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 ll the tringle right tringle. In right tringle, the two sides of the tringle tht form the right ngle re lled the legs of the right tringle, nd the side of the tringle whih is opposite to the right ngle is lled the hypotenuse of the tringle. leg hypotenuse right ngle leg Right Tringle We re now redy to stte: The Pythgoren Theorem. In tringle whih hs sides of lengths, nd, if the sides of length nd form right ngle, then side lengths stisfy the following eqution: 2 + 2 = 2. In words, the Pythgoren Theorem sys The squre of the hypotenuse is equl to the sum of the squres of the other two sides.
28 There re mny different explntions of why the Pythgoren Theorem is true. These explntions re lled proofs of the Pythgoren Theorem nd we will give them more ttention lter in this lesson. 1 Here we fous on pplitions of the Pythgoren Theorem. We now present three smple pplitions of the Pythgoren Theorem. In eh pplition, we will use the Pythgoren Theorem to find the unknown side length. When performing these lultions, we will py reful ttention to the units in whih the given dt is presented, nd we will oey the two Clultion Rules stted in Lesson 1. Applition 1. 4.4 m =? 3.3 m In the tringle shown ove, the lengths of two sides re given to the nerest tenth of meter. In this sitution Clultion Rule 2 tells us tht our finl nswer for side length should e expressed to the nerest tenth of meter. Aording to the Pythgoren Theorem: Hene, Using poket lultor, we find tht 2 = (3.3) 2 + (4.4) 2. = (3.3) 2 + (4.4) 2. (3.3) 2 + (4.4) 2 = 5.5 meters. In this lultion, we re fortunte. The nswer produed y the lultor requires no rounding. Hene, the solution to this prolem is: 1 Here re four wesites tht present vriety of proofs of the Pythgoren Theorem: http://www.usn.edu/mthdept/mdm/pyth.html http://www.ut-the-knot.org/pythgors/index.shtml http://www.ut-the-knot.org/pythgors/pythlttie.shtml http://www.m.org/mthlnd/mthtrek_11_27_00.html The first site displys n esily understood nimted proof. The seond site ontins 43 different proofs. The third site views the proof in the first site from more sophistited perspetive tht revels the proof to e one of fmily of similr proofs. The fourth site gives rief history of the Theorem, speultion out how it ws disovered, nd referenes to other interesting sites.
29 = 5.5 m to the nerest tenth of meter. Applition 2. =? 19 in 8 in In the tringle shown ove, the lengths of two sides re given to the nerest inh. Thus, ording to Clultion Rule 2, our finl nswer for side length should e expressed to the nerest inh. Here, the Pythgoren Theorem implies tht Hene, Therefore, Using poket lultor, we find tht 8 2 + 2 = 19 2. 2 = 19 2 8 2. = 19 2 " 8 2. 19 2 " 8 2 = 17.23368794. We round to the nerest inh, to otin the solution to this prolem: = 17 in to the nerest inh. Applition 3. =? = 1342 miles = 1980 miles to the nerest 10 miles In the tringle pitured ove, the lest urte mesurement is in units of 10 miles. Therefore, Clultion Rule 2 stipultes tht our finl nswer should e in units of 10 miles (even though one of the lengths is given in 1 mile units). The Pythgoren Theorem implies tht 2 + 1342 2 = 1980 2.
