Chapter 10 Geometry: Angles, Triangles and Distance

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1 hpter 10 Geometry: ngles, Tringles nd Distne In setion 1 we egin y gthering together fts out ngles nd tringles tht hve lredy een disussed in previous grdes. This time the ide is to se student understnding of these fts on the trnsformtionl geometry introdued in the preeding hpter. s efore, here the ojetive is to give students n informl nd intuitive understnding of these fts out ngles nd tringles; ll this mteril will e resumed in Seondry Mthemtis in more forml nd logilly onsistent exposition. Setion 2 is out stndrd 8G6: Explin proof of the Pythgoren Theorem nd its onverse. The lnguge of this stndrd is very preise: it does not sy Prove... ut it sys Explin proof of..., suggesting tht the point for students is to rtiulte their understnding of the theorem; not to demonstrte skill in reiting forml proof. lthough Mthemtis tends to e quite rigorous in the onstrution of forml proofs, we know through experiene, tht informl, intuitive understnding of the why of proof lwys preedes its rtiultion. Strting with this point of view, the student is guided through pprohes to the Pythgoren theorem tht mke it elievle, insted of forml rguments. In turn, the student should e etter le to explin the resoning ehind the Pythgoren theorem, thn to provide it in form tht exhiits form over grsp. In hpter 7, in the study of tilted squres, this text suggests tht y repling speifi numers y generi ones, we get the Pythgoren theorem. We strt this setion y turning this suggestion into n explntion of proof, nd we give one other wy of seeing tht this is true. There re mny; see, for exmple jwilson.oe.ug.edu/emt668/emt668.student.folders/hedngel/essy1/pythgoren.html The onverse of the Pythgoren theorem sttes this: if = 2, where,, re the lengths of the sides of tringle, then the tringle is right tringle, nd the right ngle is tht opposite the side with length. The Euliden proof of this sttement is n pplition of S S S for tringles. lthough students plyed with S S S in hpter 9, here we wnt more intuitive nd dynmi understnding. Our purpose here is to enourge thinking out dynmis, whih eomes entrl tool in lter mthemtis. We look t the olletion of tringles with two side lengths nd with. s the ngle t grows from very tiny to very ner stright ngle, the length of its opposing side stedily inreses. It strts out very ner, nd ends up very ner +. There is only on tringle in this sequene where is preisely 2. In the finl setion, we use the Pythgoren Theorem to lulte distnes etween points in oordinte plne, This is wht the relevnt stndrd sks: it does not sk tht students know the distne formul. The gol here is tht students understnd the proess to lulte distnes: this proess involves right ngles nd the Pythgoren theorem nd students re to understnd tht involvement. onentrtion on the formul perverts this ojetive. 8MF10-1

2 Setion 10.1 ngles nd Tringles Use informl rguments to understnd si fts out the ngle sum nd exterior ngle o tringles, out the ngles reted when prllel lines re ut y trnsversl, nd the ngle-ngle riterion for similrity of tringles. 8G5 In this setion, we ontinue the theme of the preeding hpter: to hieve geometri intuition through explortion. We strt with geometri fts tht students lerned in 7th grde or erlier, exploring them from the point of view of rigid motions nd diltions. (1). Vertil ngles t the point of intersetion of two lines hve the sme mesure. The mening of vertil is in the sense of vertex. Thus, in Figure 1, the ngles t V with rrows re vertil ngles, s is the pir t V without rrows. V D Figure 1 Now rottion with vertex V through stright ngle (180 ) tkes the line into itself. More speifilly, it tkes segment V to V nd V to VD, nd so rries V to VD,. This rottion therefore shows tht the ngles V nd VD re ongruent, nd thushve the sme mesure. The trditionl rgument (nd tht whih ppers in grde 7) is this: oth ngles V nd VD re supplementry to V (rell tht two ngles re supplementry if they dd to stright ngle), nd therefore must hve equl mesure. However, in grde 8 we wnt to understnd mesure equlity in terms of ongruene, nd ongruene in terms of rigid motions. (2). If if two lines re prllel, nd third line L uts ross oth, then orresponding ngles t the points of intersetion hve the sme mesure. L P L P L Figure 2 In Figure 2, the two prllel lines re L nd L, nd the orresponding ngles re s mrked t P nd P. The 8MF10-2

