A Study on the Properties of Rational Triangles
|
|
- Bernadette Gardner
- 5 years ago
- Views:
Transcription
1 Interntionl Journl of Mthemtis Reserh. ISSN Volume 6, Numer (04), pp. 8-9 Interntionl Reserh Pulition House Study on the Properties of Rtionl Tringles M. Q. lm, M.R. Hssn nd 3 M.. Hssn eprtment of Mthemtis, Mhil ollege, rhiy, 830, Indi ssoite professor in Mthemtis, S. M. ollege, hglpur-800, Indi 3 Mehnil Engineer, H. M. S. I. T. Tumkur, Krntk, 5704, Indi E-mil: hssnsm@mil.om strt In this mnusript n ttempt hs een mde to ollorte geometry nd rtionl numer system. In support of this some properties of rtionl tringles hve een disussed nd some theorems hve een estlished following the geometril onept of rhme Gupt ( 598 ), het nd Viet et. Keywords: Opposite Prity, Primitive tringle, Rtionl tringle, Juxtposition nd Olique tringle.. Introdution Let us define some key terms to mke the mnusript more understndle. The definitions re s follows: Rtionl tringle: - tringle with rtionl sides nd rtionl re is lled rtionl tringle. Juxtposition: - The proess of omintion of two right tringles with ommon leg is lled Juxtposition. The omined tringle is lled olique tringle. Exmple: - Let & e two right tringles with ommon leg nd Right <ed t, then the omined tringle is the olique tringle.
2 8 M. Q. lm, M.R. Hssn nd M.. Hssn Olique Tringle: fter juxtposition of two right tringles, the omined tringle is lled olique tringle. Opposite prity: Two reltively prime numers re sid to e of opposite prity if one of them is odd nd the other is even. Ex.: & 3 re of opposite prity, 3 & 4 re of opposite prity. Primitive tringle: right tringle with reltively prime integrl sides is lled primitive tringle. Ex.: Pythgoren triplet (3, 4, 5) represents primitive right tringle s (3, 4, 5) =. Setion: - Erly works out rtionl tringles rhme Gupt (.598-) hose ny three positive rtionl numers,, nd estlished tht,, nd re the side of n olique tringle whose ltitude nd re re rtionl, where the olique tringle is formed y juxtposition of two right tringles & with the ommon height =. + = Olique tringle Juxtposition
3 Study on the Properties of Rtionl Tringles 83. G. het [] introdued two methods to solve rtionl tringles s follows: Method : In the first method het onsidered right tringle with the sides 0, 8, 6 nd = N suh tht N 8, where is rtionl. From the properties of tringles, < is ute or otuse ording s or i. e. 6N 8 or 6N 8 se : When N 3 / 3 is ute if N 3 / 3 & is otuseif N 3 / 3. ute N N Let us hoose positive rtionl numer x suh tht 8 8 xn 6x then, N 3 / 3 x x x 0 x ( s x 0) () Ex.: Let x = 5/, then from () N 60 / Thus the sides of the tringles with ute nd ltitude 8 re 0, 6+N, 8 N i.e., 0, 86/ nd 3/ re the rtionl sides. se : When N 3 / 3
4 84 M. Q. lm, M.R. Hssn nd M.. Hssn N Similr to se () ` N 8 8 xn 6x N 3 / 3 x x x 0 x ( s x 0) Ex: Let x 3 /, then from () N 8 (8 3 N / ) Thus the sides of the tringle with otuse nd ltitude = 8 re ,6, 8 i. e. 0,6 / 5,04 / re rtionl sides. Method : In the seond method het used the method of juxtposition of two rtionl right tringles with ommon height. He onsidered = nd other two numers x nd y suh tht x ny squre nd y ny squre Thus x & y 6, 6 0 Using (m -n ) + (mn) = (m +n ) Hene one n find the rtionl tringle y juxtposing s 37, 0, = 5 with ltitude. Similrly y juxtposing the tringles & 9 5 or & 5 3, We get the rtionl tringles (37, 5, 35+9) or (37, 3, 35+5).
