PYTHAGORAS THEOREM 8YEARS. A guide for teachers  Years 8 9. The Improving Mathematics Education in Schools (TIMES) Project


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1 The Improving Mthemtis Edution in Shools (TIMES) Projet PYTHGORS THEOREM guide for tehers  Yers 8 9 MESUREMENT ND GEOMETRY Module 15 June YERS 9
2 Pythgors theorem (Mesurement nd Geometry: Module 15) For tehers of Primry nd Seondry Mthemtis 510 over design, Lyout design nd Typesetting y lire Ho The Improving Mthemtis Edution in Shools (TIMES) Projet ws funded y the ustrlin Government Deprtment of Edution, Employment nd Workple Reltions. The views expressed here re those of the uthor nd do not neessrily represent the views of the ustrlin Government Deprtment of Edution, Employment nd Workple Reltions.. The University of Melourne on ehlf of the interntionl entre of Exellene for Edution in Mthemtis (IE EM), the edution division of the ustrlin Mthemtil Sienes Institute (MSI), 2010 (exept where otherwise indited). This work is liensed under the retive ommons ttriutionnonommerilnoderivs 3.0 Unported Liense.
3 The Improving Mthemtis Edution in Shools (TIMES) Projet PYTHGORS THEOREM guide for tehers  Yers 8 9 MESUREMENT ND GEOMETRY Module 15 June 2011 Peter rown Mihel Evns Dvid Hunt Jnine MIntosh ill Pender Jqui Rmgge YERS 8 9
4 {4} The Improving Mthemtis Edution in Shools (TIMES) Projet PYTHGORS THEOREM SSUMED KNOWLEDGE Fmilirity with mesurement of lengths, ngles nd re. si knowledge of ongruene nd similrity. Fmilirity with simple geometri proofs. Simple geometri onstrutions. Ftoristion of whole numers. Simple surd nottion. MOTIVTION Is there simple reltionship etween the length of the sides of tringle? prt from the ft tht the sum of ny two sides is greter thn the third, there is, in generl, no simple reltionship etween the three sides of tringle. mong the set of ll tringles, there is speil lss, known s rightngled tringles or right tringles tht ontin right ngle. The longest side in rightngled tringle is lled the hypotenuse. The word is onneted with Greek word mening to streth euse the nient Egyptins disovered tht if you tke piee of rope, mrk off 3 units, then 4 units nd then 5 units, this n e strethed to form tringle tht ontins right ngle. This ws very useful to the Egyptin uilders. This rises ll sorts of questions. Wht is so speil out the lengths 3, 4 nd 5? re there other sets of numers with this property? Is there simple reltionship etween the lengths of the sides in rightngled tringle? Given the lengths of the sides of tringle, n we tell whether or not the tringle is right ngled?
5 The Improving Mthemtis Edution in Shools (TIMES) Projet {5} Most dults rememer the mthemtil formul 2 = or perhps the squre on the hypotenuse is the sum of the squres on the other two sides. The first version uses n implied stndrd nottion, the seond version uses rhi lnguge ut oth re Pythgors theorem. This theorem enles us to nswer the questions rised in the previous prgrph. The disovery of Pythgors theorem led the Greeks to prove the existene of numers tht ould not e expressed s rtionl numers. For exmple, tking the two shorter sides of right tringle to e 1 nd 1, we re led to hypotenuse of length 2, whih is not rtionl numer. This used the Greeks no end of troule nd led eventully to the disovery of the rel numer system. This will e disussed riefly in this module ut will e developed further in lter module, The Rel Numers. Triples of integers suh s (3, 4, 5) nd (5, 12, 13) whih our s the side lengths of right ngled tringles re of gret interest in oth geometry nd numer theory they re lled Pythgoren triples. We find ll of them in this module. Pythgors theorem is used in determining the distne etween two points in oth two nd three dimensionl spe. How this is done is outlined in the Links Forwrd setion of this module. Pythgors theorem n e generlised to the osine rule nd used to estlish Heron s formul for the re of tringle. oth of these re disussed in the Links Forwrd setion.
