The Pythgoren Theorem Pythgors ws Greek mthemtiin nd philosopher, orn on the islnd of Smos (. 58 BC). He founded numer of shools, one in prtiulr in town in southern Itly lled Crotone, whose memers eventully eme known s the Pythgorens. The inner irle t the shool, the Mthemtikoi, lived t the shool, rid themselves of ll personl possessions, were vegetrins, nd oserved strit vow of silene. They studied mthemtis, philosophy, nd musi, nd held the elief tht numers onstitute the true nture of things, giving numers mystil or even spiritul qulity. Tody, nothing is known of Pythgors s writings, perhps due to the serey nd silene of the Pythgoren soiety. However, one of the most fmous theorems in ll of Pythgors. mthemtis does er his nme, the Pythgoren Theorem. Pythgoren Theorem. Let represent the length of the hypotenuse, the side of right tringle diretly opposite the right ngle ( right ngle mesures 90 )of the tringle. The remining sides of the right tringle re lled the legs of the right tringle, whose lengths re designted y the letters nd. The reltionship involving the legs nd hypotenuse of the right tringle, given y + =, (1) is lled the Pythgoren Theorem. Note tht the Pythgoren Theorem n only e pplied to right tringles. Let s look t simple pplition of the Pythgoren Theorem (1). Exmple. Given tht the length of one leg of right tringle is 4 entimeters nd the hypotenuse hs length 8 entimeters, find the length of the seond leg. Let s egin y skething nd leling right tringle with the given informtion. We will let x represent the length of the missing leg. 15 Copyrighted mteril. See: http://msenux.redwoods.edu/intalgtext/ 505
8m 4m Figure 1. it esier. x A sketh mkes things Here is n importnt piee of dvie. Tip 3. The hypotenuse is the longest side of the right tringle. It is loted diretly opposite the right ngle of the tringle. Most importntly, it is the quntity tht is isolted y itself in the Pythgoren Theorem. + = Alwys isolte the quntity representing the hypotenuse on one side of the eqution. The legs go on the other side of the eqution. So, tking the tip to hert, nd noting the lengths of the legs nd hypotenuse in Figure 1, wewrite 4 + x =8. Squre, then isolte x on one side of the eqution. 16 + x =64 x =48 Normlly, we would tke plus or minus the squre root in solving this eqution, ut x represents the length of leg, whih must e positive numer. Hene, we tke just the positive squre root of 48. x = 48 Of ourse, ple your nswer in simple rdil form. x = 16 3 x =4 3 If need e, you n use your grphing lultor to pproximte this length. To the nerest hundredth of entimeter, x 6.93 entimeters. 506
Proof of the Pythgoren Theorem It is not known whether Pythgors ws the first to provide proof of the Pythgoren Theorem. Mny mthemtil historins think not. Indeed, it is not even known if Pythgors rfted proof of the theorem tht ers his nme, let lone ws the first to provide proof. There is evidene tht the nient Bylonins were wre of the Pythgoren Theorem over 1000 yers efore the time of Pythgors. A ly tlet, now referred to s Plimpton 3 (see Figure ), ontins exmples of Pythgoren Triples, sets of three numers tht stisfy the Pythgoren Theorem (suh s 3, 4, 5). Figure. A photogrph of Plimpton 3. One of the erliest reorded proofs of the Pythgoren Theorem dtes from the Hn dynsty (06 BC to AD 0), nd is reorded in the Chou Pei Sun Ching (see Figure 3). You n see tht this figure speifilly ddresses the se of the 3, 4, 5 right tringle. Mthemtil historins re divided s to whether or not the imge ws ment to e prt of generl proof or ws just devised to ddress this speifi se. There is lso disgreement over whether the proof ws provided y more modern ommenttor or dtes k further in time. 507