30 Notie tht we do not round the numer 1342 to 1340 even though our finl nswer must e in units of 10 miles. Here we re oeying Clultion Rule 1 whih tells us tht to preserve ury we should not do ny rounding until the end of the lultion. Hene, Using poket lultor, we find tht = 1980 2 "1342 2. 1980 2 "1342 2 = 1455.828287. We round to the nerest 10 miles to otin the solution to this prolem: = 1460 miles to the nerest 10 miles. Ativity 1. Different groups should work on prolems ) nd ) elow nd report their solutions to the lss. The lss should disuss ny disrepnies in the solutions. In eh prolem, find the unknown side length. Be sure to oey the two Clultion Rules. ) ) 14.03 m =? =? 14 in 8.6 m 9.2 in Next we present n illustrtion of how the Clultion Rules sve us from error. Suppose we re given right tringle in whih oth legs mesure.1 m to the nerest tenth of entimeter. Let us lulte the length of the hypotenuse of this right tringle. Clultion Rule 2 instruts us to ompute the length of the hypotenuse in units of tenths of entimeter. Aording to the Pythgoren Theorem the length of the hypotenuse is (.1) 2 + (.1) 2 =.01+.01 =.02 =.1414? 0.1 m 0.1 m
31 Hene, rounded to the nerest tenth of entimeter, the length of the hypotenuse of this right tringle is.1 m. During this omputtion, we oeyed oth Clultions Rules 1 nd 2. We didn t round during the lultion, ut we rounded t the end of the lultion to mke our nswer hve the sme degree of ury s the dt given t the eginning of the prolem. Suppose we hd disoeyed Clultion Rule 1 y rounding the quntity.02 to the nerest tenth of entimeter efore we ompleted the lultion y tking squre root. This would hnge.02 to.0. Then the lultion would eome: (.1) 2 + (.1) 2 =.01+.01 =.02.0 = 0. We would thus rrive t the onlusion tht the length of the hypotenuse of this right tringle is 0 m to the nerest tenth of entimeter. Sine the legs of this tringle re.1 m eh nd sine the hypotenuse of right tringle is lwys longer thn its legs, this is n surd onlusion. Morl: Don t rek the two Clultion Rules. Notie tht we hve not used the onept of signifint figures in the preeding pplitions of the Pythgoren Theorem. In Lesson 1 we hd dvised the use of signifint figures in lultions involving multiplition or division prtiulrly when input dt is expressed in different units. Pythgoren Theorem lultions involve squring whih is form of multiplition. So it is not unresonle to think tht it would e pproprite to use signifint figures in these lultions. However, the method of signifint figures is only rule of thum to void inury whih doesn t lwys produe optimlly urte results. It turns out tht Pythgoren Theorem pplitions re lultions in whih the method of signifint figures sometimes retes inury. So we dvise not using signifint figures in these kinds of lultions. We illustrte how the use of signifint figures n introdue inury y reexmining Applition 2 ove. In pplition 2 we onsider right tringle in whih the hypotenuse hs length 19 inhes nd one leg hs length 8 inhes, oth to the nerest inh. We wnt to estimte the length of the other leg. Clultion yields = =? 19 in 8 in 17.23368794, figure whih must e rounded ppropritely. Beuse the input dt for this prolem ws given to the nerest inh, we rounded to the nerest inh to get =
32 17 inhes. However, if we were to use the method of signifint figures, then euse one of the input dt 8 inhes hs only one signifint figure, we would onsequently round to one signifint figure, otining = 20 inhes. This estimte is oviously inorret euse the length of leg must lwys e less thn the length of the hypotenuse whih is 19 inhes. To get more preise ide out the inury of the estimte = 20 inhes, we perform n intervl nlysis on this prolem. Let represent the length of the leg whih is 8 inhes to the nerest inh, nd let represent the length of the hypotenuse 19 inhes to the nerest inh. Then nd tully lie in the following rnges: 7.5 8.5 nd 18.5 19.5. Sine = 2 " 2, then the smllest possile vlue of is nd the lrgest possile vlue of is Thus, lies in the rnge = 18.5 2 " 8.5 2 = 16.43167673 = 19.5 2 " 7.5 2 = 18.000 16.43167673 18.000. This rnge omfortly ontins our originl estimte, = 17. However, the estimte = 20 otin y using signifint figures lies well outside this rnge. The length of this rnge is 18 16.43167673 = 1.56832327. Hene, the estimte = 20 lies frther ove the upper end, 18, of the rnge thn the entire length of the rnge! Thus, the use of signifint figures in this lultion introdues signifint mount of inury. Morl: Don t use signifint figures in Pythgoren Theorem lultions. A Brief Review of Similr Tringles Two tringles T nd T re similr if there is orrespondene etween the sides of T nd the sides of T nd there is rel numer r > 0 lled sle ftor suh tht the length of eh side of T is equl to r times the length of the orresponding side of T. In the figure on the next pge, suppose tht tringle T with side lengths, nd is similr to tringle T with side lengths, nd so tht the side of T of length orresponds to the side of T of length, the side of T of length orresponds to the side of T of length nd the side of T of length orresponds to the side of T of length. Then there is sle ftor r > 0 suh tht = r, = r nd = r.
