21 lilavati ii (mss)

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NPTEL COURSE ON MATHEMATICS IN I NDIA : F ROM V EDIC PERIOD TO MODERN TIMES Lecture 21 ıl¯avat¯ ı of Bh¯ askar¯ ac¯ arya 2 M. S. Sriram University of Madras, Chennai.

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21 Lilavati II (MSS)

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Page 1: 21 Lilavati II (MSS)

NPTEL COURSE ON

MATHEMATICS IN INDIA:FROM VEDIC PERIOD TO MODERN TIMES

Lecture 21

Lılavatı of Bhaskaracarya 2

M. S. SriramUniversity of Madras, Chennai.

Page 2: 21 Lilavati II (MSS)

Outline

◮ Applications of right triangles, continued.◮ Sucı probems◮ Construction of a quadrilateral: Discussion on earlier

confusions◮ To find the second diagonal, given the four sides and a

diagonal of a quadrilateral◮ Cyclic quadrilaterals◮ Value of π, area of a circle, surface area of a sphere, volume

of a sphere

Page 3: 21 Lilavati II (MSS)

A rational right triangleVerse 140:

As in Brahmasphut.asiddhanta: Side = a, upright =12

(

a2

n − n)

, hypotenuse

=12

(

a2

n + n)

. It is obviously true. But how to get this?

Gan. esa Daivajna explains in his Buddhivilasinı commentary.

Let side = a. Let hypotenuse - upright = n. Let upright = x .

∴ a2 + x2 = (x + n)2 = x2 + n2 + 2xn

∴ x =a2 − n2

2n=

12

(

a2

n− n

)

.

Hypotenuse = x + n =12

(

a2

n+ n

)

.

Or given the hypotenuse a, and an assumed number n. Then by a similar rule

in Verse 142, the upright is2an

n2 + 1, and the side is found to be

n × upright - Hypotenuse =2an2

n2 + 1− a =

a(n2 − 1)(n2 + 1)

.

Page 4: 21 Lilavati II (MSS)

Bamboo problem

Suppose a bamboo of height a , standing vertically, is broken atheight x , and the tip falls to the ground at a distance b from theroot of the bamboo. A right triangle is now formed with the sideb, upright x and hypotenuse h = a − x . Verse 147 states that theupright , x = 1

2(a − b2

a ) and the hypotenuse , h = 12(a + b2

a ) :vMa;Za;a;g{a;mUa;l+.a;nta:=+BUa;Æa;ma;va;ga;eRa vMa;Za;ea:;dÄâx ;ta;~tea;na :pxa;Ta;gyua;ta;ea;na;Ea ÁvMa;Za;ea ta;d;DeRa Ba;va;taH kÒ +:mea;Na vMa;Za;~ya Ka;Nqe (rua;�a;ta;k+:ea;�a;f:�+.pea Á Á

Page 5: 21 Lilavati II (MSS)

Bamboo problem

“The square of the ground intercepted between the root andthe tip is divided by the (length of the ) bamboo, and thequotient severally added to , and subtracted from , thebamboo: the halves (of the sum and difference) will be thetwo portions of it representing hypotenuse and upright.”

Hereh + x = a.

Alsoh2 = x2 + b2.

∴ h2 − x2 = b2.

∴ h − x =h2 − x2

h + x=

b2

a.

So, from Sankraman. a, upright x = 12

(

a − b2

a

)

, Hypotenuse, h

= 12

(

a + b2

a

)

. This is essentially Ganes.a’s explanation of the result.

Page 6: 21 Lilavati II (MSS)

Example

Example in Verse 148.ya;a;d .sa;ma;Bua;a;va :vea;NuaH d;nta;pa;a; a;Na;pra;ma;a;Na;eaga;Na;k! :pa;va;na;vea;ga;a;de ;k+:de ;Zea .sa Ba;çÉîå+;aH ÁBua;a;va nxa;pa;Æa;ma;ta;h;~tea;Svea;va l+.çÉîå+;a;~ta;d;g{Mak+:Ta;ya k+:�a;ta;Sua mUa;l+.a;de ;Sa Ba;çÉîå+;aH k+.=e +Sua Á Á 156 Á Á“If a bamboo, measuring 32 cubits and standing uponlevel ground, be broken in one place by the force of thewind and the tip of it meets the ground at 16 cubits; saymathematician at how many cubits from the root it isbroken.”