5 Study on the Properties of Rtionl Tringles 85 F. Viet s Method: F. Viet [3] otined rtionl tringle y juxtposing two right tringles s follows: He onsidered rtionl right tringle with right legs nd d nd hypotenuse z nd seond right tringle hving ltitude = (f + d), then the hypotenuse nd se of the seond tringle re & f d f d respetively. Now multiplying the sides of the first tringle y nd the sides of the seond tringle y d then the two right tringles eome of ommon ltitude d. Thus y Juxtposition of two resulting right tringles with ommon ttitude we otined the rtionl olique tringle with sides z., d[( f d ) ], d[( f d ) ]. For Ex, tking = 3, d = 4, z = 5, f = 6, so tht = 60nd ignoring the proportionl ftor 4, we get the rtionl tringle of sides 75, 09, 36 with ltitude 60. Setion (present works out rtionl tringles) In this setion we hve estlished some new theorems pplying the onept of juxtposition of two right tringles nd solved some olique tringles. Theorem () Sttement: In ny tringle with rtionl sides,, nd rtionl re. [( ps qr)( pr qs)] p q r s : : : :, pqrs pq rs where p, q, r, nd s re some rtionl numers nd every pir of sides re in the rtio u v of two numers of the form. uv Proof: p q pq r s rs p q pq r s rs onsidering two right tringles nd with sides
6 86 M. Q. lm, M.R. Hssn nd M.. Hssn p q p q r s r s, nd,, pq pq rs rs with ommon ltitude =. Now juxtposing these two tringles we get the olique tringle, whose sides re [( ps qr)( pr qs)] p q r s,,, pqrs pq rs where the upper nd lower sign is to e tken ording s the tringles nd do not or do overlp. Moreover we n esily verify tht r s x y : : for x ps qr, y pr qs, rs xy So tht x y ( p q )( r s ) Theorem : Sttement: There exit three tringles with integrl sides nd re hving equl perimeters nd res in the rtio of : :. Proof: Let the three required tringles F, F, F stnding on the sme se nd of ommon ltitude EF= t F t E Tking p, q, r, s F t F t F t F t p q r s p q r s Then E t, E t, E t, E t. p q r s
7 Study on the Properties of Rtionl Tringles 87 p q q r E E t, E E t p q q r r s nd E E t. r s From the sttement, ll the three tringles hve sme perimeters so F F F F F F p q s r q r () q p, q r () & r q s r (3) & p r s q (4) gin from the sttement, F F F [s height is ommon to ll the tringles] t p q t q r t r s p q q r r s rp p r p r q s s q (5) rp s p r p r q p r (6) s rp (s p r ) q q p, s r (7) From (5) nd (7) we hve r q q q s q r r q r (8)
8 88 M. Q. lm, M.R. Hssn nd M.. Hssn Eliminting p, q nd s from (), (7) nd (8), we get r r r r r r r 4 d d r de 0 (9) where, d, e Now hoosing integrl vlues of,, so tht the iqudrti in r gives rtionl root, then we n esily see tht p, q & s re rtionls. Thus we otined the required tringles. Exmple: For exmple tking =, = 7, = 5, we get the rtionl root r = 5/3. Thus 5 p, q, s F t, F t, F t, F t, Tking t = 40, we get F 54, F 55, F 476, F 4, 6, 9, 95 The perimeters of the tringle re sme, whih is equl to 09. Prolem () If the sides nd re of the tringle re integers, then the re is divisile y 6. Proof: If p, q, r, s ll re integers then in ny rtionl tringle, the sides re proportionl to the numers ( ps qr)( pr qs) p q r s,, pqrs pq rs Thus the sides of the rtionl tringles re tken s = (ps+qr)(pr-qs), =rs(p +q ), =pq(r +s ) Let t e the perimeter of the rtionl tringle, then t. ). The re of the tringle is given y t t t t pqrs ps qr pr qs (0) If p, q, r, s re ll odd integers or t lest one of them is even then 0 (mod If one of p, q, r, s is odd then 0 (mod 3). Now suppose tht none of p, q, r, s is divisile y 3, then
9 Study on the Properties of Rtionl Tringles 89 ( ps qr)( pr qs) rs( p q ) pq( r s ) rs pq mod (3) ( ps qr)( ps qr) 0(mod 3) euse the squre of ny integer whih is not divisile y 3, is ongruent to (mod 3). Hene it follows tht pqrs ( ps qr)( ps qr) 0 (mod 3). Then, 0mod 6. Theorem (3) Sttement: There exit tringles whose sides re onseutive integers. Proof: Let x, x, x + e the sides of rtionl tringles, then the re is given y where t t t t t 3x hlf of the perimeter of the tringle. 3 x x 4 4 Sine the re of the tringle is rtionl hene we must hve x 3y 4 () Here x nd y oth nnot e odd simultneously sine 4 3 (mod 8) is impossile. lso x nd y nnot e of opposite prity sine (mod 4), 4 3 (mod 4) eh of whih is impossile hene x nd y oth must e even. Let us write x = u & y = v then from () we get u 3v () Whih is Pell s eqution nd whose fundmentl solution is u =, v =. Hene ll solutions of Pell s eqution () re given y 3 3 r u v, r,,3,... Thus ( u, v) (,),(7, 4)... nd the required tringles re (3, 4, 5), (3, 4, 5) Prolem () The tringle with the ltitude nd sides 3, 4 5 is the only one tringle in whih the ltitude nd sides re onseutive integer. Proof: Let e the ltitude nd +, +, + 3 e the three sides of tringles.