3 trnsltion tht tkes the point P to the point P tkes the line L to the line L euse trnsltion tkes line to nother one prllel to it, nd y the hypothesis, L is the line through P prllel to L. Sine trnsltions lso preserve the mesure of ngles, the orresponding ngles s mrked (t P nd P ) hve the sme mesure. Now, euse of (1) ove, tht opposing ngles t vertex re of equl mesure, we n onlude tht the ngle denoted y the dshed r is lso equl to the ngles denoted y the solid r. This figure lso demonstrtes the onverse sttement: (3). Given two lines, if third line L uts ross oth so tht orresponding ngles re equl, then the two lines re prllel. To show this, we gin drw Figure 2, ut now the hypothesis is tht the mrked ngles t P nd P hve the sme mesure. Sine trnsltions preserve the mesure of ngles, the trnsltion of P to P tkes the ngle P P to P, nd so the imge of L hs to ontin the ry P, nd so is the line L. Sine the line L is the imge of L under trnsltion, these lines re prllel. (4). The sum of the interior ngles of tringle is stright ngle. In seventh grde, students sw this to e true y drwing n ritrry tringle, utting out the ngles t the verties, nd putting them t the sme vertex. Every replition of this experiment produes stright ngle. This experiment is onvining tht the sttement is true, ut does not tell us why it is true. We hve two rguments to show why it is true. The first hs the dvntge tht it uses onstrution with whih the student is fmilir (tht to find the re of tringle) nd thus reinfores tht ide. The seond hs the dvntge tht it n e generlized to polygons with more sides. First, drw tringle with horizontl se (the tringle with solid sides in Figure 3). Rotte opy of the tringle round the vertex, nd then trnslte the new tringle upwrds to get the result shown in Figure 3, in whih the tringle with dshed sides is the new position of the opied tringle. We hve indited the orresponding ngles with the greek letters α, β, γ. Sine the ngles nd hve the sme mesure (β), the lines nd re prllel. Sine they re prllel, the ngles nd E hve the sme mesure Now look t the point = : the ngles α, β, γ. form stright ngle. = β α γ α = n lterntive rgument is sed on Figure 4 elow: β Figure 3 γ γ γ α α β β Figure 4 In this figure we hve nmed the exterior ngles of the tringle, α, β, γ, eh of whih is outside the tringle formed y the extension of the side of the tringle on the right. If we were to wlk round the perimeter of the 8MF10-3

4 tringle, strting nd ending t looking in the diretion of, we would rotte our line of vision y full irle, 360. s this is the sum of the exterior ngles, we hve α + β + γ = 360. ut eh ngle in this expression is supplementry to the orresponding ngle of the tringle, tht is, the sum of the mesures of the ngle is 180. So, the ove eqution eomes from whih we get α + β + γ = 180. (180 α) + (180 β) + (180 γ) = 360, We n generlize the seond rgument to polygons with more sides. onsider the qudrilterl in Figure 5. D δ γ α β Figure 5 y the sme resoning s for the tringle, the sum of the exterior ngles of the qudrilterl is lso 360 nd the sum of eh interior ngle nd its exterior ngle is 180. ut now there re four ngles, so we end up with the eqution (180 α) + (180 β) + (180 γ) δ) = 360, or 720 (α + β + γ + δ) = 360. (5). The sum of the interior ngles of qudrilterl is 360. n you now show tht, for five sided polygon, the sum of the interior ngles of qudrilterl is 540? n you go from there to the formul for generl polygon? (6). If two tringles re similr, then the rtios of the lengths of orresponding sides is the sme, nd orresponding ngles hve the sme mesure. (7). Given two tringles, if we n lel the verties so tht orresponding ngles hve the sme mesure, then the tringles re similr. Figure 6 We sw in hpter 9 why (6) is true. Let us look more losely t sttement (7). 8MF10-4