However, Figure 3 does suggest pth we might tke on the rod to proof of the Pythgoren Theorem. Strt with n ritrry right tringle hving legs of lengths nd, nd hypotenuse hving length, sshowninfigure 4(). Next, mke four opies of the tringle shown in Figure 4(), then rotte nd trnslte them into ple s shown in Figure 4(). Note tht this forms ig squre tht is units on side. Figure 3. A figure from the Chou Pei Sun Ching. Further, the position of the tringles in Figure 4() llows for the formtion of smller, unshded squre in the middle of the lrger squre. It is not hrd to lulte the length of the side of this smller squre. Simply sutrt the length of the smller leg from the lrger leg of the originl tringle. Thus, the side of the smller squre hs length. () Figure 4. () Proof of the Pythgoren Theorem. Now, we will lulte the re of the lrge squre in Figure 4() in two seprte wys. First, the lrge squre in Figure 4() hs side of length. Therefore, the re of the lrge squre is Are =. Seondly, the lrge squre in Figure 4()ismdeupof4tringlesofthesmesize nd one smller squre hving side of length. We n lulte the re of the lrge squre y summing the re of the 4 tringles nd the smller squre. 508
1. The re of the smller squre is ( ).. The re of eh tringle is /. Hene, the re of four tringles of equl size is four times this numer; i.e., 4(/). Thus, the re of the lrge squre is Are = Are of smll squre + 4 Are of tringle ( ) =( ) +4. We lulted the re of the lrger squre twie. The first time we got ;the seond time we got ( ) +4(/). Therefore, these two quntities must e equl. ( ) =( ) +4 Expnd the inomil nd simplify. Tht is, = + + = + nd the Pythgoren Theorem is proven. + =, Applitions of the Pythgoren Theorem In this setion we will look t few pplitions of the Pythgoren Theorem, one of the most pplied theorems in ll of mthemtis. Just sk your lol rpenter. The nient Egyptins would tke rope with 1 eqully sped knots like tht shown in Figure 5, nd use it to squre orners of their uildings. The tool ws instrumentl in the onstrution of the pyrmids. The Pythgoren theorem is lso useful in surveying, rtogrphy, nd nvigtion, to nme few possiilities. Let s look t few exmples of the Pythgoren Theorem in tion. Figure 5. A si 3-4-5 right tringle for squring orners. Exmple 4. One leg of right tringle is 7 meters longer thn the other leg. The length of the hypotenuse is 13 meters. Find the lengths of ll sides of the right tringle. Let x represent the length of one leg of the right tringle. Beuse the seond leg is 7 meters longer thn the first leg, the length of the seond leg n e represented y the expression x +7, s shown in Figure 6, where we ve lso leled the length of the hypotenuse (13 meters). 509
x +7 x 13 Figure 6. The seond leg is 7 meters longer thn the first. Rememer to isolte the length of the hypotenuse on one side of the eqution representing the Pythgoren Theorem. Tht is, x +(x +7) =13. Note tht the legs go on one side of the eqution, the hypotenuse on the other. Squre nd simplify. Rememer to use the squring inomil pttern. x + x +14x + 49 = 169 x +14x + 49 = 169 This eqution is nonliner, so mke one side zero y sutrting 169 from oth sides of the eqution. x +14x +49 169 = 0 x +14x 10 = 0 Note tht eh term on the left-hnd side of the eqution is divisile y. Divide oth sides of the eqution y. x +7x 60 = 0 Let s use the qudrti formul with =1, =7,nd = 60. x = ± 4 = 7 ± 7 4(1)( 60) (1) Simplify. x = 7 ± 89 Note tht 89 is perfet squre (17 = 89). Thus, Thus,wehvetwosolutions, x = 7 ± 17. x =5 or x = 1. 510
Beuse length must e positive numer, we eliminte 1 from onsidertion. Thus, thelengthofthefirstlegisx = 5 meters. The length of the seond leg is x +7,or1 meters. Chek. Cheking is n esy mtter. The legs re 5 nd 1 meters, respetively, nd the hypotenuse is 13 meters. Note tht the seond leg is 7 meters longer thn the first. Also, whih is the squre of 13. 5 +1 = 5 + 144 = 169, The integrl sides of the tringle in the previous exmple, 5, 1, nd 13, re n exmple of Pythgoren Triple. Pythgoren Triple. A set of positive integers,, nd, is lled Pythgoren Triple if they stisfy the Pythgoren Theorem; tht is, if + =. If the gretest ommon ftor of,, nd is 1, then the triple (,, ) is lled primitive Pythgoren Triple. Thus, for exmple, the Pythgoren Triple (5, 1, 13) is primitive. Let s look t nother exmple. Exmple 5. If (,, ) is Pythgoren Triple, show tht ny positive integrl multiple is lso Pythgoren Triple. Thus, if the positive integers (,, ) is Pythgoren Triple, we must show tht (k, k, k), where k is positive integer, is lso Pythgoren Triple. However, we know tht Multiply oth sides of this eqution y k. This lst result n e written + =. k + k = k (k) +(k) =(k). Hene, (k, k, k) is Pythgoren Triple. Hene, euse (3, 4, 5) is Pythgoren Triple, you n doule everything to get nother triple (6, 8, 10). Note tht 6 +8 =10 is esily heked. Similrly, tripling gives nother triple (9, 1, 15), nd so on. 511