6 {6} The Improving Mthemtis Edution in Shools (TIMES) Projet ONTENT STNDRD NOTTION Let e tringle. We my write. Then y onvention, is length of the intervl. We lso tlk out ngle or for. So is the length of the side opposite ngle. Using this nottion we n suintly stte Pythgors theorem nd two of the most importnt theorems in trigonometry, the sine rule nd the osine rule. The sine rule nd osine rule re estlished in the Links Forwrd setion. RIGHTNGLED TRINGLES mong the set of ll tringles there is speil lss known s rightngled tringles or right tringles. right tringle hs one ngle, right ngle. The side opposite the right ngle is lled the hypotenuse. It is the longest side of the tringle hypotenuse We lso tlk out the shorter sides of rightngled tringle. Let us use the stndrd nottion desried ove nd ssume = 90. If nd re fixed then is determined. lso <, < nd < +. To prove is determined note tht D DFE (SS), so = y (See the module, ongruene) y F E
7 The Improving Mthemtis Edution in Shools (TIMES) Projet {7} THE THEOREM tringle with sides 3 m, 4 m, 5 m is rightngled tringle. Similrly, if we drw right ngled tringle with shorter sides 5 m, 12 m nd mesure the third side, we find tht the hypotenuse hs length lose to 13 m. To understnd the key ide ehind Pythgors theorem, we need to look t the squres of these numers. You n see tht in 3, 4, 5 tringle, = 25 or = 5 2 nd in the 5, 12, 13 tringle, = 169 or = We stte Pythgors theorem: The squre of the hypotenuse of rightngled tringle is equl to the sum of the squres of the lengths of the other two sides. In symols 2 = EXMPLE Find the length of the hypotenuse in the right tringle opposite. 12 x SOLUTION 16 Let x e the length of the hypotenuse. Then y Pythgors theorem, x 2 = = 400. So x = 20. Proof of the theorem mthemtil theorem is logil sttement, If p then q where p nd q re luses involving mthemtil ides. The onverse of If p then q is the sttement, If q then p. The onverse my or my not e true ut ertinty needs seprte proof. onverse of Pythgors theorem: If 2 = then is right ngle. There re mny proofs of Pythgors theorem. Proof 1 of Pythgors theorem For ese of presenttion let = 1 e the re of the right ngled tringle with 2 right ngle t.
8 {8} The Improving Mthemtis Edution in Shools (TIMES) Projet The two digrms show squre of side length + divided up into vrious squres nd tringles ongruent to. From the left hnd digrm ( + ) 2 = (1) From the right hnd digrm ( + ) 2 = (2) ompring the two equtions we otin 2 = nd the theorem is proved. Severl other proofs of Pythgors theorem re given in the ppendix. EXERISE 1 Find the hypotenuse of the rightngled tringles whose other sides re: 5, 12 9, 12 35, 12 d 15, 8 e 15, 20 f 15, 112 Note: lerly one n use lultor nd redue eh of the ove lultions to hlf dozen keystrokes. This leds to no insights t ll. s suggestion, if perfet squre is etween 4900 nd 6400 then the numer is etween 70 nd 80. If the lst digit of the squre is 1 then the numer ends in 1 or 9, et. pplitions of Pythgors theorem EXMPLE retngle hs length 8 m nd digonl 17 m. Wht is its width? SOLUTION Let e the width, mesured in m. Then 17 2 = (Pythgors theorem) 289 = = = 225, so = 15. The width of the retngle is 15 m. 17 m m 8 m
9 The Improving Mthemtis Edution in Shools (TIMES) Projet {9} EXERISE 2 ldder of length 410 m is lening ginst wll. It touhes the wll 400 m ove the ground. Wht is the distne etween the foot of the ldder nd the wll? 410 m 400 m The onverse theorem We now ome to the question: Given the lengths of the sides of tringle, n we tell whether or not the tringle is right ngled? This is nswered using the onverse of Pythgors theorem. The onverse theorem sys: If = 2 then the tringle is right ngled (with right ngle t ). Thus, for exmple, tringle with sides 20, 21, 29 is right ngled sine = = 841 = 29 2 The ovious question, whih we shll nswer lter in this module, is n we find ll suh Pythgoren triples of whole numers? We shll give two proofs of the onverse rther different in nture. However, oth use the theorem itself in the proof! This does not often hppen in elementry mthemtis ut is quite ommon in more dvned topis. First proof of the onverse We ssume 2 = onstrut seond tringle DEF E with EFD = 90, EF = nd DF =. Then, y Pythgors theorem, ED 2 = ut 2 = so ED =. Hene F D DEF (SSS). So, = 90 sine EFD = 90 nd the onverse is proved.