10 90 M. Q. lm, M.R. Hssn nd M.. Hssn x 3 If the perpendiulr is drwn from to the side + 3 nd x is the se of one of the right tringle formed fter drwing perpendiulr, then (I.) x, 3 x For other perpendiulrs, we hve (II) x, x 3 (III) x 3, x Sutrting the result of (I.), we get (mod ( - 3)) whih is impossile Similrly (II) gives 5 0 (mod( )) whih is impossile only when =. Now sustituting the two equtions of se (III), we get x ( ) 4 x 6 (Negtive x is indmissile) Thus from the first Eqution, we get 3 3 Hene the theorem follows. Theorem (4) Sttement: If in ny primitive rtionl tringle, two sides re odd, the lst side > nd the differene etween the sum of two smller sides nd the lrger sides is not unity. Proof: Let,, e the integrl sides of tringle with (,, ) = nd let < <. (I.) Let us first prove two of,, re odd. The re of the tringle is given y t t t t, where t
11 Study on the Properties of Rtionl Tringles 9 6 ( ) ( ) (3) (mod 4) if, nd re even (mod 4) if, even nd odd (mod 4) if, even nd odd. Hene two of,, must e odd. (II) Now let us prove tht > We hve & s M GM In one of the ove two inequlity ( ) Hene / Similrly from the seond inequlity / Thus (dding two results) 4 5 If =, the lest vlue of = nd lest vlue of = 3 nd so whih is 4 impossile. 7 If =, the lest vlue of = 3 nd lest vlue of = 4 nd whih is 4 impossile. 9 If = 3, the lest vlue of = 4 nd 3 whih is impossile. 4 Hene, >. (III) Here to prove tht. If possile, suppose + =, then from (3) 6 6 I (4) The reltion (4) does not hold euse left side of (4) is even wheres right side is odd. Thus + = is impossile. Hene, +. onlusion: ll the propositions nd theorems re the tools for numer system nd for the theory of numers. With the help of these propositions nd theorems mny unsolved prolems of rtionl tringles n e solved.
12 9 M. Q. lm, M.R. Hssn nd M.. Hssn Referenes [] rhme-sphut-sidhnt, h., lger with rith nd Mensurtion, from the Snskrit of rhme, Gupt nd hsr, trnslted y H.T. olerooke, LONON, (87) p; 306 [] L.E. ikson: History of the theory of numers, helse Pulishing ompny, New York (95) [3] H. renport: Higher rithmeti, Hithinson, LONON (95)
Section 1.3 Triangles
Se 1.3 Tringles 21 Setion 1.3 Tringles LELING TRINGLE The line segments tht form tringle re lled the sides of the tringle. Eh pir of sides forms n ngle, lled n interior ngle, nd eh tringle hs three interior
More informationQUADRATIC EQUATION. Contents
QUADRATIC EQUATION Contents Topi Pge No. Theory 0-04 Exerise - 05-09 Exerise - 09-3 Exerise - 3 4-5 Exerise - 4 6 Answer Key 7-8 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,
More information1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the
More informationComparing the Pre-image and Image of a Dilation
hpter Summry Key Terms Postultes nd Theorems similr tringles (.1) inluded ngle (.2) inluded side (.2) geometri men (.) indiret mesurement (.6) ngle-ngle Similrity Theorem (.2) Side-Side-Side Similrity
More informationProbability. b a b. a b 32.
Proility If n event n hppen in '' wys nd fil in '' wys, nd eh of these wys is eqully likely, then proility or the hne, or its hppening is, nd tht of its filing is eg, If in lottery there re prizes nd lnks,
More informationIntermediate Math Circles Wednesday 17 October 2012 Geometry II: Side Lengths
Intermedite Mth Cirles Wednesdy 17 Otoer 01 Geometry II: Side Lengths Lst week we disussed vrious ngle properties. As we progressed through the evening, we proved mny results. This week, we will look t
More informationQUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP
QUADRATIC EQUATION EXERCISE - 0 CHECK YOUR GRASP. Sine sum of oeffiients 0. Hint : It's one root is nd other root is 8 nd 5 5. tn other root 9. q 4p 0 q p q p, q 4 p,,, 4 Hene 7 vlues of (p, q) 7 equtions
More informationGeometry of the Circle - Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272.
Geometry of the irle - hords nd ngles Geometry of the irle hord nd ngles urriulum Redy MMG: 272 www.mthletis.om hords nd ngles HRS N NGLES The irle is si shpe nd so it n e found lmost nywhere. This setion
More informationProportions: A ratio is the quotient of two numbers. For example, 2 3
Proportions: rtio is the quotient of two numers. For exmple, 2 3 is rtio of 2 n 3. n equlity of two rtios is proportion. For exmple, 3 7 = 15 is proportion. 45 If two sets of numers (none of whih is 0)
More informationNumbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point
GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply
More information12.4 Similarity in Right Triangles
Nme lss Dte 12.4 Similrit in Right Tringles Essentil Question: How does the ltitude to the hpotenuse of right tringle help ou use similr right tringles to solve prolems? Eplore Identifing Similrit in Right
More informationIntroduction to Olympiad Inequalities
Introdution to Olympid Inequlities Edutionl Studies Progrm HSSP Msshusetts Institute of Tehnology Snj Simonovikj Spring 207 Contents Wrm up nd Am-Gm inequlity 2. Elementry inequlities......................
