABP

Also, BP^ = BC^ - PC * from BPC

Therefore, AB^ - AP^ = BC^ - PC^

Therefore, AB' - BC^ = AP= - PC^

i.e., the difference of squares of these two sides is equal to the difference

of squares of their projections on the base. Condition of similarity of right

angled triangles given in Yuktibhasa is given below:'^

(1)Two right angled triangles are similar if the hypotenuse {kamam)

and one side {bhuja) of one triangle is parallel to the hypotenuse and

one side of the other respectively or one side (koti) and hypotenuse *

of one triangle is perpendicular to one side (koti) and hypotenuse of

the other triangle respectively.

(2) The three sides of one triangle are perpendicular to the three sides

of the other respectively.

(3) Two triangles are similar if the three sides of one triangle are parallel

to the three sides of the other triangle respectively.

From this it is clear that ancient people in India were awarw of the

basic axioms of trigonometry. They used these properties for the

measurement of the rafters etc. in the construction of roofs.

Another property seen in Yuktibhasa is that if two sides of a

triangle are equal, the perpendicular from the vertex bisects the base. If

they are unequal, the foot of the perpendicular is nearer the shorter side. ®''

"^ Ibid., ch-6, p.89

//7/J.,ch-7,p.l44.

141

This property was indicated in the process of inscribing a hexagon

of sides equal to the radius within a circle. We can prove this using the

simple properties of trigonometry.

A

B M C

Let ABC be a triangle in which AB = AC. Let M be the foot of the

perpendicular from A on BC. ClearlyWBM and AACM are right angled

triangles. By the theorem of hypotenuse

BM2 = AB»-AM2

MC* = A C - AM*

If AB= AC, then clearly BM* = MC^

Therefore, BM = MC.

If AB < AC, then BM < MC. ie, the foot of the perpendicular M is nearer to

the shorter side AB.

Another property given in the process of inscribing a hexagon

within a circle is that six chords of length equal to the radius can be placed

in order inside a circle, i.e., the side length of the hexagon inscribed within

a circle is equal to the radius of the circle.

We can give a simple proof for this.

142

D

Consider AOBC, where OB = OC = radius of the circle.

Clearly <BOC = 360/6 = 60.

Since AOBC is an isosceles triangle, we have the opposite angles of the

equal sides equal.

Therefore, <B + <C = 180 - <BOC

= 180-60=120

Therefore, <B = <C = 60.

Therefore, AOBC is an equilateral triangle and therefore BC = OB = OC.

Therefore, BC is the radius of the circle.

Yuktibhasa gives the area of the triangle as half of the product of the

base and altitude. ®® There is also given a proof for this.

"•" JhiJ., ch-7, p.222.

143

B C S O R

Let ABC be a triangle. Let M be the mid point of AB, and N be the

mid point of AC. Draw a line through M perpendicular to BC and a line

through N perpendicular to BC. Choose points P and Q on a line parallel to

BC through A, such that PQRS is a rectangle. Let S and R be the feet of

the perpendiculars on BC respectively. Clearly triangles PAM and SBM are

similar, (three sides parallel). Also BM = AM Therefore, corresponding

sides are equal. So these two triangles are equal triangles. Similarly, AAQN

= A CRN. Cut AMBS and paste it at the place of APAM with the vertices B

at the position of A and S at the position of P. Similarly, cut A RCN and

paste it at the place ofA QAN with C at the position of A and R at the

position of Q. We obtain a rectangle PQRS.

Therefore, Area of AABC = Area of n PQRS = PS x RS = AG x RS.

The triangles ABMS and ABAC are similar. (Since three sides are parallel.)

Also,

BM = V2 BA.

Therefore, BS = Vz BO

Similarly, CR = V^ CO

Therefore, ^S - CR = 'A (BO + CO)

144

(BC-SR) = 72 BC

.•.SR = y2BC

So the Area of the AABC = AO x RS = AO x 72 BC = Vi BC AO

= V2 base X altitude

Another reference of the triangle inscribed in a semi-circle is that the

chords of any two arcs of a semi-circle are mutually perpendicular.^®^ In a

triangle, the product of the two sides divided by the diameter of the

circumcircle gives the altitude.^^°

We can prove this using some formulae of trigonometry.

The altitude of the AABC is CP, where P is the foot of the perpendicular

from C on AB. From the right angled triangle ACAP.Sin A=CP/AC.

Therefore, CP=AC SinA = b SinA

But from the trigonometric formula a/SinA = b/SinB = c/SinC = 2R where

a=BC, b= AC, c= AB and R is the radius of the circumcircle.

Therefore, a/2R = SinA

'•'' Ihid., ch-7, p.229.