Page 7: 21 Lilavati II (MSS)

Bamboo problem: Solution

b=16

x

a=32

a-x

2012

20

x =12

(

a − b2

a

)

∴ x =12

(

32 − 25632

)

=12(32 − 8) = 12

h = a − x = 32 − 12 = 20

Page 8: 21 Lilavati II (MSS)

Snake-Peacock problem

Snake -Peacock Problem in Verse 150 :A;�/////�a;~ta .~ta;}Ba;ta;le ;a;ba;lM ta;du ;pa;�a:= kÒ +:a;q+.a; a;Za;Ka;Nq� +.a ;�//////�a;~Ta;taH.~ta;}Ba;ea na;nd;k+.=+ea;�/////�a;. C" +.ta;�///�a;~:�a;gua; a;Na;ta;~ta;}Ba;pra;ma;a;Na;a;nta:=e Ádx ;�õÅ ;a;a;hM ;a;ba;l+.ma;a;v.ra:$a;nta;ma;pa;ta;t,a ;�a;ta;yRa;k, .sa ta;~ya;ea;pa;�a:=;Æa;[a;prMa b.rUa;a;h ta;ya;ea;a;bRa;l+.a;t,a k+:�a;ta;Æa;ma;tea .sa;a;}yea;na ga;tya;ea;yRua;�a;taH Á Á“A snake’s hole is at the foot of a pillar, 9 cubits high,and a peacock is perched on its summit. Seeing asnake at the distance of thrice the pillar gliding towardshis hole, he pounces obliquely upon him. Say quickly athow many cubits from the snakes’s hole they meet,both proceeding an equal distance.”

Page 9: 21 Lilavati II (MSS)

Snake-Peacock problem: Solution

9

B

A

x

27-x

D C

27-x

Snake - Peacock problem

AB = 9 cubits, AC = 3 × 9 = 27 cubits. Let AD = x ,CD = 27 − x = BD. Now

BD2 − AD2 = AB2

∴ (27 − x)2 − x2 = 92

∴ (27 − x + x)(27 − x − x) = 81

∴ 27 · (27 − 2x) = 81

∴ 27 − 2x =8127

= 3

∴ 2x = 27 − 3 = 24

∴ x = 12

Snake, Peacock meet at distance x = 12, from the hole.

Page 10: 21 Lilavati II (MSS)

Lotus problemLotus probem in Verse 154 :.sa;Kea :pa;�íå Å+a;ta;n}å.a:êêÁ*.a;na;~Ta;a;na;ma;DyMa Bua:$aH k+:ea;�a;f;k+:Na;Ra;nta:=M :pa;�íåÅ+a;dx ;Zya;m,a Ána;l:H k+:ea;�a;f:=e +ta;�///////�a;n}å.a;tMa .~ya;a;dùÅ;a;ta;ea.Y;}Ba;ea va;dE ;vMa .sa;ma;a;n�a;a;ya :pa;a;n�a;a;ya;ma;a;na;m,a Á Á154 Á Á

“Friend, the space between the lotus (as it stood) and the spotwhere it submerged, is the side. The lotus as seen (above water) isthe difference between the hypotenuse and upright. The stalk is theupright, for the depth of water is measued by it. say, what the depthof the water is.”

B

A

D

a+d

ba

d

C

Lotus Problem

Page 11: 21 Lilavati II (MSS)

Lotus problem: Solution

AC : original position of the lotus = a + d . BC = a : portionabove water. AB = d : portion inside water = Depth of water.Due to wind the lotus is swept and assumes the positionAD = a + d . (So it is just submerged at D). Suppose BD = b isgiven. Find d (Depth of water).

Solution :

(a + d)2 = d2 + b2, or d =b2 − a2

2a.

Example in Verse 153.

a = 1 span = 12 cubit. b = 2 cubits. d =

22 − 1/4

2 · 12

=154.