10 {10} The Improving Mthemtis Edution in Shools (TIMES) Projet Seond proof of the onverse We ssume 2 = Drop the perpendiulr from to, ssume D is etween nd. x lerly x < nd y < so D y 2 = x 2 + y 2 (Pythgors theorem) < = 2 This is ontrdition, so = D or D is to the left of on the line. EXERISE 3 Work out the detils of the proof when D is to the left of on the line. EXMPLE Whih of the tringles elow re rightngled tringles? Nme the right ngle in eh se. 14 m D 36 m 32 m 27 m 39 m E 15 m F SOLUTION Tringle is not rightngled tringle sine Tringle DEF is rightngled tringle sine = F is the right ngle IRRTIONL NUMERS onsider the sequene 1, 2, 3, 4, 5, This sequene of positive rel numers is stritly inresing nd n is whole numer if nd only if n is perfet squre suh s 36 or 49. The sequene tends to infinity, tht is, there is no upper ound for n.
11 The Improving Mthemtis Edution in Shools (TIMES) Projet {11} rightngled tringle with equl side lengths is n isoseles tringle. Hene the ngles re 45, 45 nd 90. If the length of side is 1 then the hypotenuse is of length 2 (sine = 2). Next we onsider the rightngled tringle with shorter sides 1 nd 2. It s hypotenuse hs length 3. We n iterte this ide otining: Using the ove onstrutions it follows tht length where n is whole numer greter thn 1, n e onstruted using just ruler nd ompss (see module, onstrutions). Sine 1 < 2 < 2, 2 is not whole numer ut perhps 2 is rtionl. This is not so, s ws disovered out 600. These ides re delt with in more detil in the module, The Rel Numers. When irrtionl numers our in prolems involving Pythgors theorem, we n either leve the nswer in symoli form, for exmple 24 (= 2 6) or pproximte the nswer using lultor, for exmple (to two deiml ples). EXMPLE Find the length, orret to 2 deiml ples, of the missing side in the right tringle opposite. x 5 SOLUTION 7 y Pythgors theorem, x = 7 2 x = 49 x 2 = 24 x = 24 = (orret to two deiml ples)
12 {12} The Improving Mthemtis Edution in Shools (TIMES) Projet EXERISE 4 rossountry runner runs 3km west, then 2km south nd then 8km est. How fr is she from her strting point? Give your nswer in kilometres nd orret to 2 deiml ples. EXERISE 5 Find the ext length of the long digonl in ue of side length 3 m. PYTHGOREN TRIDS Three whole numers tht re the lengths of the sides of rightngled tringle re lled Pythgoren Trid or Pythgoren Triple. Thus, {3, 4, 5} is Pythgoren Trid. The formul for how to generte suh triples ws known y out This is proved y ly tlet (Plimpton 322) whih ontins fifteen different triples inluding (1679, 2400, 2929). The tlet is dted to With your lultor hek this is Pythgoren triple. This ws oviously not found y hne! Strting with (3, 4, 5) we n find or onstrut infinitely mny suh triples y tking integer multiples: (3, 4, 5), (6, 8, 10), (9, 12, 15),.. onsider triple (,, ) of positive whole numers with = 2. If nd hve ommon ftor then it lso divides. So useful definition is tht the Pythgoren triple (,, ) is primitive if, HF (, ) = HF (, ) = HF (, ) = 1 tht is, the highest ommon ftor of nd is 1, et. If we n find ll primitive Pythgoren triples then we n find ll triples y simply tking whole numer multiples of the primitive triples. There re vrious fmilies of exmples. onsider the identity: (n + 1) 2 n 2 = 2n + 1 So if 2n + 1 is perfet squre then we n onstrut primitive triple ( 2n + 1, n, n + 1). In this wy, tking 2n + 1 = 9, 25, 49, 81,.. we otin triples: (3, 4, 5); (5, 12, 13); (7, 24, 25); (9, 40, 41),. It is possile to list ll primitive triples. One form of this lssifition is in the following theorem. We shll prove it using some elementry numer theory inluding the use of the fundmentl theorem of rithmeti nd the use of the HF. The symol is used for divides extly into. The result gives formul for ll primitive Pythgoren Trids.