In Exmple 5, we showed tht (5, 1, 13) ws triple, so we n tke multiples to generte other Pythgoren Triples, suh s (10, 4, 6) or (15, 36, 39), nd so on. Formule for generting Pythgoren Triples hve een know sine ntiquity. Exmple 6. The following formul for generting Pythgoren Triples ws pulished in Eulid s (35 65 BC) Elements, one of the most suessful textooks in the history of mthemtis. If m nd n re positive integers with m>n,show genertes Pythgoren Triples. = m n, =mn, = m + n, We need only show tht the formule for,, nd stisfy the Pythgoren Theorem. With tht is mind, let s first ompute +. + =(m n ) +(mn) = m 4 m n + n 4 +4m n = m 4 +m n + n 4 (7) On the other hnd, =(m + n ) = m 4 +m n + n 4. Hene, + =, nd the expressions for,, nd form Pythgoren Triple. It is oth interesting nd fun to generte Pythgoren Triples with the formule from Exmple 6. Choose m =4ndn =,then = m n =(4) () =1, =mn = (4)() = 16, = m + n =(4) +() =0. It is esy to hek tht the triple (1, 16, 0) will stisfy 1 +16 =0. Indeed, note tht this triple is multiple of the si (3, 4, 5) triple, so it must lso e Pythgoren Triple. It n lso e shown tht if m nd n re reltively prime, nd re not oth odd or oth even, then the formule in Exmple 6 will generte primitive Pythgoren Triple. For exmple, hoose m =5ndn =. Note tht the gretest ommon divisor of m =5ndn = is one, so m nd n re reltively prime. Moreover, m is odd while n is even. These vlues of m nd n generte = m n =(5) () =1, =mn = (5)() = 0, = m + n =(5) +() =9. 51
Note tht 1 +0 = 441 + 400 = 841 =9. Hene, (1, 0, 9) is Pythgoren Triple. Moreover, the gretest ommon divisor of 1, 0, nd 9 is one, so (1, 0, 9) is primitive. The prtil pplitions of the Pythgoren Theorem re numerous. Exmple 8. A pinter lens 0 foot ldder ginst the wll of house. The se of the ldder is on level ground 5 feet from the wll of the house. How high up the wll of the house will the ldder reh? Consider the tringle in Figure 7. The hypotenuse of the tringle represents the ldder nd hs length 0 feet. The se of the tringle represents the distne of the se of the ldder from the wll of the house nd is 5 feet in length. The vertil leg of the tringle is the distne the ldder rehes up the wll nd the quntity we wish to determine. 0 h Applying the Pythgoren Theorem, 5 Figure 7. A ldder lens ginst the wll of house. 5 + h =0. Agin, note tht the squre of the length of the hypotenuse is the quntity tht is isolted on one side of the eqution. Next, squre, then isolte the term ontining h on one side of the eqution y sutrting 5 from oth sides of the resulting eqution. 5 + h = 400 h = 375 513
We need only extrt the positive squre root. h = 375 We ould ple the solution in simple form, tht is, h =5 15, ut the nture of the prolem wrrnts deiml pproximtion. Using lultor nd rounding to the nerest tenth of foot, h 19.4. Thus, the ldder rehes out 19.4 feet up the wll. TheDistneFormul We often need to lulte the distne etween two points P nd Q in the plne. Indeed, this is suh frequently reurring need, we d like to develop formul tht will quikly lulte the distne etween the given points P nd Q. Suh formul is the gol of this lst setion. Let P (x 1,y 1 ) nd Q(x,y ) e two ritrry points in the plne, s shown in Figure 8() nd let d represent the distne etween the two points. y y Q(x,y ) Q(x,y ) d x d y y 1 x P (x 1,y 1 ) P (x 1,y 1 ) x x 1 R(x,y 1 ) () () Figure 8. Finding the distne etween the points P nd Q. To find the distne d, first drw the right tringle PQR, with legs prllel to the xes, s shown in Figure 8(). Next, we need to find the lengths of the legs of the right tringle PQR. The distne etween P nd R is found y sutrting the x oordinte of P from the x-oordinte of R nd tking the solute vlue of the result. Tht is, the distne etween P nd R is x x 1. The distne etween R nd Q is found y sutrting the y-oordinte of R from the y-oordinte of Q nd tking the solute vlue of the result. Tht is, the distne etween R nd Q is y y 1. 514