More informationPAIR OF LINEAR EQUATIONS IN TWO VARIABLES
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES. Two liner equtions in the sme two vriles re lled pir of liner equtions in two vriles. The most generl form of pir of liner equtions is x + y + 0 x + y + 0 where,,,,,,
More informationLESSON 11: TRIANGLE FORMULAE
. THE SEMIPERIMETER OF TRINGLE LESSON : TRINGLE FORMULE In wht follows, will hve sides, nd, nd these will e opposite ngles, nd respetively. y the tringle inequlity, nd..() So ll of, & re positive rel numers.
More informationApril 8, 2017 Math 9. Geometry. Solving vector problems. Problem. Prove that if vectors and satisfy, then.
pril 8, 2017 Mth 9 Geometry Solving vetor prolems Prolem Prove tht if vetors nd stisfy, then Solution 1 onsider the vetor ddition prllelogrm shown in the Figure Sine its digonls hve equl length,, the prllelogrm
More informationTHE PYTHAGOREAN THEOREM
THE PYTHAGOREAN THEOREM The Pythgoren Theorem is one of the most well-known nd widely used theorems in mthemtis. We will first look t n informl investigtion of the Pythgoren Theorem, nd then pply this
More informationMath Lesson 4-5 The Law of Cosines
Mth-1060 Lesson 4-5 The Lw of osines Solve using Lw of Sines. 1 17 11 5 15 13 SS SSS Every pir of loops will hve unknowns. Every pir of loops will hve unknowns. We need nother eqution. h Drop nd ltitude
More information9.1 Day 1 Warm Up. Solve the equation = x x 2 = March 1, 2017 Geometry 9.1 The Pythagorean Theorem 1
9.1 Dy 1 Wrm Up Solve the eqution. 1. 4 2 + 3 2 = x 2 2. 13 2 + x 2 = 25 2 3. 3 2 2 + x 2 = 5 2 2 4. 5 2 + x 2 = 12 2 Mrh 1, 2017 Geometry 9.1 The Pythgoren Theorem 1 9.1 Dy 2 Wrm Up Use the Pythgoren
More informationm m m m m m m m P m P m ( ) m m P( ) ( ). The o-ordinte of the point P( ) dividing the line segment joining the two points ( ) nd ( ) eternll in the r
CO-ORDINTE GEOMETR II I Qudrnt Qudrnt (-.+) (++) X X - - - 0 - III IV Qudrnt - Qudrnt (--) - (+-) Region CRTESIN CO-ORDINTE SSTEM : Retngulr Co-ordinte Sstem : Let X' OX nd 'O e two mutull perpendiulr
More information= x x 2 = 25 2
9.1 Wrm Up Solve the eqution. 1. 4 2 + 3 2 = x 2 2. 13 2 + x 2 = 25 2 3. 3 2 2 + x 2 = 5 2 2 4. 5 2 + x 2 = 12 2 Mrh 7, 2016 Geometry 9.1 The Pythgoren Theorem 1 Geometry 9.1 The Pythgoren Theorem 9.1
More information2. There are an infinite number of possible triangles, all similar, with three given angles whose sum is 180.
SECTION 8-1 11 CHAPTER 8 Setion 8 1. There re n infinite numer of possile tringles, ll similr, with three given ngles whose sum is 180. 4. If two ngles α nd β of tringle re known, the third ngle n e found
More informationNon Right Angled Triangles
Non Right ngled Tringles Non Right ngled Tringles urriulum Redy www.mthletis.om Non Right ngled Tringles NON RIGHT NGLED TRINGLES sin i, os i nd tn i re lso useful in non-right ngled tringles. This unit
More informationFactorising FACTORISING.
Ftorising FACTORISING www.mthletis.om.u Ftorising FACTORISING Ftorising is the opposite of expning. It is the proess of putting expressions into rkets rther thn expning them out. In this setion you will
More informationProject 6: Minigoals Towards Simplifying and Rewriting Expressions
MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy
More informationChapter 8 Roots and Radicals
Chpter 8 Roots nd Rdils 7 ROOTS AND RADICALS 8 Figure 8. Grphene is n inredily strong nd flexile mteril mde from ron. It n lso ondut eletriity. Notie the hexgonl grid pttern. (redit: AlexnderAIUS / Wikimedi
More informationAP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals
AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into
More informationMaintaining Mathematical Proficiency
Nme Dte hpter 9 Mintining Mthemtil Profiieny Simplify the epression. 1. 500. 189 3. 5 4. 4 3 5. 11 5 6. 8 Solve the proportion. 9 3 14 7. = 8. = 9. 1 7 5 4 = 4 10. 0 6 = 11. 7 4 10 = 1. 5 9 15 3 = 5 +
More informationTrigonometry and Constructive Geometry
Trigonometry nd Construtive Geometry Trining prolems for M2 2018 term 1 Ted Szylowie tedszy@gmil.om 1 Leling geometril figures 1. Prtie writing Greek letters. αβγδɛθλµπψ 2. Lel the sides, ngles nd verties
More informationCHENG Chun Chor Litwin The Hong Kong Institute of Education
PE-hing Mi terntionl onferene IV: novtion of Mthemtis Tehing nd Lerning through Lesson Study- onnetion etween ssessment nd Sujet Mtter HENG hun hor Litwin The Hong Kong stitute of Edution Report on using
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationLesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem.