Page 12: 21 Lilavati II (MSS)

Apes problemvxa;[a;a:;dÄâ ;~ta;Za;ta;ea;. C" +.ya;a;. C+.ta;yua;gea va;a;pya;Ma k+:a;paH k+:ea.Y;pya;ga;a;t,ao+ññ*� +:a;ya;a;Ta :pa:=+ea dÒu +tMa (rua;�a;ta;pa;Ta;a;t,a :pra;ea:ññ*� +:a;ya ;a;k+:�a:úãÁ*.a;d, dÒu +ya;a;t,a Á.$a;a;tEa;vMa .sa;ma;ta;a ta;ya;ea;yRua;�a;ta:=+a;pa :pra;ea:ññ*� +:a;ya;ma;a;nMa ;a;k+:ya;d,;a;va;dõ ;n,a :vea;�//////�a;tsa :pa;�a:=+(ra;ma;ea.Y;�/////�a;~ta ga; a;Na;tea ;Æa;[a;prMa ta;d;a;. a;[va mea Á Á“From a tree, a hundred cubits high, an ape descendedand went to a pond, two hundred cubits distant; whileanother ape, vaulting to some height off the tree,proceeded with a velocity diagonally to the same spot.If the space traveled by them be equal, tell me quickly,learned man, the height of the leap, if thou havediligently studied calculation.”

Solution given in (earlier) Verse 154.

Page 13: 21 Lilavati II (MSS)

Solution of Apes problem;a;dõ ;�a;na.Èåî ÁÁ*+;ta;a;l+.ea;�/////�a;. C" +.�a;ta;sMa;yua;tMa ya;t,a .sa:=+ea.Y;nta:=M .tea;na ;a;va;Ba;a;Æa:$a;ta;a;ya;aH Áta;a;l+.ea;�/////�a;. C" +.tea;~ta;a;l+.sa:=+ea;nta:=:Èåî ÁÁ*+�a;a o+ññ*� +:a;ya;ma;a;nMa Ka;lu l+.Bya;tea ta;t,a Á Á“The height of the tree multiplied by its distance fromthe pond, is divided by twice the height of the tree,added to the space between the tree and the pond: thequotient will the measure of the leap.”

B

D

a

A

x

Cb

Ape Problem

Page 14: 21 Lilavati II (MSS)

Solution of Apes problemThis is essentially the same as the ”Two ascetics problem” inBrahmasphut.asiddhanta.

BD = a = Height of the tree; BC = b = Distance between treeand pond. Let the leap be x = AD

Given that

BD + BC = a + b = AD + AC = x + AC = x +√

(a + x)2 + b2.

∴ x +√

(a + x)2 + b2 = a + b

∴ (a + x)2 + b2 = (a + b − x)2 = x2 + (a + b)2 − 2x(a + b)

∴ a2 + b2 + 2ax + x2 = x2 + (a + b)2 − 2x(a + b) = x2 + a2

+b2 + 2ab − 2x(a + b).

∴ x =ab

2a + b

Page 15: 21 Lilavati II (MSS)

Two bamboo pillars: Segments and Perpendicular

Verse 159.

A

B

a

Cx1

D

bE

x2F

p

d

Two bamboos and perpendicular at the junction

Two bamboos of heights AB = a,CD = b. Distance betweenthem = AC = d . Top of each bamboo joined by string to thebottom of the other. p = EF is the perpendicular from theintersection of the strings at E to the base AC at F . Find p andthe segments AF = x1,FC = x2.

Page 16: 21 Lilavati II (MSS)

Segments and Perpendicular

From similar triangles,EFAB

=pa=

FCAC

=x2

x1 + x2=

x2

d

Similarly,EFCD

=pb=

AFAC

=x1

x1 + x2=

x1

d

∴ p(

1a+

1b

)

=x1x2

d=

dd

= 1

.

∴ p =ab

a + b

Also x1 =pdb

=ad

(a + b). Similarly, x2 =

bd(a + b)

.

Page 17: 21 Lilavati II (MSS)

Example

Example Verse 160.

15

10

6

3 2

5

An example

Given Bamboos; 15, 10. Distance: 5. Then p =15 × 1015 + 10

= 6.

x1 = 3, x2 = 2.