13 The Improving Mthemtis Edution in Shools (TIMES) Projet {13} Theorem If = 2 nd (,, ) is primitive trid then = p 2 q 2, = 2pq nd = p 2 + q 2 where the HF of p nd q is 1 nd p nd q re not oth odd. Proof t lest one of, nd is odd sine the trid is primitive. The squre of whole numer is either multiple of 4 or one more thn multiple of 4, hene nd nnot oth e odd. So we my ssume is odd, is even nd is odd. 2 = so 2 = 2 2 = ( )( + ) Let d e the HF of nd +, so d nd d + so d 2 nd d 2 ut nd re oprime so d = 1 or 2 ut nd + re even. So d = 2 Hene we hve 2 nd + 2 re oprime integers. ut 2 4 = so + 2 is squre s is 2. Set + 2 = p2, 2 = q2 then = p 2 + q 2, = p 2 q 2 nd 2 4 = p2 q 2 or = 2pq. Finlly if p nd q re odd then nd re even whih is not the se. So the theorem is proved. EXMPLE = 3, = 4 nd = 5 so the triple (3, 4, 5) orresponds to (p, q) = (2, 1) EXERISE 6 Investigte: (p, q) = (2, 1), (3, 2), (4, 3),.. (p, q) = (4, 1), (5, 2), (6, 3), (7, 4),.. (p, q) = (2, 1), (4, 1), (6, 1), (8, 1),..
14 {14} The Improving Mthemtis Edution in Shools (TIMES) Projet EXERISE 7 Find the vlues of p nd q orresponding to the triple (1679, 2400, 2929) from the ly tlet Plimpton 322. The lrgest triple on Plimpton 322 is (12 709, , ) find p nd q in this se s well. LINKS FORWRD THE DISTNE FORMUL IN R 2 (THE OORDINTE PLNE) In two dimensionl oordinte geometry perhps the most si question is Wht is the distne etween two points nd with oordintes (x 1, y 1 ) nd (x 2, y 2 )? y (x 2, y 2 ) Suppose tht (x 1, y 1 ) nd (x 2, y 2 ) re two points in the plne. onsider the rightngled tringle X where X is the point (x 2, y 1 ). Then (x 1, y 1 ) X(x 2, y 1 ) x X = x 2 x 1 or x 1 x 2 nd X = y 2 y 1 or y 1 y 2 depending on the reltive positions of nd. y Pythgors theorem 2 = X 2 + X 2 = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 Therefore = = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 DISTNES IN THREEDIMENSIONL SPE The distne d from (0, 0) to (x, y) in the oordinte plne stisfies d 2 = x 2 + y 2. We n extend oordinte geometry to 3dimensions y hoosing point O lled the origin nd hoosing three lines through O ll perpendiulr to eh other. We ll these lines the xxis, the yxis nd the zxis. z y P It is possile to go from O to ny point P y moving units long the xxis, then units prllel to the yxis nd then units prllel to the zxis. We sy the oordintes of the point P re (,, ). O x
15 The Improving Mthemtis Edution in Shools (TIMES) Projet {15} gin si question is Wht is the distne OP? The nswer is OP 2 = To see this, let = (, 0, 0) nd = (,, 0). The tringle O is rightngled t. Hene O 2 = O = Next OP is right ngled t so OP 2 = O 2 + P 2 = IRLES IN THE PLNE, ENTRE THE ORIGIN irle is the pth tred out y point moving fixed distne from fixed point lled the entre. First suppose we drw irle in the rtesin plne entre the origin nd rdius 1 nd suppose (x, y) is on this irle. y 1 1 (x, y) r r (x, y) 1 r x Then y Pythgors theorem (or the distne formul in R 2 ) x 2 + y 2 = 1 2 onversely, if x 2 + y 2 = 1 then the point (x, y) lies on the irle of rdius 1. Similrly, if x 2 + y 2 = r 2 then (x, y) lies on the irle, entre the origin, of rdius r nd onversely ll points on this irle stisfy the eqution. PYTHGORS THEOREM IN TRIGONOMETRY onsider the rightngled tringle, with = 90 nd = Now y definition os = nd sin =. y Pythgors theorem, = 2. Therefore = 1. Hene, one of the most fundmentl identities in trigonometry. os 2 + sin 2 = 1 There re 2 importnt formuls linking the side lengths of tringle nd the ngles of the tringle.