27 Lesson 2: The Pythgoren Theorem nd Similr Tringles A Brief Review of the Pythgoren Theorem. Rell tht n ngle whih mesures 90º is lled right ngle. If one of the ngles of tringle is right ngle, then we
More informationSimilarity and Congruence
Similrity nd ongruence urriculum Redy MMG: 201, 220, 221, 243, 244 www.mthletics.com SIMILRITY N ONGRUN If two shpes re congruent, it mens thy re equl in every wy ll their corresponding sides nd ngles
More informationMatrix Algebra. Matrix Addition, Scalar Multiplication and Transposition. Linear Algebra I 24
Mtrix lger Mtrix ddition, Sclr Multipliction nd rnsposition Mtrix lger Section.. Mtrix ddition, Sclr Multipliction nd rnsposition rectngulr rry of numers is clled mtrix ( the plurl is mtrices ) nd the
More information5. Every rational number have either terminating or repeating (recurring) decimal representation.
CHAPTER NUMBER SYSTEMS Points to Rememer :. Numer used for ounting,,,,... re known s Nturl numers.. All nturl numers together with zero i.e. 0,,,,,... re known s whole numers.. All nturl numers, zero nd
More information1.3 SCALARS AND VECTORS
Bridge Course Phy I PUC 24 1.3 SCLRS ND VECTORS Introdution: Physis is the study of nturl phenomen. The study of ny nturl phenomenon involves mesurements. For exmple, the distne etween the plnet erth nd
More informationLinear Algebra Introduction
Introdution Wht is Liner Alger out? Liner Alger is rnh of mthemtis whih emerged yers k nd ws one of the pioneer rnhes of mthemtis Though, initilly it strted with solving of the simple liner eqution x +
More informationp-adic Egyptian Fractions
p-adic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Set-up 3 4 p-greedy Algorithm 5 5 p-egyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction
More informationFarey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University
U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions
More informationOn Implicative and Strong Implicative Filters of Lattice Wajsberg Algebras
Glol Journl of Mthemtil Sienes: Theory nd Prtil. ISSN 974-32 Volume 9, Numer 3 (27), pp. 387-397 Interntionl Reserh Pulition House http://www.irphouse.om On Implitive nd Strong Implitive Filters of Lttie
More informationChapter Gauss Quadrature Rule of Integration
Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to
More information( ) { } [ ] { } [ ) { } ( ] { }
Mth 65 Prelulus Review Properties of Inequlities 1. > nd > >. > + > +. > nd > 0 > 4. > nd < 0 < Asolute Vlue, if 0, if < 0 Properties of Asolute Vlue > 0 1. < < > or
More informationBridging the gap: GCSE AS Level
Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions
More informationTrigonometry Revision Sheet Q5 of Paper 2
Trigonometry Revision Sheet Q of Pper The Bsis - The Trigonometry setion is ll out tringles. We will normlly e given some of the sides or ngles of tringle nd we use formule nd rules to find the others.
More information1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE
ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check
More informationMATH 573 FINAL EXAM. May 30, 2007
MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems.