5 = Figure 6 Figure 6 shows possile onfigurtion of the two tringles. y trnsltion, we n ple point on top of pint to get Figure 6. Now move the smller tringle y rottion with enter =, so tht the point lnds on the segment. Sine the ngles nd hve the sme mesure, the rottion must move line segment so tht it lies on. Now, sine nd hve the sme mesure, the line segments nd must e prllel (y Proposition 2). Now the diltion with enter tht puts point on, puts tringle onto tringle, so they re similr. The rgument is not fully ompleted, for the onfigurtion of Figure 6 is not the only possiility. In FIgure 6, is etween nd if we trverse the edge of the tringle in the lokwise diretion, nd the sme is true for. However, this is not true in the onfigurtion of Figure 7. Now how do you find the desired similrity trnsformtion? Figure 7 gin, we trnslte so tht the points nd oinide. ut now, the rottion tht puts the line segment on the sme line s doesn t led to the onfigurtion of Figure 6 nd the smller tringle nnot e rotted so tht orresponding sides lie on the sme ry. ut this is fixed y refleting the smller tringle in the line ontining the segment nd, nd now we re in the onfigurtion of Figure 6. Indeed, we ould hve strted with refletion of the smll tringle in the line through, nd then followed the originl rgument. Hve we overed ll ses? The nswer is yes: the differene etween Figure 6 nd tht of Figure 7 is tht of orienttion. So, if we strt gin with the two tringles nd with orresponding ngles of the sme mesure, then we should first sk: is the orienttion the sme s the orienttion (oth lokwise or oth ounter lokwise)? If so, we re in the se of Figure 6. If not, fter refletion in ny side of tringle puts us in the se of Figure 7. Setion 10.2 The Pythgoren Theorem. Explin proof of the Pythgoren Theorem nd its onverse. 8G6 In hpter 7 we onstruted tilted squres of side length, whose re 2 is speifi integer. If is not n integer, we hve oserved tht it nnot e expressed s rtionl numer ( quotient of integers). t the end of the disussion we mentioned tht these speifi exmples generlized to generl theorem (known s the Pythgoren 8MF10-5

6 theorem) relting the lengths of the sides of right tringle. This mthemtil ft is nmed fter sixth entury E mthemtil soiety (presumed to e led y someone nmed Pythgors). It is ler tht this ws known to muh erlier iviliztions: the written reord shows it eing used y the Egyptins for lnd mesurements, nd n nient hinese doument even illustrtes proof. ut the Pythgoren Soiety ws given the redit for this y third entury E Greek mthemtiins. The Pythgoren Soiety is lso redited with the disovery of onstrutile line segments whose length nnot e represented y quotient of integers (in prtiulr, the hypotenuse of tringle whose legs re oth the sme integer. ). The legend is tht the disoverer of this ft ws srified y the Pythgorens. We mention this only to highlight how muh the pproh to mthemtis hs hnged in 2500 yers; in prtiulr this ft, onsidered unfortunte then, is now ppreited s ornerstone of the ttempt to fully understnd the onept of numer nd its reltionship to geometry. Let s pik up with the disussion t the end of hpter 7, setion 1. There we pled our tilted squres in oordinte plne so s to e le to more esily see the reltionship etween the res of the squres nd its ssoited tringles. Here, in order to stress tht the understnding of the Pythgoren theorem does not involved oordintes, we look t those hpter 7 rguments in oordinte free plne. For two positive numers, nd, onstrut the squre of side length +. This is the squre ounded y solid lines in Figure 8, with the division points etween the lengths nd mrked on eh side. Drw the figure joining these points - this is the squre with dshed sides in Figure 8. In hpter 7 we oserved tht this is squre; now, let us hek tht this is so. First of ll, tringle I is ongruent to tringle II: trnslte tringle I horizontlly so tht the side leled lies on the side of tringle II leled. Now rotte tringle I (in its new position) y 90. Then the sides leled nd oinide nd in the ft the tringles oinide, so re ongruent. In the sme wy we n show tht tringe II is ongruent to tringle III, nd III is ongrurent to IV. I III IV II Figure 8 We onlude tht the dshed lines re ll of the sme length, telling us tht the dshed figure is rhomus. To e squre, we hve to show tht one of the orner ngles is right ngle. Lets onentrte on the vertex mrked, where the ngle is mrked γ. We see tht α + γ + β = 180. ut the sum of the non-right tringes of right tringe is 90, nd these re, y the ongruene of tringles I nd IV, the sme s the ngles mrked α nd β; tht is So, it follows tht γ = 90. α + β = 90. In hpter 7 we used these fts to lulte the re of the tilted squre, whih is 2, where is the length of the hypotenuse of the four tringles. y reonfiguring the piture s in Figure 9, we n show tht this is lso This is known s the hinese proof of the Pythgoren theorem, nd reords show tht it preedes the Pythgoren Soiety y lose to 1000 yers. 8MF10-6