We n now use the Pythgoren Theorem to lulte d. Thus, However, for ny rel numer, d =( x x 1 ) +( y y 1 ). ( ) = = =, euse is nonnegtive. Hene, ( x x 1 ) =(x x 1 ) nd ( y y 1 ) =(y y 1 ) nd we n write d =(x x 1 ) +(y y 1 ). Tking the positive squre root leds to the Distne Formul. The Distne Formul. Let P (x 1,y 1 )ndq(x,y ) e two ritrry points in the plne. The distne d etween the points P nd Q is given y the formul d = (x x 1 ) +(y y 1 ). (9) The diretion of sutrtion is unimportnt. Beuse you squre the result of the sutrtion, you get the sme response regrdless of the diretion of sutrtion (e.g. (5 ) =( 5) ). Thus, it doesn t mtter whih point you designte s the point P, nor does it mtter whih point you designte s the point Q. Simply sutrt x- oordintes nd squre, sutrt y-oordintes nd squre, dd, then tke the squre root. Let s look t n exmple. Exmple 10. Find the distne etween the points P ( 4, ) nd Q(4, 4). It helps the intuition if we drw piture, s we hve in Figure 9. One n now tke ompss nd open it to the distne etween points P nd Q. Then you n ple your ompss on the horizontl xis (or ny horizontl gridline) to estimte the distne etween the points P nd Q. We did tht on our grph pper nd estimte the distne d 10. y Q(4, 4) d x P ( 4, ) Figure 9. Guging the distne etween P ( 4, ) nd Q(4, 4). 515
Let s now use the distne formul to otin n ext vlue for the distne d. With (x 1,y 1 )=P ( 4, ) nd (x,y )=Q(4, 4), d = (x x 1 ) +(y y 1 ) = (4 ( 4)) +(4 ( )) = 8 +6 = 64 + 36 = 100 =10. It s not often tht your ext result grees with your pproximtion, so never worry if you re off y just little it. 516
Exerises In Exerises 1-8, stte whether or not the given triple is Pythgoren Triple. Give reson for your nswer. 10. 1. (8, 15, 17). (7, 4, 5) 3. (8, 9, 17) 4. (4, 9, 13) 11. 5. (1, 35, 37) 6. (1, 17, 9) 7. (11, 17, 8) 4 8 8. (11, 60, 61) In Exerises 9-16, set up n eqution to model the prolem onstrints nd solve. Use your nswer to find the missing side of the given right tringle. Inlude sketh with your solution nd hek your result. 9. 1. 13. 10 1 3 3 16 Copyrighted mteril. See: http://msenux.redwoods.edu/intalgtext/ 517
14. legs. 1 4 3 0. One leg of right tringle is 3 feet longer thn 3 times the length of the first leg. The length of the hypotenuse is 5 feet. Find the lengths of the legs. 15. 1. Pythgors is redited with the following formule tht n e used to generte Pythgoren Triples. 16. 5 10 = m = m 1, = m +1 8 8 Use the tehnique of Exmple 6 to demonstrte tht the formule given ove will generte Pythgoren Triples, provided tht m is n odd positive integer lrger thn one. Seondly, generte t lest 3 instnes of Pythgoren Triples with Pythgors s formul. In Exerises 17-0, set up n eqution tht models the prolem onstrints. Solve the eqution nd use the result to nswer the question. Look k nd hek your result. 17. The legs of right tringle re onseutive positive integers. The hypotenuse hs length 5. Wht re the lengths of the legs? 18. The legs of right tringle re onseutive even integers. The hypotenuse hs length 10. Wht re the lengths of the legs?. Plto (380 BC) is redited with the following formule tht n e used to generte Pythgoren Triples. =m = m 1, = m +1 Use the tehnique of Exmple 6 to demonstrte tht the formule given ove will generte Pythgoren Triples, provided tht m is positive integer lrger thn 1. Seondly, generte t lest 3 instnes of Pythgoren Triples with Plto s formul. 19. One leg of right tringle is 1 entimeter less thn twie the length of the first leg. If the length of the hypotenuse is 17 entimeters, find the lengths of the 518