™//)/J., ch-7, p.23l

145

Therefore, CP = b SinA = b x a/2R = ab/2R

ie, altitude = product of the two sides / diameter of the circumcircle.

Another property given in Yuktibhasa about the cyclic quadrilateral is

that the sum of the product of the opposite sides of a cyclic quadrilateral is

equal to the product of the diagonals.^^^

6.8 SOME MATHEMATICAL CALCULATIONS IN B A K S H A L T

MANUSCRIPT

The Bakshali Manuscript is a mathematical work found at Bakshaii,

a village of Peshawar district of the North West Frontier Province of India in

1881. It was written on brich-bark, and each layer of the bark Is white or

pinky white on the outer side, but is a reddish or yellowish buff on the inner

side. The language used in this text is an irregular Sanskrit. The beginning

and end of the manuscript was lost. The author and name of the work were

unknown. It contains problems involving systems of linear equations,

indeterminate equations of the second degree, arithmetical progressions,

quadratic equations, approximate values of square roots and so on. It

contains a large variety of problems relating to our daily life.

' " Ibid., ch-7,p.233

146

This consists of Sutras (rules) and examples. There is given a

method of extracting square roots in this text. The method gives the first

approximation for the square root of

Q=A» + b is A+(b/2A)=qi

The first error qi*-Q=(b/2A) '' = e^.

The second approximation for the square root of Q is

VQ = V (qi» - ei) = qi-(ei/2qi)

=[A+(b/2A)] - (b'/4A^)/(2[A+(b/2A)])

=[A+(b/2A)] -(b'/4A2)[A/(2A^+b)]

= A+ (b/2) [(1 /A)-(b/4A^+2Ab)]

=A +[(4A^b+b2)/(8A='+4Ab)]

The second en-or 62 = (ei /2qi)* ^

There are given some examples.

V41 =V(36+5), qi = 6+5/12. 61=45/144, q2=6 + 745/1848

V481= V(21» + 40)(, qi=21 40/42, ei/8 = 1600/14112, q2=21+ 9020/9681

We can compare these with modern values

Q qi q2 q (approximately)

41 6.41667 6.40313 6.403124

481 21.9524 21.9307 21.9317122

The first one is correct upto four decimal places and the second one

is correct upto two decimal places.

' ^ KAYE G.R, ""The Bakshali Manuscript, A Study in mediaeval mathematics", Cosmo Publications, New Delhi,l98l, p.30.

147

6.9 GENERAL TRIGNOMETRICAL RESULTS

There is a reference of the expression for the circumradius in

Yuktibhasa^^^ which says that the product of two chords divided by the

diameter is equal to the altitude from the point of intersection of the chords

to the line joining their ends^ '*. The theorem "Perpendicular bisectors of the

sides of a triangle are concurrent " is proved for the equilateral triangle

under Aryabhatiya Ganitapada 6^^^

From the fourth Century A.D Indian astronomers gave very large

importance to the construction of Jya^table (Sine - table). It was necessary

to calculate the position of planets accurately.

C

The trignometrical functions Jya^ (PM), Kojya (OM) for a small arc PB is

defined as

Jya A =PM = R SinA, Kojya = OM=R CosA

There is a reference of the value of Jya"30°(Sin 30°) and Jya~60°(Sin 60°)

in Pancasiddhantika^^^ without giving their derivation. They are as follows:

' " Ibid, pp.244-246. ''" /bid, p.23\.

'^'Aryabhatiya-Ganitapada, TrivanJrumSanskritSeries,]930,versus 14-16 '/2

(Ref. T.A Saraswathy, "Development ofMathamatical ideas in India", IJHS, Volume 4, Nov-1969)

''"' Pancasiddhantika,Chapter 4 V-2 (Ref. AmulyaKumar Bag, "SINE TABLE IN ANCIEN I INDIA", National Institute of Science of lndia,IJHS.Vol-4,l969.)

148

"Square of the radius R is to be defined constant. The one-fourth part of

that is the square of the Aries. The square root of the two quantities- the

square of the Aries and the Aries lessened from the constant are the Jyas

of 30° and 60° respectively"

ievyya30°=A/(R2/4)=R/2

Jya 60°=V[RMR^/4)] =(V3/2)R.

Another result is^^ "Twice of any desired arc is subtracted from 90°,

the Jyaoi the remainder is subtracted from the radius. The square root of

the result multiplied by half the radius is the Jya of that arc. By deducing

that square from the constant R , the square of the Kojya is obtained,

i.e. 1) (Jya A) ^ = R/2 [R-Jya(Tr/2 - 2A)]

(Jya A) ^ = R/2 [R-Kojya 2A] and

2) R2-(JyaA)=^ = [Jya{TT/2 -A) ]^

i.e. R^-(Jya A) 2 = {Kojya A) ^

or (Jya A) ^ + {Kojya A) ^ = R^

These two results arc equivalent to the identities SinA^ =(1/2) [1-Cos2A]

and Sin='A+Cos='A=1

Mahavira^^^ in the 9* century refers to the circle inscribed in

triangles and in quadrilaterals giving a formula for radius as radius = area /

half the perimeter.