33 T T We oserve tht in this sitution, the similrity of tringles T nd T n e expressed without expliitly mentioning the sle ftor r. This is euse the rtios ", " nd re ll equl to the sle ftor r. Hene the three equtions " = r, = r nd = r n e expressed without mentioning the sle ftor r y writing the equtions " = " = ". Furthermore, y ross-multiplying these equtions n e trnsformed into the equtions = ", = " nd = ". Thus, the similrity of tringles T nd T n e expressed y ny one of the following three different fmilies of similrity equtions: 1) = r, = r nd = r. 2) " = " = ". 3) = ", = " nd = ". In two of these fmilies of similrity equtions, the sle ftor doesn t pper. The importnt lesson for student onfronted with prolem involving similr tringles is: the student should use the fmily of similrity equtions tht provides the most help with the solving the prolem t hnd. The knk of hoosing the most helpful fmily of similrity equtions to solve prtiulr similr tringles prolem is prt of the rt of prolem solving. We mke seond oservtion out this sitution. The lengths, nd of the sides of tringle T must ll e expressed in the sme units. Similrly, the lengths, nd of the sides of tringle T must ll e expressed in the sme units. However,
34, nd n e expressed in different units from, nd! The three fmilies of similrity equtions will vlidly express the similrity of tringles T nd T even if the side lengths of T re mesured in different units thn the side lengths of T. There is no need to onvert the side lengths of T nd T to ommon units of mesure to use the similrity equtions. We will solve prolem elow whih illustrtes this oservtion; in this prolem the side lengths of T re expressed in inhes, the side lengths of T re expressed in entimeters, nd the prolem is solved without ever onverting to ommon unit of mesure. Aording to our definition, the similrity of two tringles depends on the rtios of side lengths of the two tringles. There is surprising nd useful geometri ft (or theorem) tht whih llows us to detet the similrity of two tringles, not from the lengths of their sides, ut insted from the mesures of their ngles. This theorem, whih we ll the Angle Theorem for Similr Tringles sys tht if two ngles in tringle T hve the sme mesure (i.e. size) s two ngles in tringle T, then tringles T nd T re similr. We stte this theorem more preisely: The Angle Theorem for Similr Tringles. Suppose T is tringle with side lengths, nd nd T is tringle with side lengths, nd. If the ngle in tringle T tht is opposite the side of length hs the sme mesure s the ngle in tringle T tht is opposite the side of length, nd the ngle in tringle T tht is opposite the side of length hs the sme mesure s the ngle in tringle T tht is opposite the side of length, then tringles T nd T re similr, nd, in prtiulr, there is sle ftor r > 0 suh tht = r, = r nd = r. (Then, of ourse, the other two fmilies of similrity equtions lso hold: " = " = " nd = ", = " nd = ".) T T We will now show how pply this theorem together with the two Clultion Rules to solve prolem. In the figure on the next pge, suppose two ngles in the tringle T on the left hve the sme mesures s two ngles in the tringle T on the
35 right (s indited). The lengths of two sides of T nd one side of T re given. Find the side length leled in tringle T in entimeters. =? 13 inhes T T 18 inhes 6.89 m Beuse two ngles in T hve the sme mesures s two ngles in T, the Angle Theorem for Similr Tringles implies tht T nd T re similr in suh wy tht the following eqution holds. Solving this eqution for, we get " 6.89 = 13 18. = 6.89 x 13 18 Using lultor, we evlute the expression on the right side of this eqution. Following Clultion Rule 1, we don t round t ny point during this lultion. For exmple, we don t lulte the frtion 13/18 nd round it efore multiplying y 6.89. This lultion gives us: = 4.9761111 Hving ompleted this lultion, we round the finl nswer following Clultion Rule 2. This lultion is n pproprite ple to use signifint figures, euse the input dt is expressed in different units (inhes nd entimeters) nd the lultion involves division. Our vlue for n e no more urte thn the lest urte input. Sine the two inputs 13 inhes nd 18 inhes hve only two signifint figures, then we should round our vlue for to hve only two signifint figures. This gives us: = 5.0 entimeters (to the nerest tenth of entimeter). Oserve tht the side lengths of T re expressed in inhes, the side lengths of T re expressed in entimeters, nd the prolem is solved without onverting the lengths to ommon unit of mesure.