7 lthough this is proof without words, here is desription of how it goes: For the originl squre of side length + is now sudivided in different onfigurtion: the ottom left orner is filled with squre of side length, nd the upper right orner, y squre of side length. I II III IV Figure 9 The rest of the ig squre of Figure 9 onsists of four tringles, so wht is outside those tringles hs re ut eh of those tringles is ongruent to the tringle of the sme lel in Figure 8. Thus wht is outside those four tringles must hve re 2. This result is: The Pythgoren Theorem: = 2 for right tringle whose leg lengths re nd nd whose hypotenuse is of length. It is instrutive to give nother proof without words of the Pythgoren theorem; this one is due to hskr, 12th entury E mthemtiin in Indi. Strt with the squre of side length (so of re 2, nd drw the right tringles of leg lengths nd, with hypotenuse side of this squre, s shown in the top left of Figure 10. Now reonfigure these tringles nd the interior squre s in the imge on the right. Sine it is reonfigurtion, it still hs re 2. ut now, y redrwing s in the lower imge, we see tht this figure onsists of two squres, one of re 2 nd the other of re 2. Figure 10 Exmple 1.. right tringle hs leg lengths 6 in nd 8 in. Wht is the length of its hypotenuse? 8MF10-7

8 Solution. Let e the length of the hypotenuse. y the Pythgoren theorem, we know tht so = = = = 100,. nother tringle hs leg lengths 20 in nd 25 in. Give n pproximte vlue for the length of its hypotenuse. Solution. 2 = = = 1025 = So, = = Sine 6 2 = 36 nd 7 2 = 49, we know tht 41 is etween 6 nd 7; proly it loser to 6. We lulte = 40.96, so it mkes good sense to use the vlue 6.4 to pproximte 41. Then the orresponding pproximte vlue of is = 32. Exmple 2.. The hypotenuse of tringle is 25 ft, nd one leg is 10 ft long. How long is the other leg? Solution. Let e the length of the other leg. We know tht = 25 2, or = 625, nd thus 2 = 525. We n then write = 525. If we wnt to pproximte tht, we first ftor 525 = 25 21, sp = 5 21 We n pproximte 21 y 4.5 (4.5 2 = 20.25). nd thus is pproximtely given y = n isoseles right tringle hs hypotenuse of length 100 m. Wht is the leg length of the tringle. Solution. Referring to to the Pythgoren theorem, we re given: = (the tringle is isoseles), nd = 100. So, we hve to solve the eqution 2 2 = Sine = 10 4, we hve to solve 2 2 = 10 4, or 2 = Write = , whih rings us to = = Sine 7 2 = 49, we n give the pproximte nswers = 10 7 = 70. Exmple 3. h h D h Figure 11 Figure 11 is tht of n isoseles right tringle,, lying on top of squre. The totl re of the figure is 1250 sq. ft. Wht is the length of the ltitude (D) of the tringle? Note: This prolem my e eyond the sope of 8th grde mthemtis, ut it still my e worth disussing, sine it illustrtes how the intertion of geometry nd lger works to solve omplex struturl prolem. Solution. Sine the tringle is isoseles, the mesure of D nd D re oth 45. The ltitude of tringle is perpendiulr to the se ; from whih we onlude tht the tringles D nd D 8MF10-8