Page 18: 21 Lilavati II (MSS)

Triangles and quadrilaterals

In Verse 161. It is stated: In any rectilinear figure, one sidecannot be greater than the sum of the other sides.

Verse 163,164: Given the segments and the perpendicular(altitude) in terms of the two sides and the third side (base). Alsoit is stated that:

Area =12× Base × Altitude

Verse 166. Case of a triangle with an obtuse angle.d;Za;sa;�a;d;Za;pra;ma;Ea Bua:$a;Ea ;a:�a;Bua:$ea ya:�a na;va;pra;ma;a ma;h� ;a ÁA;ba;Dea va;d l+.}ba;kM ta;Ta;a ga; a;Na;tMa ga; a;Na;ta;k+:a;Zua ta:�a mea Á Á 166 Á Á“In a triangle, wherein the sides measure ten andseventeen , and the base nine, tell me promptly, expertmathematician, the segments, perpendicular, and area.”

Page 19: 21 Lilavati II (MSS)

Negative seqment

Here, the quotient b2−c2

2a is 21. This cannot be subtracted from

the base; wherefore the base is subtracted from it : (A;nea;naBUa:�+.na;a na .~ya;a;t,a A;sma;a;de ;va BUa:=+pa;n�a;a;ta;a ) . (One of ) the segmentis negative , that is to say, in the contrary direction.

(�+.Na;ga;ta;a.Y;ba;a;Da;a ;a;d;gvEa;pa:=� +a;tyea;nea;tya;TRaH ). The two segments arefound 15 and 6(negative) . Perpendicular is 8 and the area is 36.

The negative segment is shown by the dotted line.

Obtuse triangle

Page 20: 21 Lilavati II (MSS)

Construction of a quadrilateral and its area

Verse 167..sa;vRa;d;ea;yRua;�a;ta;d;lM .. a;tuaH+�//////�a;~Ta;tMa ba;a;hu ;Æa;Ba;a;vRa:=+a;h;tMa .. a ta:;dÄâ ;teaH ÁmUa;l+.ma;~å.Pu +.f;P+.lM :pra:$a;a;ya;tea, .~å.pa;�;mea;va;mua;a;d;tMa ;a:�a;ba;a;hu ;ke Á Á 167 Á Á“Half the sum of all the sides is set down in four placesand the sides are severally subtracted. The remaindersmultiplied together, the square root of the product is thearea, inexact for quadrilateral, but pronounced exact fortriangle.”

Bhaskara, Area of quadrilateral,

A =√

(s − a)(s − b)(s − c)(s − d), s =(a + b + c + d)

2

is “Inexact” in a quadrilateral, but exact for a triangle (by puttingone of the sides =0).

Page 21: 21 Lilavati II (MSS)

Construction of a quadrilateral

Verse 169-172:.. a;tua;BRua:$a;~ya;a.Y;�a;na;ya;ta;Ea ;a;h k+:Na;ERa k+:TMata;ta;ea.Y;�/////////�a;sma;n,a ;�a;na;ya;tMa :P+.lM .~ya;a;t,a Á:pra;sa;a;�a;Da;ta;Ea ta;. C" +.va;Na;Ea ya;d;a dõ ;Ea .~va;k+:�//////�a;�pa;ta;tva;a;a;d;ta:=;�a na .~taH Á Á 169 Á Á.tea;Svea;va ba;a;hu ;Sva;pa:=+Ea .. a k+:Na;Ra;va;nea;k+:Da;a [ea:�a;P+.lM ta;ta;(ãÉa Á Á 170 Á Ál+.}ba;ya;eaH k+:NRa;ya;ea;nERa;kM .sa;mua;a;�+Zya;a;pa:=+a;n,a k+:Ta;m,a Á:pxa;. C+.tya;�a;na;ya;ta;tvea.Y;a;pa ;�a;na;ya;tMa .. a;a;a;pa ta;tP+.l+.m,a Á Á 171 Á Á.sa :pxa;. C+.k;H ;a;pa;Za;a;. a;ea va;a ga;Na;k+:ea ;�a;na;ta:=+Ma ta;taH Áya;ea na :vea; a:�a .. a;tua;ba;Ra;h;Ea [ea:�a;~ya;a.Y;�a;na;ya;ta;Ma ;�//////�a;~Ta;�a;ta;m,a Á Á 172 Á Á“Since the diagonals of a quadrilateral are indeterminate howshould the area be in this case, determinate? The diagonalsfound as assumed by the ancients do not answer in anothercase. with the same sides, there are other diagonals; andthe area of the quadrilateral is accordingly manifold.