36 Ativity 2. Eh group should work on the prolem stted elow nd report its solution to the lss. The lss should disuss ny disrepnies in the solutions. Be sure to oey the two Clultion Rules. As indited in the figure elow, two ngles in the tringle T on the left hve the sme mesures s two ngles in the tringle T on the right. The lengths of one side of T nd two sides of T re given. Find the side length leled in tringle T in inhes. 1.3 feet 1.83 feet 6.45 inhes =? T T
37 A Proof of the Pythgoren Theorem Ativity 3. Groups should work on the following prolem. This tivity is designed to led you to proof of the Pythgoren Theorem. Suppose you re given the tringle shown here whih hs sides of length, nd suh tht the sides of length nd form right ngle. The following tivity is designed to led you to proof of the eqution 2 + 2 = 2. Below re two squres of the sme re, eh with sides of length +. We ll the squre on the left Big Squre 1 nd the squre on the right Big Squre 2. Big Squre 1 Big Squre 2 On the next pge re 11 figures: 8 opies of the ove tringle together with 3 squres, one of whih hs sides of length, nother of whih hs sides of length, nd the third of whih hs sides of length. Photoopy this pge onto stiffer pper if possile. Cut out the 11 figures in the photoopy nd rrnge them so tht they extly over the Big Squres 1 nd 2 without overlpping. Furthermore, over Big Squre 1
38 with 4 tringles nd the two squres of side lengths nd, nd over Big Squre 2 with 4 tringles nd the squre of side length. Compre the sum of the res of the figures overing Big Squre 1 to the sum of the res of the figures overing Big Squre 2. From this omprison, prove the eqution 2 + 2 = 2.