9 re ongruent. Sine we wnt to find the length of D, let s denote tht numer y h. This gives us the full lelling of Figure 11. The informtion given to us is out re, so let s do n re lultion. Sine the length of side of the squre is 2h, its re is 4h 2. The re of the tringle on top of the squre (one-hlf se times ltitude) is 1 2 (2h)(h) = h2. So, the re of the entire figure is 4h 2 + h 2 = 5h 2, nd we re given tht tht is 1250 sq. ft. We then hve 5h 2 = 1250 so h 2 = 250 = nd h = Now the Pythgoren theorem desries reltion mong the lengths of the sides of right tringle; it is lso true tht this reltion desries right tringle. This is wht is lled the onverse of the Pythgoren theorem. The ide of onverse is importnt in mthemtis. Most theorems of mthemtis re of the form : under ertin onditions we must hve speifi onlusion. The onverse sks: if the onlusion is oserved, does tht men tht the given onditions hold, tht is, does the eqution = 2 relting the sides of tringle tell us tht the tringle s right tringle? onverse of the Pythgoren Theorem: For tringle with side lengths,, if = 2, then is right ngle. In order to see why this is true, we show how to drw ll tringles with two side lengths nd. Let s suppose tht. On horizontl line, drw line segment of length. Now drw the semiirle whose enter is nd whose rdius is (see Figure 12). Then, ny tringle with two side lengths nd is ongruent to tringle with one side, nd the other side the line segment from to point on the irle. Figure 12 Now, s the line segment is rotted round the point, the length of the line segment ontinully inreses. When is vertil, we hve the right tringle, for whih the length = We onlude tht for ny tringle with side lengths nd, the length of the third side is either less thn (tringle is ute), or greter thn (tringle is otuse) exept for the right tringle(where the segment is vertil). Exmple 4. Drw irle nd its horizontl dimeter ( in Figure 13). Pik point on the irle. Verify y mesurement tht tringle is right tringle. Solution. For the prtiulr tringle the mesures of the side lengths, up to nerest millimeter re: = 44 mm, = 18 mm, = 40 mm. Now, lulte: = = 1924, nd 2 = This is pretty lose. If ll students in the lss get this lose, ll with different figures, then tht is sustntil sttistil evidene for the lim tht the tringle is lwys right tringle. 8MF10-9

10 Figure 13 Setion 10.3 pplitions of the Pythgoren Theorem. pply the Pythgoren Theorem to determine unknown side lengths in right tringles in rel-world nd mthemtil prolems in two nd three dimensions. 8G7 Exmple 5. Wht is the length of the digonl of retngle of side lengths 1 inh nd 4 inhes? Solution. The digonl is the hypotenuse of right tringle of side lengths 1 nd 4, so is of length = 17. Exmple 6. Suppose we doule the lengths of the legs of right tringle. y wht ftor does the length of the digonl hnge, nd y wht ftor does the re hnge? Solution. This sitution is illustrted in Figure 14, where the tringles hve een moved y rigid motions so tht they hve legs tht re horizontl nd vertil, nd they hve the vertex in ommon. ut now we n see tht the diltion with enter tht moves to puts the smller tringle on top of the lrger one. The ftor of this diltion s 2. Thus ll length hnge y the ftor 2, nd re hnges y the ftor 2 2 = 4. Exmple 7. Figure 14 n 18 ft ldder is lening ginst wll, with the se of the ldder 8 feet wy from the se of the wll pproximtely how high up the wll is the top of the ldder? Solution. The sitution is visulized in Figure 15. The onfigurtion is right tringle with hy- 8MF10-10

11 potenuse (the ldder) of length 18 feet, the se of length 8 feet, nd the other leg of length h. y the Pythgoren theorem, we hve h = 18 2 or h = 324. Then h 2 = 260. Sine we just wnt n pproximte nswer, we look for the integer whose squre is lose to 260: tht would e 16 (16 2 = 256). So, the top of the ldder hits the wll out 16 feet ove the ground. 18 h 8 Figure 15 Exmple 8. room is in the shpe of retngle of width 12 feet, length 20 feet, nd height 8 feet. Wht is the distne from one orner of the floor (point in the figure) to the opposite orner on the eiling? Figure Solution. In Figure 16, we wnt to find the length of the dshed line from to. Now, the dsh-dot line on the floor of the room is the hypotenuse of right tringle of leg lengths 12 ft nd 20 ft. So, its length is = The length whose mesure we wnt is the hypotenuse of tringle whose leg lengths re 23.3 nd 8 feet. Using the Pythgoren theorem gin we onlude tht the mesure of the line segment in whih we re interested () is = 24.64; sine our originl dt were given in feet, the nswer: 25 ft. should suffie. Exmple 9. Wht is the length of the longest line segment in the unit ue? Solution. We n use the sme figure s in the preeding prolem, tking tht to e the unit ue. 8MF10-11