Page 22: 21 Lilavati II (MSS)

Construction of a quadrilateral

For in a quadrilateral, opposite angles being made toapproach, contract their diagonal as they advanceinwards. While the other angles receding outwardslengthen the diagonal. Therefore it is said that with thesame sides there are other diagonals.

How can person, neither specifying one of theperpendicular nor the either of the diagonals, ask therest? Or how can he demand a determinate area, whilethey are indefinite?

Such a questioner is a blundering devil(pisaca). Stillmore so is he, who answers the question. For heconsiders not the indefinite nature of the lines in aquadrilateral figure.”

Page 23: 21 Lilavati II (MSS)

Construction of a quadrilateral

A

d

D c C

b

Ba2

p

a1

a

A quadrilateral

AB,AD,CD,CB are the sides. To specify D, we should be givenp (or equivalently BAD) or DB = D1: one of the diagonals. Laterif d ,D1,a are known, p can be found. Similarly, if p,d are known,

a1,a2 are known, and D1 =√

p2 + a22.

Page 24: 21 Lilavati II (MSS)

Finding the second diagonal

Determination of the second diagonal given the four sides and one diagonal inVerses 181-182 :I+.�;ea.Y:�a k+:NRaH :pra;Ta;mMa :pra;k+:�pyaH �ya;~åò:ea tua k+:Na;eRa;Ba;ya;taH ;�//////�a;~Ta;tea yea Ák+:N a ta;ya;eaH [ma;a;Æa;ma;ta:=+Ea .. a ba;a;hU :pra;k+:�pya l+.}ba;a;va;ba;Dea :pra;sa;a;Dyea Á Á181 Á ÁA;a;ba;a;Da;ya;ea;=e +k+:k+:ku +:p~Ta;ya;eaH ya;t,a .~ya;a;d;nta:=M ta;tkx +:�a;ta;sMa;yua;ta;~ya Ál+.}bEa;k�+.a;va;gRa;~ya :pa;dM ;a;dõ ;t�a;a;yaH k+:Na;eRa Ba;vea;t,a .sa;vRa;. a;tua;BRua:$ea;Sua Á Á 182 Á Á

“In the figure, first a diagonal is assumed. In the two trianglessituated on each side of the diagonal, tis diagonal is madethe base of each; and the other sides are given; theperpendiculars and segments must be found. Then thesquare of the difference of two segments on the same sidebeing added to the square of the sum of the perpendiculars,the square root of the sum of those squares will be thesecond diagonal in all quadrilaterals.”

Page 25: 21 Lilavati II (MSS)

Determination of the second diagonal

A

D c

C

B

bd

D1

a

p2

H’

G

H

p1

Determination of the second diagonal

In the figure, BD is a diagonal. To find the other diagonal AC. p1, p2

are perpendicular to the first diagonal BD = D1 can be determined.DH = D1c and HB = D1b: segments associated with p1. DH ′ = D1d

and H ′B = D1a: segments associated with p2. Extend AH ′ to G suchthat H ′G = CH = p1. Clearly, CG is perpendicular to H ′G.

CG = H ′H = DH − DH ′ = D1c − D1d

. Then the second diagonal AC = D2 is given by

AC2 = AG2 + CG2 = (p1 + p2)2 + (D1c − D1d )

2

Page 26: 21 Lilavati II (MSS)

Example

Example: d = 68, a = 75, c = 51,b = 40,D1 = 77..

DH = D1c =D1 +

(c2 − b2)

D1

2=

77 +(512 − 402)

772

=77 +

91 × 1177

2=

77 + 132

= 45

DH ′ =D1 −

(a2 − d2)

D1

2=

77 − (752 − 682)

772

=77 − 143 × 7

772

=77 − 13

2= 32.