The point of intersection of the Xr bisectors of the sides is the incentre.

' " Pancasiddhantika, Chapter 4 V-2 (Ref. AmulyaKumar Bag, "SINE TABLE IN ANCIENT INDIA", National Institute of Science of India,IJHS.Vol-4,1969.)

' '* T.A.Saraswathy, Develupmeni of Mathematical ideas in India, IJHS, May-Nov. 1969, p.70.

149

Total area of A ABC = area of AQOR + Area of BPOR + Area of CPOQ

.-. Area of AQOR = Area of A A R O + Area of A AOQ

= 72 R O X AR + 72 OQ x AQ

= 72 r (AR + AQ)

B P C

.-.Area of AABC = Ya r (AR+ AQ + BR+ BP + CQ +CP) = 72 r (a +b+ c)

Sulbakaras were aware that the volume of the regular solid with opposite

faces parallel, is the area of the base multiplied by the height.

Brahmagupta^'^^ knows the volume of a pyramid is 1/3' ' of the volume of a

prism with the same height and base. It was the astronomical need of

knowing the circumference, chords and arc of the circle accurately. The

main reason for this achievement was their great interest in series.

6.10 KNOWLEDGE OF GEOMETRY IN INDUS CIVILIZATION

It is clear from the archaeological investigations that the geometry of

Mohenjo Daro and Harappa was well developed. They^^ used a slip of

shell marked with division of 0.67 cm. The error of the scaling is only

0.00762 cm. From the measurement of staircases of buildings of Mohenjo

Daro, it is evident that the fundamental operations like sum subtraction,

" ' T A Saraswathy, Mahaviras trealmcnt of series. Journal of Kanchi University 1,1962, p.43 (Ref: Development of Mathematical ideas in India , T. A Saraswathy, Indian Journal of History Of Science, May-Nov. 1969 )

"'° K.P Kulkami, "Geometry as known to the people of Indus civilization", IJHS Vol-13,1978

150

multiplication and division were known to them. The bricks used for

construction of wells are of trapezium shape. This is because the thickness

of the joints between bricks should not be increasing with the increase in

the radius of the circular wall of the well. This is an example of a good

engineering practice.

6.10.1 Knowledge of Pythagoras^®^

The dimensions of the great bath are given as 11.99 meters long on

the west side and 11.96 meters, long on the east side 6.98 meters broad

on the south side and 6.87 meters broad on the north side. The difference

between the lengths of the opposite sides is 2.54 cm and 10.9 cm

respectively. The difference is very small compared to the other layouts of

rec^ngle brick works. This is an example of their knowledge of drawing a

right angled triangle on the ground. From the well laid out street plans and

from the arrangement of buildings, it is evident that drawing of similar

figures was known to the people of the Indus civilization.

The experiments conducted in Sindh, Baloochistan and Panjab

give knowledge about a civilization 4000 or 5000 years ago. More than

500 seals were obtained. These seals are made of sand and clay and ot

small in shape. Beautiful pictures were sculptured on this.

"" Ibid.,

151

6.10.2 Knowledge about the Circle

The most popular design on pots is composed of a series of

intersecting circles^^^. We can see a drawing of a copper vessel and a

circle containing a flower with four petals such that all circular parts of the

copper vessel are concentric. Some circular stone rings with concentric

rings marked on them were found at Mohenjo Daro of height 24.892 cms to

28.448 cms.^^^

From these we can assume that they knew about the construction of

a circle, how to draw a perpendicular to a given line at a given point,

bisecting a given angle and so on.

The scales^®^ and measures used in Sindhu river ci\'ilization

(culture) are very accurate. They used burnt bricks for construction; and

their measures are 11x5.5x2.5 inches and 9.2x4.5x2.2 inches. Modern

science agrees that these measures are very suitable for the construction

of walls.

'"^ Mackey, B. (1948) '"Early Indus Civilisation", Luzac and Co Limited,London,2™' Edition

(Ref. K.P Kulkami, "Geometry as known to the people of Indus civilization", IJHS Vol-13,1978)

'*' K.P Kulkami, ""Geometry as known to the people of Indus civilization'", IJHS Vol-13,1978

"'' Dr. P.V. Ouseph, "Vaslusastram oru sumagrapadanam" Published by D.C Books, Kottayam.