12 Then the length of the digonl on the ottom fe is = 2 units, nd the length of the digonl is ( 2) 2 = 3. These exmples show tht we n use the Pythgoren theorem to find lengths of line segments in spe. Given the points nd, drw the retngulr prism with sides prllel to the oordinte plnes tht hs nd s dimetrilly opposite verties (refer to Figure 16). Then, s in exmple 8, the distne etween nd is the squre root of the sum of the lengths of the sides. Exmple 10. Wht is the length of the longest line segment in ox of width 10, length 16 nd height 8? Length = = = 428 = = inhes, or little more tht 20 inhes. Exmple 11. In the movie Despile Me, n infltle model of The Gret Pyrmid of Giz in Egypt ws reted y Vetor to trik people into thinking tht the tul pyrmid hd not een stolen. When inflted, the flse Gret Pyrmid hd squre se of side length 100 m. nd the height of one of the side tringles ws 230 m. Wht is the volume of gs tht ws used to fully inflte the fke Pyrmid? h 230m 100m 50m Figure 17 Solution. The sitution is depited in Figure 17. Now, we know tht the formul for the volume of pyrmid is 1 3 h, whee is the re of the se nd h is the height of the pyrmid (the distne from the se to the pex, denoted y h in the figure). Sine the se is squre of side length 100 m., its re is 10 4 m 2. To lulte the height, we oserve (sine the pex of the pyrmid is diretly ove the enter of the se), tht h is leg of right tringle whose other leg is 50 m. nd whose hypotenuse hs length 230 m. y the Pythgoren theorem h = lulting we find h 2 = = = Tking squre roots, we hve h = 225 pproximtely. Then, the volume of the pyrmid is Volume = 1 3 (104 )( ) = = 75, 000 m 3. Setion 10.4 The Distne etween Two Points pply the Pythgoren theorem to find the distne etween two points. 8G8. 8MF10-12

13 For ny two points P nd Q, the distne etween P nd Q is the length of the line segment PQ. We n pproximte the distne etween two points y mesuring with ruler, nd if we re looking t sle drwing, we will hve to use the sle onversion. If the two points re on oordinte plne, we n find the distne etween the points using the oordintes y pplying the Pythgoren theorem. The following sequene of exmples demonstrtes this method, strting with stright mesurement. Exmple 12.. Using ruler, estimte the distne etween eh of the three points P, Q nd R on Figure 18. Solution. The mesurements I get re PQ = 39 mm; PR = 39 mm nd QR = 41 mm. Of ourse, the tul mesures one gets will depend upon the disply of the figure.. Now mesure the horizontl nd vertil line segments (the dshed line segments in the figure), to onfirm the Pythgoren theorem. Solution. The horizontl line is 30 mm, nd the vertil line is 27 mm. Now = = 1629, nd 39 2 = This pproximte vlues re lose enough to onfirm the distne lultion, nd ttriute the disrepny to minor impreision in mesurement. Q P Exmple 13. Figure 18 In the ompnying mp of the northest United Sttes, one inh represents 100 mies. Using ruler nd the sle on the mp, lulte the distne etween R. Pittsurgh nd Providene. Providene nd onord. Pittsurgh nd onord d. Now test whether or not the diretions Providene Pittsurgh nd Providene onord re t right ngles. Solution. fter printing out the mp, we mesured the distnes with ruler, nd found. Pittsurgh to Providene: Four nd n eighth inhes, or miles;. Providene to onord: 15/16 of n inh, or miles; 8MF10-13