16 {16} The Improving Mthemtis Edution in Shools (TIMES) Projet Theorem The osine rule Let e tringle with n ute ngle t. Then 2 = os Proof Suppose the ltitude from hs length h nd divides into intervls of length x nd y y Pythgors theorem h 2 = h 2 + x 2 nd 2 = h 2 + y 2 x y lso = x + y so, eliminting h = y 2 x 2 2 = (x + y) 2 + y 2 x 2 = x 2 2xy = x(x + y) Now x = os nd x + y = so 2 = os EXERISE 8 Show tht the osine rule is still true when is otuse. Theorem The sine rule sin = sin = sin EXERISE 9 Write down expressions for sin nd sin nd hene prove the sine rule. h
17 The Improving Mthemtis Edution in Shools (TIMES) Projet {17} POLLONIUS THEOREM ND HERON S FORMUL Eulid s Elements ws written out 300. s disussed elsewhere in these modules this mzing set of thirteen ooks olleted together most of the geometry nd numer theory known t tht time. During the next entury pollonius nd rhimedes developed mthemtis onsiderly. pollonius is est rememered for his study of ellipses, prols nd hyperols. rhimedes is often rnked s one of the most importnt mthemtiins of ll time. He rried out numer of lultions, whih ntiipted ides from integrl lulus. In this setion we disuss Heron s formul tht sholrs elieve ws disovered y Pythgors. We shll prove pollonius theorem nd Heron s formul whih oth follow from Pythgors theorem using lger. pollonius theorem Suppose is ny tringle, = 2x nd m is the length of the medin from to then = 2x 2 + 2m 2 Proof Let length of the ltitude E e h. lso let E = t, ED = s. h m t E s D x lerly s + t = x nd there re three rightngled tringles so m 2 = h 2 + s 2 2 = h 2 + (s + x) 2 2 = h 2 + t 2 Tking into ount the formul to e proved we onsider m 2 = h 2 + (s + x) 2 + h 2 + t 2 2h 2 2s 2 = s 2 + 2sx + x 2 + t 2 2s 2 = x 2 + 2sx + t 2 s 2 = x 2 + 2sx + (t + s)(t s) = x 2 + 2sx + (t s)x = x 2 + sx + tx = 2x 2 Tht is, = 2m 2 + 2x 2 In the ove digrm we sumed then is ute nd E is etween nd. The other ses n e delt with similrly.