39 Homework Prolem 1. This prolem hs prts ) through k). In prts ) through h), find the unknown side lengths. Be sure to oey the two Clultion Rules in ll prts of the prolem. ) ) 15.0 m =? =? 5.05 in 8.0 m 3.03 in ) d).01 km.02 km =? =? = 25 m = 60 m to the nerest 10m e) f) 3.5 ft 3.70 ft 214.37 m x =? z =? 365.2 m g) h) =?.5 m 55.00 mi =?.16 m 23.000 mi
40 i) The distne from the enter of the Erth to the enter of the Sun is 93,000,000 miles to the nerest 100,000 miles. The rdius of the Erth is 4000 miles to the nerest 100 miles. P is point on the surfe of the Erth with the property tht the line joining the enter of the Erth to P is perpendiulr to the line joining the enters of the Erth nd the Sun. How fr is P from the enter of the Sun? P 4000 mi? Sun Erth 93,000,000 mi j) A ldder lens ginst wll. The foot of the ldder is 3 feet 5 1 / 4 inhes from the ottom of the wll to the nerest qurter of n inh. The ldder is 12 feet long to the nerest inh. How high ove the ground is the top of the ldder? k) On piee of grph pper drw the x nd y xes nd lel units on these xes so tht you n plot the points A = (17,2), B = (2,19) nd C = (3,11) on the piee of pper. Clulte the perimeter of the tringle with verties A, B nd C to the nerest tenth of unit. Homework Prolem 2. In prts ) nd ) of this prolem, two ngles in the tringle on the left hve the sme mesures s two ngles in the tringle on the right (s indited). Find the unknown side lengths. Be sure to oey the two Clultion Rules. ) 2.7 in 237 mi =? 320 mi
41 ) 75 feet =? 98 feet.0583 m Homework Prolem 3. ) Refer to Ativity 3. Write out nd hnd in detiled nd reful explntion (i.e., proof) of why 2 + 2 = 2. You my use fts out res of squres nd tringles in your proof. ) Adpt this rgument to the tringle with sides of length x, y nd z shown elow. In other words, drw two squres of side x + y nd sudivide them into squres nd tringles so tht the proof desried ove for the tringle with side lengths, nd n e modified to yield proof tht x 2 + y 2 = z 2. z y x Homework Prolem 4. The gol of this prolem is lso to led you to proof of the Pythgoren Theorem. This proof is different from the one presented in Ativity 3. Agin suppose you re given the tringle shown here whih hs sides of length, nd suh tht the sides of length nd form right ngle.
42 ) In the figure elow, we eret three squres on the sides of this tringle. One squre hs side length, the seond hs side length, nd the third hs side length. The squre of side length is sudivided y dotted lines. Photoopy this digrm onto stiffer pper if possile. Cut out the squre of side length nd ut it prt long the dotted lines to otin five piees. Show tht the five piees n e rrnged to over the two squres of side lengths nd with no overlp. Argue tht this proves 2 + 2 = 2. ) Adpt this rgument to the tringle with sides of length x, y nd z shown elow. In other words, eret 3 squres on the sides of this tringle nd show how to sudivide the squre of side length z into five piees tht will extly over the two squres of side lengths x nd y. z y x
43 Homework Prolem 5. This prolem leds to yet nother proof of the Pythgoren Theorem. This proof is similr to the previous one. Agin strt with the tringle shown here with side lengths, nd nd right ngle etween the sides of length nd. ) Agin, in the figure elow, we eret three squres on the sides of this tringle. This time the ig squre with side length is sudivided y four dotted line segments. These dotted line segments re determined y the rule tht they re either horizontl or vertil nd one endpoint of eh dotted line segment is the midpoint of side of the ig squre. Agin, photoopy this digrm onto stiffer pper if possile, nd ut the ig squre prt long the dotted line segments to otin five piees, nd show tht the five piees n e rrnged to over the two smller squres with no overlp. Argue tht this proves 2 + 2 = 2.
44 ) Adpt this rgument to the tringle with sides of length x, y nd z shown elow. z y x Homework Prolem 6. The gol of this prolem is to led you to proof of the Pythgoren Theorem tht ws known to the nient Chinese out 1000 BC. (Pythgors lived out 500 BC.) Suppose you re given the tringle T shown here whih hs sides of length, nd suh tht the sides of length nd form right ngle. T ) Consider the following two figures. The figure on the left is squre whih hs sides of length. The figure on the right is the union of two squres, one of whih hs sides of length nd the other of whih hs sides of length. Eh figure is sudivided y dotted lines into 5 piees: 4 opies of the originl tringle T nd squre. Compre the sum of the res of the piees mking up the squre on the left one with the sum of the res of the piees mking up the figure on the right. Try to derive the eqution 2 + 2 = 2. (Hint: Eh figure ontins 4 tringles nd squre. How long re the sides of these squres?) T T T T T T T T
45 ) Adpt this rgument to the tringle with sides of length x, y nd z shown elow. z y x
46