DH ′ = D1d = 32,D1c − D1d = 13.

.p1 =

b2 − D21b =

402 − 322 = 8√

52 − 42 = 24

Similarly, p2 = 60.

∴ AC2 = (p1 + p2)2 + (D1c − D1d)

2 = 842 + 132 = 7225 = 852

∴ AC = 85 (852 = (84+1)2 = 842+2·84+1 = 842+169 = 842+132)

Page 27: 21 Lilavati II (MSS)

Construction of a cyclic quadrilateral

Bhaskara gives the method to construct a cyclic quadrilateralbeginning with two right triangles in Verses 191-192 : Same asthe construction in Brahmasphut.asiddhanta.

c1a1

b1 b2

a2 →c2

b1a2

a1a2c1a2

b1b2

a1b2

c1b2c2a1

a2a1

b2a1 b2b1

a2b1c2b1

Triangles used to construct a cyclic quadrilateral

Page 28: 21 Lilavati II (MSS)

Construction of a cyclic quadrilateral

c1a2

a1a2

c2b1

a2b1

c1b2

b1b2

a1b2

c2a1

A cyclic quadrilateral constructed from 4 right triangles

Page 29: 21 Lilavati II (MSS)

Another cyclic quadrilateral with same sidesAnother cyclic quadrilateral can be constructed from the four right triangles, byinterchanging the sides c2b1 and c1b2:

D

A

b1a2

c1b2

b1b2

B

C

c2b1

a1b2

a2b1

a1a2

Another cyclic quadrilateral from the same 4 right triangles

Here one diagonal AC = a1b2 + a2b1.

(other diagonal)2 = (a1a2 + b1b2)2 + (b1a2 − a1b2)

2

= (a21 + b2

1)(a22 + b2

2)

= c21c2

2 .

∴ Other diagonalBD = c1c2.

We may also note that in this construction, the diagonal BD is the diammeterof the circle.

Page 30: 21 Lilavati II (MSS)

Circle

After triangles and quadrilaterals, Bhaskara discusses Circles .He has many new things to say regarding circles.

Verse 201.v.ya;a;sea Ba;na;nd;a;�a;çÉîå+;a;h;tea ;a;va;Ba;�eKa;ba;a;Na;sUa;yERaH :pa;�a:=+�a;DaH .sua;sUa;[maH Ádõ ;a;a;vMa;Za;�a;ta.éÈåîÅéé:ae ;a;va;h;tea.Y;Ta ZEa;lE :H.~TUa;l+.ea.Y;Ta;va;a .~ya;a;d, v.ya;va;h;a:=+ya;ea;gyaH Á Á 207 Á Á“When the diameter of a circle is multiplied by 3927 anddivided by 1250, the quotient is nearly thecircumference: or multiplied by 22 and divided by 7, it isthe gross circumference adapted to the practice”.

Page 31: 21 Lilavati II (MSS)

Ratio of circumference and diameter

π =Circumference

Diameter≈︸︷︷︸

‘Near’

39271250

= 3.1416

Rough value: π =227.

.In his Buddhivilasini, Gan. esa Daivajna explains the value 3927

1250 forπ.

Page 32: 21 Lilavati II (MSS)

Inscribed hexagon

h

r2r

a12

Side of an inscribed hexagon = r . Circumference ≈ 6r .

Page 33: 21 Lilavati II (MSS)

Inscribed dodecagonSide of an inscribed dodecagon (12 sides), a12 is obtained as follows:

(r − h)2 +( r

2

)2= r2

.

∴ r2 + h2 − 2rh +r2

4= r2

.

∴ h2 − 2rh +r2

4= 0

∴ h = r − 12

4r2 − r2 = r −√

32

r .

a212 = h2 +

r2

4.

∴ a212 = r2

{

(

1 −√

32

)2

+14

}

= r2{2 −√

3}.

∴ a12 = r√

2 −√

3 = d

12−

√3

4≈ d

√669.87100

.

Gan. esa gives a12 ≈ d√

673100 .