18 {18} The Improving Mthemtis Edution in Shools (TIMES) Projet EXERISE 10 Use the osine rule to write m 2 in D nd 2 in. Dedue pollonius theorem in ouple of steps. Heron s formul This is n mzing formul expressing the re of tringle in terms of its side lengths. To write this in its stndrd form onsider with side lengths, nd. We let 2s = + + nd let e the re of (s is the semiperimeter of the tringle). Then 2 = s (s ) (s ) (s ) We give proof tht uses only Pythgors theorem, the formul for the re of tringle nd some lger. Proof Let h e the length of the ltitude from to whih divides into intervls of length x nd y. Then x + y =. The re of tringle is hlf se times height so h = 1 2 h nd 4 2 = h 2 2 (1) x y Pythgors theorem gives 2 = h 2 + x 2 (2) 2 = h 2 + y 2 = h 2 + ( x) 2 (3) We must eliminte h nd x from equtions (1), (2) nd (3). This is nontrivil! (2) (3) 2 2 = h 2 + x 2 h 2 ( x) = 2x 2 2x = (4) Next (2) gives h 2 = 2 x 2 sustitute into (1) 4 2 = 2 ( 2 x 2 ) 16 2 = x 2 (5)
19 The Improving Mthemtis Edution in Shools (TIMES) Projet {19} Squre (4) nd sustitute into (5) 16 2 = (2) 2 ( ) 2 This is the differene of two squres, so 16 2 = ( )( ) = (( + ) 2 2 )( 2 ( ) 2 ) = ( + + )( + )( + )( + ) = 2s(2s 2)(2s 2)(2s 2) So 2 = s(s )(s )(s ) EXERISE 11 Find the res of the tringles with side lengths: 13, 14, 15 13, 20, 21 10, 17, 21 d 51, 52, 53 HISTORY s outlined ove, the theorem, nmed fter the sixth entury Greek philosopher nd mthemtiin Pythgors, is rguly the most importnt elementry theorem in mthemtis, sine its onsequenes nd generlistions hve wide rnging pplitions. It is often diffiult to determine vi historil soures how long ertin fts hve een known. However, in the se of Pythgors theorem there is ylonin tlet, known s Plimpton 322, tht dtes from out This tlet lists fifteen Pythgoren triples inluding (3, 4, 5), (28, 45, 53) nd (65, 72, 97). It does not inlude (5, 12, 13) or (8, 15, 17) ut it does inlude (12 709, , )! The fifteen triples orrespond (very roughly) to ngles etween 30 nd 45 in the rightngled tringle. The ylonin numer system is se 60 nd ll of the even sides re of the form presumly to filitte lultions in se 60. Most historil douments re found s frgments nd one ould ll this the Rosett Stone of mthemtis. Whihever interprettion of the purpose of Plimpton 322 is orret, nd there re severl, it is ler tht oth Pythgors theorem nd how to onstrut Pythgoren Triples ws known well efore The nture of mthemtis egn to hnge out 600. This ws losely linked to the rise of the Greek ity sttes. There ws onstnt trde nd hene ides spred freely from the erlier ivilistions of Egypt nd yloni. Most of the history is lost forever, ut trdition hs it the Thles, Pythgors nd their students, were responsile for developing mny of the key ides in prtiulr the need to prove theorems! Wht we do know is wht ws known t out 300. This is euse Eulid of lexndri wrote his thirteen volume ook the Elements. ontrry to populr elief, this ook is y no mens solely out geometry.
20 {20} The Improving Mthemtis Edution in Shools (TIMES) Projet ook 1 of the Elements is on geometry nd ttempts to set geometry on sound logil sis y giving some twentythree definitions nd lists five postultes nd five ommon notions. This xiomti pproh, lthough flwed nd inomplete gve logil pproh to the study of geometry whih ws entrl prt of lssil edution right up to the twentieth entury. The imperfetions of Eulid were not fixed until 1900 when Dvid Hilert gve modern orret system of xioms. In ook 1 of Eulid numer of theorems re proved suh s the wellknown result tht in n isoseles tringle the se ngles re equl. The finl theorem, Proposition 147, is Pythgors theorem. The proof given is not the esiest known t the time, ut uses only ongruene nd other results proved in ook 1. Eulid s Elements re very sophistited. PPENDIX OTHER PROOFS OF PYTHGORS THEOREM There re hundreds of proofs of Pythgors theorem one ttriuted to Npoleon nd one ttriuted to 19 th entury US president! We shll present few more inluding Eulid s proof. Seond proof We tke the seond digrm from the first proof. is the re of the tringle so = 1 2. The lrge squre is + y + so ( + ) 2 = = or 2 = Third proof We ssume >. The side length of the inner squre is Hene ( ) = = = 2 nd we hve nother proof of Pythgors theorem.