More informationTorsion in Groups of Integral Triangles
Advnces in Pure Mthemtics, 01,, 116-10 http://dxdoiorg/1046/pm011015 Pulished Online Jnury 01 (http://wwwscirporg/journl/pm) Torsion in Groups of Integrl Tringles Will Murry Deprtment of Mthemtics nd Sttistics,
More informationSTRAND J: TRANSFORMATIONS, VECTORS and MATRICES
Mthemtics SKE: STRN J STRN J: TRNSFORMTIONS, VETORS nd MTRIES J3 Vectors Text ontents Section J3.1 Vectors nd Sclrs * J3. Vectors nd Geometry Mthemtics SKE: STRN J J3 Vectors J3.1 Vectors nd Sclrs Vectors
More informationMATHEMATICS AND STATISTICS 1.6
MTHMTIS N STTISTIS 1.6 pply geometri resoning in solving prolems ternlly ssessed 4 redits S 91031 inding unknown ngles When finding the size of unknown ngles in figure, t lest two steps of resoning will
More information#A42 INTEGERS 11 (2011) ON THE CONDITIONED BINOMIAL COEFFICIENTS
#A42 INTEGERS 11 (2011 ON THE CONDITIONED BINOMIAL COEFFICIENTS Liqun To Shool of Mthemtil Sienes, Luoyng Norml University, Luoyng, Chin lqto@lynuedun Reeived: 12/24/10, Revised: 5/11/11, Aepted: 5/16/11,
More information6.2 The Pythagorean Theorems
PythgorenTheorems20052006.nb 1 6.2 The Pythgoren Theorems One of the best known theorems in geometry (nd ll of mthemtics for tht mtter) is the Pythgoren Theorem. You hve probbly lredy worked with this
More informationTHREE DIMENSIONAL GEOMETRY
MD THREE DIMENSIONAL GEOMETRY CA CB C Coordintes of point in spe There re infinite numer of points in spe We wnt to identif eh nd ever point of spe with the help of three mutull perpendiulr oordintes es
More informationTOPIC: LINEAR ALGEBRA MATRICES
Interntionl Blurete LECTUE NOTES for FUTHE MATHEMATICS Dr TOPIC: LINEA ALGEBA MATICES. DEFINITION OF A MATIX MATIX OPEATIONS.. THE DETEMINANT deta THE INVESE A -... SYSTEMS OF LINEA EQUATIONS. 8. THE AUGMENTED
More informationMatrices SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics (c) 1. Definition of a Matrix
tries Definition of tri mtri is regulr rry of numers enlosed inside rkets SCHOOL OF ENGINEERING & UIL ENVIRONEN Emple he following re ll mtries: ), ) 9, themtis ), d) tries Definition of tri Size of tri
More informationHS Pre-Algebra Notes Unit 9: Roots, Real Numbers and The Pythagorean Theorem
HS Pre-Alger Notes Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Roots nd Cue Roots Syllus Ojetive 5.4: The student will find or pproximte squre roots of numers to 4. CCSS 8.EE.-: Evlute squre roots
More information6.5 Improper integrals
Eerpt from "Clulus" 3 AoPS In. www.rtofprolemsolving.om 6.5. IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y =
More informationIn right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.
Mth 3329-Uniform Geometries Leture 06 1. Review of trigonometry While we re looking t Eulid s Elements, I d like to look t some si trigonometry. Figure 1. The Pythgoren theorem sttes tht if = 90, then
More information50 AMC Lectures Problem Book 2 (36) Substitution Method
0 AMC Letures Prolem Book Sustitution Metho PROBLEMS Prolem : Solve for rel : 9 + 99 + 9 = Prolem : Solve for rel : 0 9 8 8 Prolem : Show tht if 8 Prolem : Show tht + + if rel numers,, n stisf + + = Prolem
More informationNON-DETERMINISTIC FSA
Tw o types of non-determinism: NON-DETERMINISTIC FS () Multiple strt-sttes; strt-sttes S Q. The lnguge L(M) ={x:x tkes M from some strt-stte to some finl-stte nd ll of x is proessed}. The string x = is
More informationComputing data with spreadsheets. Enter the following into the corresponding cells: A1: n B1: triangle C1: sqrt
Computing dt with spredsheets Exmple: Computing tringulr numers nd their squre roots. Rell, we showed 1 ` 2 ` `n npn ` 1q{2. Enter the following into the orresponding ells: A1: n B1: tringle C1: sqrt A2:
More informationAlgebra 2 Semester 1 Practice Final
Alger 2 Semester Prtie Finl Multiple Choie Ientify the hoie tht est ompletes the sttement or nswers the question. To whih set of numers oes the numer elong?. 2 5 integers rtionl numers irrtionl numers
More informationGM1 Consolidation Worksheet
Cmridge Essentils Mthemtis Core 8 GM1 Consolidtion Worksheet GM1 Consolidtion Worksheet 1 Clulte the size of eh ngle mrked y letter. Give resons for your nswers. or exmple, ngles on stright line dd up
More information03. Early Greeks & Aristotle
03. Erly Greeks & Aristotle I. Erly Greeks Topis I. Erly Greeks II. The Method of Exhustion III. Aristotle. Anximnder (. 60 B.C.) to peiron - the unlimited, unounded - fundmentl sustne of relity - underlying
More informationThe Legacy of Pythagoras Theorem
Prol Volue 39, Issue 1(2003) The Legy of Pythgors Theore Peter G.rown 1 When sked wht thetil result they reeer fro High Shool, the verge person would proly reply with Pythgors Theore. In ny right tringle,
More informationPart 4. Integration (with Proofs)
Prt 4. Integrtion (with Proofs) 4.1 Definition Definition A prtition P of [, b] is finite set of points {x 0, x 1,..., x n } with = x 0 < x 1
More informationChapter 1: Fundamentals
Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,
More informationset is not closed under matrix [ multiplication, ] and does not form a group.
Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed
More informationCS 491G Combinatorial Optimization Lecture Notes
CS 491G Comintoril Optimiztion Leture Notes Dvi Owen July 30, August 1 1 Mthings Figure 1: two possile mthings in simple grph. Definition 1 Given grph G = V, E, mthing is olletion of eges M suh tht e i,
More informationTwo Triads of Congruent Circles from Reflections
Forum Geometriorum Volume 8 (2008) 7 12. FRUM GEM SSN 1534-1178 Two Trids of ongruent irles from Refletions Qung Tun ui strt. Given tringle, we onstrut two trids of ongruent irles through the verties,
More informationLesson-5 PROPERTIES AND SOLUTIONS OF TRIANGLES
Leon-5 PROPERTIES ND SOLUTIONS OF TRINGLES Reltion etween the ide nd trigonometri rtio of the ngle of tringle In ny tringle, the ide, oppoite to the ngle, i denoted y ; the ide nd, oppoite to the ngle
More informationProving the Pythagorean Theorem. Proving the Pythagorean Theorem and the Converse of the Pythagorean Theorem
.5 Proving the Pythgoren Theorem Proving the Pythgoren Theorem nd the Converse of the Pythgoren Theorem Lerning Gols In this lesson, you will: Prove the Pythgoren Theorem using similr tringles. Prove the
More informationPYTHAGORAS THEOREM,TRIGONOMETRY,BEARINGS AND THREE DIMENSIONAL PROBLEMS
PYTHGORS THEOREM,TRIGONOMETRY,ERINGS ND THREE DIMENSIONL PROLEMS 1.1 PYTHGORS THEOREM: 1. The Pythgors Theorem sttes tht the squre of the hypotenuse is equl to the sum of the squres of the other two sides
More informationPROPERTIES OF TRIANGLES
PROPERTIES OF TRINGLES. RELTION RETWEEN SIDES ND NGLES OF TRINGLE:. tringle onsists of three sides nd three ngles lled elements of the tringle. In ny tringle,,, denotes the ngles of the tringle t the verties.
More informationPYTHAGORAS THEOREM WHAT S IN CHAPTER 1? IN THIS CHAPTER YOU WILL:
PYTHAGORAS THEOREM 1 WHAT S IN CHAPTER 1? 1 01 Squres, squre roots nd surds 1 02 Pythgors theorem 1 03 Finding the hypotenuse 1 04 Finding shorter side 1 05 Mixed prolems 1 06 Testing for right-ngled tringles
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationm A 1 1 A ! and AC 6
REVIEW SET A Using sle of m represents units, sketh vetor to represent: NON-CALCULATOR n eroplne tking off t n ngle of 8 ± to runw with speed of 6 ms displement of m in north-esterl diretion. Simplif:
More informationR(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of
Higher Mthemtics Ojective Test Prctice ook The digrm shows sketch of prt of the grph of f ( ). The digrm shows sketch of the cuic f ( ). R(, 8) f ( ) f ( ) P(, ) Q(, ) S(, ) Wht re the domin nd rnge of
More informationMAT 403 NOTES 4. f + f =
MAT 403 NOTES 4 1. Fundmentl Theorem o Clulus We will proo more generl version o the FTC thn the textook. But just like the textook, we strt with the ollowing proposition. Let R[, ] e the set o Riemnn
More informationPythagoras theorem and surds
HPTER Mesurement nd Geometry Pythgors theorem nd surds In IE-EM Mthemtis Yer 8, you lernt out the remrkle reltionship etween the lengths of the sides of right-ngled tringle. This result is known s Pythgors
More informationSpecial Numbers, Factors and Multiples
Specil s, nd Student Book - Series H- + 3 + 5 = 9 = 3 Mthletics Instnt Workooks Copyright Student Book - Series H Contents Topics Topic - Odd, even, prime nd composite numers Topic - Divisiility tests
More informationSymmetrical Components 1
Symmetril Components. Introdution These notes should e red together with Setion. of your text. When performing stedy-stte nlysis of high voltge trnsmission systems, we mke use of the per-phse equivlent
More informationIndividual Group. Individual Events I1 If 4 a = 25 b 1 1. = 10, find the value of.
Answers: (000-0 HKMO Het Events) Creted y: Mr. Frnis Hung Lst udted: July 0 00-0 33 3 7 7 5 Individul 6 7 7 3.5 75 9 9 0 36 00-0 Grou 60 36 3 0 5 6 7 7 0 9 3 0 Individul Events I If = 5 = 0, find the vlue
More information1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE
ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check
More information2.1 ANGLES AND THEIR MEASURE. y I
.1 ANGLES AND THEIR MEASURE Given two interseting lines or line segments, the mount of rottion out the point of intersetion (the vertex) required to ring one into orrespondene with the other is lled the
More informationPolynomials. Polynomials. Curriculum Ready ACMNA:
Polynomils Polynomils Curriulum Redy ACMNA: 66 www.mthletis.om Polynomils POLYNOMIALS A polynomil is mthemtil expression with one vrile whose powers re neither negtive nor frtions. The power in eh expression
More informationTriangles The following examples explore aspects of triangles:
Tringles The following exmples explore spects of tringles: xmple 1: ltitude of right ngled tringle + xmple : tringle ltitude of the symmetricl ltitude of n isosceles x x - 4 +x xmple 3: ltitude of the
More information15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions )
- TRIGONOMETRY Pge P ( ) In tringle PQR, R =. If tn b c = 0, 0, then Q nd tn re the roots of the eqution = b c c = b b = c b = c [ AIEEE 00 ] ( ) In tringle ABC, let C =. If r is the inrdius nd R is the
More informationUniversity of Sioux Falls. MAT204/205 Calculus I/II
University of Sioux Flls MAT204/205 Clulus I/II Conepts ddressed: Clulus Textook: Thoms Clulus, 11 th ed., Weir, Hss, Giordno 1. Use stndrd differentition nd integrtion tehniques. Differentition tehniques
More informationSimilar Right Triangles
Geometry V1.noteook Ferury 09, 2012 Similr Right Tringles Cn I identify similr tringles in right tringle with the ltitude? Cn I identify the proportions in right tringles? Cn I use the geometri mens theorems
More information378 Relations Solutions for Chapter 16. Section 16.1 Exercises. 3. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses on A.