∴ Circumference ≈ 12a12 ≈(

12√

673100

)

× d .

Page 34: 21 Lilavati II (MSS)

Better value of π

In this manner, he says if we have a polygon of 384 sides.(384 = 12 × 32 = 12 × 25).

Circumference ≈(√

98683100

)

× d .

∴ π ≈√

98683100

≈ 39271250

.

(check:√

98683 × 1250 ≈ 392673 ≈ 392700).

Page 35: 21 Lilavati II (MSS)

Area of a circle, Surface area and Volume of a Sphere

Area of a circle, Surface area and Volune of a Sphere in Verse203 :vxa:�a;[ea:�ea :pa;�a:=+�a;Da;gua; a;Na;ta;v.ya;a;sa;pa;a;dH :P+.lM ya;t,a[ua;NNMa :vea;dE :�+:pa;�a:= :pa;�a:=+taH k+:ndu ;k+:~yea;va .$a;a;l+.m,a Ága;ea;l+.~yEa;vMa ta;d;a;pa .. a :P+.lM :pxa;�;$Ma v.ya;a;sa;�a;na.ÈåîÁÁ*+M:Sa;a:ñÂÅÅå*.+:BRa;�M Ba;va;�a;ta ;�a;na;ya;tMa ga;ea;l+.ga;BeRa ;Ga;na;a;K.ya;m,a Á Á 203 Á Á

“In a circle, a quarter of the diameter multiplied by thecircumference is the area. That multiplied by four is thenet all around the ball.This content of the surface of thesphere, multiplied by the diameter and divided by six, isthe precise solid, termed cubic, content within thesphere. ”

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Area etc.

Area of a circle =Circumference × Diameter

4=

πd2

4.

Surface area of a sphere = Circumference × Diameter= πd2 = 4πr2.

Volume of a sphere = Circumference × (Diameter)2

6 =43πr3.

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Gan. esa’s derivation for the area of a circle

→ ...

...

Combine

... ... r =Diameter

2

Rectangle with sides:Circumference

2and

diameter2

.

∴ Area =14× Circumference × Diameter.

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Surface area of a sphereSurface area of a sphere. Siddhantas. iroman. i - Goladhyaya - Vasana.

R sinθ∆θ

θ

Divide hemisphere into 24 strips.

Area of strip = 2πR sin θR∆θ.

R∆θ =π · R2 · 24

θi = i × π

2 × 24R sin θi → Given in table.

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Area of hemisphere

∴ Area of hemisphere = 2π(∑

R sin θi

)

(R∆θ).

Bhaskara carries out the sum explicitly using the 24 Rsine valuesfrom the table and reports the measure of the hemisphere to be2πR2. (Actually, half the product of the circumference and thediameter). Yuktibhas. a proves that the area of the hemisphere is2πR2 by integration: essentially

∫π

20

sin θdθ = cos θ|π

20

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Volume of a sphere

→∆A

Volume of the cone bit (Fig.68) = 13 Area × height = 1

3∆A × R.Then Volume of sphere = 1

3× Area of sphere ×R = 13 × 4πR3.

Relation among chord, sara and diameter, as inBrahmasphut.asiddhanta and Gan. itasarasangraha.

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References

1.H.T.Colebrooke, Algebra with Arithmetic and Mensuration fromthe Sanskrit of Brahmagupta and Bhaskara, London 1817; Rep.Sharada Publishing House, New Delhi, 2006.

2. Bhaskaracarya’s Lılavatı with Colebrooke’s translation andnotes by H.C.Banerji, Second Edition,The Book Company,Calcutta, 1927; Rep. Asian Educational Series, New Delhi, 1993.

3.Lılavatı of Bhasakaracarya II: Ed. with Bhaskara’s Vasana andBuddhivilasanı of Gan. esa Daivajna by V.G.Apte, 2 Vols., Pune,1937.

4.Lılavatı of Bhasakaracarya , A Treatise of Mathematics of VedicTradition, Tr. by K.S.Patwardhan, S.M.Naimpally and S.L.Singh,Motilal Banarsidass, Delhi, 2001.

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Thanks!

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