14 . Pittsurg to Providene: Four nd qurter inhes, or 425 miles. d. Sine we re only pproximting these distnes, we round to integer vlues nd then hek the Pythgoren formul to see how lose we ome to hve oth sides equl Let us denote these distnes y the orresponding letters,,, so tht = 413, = 94, = 425. We now lulte the omponents of the Pythgoren formul: 2 = 170, 569, 2 = 8, 836, so = 179, 405 ; 2 = 180, 625. The error, 1220, is well within one perent of 2, so this ngle n e tken to e right ngle. Exmple 14. On oordinte plne, lote the points P(3, 2) nd Q(7, 5) nd estimte the distne etween P nd Q. Now drw the horizontl line strting t P nd the vertil line strting t Q nd let R e the point of intersetion. lulte the length of PQ using the Pythgoren theorem. Solution. First of ll, we know the oordintes for R: (7,2). So the length of PR is 4, nd the length of QR is 3. y the Pythgoren theorem, the length of PQ is = 5. The mesurement with ruler should onfirm tht. Exmple 15. Find the distne etween eh pir of these three points on the oordinte plne: P( 3, 2), Q(7, 7) nd R(2, 4). 8MF10-14

15 (7, 7) ( 3, 2) (2, 4) Figure 19 Solution. In Figure 19 hve drwn the points nd represented the line joining them y dotted lines. To lulte the lengths of these line segments, we onsider the right tringle with horizontl nd vertil legs nd PQ s hypotenuse. The length of the horizontl leg is 7-(-3) = 10, nd tht of the vertil leg is 7-2 = 5. So PQ = = 5 2 ( ) = 5 5. For the other two lengths, use the slope tringles s shown nd perform the sme lultion: QR = = = 146, PR = = 61. These exmples show us tht the distne etween two points in the plne n e lulted using the Pythgoren theorem, sine the slope tringle with hypotenuse the line segment joining the two points is right ngle. This n e stted s formul, using symols for the oordintes of the two points, ut it is est if students understnd the protool nd the resoning ehind it, nd y no mens should memorize the formul. Nevertheless, for ompleteness, here it is. The oordinte Distne Formul: Given points P : (x 0, y 0 ), Q(x 1, y 1 ) PQ = (x 1 x 0 ) 2 + (y 1 y 0 ) 2. Exmple 16. Figure 20 is photogrph, y Jn-Pieter Np, of the Mount romo volno on the islnd Jv of Indonesi tken on July MF10-15

16 Figure 20 From the ottom of the volno (in our line of vision) to the top is 6000 feet. Given tht informtion, y mesuring with rule, find out how long the visile prt of the left slope is, nd how high the plume of smoke is. Solution. Using ruler, we find tht the height of the imge of the volno is 12 mm, the length of the left slope is 21 mm, nd the plume is 10 mm high. Now we re given the informtion tht the visile prt of the volno is 5000 feet, nd tht is represented in the imge y 12 mm. Thus the sle of this photo (t the volno) is 12 mm : 6000 feet, or 1 mm : 500 feet. Then the slope is = 10, 500 feet, nd the plume is = 5000 feet high. Exmple f t? f t 12 f t Figure 21 In my kyrd I pln to uild retngulr shed tht is 12 y 20, with peked roof, s shown in Figure 21. The pek of the roof is 3 ove the eiling of the shed. How long do I hve to ut the roof ems? Exmple 18. 8MF10-16

17 Figure 22 is detiled mp of prt of the Highline Tril, ourtesy of Using the sle, find the distne from Kings Pek to Dedhorse Pss s the row flies. Now find the length of the tril etween these points. In oth ses, just mesure distnes long horizontl nd vertil lens nd use the Pythgoren theorem. Mesuring with ruler on the sle t the ottom, we Figure 22 find tht the sle is 20 mm : 5 km, or 4 mm/km. Mesuring the distne from King s Pek to Dedhorse Pss, we get 108 mm. Then the tul distne in km is 1 km = 27 km. 4 mm Now, to find the length of the tril, you will hve to mesure eh stright length individully nd dd the mesurements. lterntively, you n go to tril.htm nd red n entertining ount of the hike. (108 mm) 8MF10-17

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