378 Reltions 16.7 Solutions for Chpter 16 Section 16.1 Exercises 1. Let A = {0,1,2,3,4,5}. Write out the reltion R tht expresses > on A. Then illustrte it with digrm. 2 1 R = { (5,4),(5,3),(5,2),(5,1),(5,0),(4,3),(4,2),(4,1),
More informationProving the Pythagorean Theorem
Proving the Pythgoren Theorem W. Bline Dowler June 30, 2010 Astrt Most people re fmilir with the formul 2 + 2 = 2. However, in most ses, this ws presented in lssroom s n solute with no ttempt t proof or
More informationSomething found at a salad bar
Nme PP Something found t sld r 4.7 Notes RIGHT TRINGLE hs extly one right ngle. To solve right tringle, you n use things like SOH-H-TO nd the Pythgoren Theorem. n OLIQUE TRINGLE hs no right ngles. To solve
More informationVECTOR ALGEBRA. Syllabus :
MV VECTOR ALGEBRA Syllus : Vetors nd Slrs, ddition of vetors, omponent of vetor, omponents of vetor in two dimensions nd three dimensionl spe, slr nd vetor produts, slr nd vetor triple produt. Einstein
More informationTrigonometric Functions
Exercise. Degrees nd Rdins Chpter Trigonometric Functions EXERCISE. Degrees nd Rdins 4. Since 45 corresponds to rdin mesure of π/4 rd, we hve: 90 = 45 corresponds to π/4 or π/ rd. 5 = 7 45 corresponds
More informationSECTION A STUDENT MATERIAL. Part 1. What and Why.?
SECTION A STUDENT MATERIAL Prt Wht nd Wh.? Student Mteril Prt Prolem n > 0 n > 0 Is the onverse true? Prolem If n is even then n is even. If n is even then n is even. Wht nd Wh? Eploring Pure Mths Are
More information3 Angle Geometry. 3.1 Measuring Angles. 1. Using a protractor, measure the marked angles.
3 ngle Geometry MEP Prtie ook S3 3.1 Mesuring ngles 1. Using protrtor, mesure the mrked ngles. () () (d) (e) (f) 2. Drw ngles with the following sizes. () 22 () 75 120 (d) 90 (e) 153 (f) 45 (g) 180 (h)
More informationH (2a, a) (u 2a) 2 (E) Show that u v 4a. Explain why this implies that u v 4a, with equality if and only u a if u v 2a.
Chpter Review 89 IGURE ol hord GH of the prol 4. G u v H (, ) (A) Use the distne formul to show tht u. (B) Show tht G nd H lie on the line m, where m ( )/( ). (C) Solve m for nd sustitute in 4, otining
More informationMathematical Proofs Table of Contents
Mthemtil Proofs Tle of Contents Proof Stnr Pge(s) Are of Trpezoi 7MG. Geometry 8.0 Are of Cirle 6MG., 9 6MG. 7MG. Geometry 8.0 Volume of Right Cirulr Cyliner 6MG. 4 7MG. Geometry 8.0 Volume of Sphere Geometry
More informationIf C = 60 and = P, find the value of P. c 2 = a 2 + b 2 2abcos 60 = a 2 + b 2 ab a 2 + b 2 = c 2 + ab. c a
Answers: (000-0 HKMO Finl Events) Creted : Mr. Frncis Hung Lst updted: 0 June 08 Individul Events I P I P I P I P 5 7 0 0 S S S S Group Events G G G G 80 00 0 c 8 c c c d d 6 d 5 d 85 Individul Event I.,
More informationarxiv: v1 [math.ca] 21 Aug 2018
rxiv:1808.07159v1 [mth.ca] 1 Aug 018 Clulus on Dul Rel Numbers Keqin Liu Deprtment of Mthemtis The University of British Columbi Vnouver, BC Cnd, V6T 1Z Augest, 018 Abstrt We present the bsi theory of
More informationDiscrete Structures Lecture 11
Introdution Good morning. In this setion we study funtions. A funtion is mpping from one set to nother set or, perhps, from one set to itself. We study the properties of funtions. A mpping my not e funtion.
More information