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Presentation by Aruna Shobhane Dinanath Jr. College, Nagpur

Bramahagupta

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Born in 598 AD in Bhinmal ( Rajasthan )

In the reign of Vaghra mukha, a great king of capa dynesty, when 500 years of Saka era had elapsed , Bramhagupta son of Jishnu at the age of 30 , composed Bramhasphutsiddhanta for the pleasure of good mathematicians and astronomers.

Ujjain School of Hindu MathematicsAuthority on Astronomy and AstrologyHead of the astronomical observatory at Ujjain

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Texts on mathematics and astronomy: Brahmasphutasiddhanta in 628 A. D.

(The Opening of the Universe)

Khandakhadyaka in 665 A. D. (literally meaning sweet toast )

Uttar Khandakhadyaka in 672 A. D (literally meaning more sweet toast ).

Works of Bramhagupta

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Brahmasphutasiddhanta The Opening of the Universe Contains 24 chapters and 1080 verese. First 10 - Astronomy

Solar and lunar eclipses planetary conjunctions positions of the planets Motion of celestial bodies and their speed Prediction of full moon day and New moon day

Chapter 11 – Dushana Criticism of different results by previous mathematicians.

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Chapter 12 –Ganita Study of Arithmetic, Algebra, Progressions ,

Permutations , Polynomials, study of Plane figures

Chapter 18 –Kuttaka or Kuttakadhyya Various forms of Quiz, Puzzels.

Chapter 19 to 24 – Solid Figures – solid Geometry Volume ,Surface Area – Cone Cylinder Other Properties.

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Kahandakhadyaka- 5 Chapters Effects of Movements of Celestial bodies on

Human life Results useful in everyday life Favorable timings of marriages , Birth Popular in Short time.

UttarKahandakhadyaka – 5 Chapters Different methods of proofs – some questions

posed in Kahandakhadyak

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Telegraphic rule for Sine function

4/)4,3,2,1,0()90,60,45,30,0sin( 0 BSS XXI -19

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The square root of the product of four factors formed by the semi-perimeter which is diminished by each side is the exact area of cyclic quadrilateral .

Area Theorem

))()()(( sSrSqSpSArea Where p, q, r, s are the sides of the cyclic quadrilateral.And S, the semi perimeter, given by

2

srqps

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))........(1()(4

)(4

12

1

2

1

180

2

1

2

1

)()(

)(

222

222

0

IACosrspqk

ASinrspqk

rsSinApqSinAk

SinCSinA

DCBDAB

rsSinCpqSinA

BDCAADBA

ABCDCyclicQuadAk

But since ABCD is a Cyclic Quadrilateral

Trigonometric Proof

April 15, 2023 10))()()((

))()()((

2

)(2

)(

2

)(

2

)(

2

)(

))()()((16

)(4

1)(4

)(4

1)(

)(2

)(

22

2

2

2

2222222

2222222

2222

2222

sSrSqSpSk

sSrSqSpSk

psrqS

srqprsqpqsrppsrqk

srqprsqpqsrppsrqk

srqprspqk

srqpACosrspq

CosArspqsrqp

CosAACosCosC

but

rsCosCsrpqCosAqp

By law of cosines

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A

B

C

D

))()(( rSqSpSSk

Heron’s Formula for Triangle with sides p, q, r

Bramhagupta’s Area Theorem

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The sums of the products of the sides about the diagonals bee both divided by each other. Multiply [the quotients obtained] by the sum of the products of the opposite sides the square roots (of the result )are diagonals.

In a cyclic quadrilateral ABCD and sides a = AB, b = BC, c = CD, and d = DA, the lengths of the diagonals p = AC and q = BD,then

Bramhaguptas expression for diagonals

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Brahmaguptan quadrilateral

Formation of cyclic quadrilateral with integer sides, integer diagonals, and integer area.

The kotis and the bhujas of two Jatyas multiplied by each others hypotenuse are the four sides in a Visama Quadrilateral.

If e, f, g and p, q, r are the sides (integral or rational ) of Jatyas i. e. Right Angled Triangles with g and r being hypotenuse , then e.r, f.r, g.p g.q are the required sides of Brahmaguptan quadrilateral.

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Construction of Brahmaguptan quadrilateral

(e, f, g ) x p ….(Triangle with sides p.e , p.f, p.g)

(e, f, g) x q …. (Triangle with sides e.q, f.q, g.q)

(p, q, r) x e …. (Triangle with sides p.e, q.e, r.e)

(p, q, r ) x f …. (Triangle with sides p.f, q.f, r.f)

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Brahmaguptan quadrilateral

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This is a note worthy contribution of Brahmagupta to pure mathematics

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All Brahmagupta quadrilaterals with sides a, b, c, d, diagonals e, f, area K, and circumradius R can be obtained by the following expressions involving rational parameters t, u, and v:

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Brahmagupta dedicated a substantial portion of his work to geometry and trigonometry. He established √10 (3.162277) as a good practical approximation for π (3.141593)

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Brahmagupta's theorem

Brahmagupta's theorem states that BM = MC.

If a cyclic quadrilateral is orthodiagonal then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.

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We need to prove that AF = FD =FM

To prove that AF = FM,

first note that the angles FAM and CBM are equal, because they are inscribed angles that intercept the same arc of the circle. Furthermore, the angles CBM and CME are both complementary to angle BCM (i.e., they add up to 90°), and are therefore equal. Finally, the angles CME and FMA are the same. Hence, AFM is an isosceles triangle, and thus the sides AF and FM are equal.

The proof that FD = FM goes similarly: the angles FDM, BCM, BME and DMF are all equal, so DFM is an isosceles triangle, so FD = FM. It follows that AF = FD,

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References:1. THE HISTORY OF MATHEMATICS AND MATHEMATICIANS OF INDIA-

BY ER. VENUGOPAL D. HERROR.Published by VIDYABHARATI KARNATAKA.

2. ANCIENT INDIAN MATHEMATICIANS –Editor Prof K. V. Krishna MurtyPublished BY Institute of Scientific Research on Vedas Hyderabad (A. P)

3. THE TEACHING OF MATHEMATICS – By Kulbir Singh Sidhu.Published by Sterling publishers Private Limited. Jalander.

4. Internet Websites on Indian Geometry.

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THANKS !!!