21 The Improving Mthemtis Edution in Shools (TIMES) Projet {21} The first three proofs re essentilly sed on ongruene of tringles, prtilly disguised s sums of res. Some proofs use similrity. One of the niest or perhps minimlist proofs omes from onsidering, simple digrm whih ontins three tringles ll similr to eh other. Fourth proof We tke n ritrry rightngled tringle with = 90 nd let D e n ltitude of the tringle of length h. D is perpendiulr to nd x + y = D is similr to D (), hene = D D = D D Tht is, = h x = y h x h D y D is similr to (), hene = D = D Tht is, x + y = h = y D is similr to (), hene = D = D Tht is, x + y = x = h From the seond group of equtions we otin (y ross multiplition) 2 = y(x + y) Similrly, from the third group we otin 2 = x(x + y) Hene = x(x + y) + y(x + y) = (x + y) 2 = 2 nd the proof is omplete. Next we shll disover Pythgors proof of his theorem. More properly it is Eulid s proof proposition 47 of Eulid s elements. 2 We drw squres of res 2, 2 nd 2 djent to the sides of the tringle. From 2 = y(x + y) = y in the previous proof 2 x y y = 2 2 From 2 = x(x + y) = x in the previous proof x = 2
22 {22} The Improving Mthemtis Edution in Shools (TIMES) Projet The re of the shded retngle is x = 2 = 2 nd the other retngle is y = 2 = 2 So we hve divided the squre, re 2, into two retngles of re 2 nd 2. This is the key ide in Eulid s proof. Fifth proof: Eulid s proof D Eulid s proof of Pythgors onsists of proving the squre of re 2 is the sme s the re of retngle. This is done y finding ongruent tringles of hlf the re of the two regions. E H Here re the detils: E is ongruent to G (SS) sine E = (sides of squre) G F I = G (sides of squre) E = G (equl to 90 + ) Hene re E = re G re DE = 2 re E (tringle nd retngle on sme se nd sme height) Similrly, re GHF = 2 re G so, re GFH = re DE = 2. In similr wy it n e shown tht re HFI = 2 nd the theorem is proved. NSWERS TO EXERISES EXERISE d 17 e 25 f 113 EXERISE 2 90 m
23 The Improving Mthemtis Edution in Shools (TIMES) Projet {23} EXERISE 3 Let D = y where D is the point on produed so tht D is perpendiulr to produed. Let D = x. ssume 2 = Using Pythgors theorem in tringle D: 2 = x 2 + y 2. Using Pythgors theorem in tringle D: 2 = x 2 + (y + ) 2. Use the three equtions to show 2y = 0 whih is ontrdition. EXERISE km EXERISE m EXERISE 6 p q p q
24 {24} The Improving Mthemtis Edution in Shools (TIMES) Projet p q EXERISE 7 p = 48 nd q = 25 nd p = 125 nd q = 54. EXERISE 8 Let D e the point on produed so tht D is perpendiulr to produed. Let D = h nd D = x. Use Pythgors theorem twie: 2 = h 2 + x 2 nd 2 = h 2 + (x + ) 2. Eliminte h nd sustitute x = os(180 ) = os to otin the result. EXERISE 9 Use h = sin = sin so sin = sin EXERISE 10 m 2 = 2 + x 2 2x os nd 2 = 2 + 4x 2 4x os. Multiply the first eqution y 2 nd sutrt the seond eqution to otin the result. EXERISE d 1170
25 The Improving Mthemtis Edution in Shools (TIMES) Projet {25}
26 The im of the Interntionl entre of Exellene for Edution in Mthemtis (IEEM) is to strengthen edution in the mthemtil sienes t ll levelsfrom shool to dvned reserh nd ontemporry pplitions in industry nd ommere. IEEM is the edution division of the ustrlin Mthemtil Sienes Institute, onsortium of 27 university mthemtis deprtments, SIRO Mthemtil nd Informtion Sienes, the ustrlin ureu of Sttistis, the ustrlin Mthemtil Soiety nd the ustrlin Mthemtis Trust. The IEEM modules re prt of The Improving Mthemtis Edution in Shools (TIMES) Projet. The modules re orgnised under the strnd titles of the ustrlin urriulum: Numer nd lger Mesurement nd Geometry Sttistis nd Proility The modules re written for tehers. Eh module ontins disussion of omponent of the mthemtis urriulum up to the end of Yer 10.
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