In Exerises 3-8, set up n eqution tht models the prolem onstrints. Solve the eqution nd use the result to nswer the question. Look k nd hek your result. 3. Fritz nd Gret re plnting 1- foot y 18-foot retngulr grden, nd re lying it out using string. They would like to know the length of digonl to mke sure tht right ngles re formed. Find the length of digonl. Approximte your nswer to within 0.1 feet. 4. Angelin nd Mrkos re plnting 0-foot y 8-foot retngulr grden, nd re lying it out using string. They would like to know the length of digonl to mke sure tht right ngles re formed. Find the length of digonl. Approximte your nswer to within 0.1 feet. 5. The se of 36-foot long guy wire is loted 16 feet from the se of the telephone pole tht it is nhoring. How high up the pole does the guy wire reh? Approximte your nswer to within 0.1 feet. 6. The se of 35-foot long guy wire is loted 10 feet from the se of the telephone pole tht it is nhoring. How high up the pole does the guy wire reh? Approximte your nswer to within 0.1 feet. 7. A stereo reeiver is in orner of 13-foot y 16-foot retngulr room. Speker wire will run under rug, digonlly, to speker in the fr orner. If 3 feet of slk is required on eh end, how long piee of wire should e purhsed? Approximte your nswer to within 0.1 feet. 8. A stereo reeiver is in orner of 10-foot y 15-foot retngulr room. Speker wire will run under rug, digonlly, to speker in the fr orner. If 4 feet of slk is required on eh end, how long piee of wire should e purhsed? Approximte your nswer to within 0.1 feet. In Exerises 9-38, use the distne formul to find the ext distne etween the given points. 9. ( 8, 9) nd (6, 6) 30. (1, 0) nd ( 9, ) 31. ( 9, 1) nd ( 8, 7) 3. (0, 9) nd (3, 1) 33. (6, 5) nd ( 9, ) 34. ( 9, 6) nd (1, 4) 35. ( 7, 7) nd ( 3, 6) 36. ( 7, 6) nd (, 4) 37. (4, 3) nd ( 9, 6) 38. ( 7, 1) nd (4, 5) In Exerises 39-4, set up n eqution tht models the prolem onstrints. Solve the eqution nd use the result to nswer the question. Look k nd hek your result. 39. Find k so tht the point (4,k)is units wy from the point (, 1). 40. Find k so ht the point (k, 1) is units wy from the point (0, 1). 519
41. Find k so tht the point (k, 1) is 17 units wy from the point (, 3). 4. Find k so tht the point ( 1,k)is 13 units wy from the point ( 4, 3). 43. Set up oordinte system on sheet of grph pper. Lel nd sle eh xis. Plot the points P (0, 5) nd Q(4, 3) on your oordinte system. ) Plot severl points tht re equidistnt from the points P nd Q on your oordinte system. Wht grph do you get if you plot ll points tht re equidistnt from the points P nd Q? Determine the eqution of the grph y exmining the resulting imge on your oordinte system. ) Use the distne formul to find the eqution of the grph of ll points tht re equidistnt from the points P nd Q. Hint: Let (x, y) represent n ritrry point on the grph of ll points equidistnt from points P nd Q. Clulte the distnes from the point (x, y) to the points P nd Q seprtely, then set them equl nd simplify the resulting eqution. Note tht this nlytil pproh should provide n eqution tht mthes tht found y the grphil pproh in prt (). 44. Set up oordinte system on sheet of grph pper. Lel nd sle eh xis. Plot the point P (0, ) nd lel it with its oordintes. Drw the line y = nd lel it with its eqution. ) Plot severl points tht re equidistnt from the point P nd the line y = onyouroordintesystem. Wht grph do you get if you plot ll points tht re equidistnt from the points P nd the line y =. ) Use the distne formul to find the eqution of the grph of ll points tht re equidistnt from the points P nd the line y =. Hint: Let (x, y) represent n ritrry point on the grph of ll points equidistnt from points P nd the line y =. Clulte the distnes from the point (x, y) to the points P nd the line y = seprtely, then set them equl nd simplify the resulting eqution. 50
45. Copy the following figure onto sheet of grph pper. Cut the piees of the first figure out with pir of sissors, then rerrnge them to form the seond figure. Explin how this proves the Pythgoren Theorem. 46. Compre this imge to the one tht follows nd explin how this proves the Pythgoren Theorem. 51
Answers 1. Yes, euse 8 +15 =17 3. No, euse 8 +9 17 5. Yes, euse 1 +35 =37 43. ) In the figure tht follows, XP = XQ. y 10 7. No, euse 11 +17 8 9. 4 11. 4 3 P (0,5) X(x,y) y=(1/)x 13. x 5 10 15. 5 3 17. Thelegshvelengths3nd4. 19. The legs hve lengths 8 nd 15 entimeters. 1. (3, 4, 5), (5, 1, 13), nd (7, 4, 5), with m = 3, 5, nd 7, respetively. 3. 1.63 ft 5. 3.5 ft 7. 6.6 ft 5 ) y =(1/)x Q(4, 3) 9. 31. 33. 35. 37. 05 37 34 = 3 6 17 50 = 5 10 39. k =3, 1. 41. k =1,3. 5