12.4 Similarity in Right Triangles
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1 Nme lss Dte 12.4 Similrit in Right Tringles Essentil Question: How does the ltitude to the hpotenuse of right tringle help ou use similr right tringles to solve prolems? Eplore Identifing Similrit in Right Tringles Mke two opies of the right tringle on piee of pper nd ut them out. Resoure Loker hoose one of the tringles. Fold the pper to find the ltitude to the hpotenuse. ut the seond tringle long the ltitude. Lel the tringles s shown Module Lesson 4 DO NOT EDIT--hnges must e mde through "File info" orretionke=nl-;- Dte
2 D Ple tringle 2 on top of tringle 1. Wht do ou notie out the ngles? The orresponding ngles re ongruent. E Wht is true of tringles 1 nd 2? How do ou know? The re ongruent the Similrit riterion. F Repet Steps 1 nd 2 for tringles 1 nd 3. Does the sme reltionship hold true for tringles 1 nd 3? Yes; the orresponding ngles re ongruent, nd the tringles re similr the Similrit riterion. Reflet 1. How re the hpotenuses of the tringles 2 nd 3 relted to tringle 1? The hpotenuses of the smller tringles re the legs of the originl tringle. 2. Wht is the reltionship etween tringles 2 nd 3? Eplin. The re similr euse tringle similrit is trnsitive. 3. When ou drw the ltitude to the hpotenuse of right tringle, wht kinds of figures re produed? Two tringles tht re similr to the originl tringle nd to eh other. 4. Suppose ou drw suh tht is right ngle nd the ltitude to the hpotenuse intersets hpotenuse _ t point P. Mth eh tringle to similr tringle. Eplin our resoning.. P. P. P P. ngles nd P re orresponding right ngles. ngle orresponds to itself.. P orresponds to P nd P orresponds to P.. ngles nd P re orresponding right ngles. ngle orresponds to itself. Module Lesson 4
3 Eplin 1 Finding Geometri Mens of Pirs of Numers onsider the proportion = where two of the numers in the proportion re the sme. The numer is the geometri men of nd. The geometri men of two positive numers is the positive squre root of their produt. So the geometri men of nd is the positive numer suh tht = or 2 =. Emple 1 Find the geometri men of the numers. 4 nd 25 Write proportion. 4_ = _ 25 Multipl oth sides the produt of the denomintors. 25 4_ = 25 _ 25 Multipl. _ 100 = _ Simplif. 100 = 2 Tke the squre root of oth sides. _ 100 = _ 2 Simplif. 9 nd 20 Write proportion. 10 = 9 _ Multipl oth sides the produt of the denomintors _ = 20 _ 20 Multipl. _ 180 = _ Simplif. 180 = 2 = _ 20 Tke the squre root of oth sides. _ 180 = _ 2 Simplif. 6 5 = Reflet 5. How n ou show tht if positive numers nd re suh tht =, then =? Multipl oth sides of the proportion, then tke the squre root of oth sides: ( _ ) = ( _ ) ; = 2 ; = 2 ; =. Your Turn Find the geometri men of the numers. If neessr, give the nswer in simplest rdil form nd nd = 144 = = 4 15 = 2 15 Module Lesson 4
4 Eplin 2 Proving the Geometri Mens Theorems In the Eplore tivit, ou disovered theorem out right tringles nd similrit. The ltitude to the hpotenuse of right tringle forms two tringles tht re similr to eh other nd to the originl tringle. Tht theorem leds to two dditionl theorems out right tringles. oth of the theorems involve geometri mens. Geometri Mens Theorems Theorem Emple Digrm The length of the ltitude to the hpotenuse of right tringle is the geometri men of the lengths of the segments of the hpotenuse. The length of leg of right tringle is the geometri men of the lengths of the hpotenuse nd the segment of the hpotenuse djent to tht leg. h 2 = or h = 2 = or = 2 = or = h Emple 2 Prove the first Geometri Mens Theorem. D Given: Right tringle with ltitude _ D Prove: D_ D = D_ D Reflet Sttements 1. with ltitude _ D 1. Given 2. D D 2. D_ D_ 3. D = D 3. Resons The ltitude to the hpotenuse of right tringle forms two tringles tht re similr to the originl tringle nd to eh other. orresponding sides of similr tringles re proportionl. 8. Disussion How n ou prove the seond Geometri Mens Theorem? Use the tringle shown in the Emple. Show tht D euse the ltitude to the hpotenuse of right tringle forms two tringles tht re similr to the originl tringle nd to eh other. Then show tht D = euse orresponding sides of similr tringles re proportionl. Then prove tht D = in the sme w. Module Lesson 4
5 Eplin 3 Using the Geometri Mens Theorems You n use the Geometri Mens Theorems to find unknown segment lengths in right tringle. Emple 3 Find the indited vlue. z 10 2 Write proportion. 2_ = _ 10 Multipl oth sides the produt of the denomintors. 10 2_ = 10 _ 10 Multipl. _ 20 = _ Simplif. 20 = 2 Tke the squre root of oth sides. 20 = 2 Simplif. 2 5 = Write proportion. Multipl oth sides the produt of the denomintors. Multipl _ = _ _ = _ = _ 12 _ 12 Simplif. 120 = Tke the squre root of oth sides. Simplif. Reflet 9. Disussion How n ou hek our nswers? Possile nswer: Use the Pthgoren Theorem. Your Turn 10. Find. z 5_ = _ 7 ; 7 ( 5_ ) = 7 ( = = 2 _ 7 ) ; 35 = 2 ; 35 = 5 7 Module Lesson 4
6 Eplin 4 Proving the Pthgoren Theorem using Similrit You hve used the Pthgoren Theorem in erlier ourses s well s in this one. There re mn, mn proofs of the Pthgoren Theorem. You will prove it now using similr right tringles. The Pthgoren Theorem In right tringle, the squre of the sum of the lengths of the legs is equl to the squre of the length of the hpotenuse. Emple 4 omplete the proof of the Pthgoren Theorem. Given: Right Prove: = 2 Prt 1 Drw the ltitude to the hpotenuse. Lel the point of intersetion X. X euse ll right ngles re ongruent. X the Refleive Propert of ongruene. So, X the Similrit riterion. X euse ll right ngles re ongruent. the Refleive Propert of ongruene. So, X the Similrit riterion. Prt 2 Let the lengths of the segments of the hpotenuse e d nd e, s shown in the figure. Use the ft tht orresponding sides of similr tringles re proportionl to write two proportions. Proportion 1: X, so _ e = _. Proportion 2: X, so _ = d. d X e Module Lesson 4
7 Prt 3 Now perform some lger to omplete the proof s follows. Multipl oth sides of Proportion1. Write the resulting eqution. 2 = e Multipl oth sides of Proportion12. Write the resulting eqution. 2 = d dding the two resulting equtions give this: = e + d Ftor the right side of the eqution: = (e + d) Finll, use the ft tht e + d = the Segment ddition Postulte to rewrite the eqution s = 2. Reflet 11. Error nlsis student used the figure in Prt 2 of the emple, nd wrote the following inorret proof of the Pthgoren Theorem. ritique the student s proof. X nd X, so X X trnsitivit of similrit. Let X = f. Sine orresponding sides of similr tringles re proportionl, e_ f = _ f d nd f 2 = ed. euse X X nd the re right tringles, 2 = e 2 + f 2 nd 2 = f 2 + d 2. dd the equtions = e f 2 + d 2 Sustitute. = e 2 + 2ed + d 2 Ftor. = (e + d) 2 Segment ddition Postulte = 2 The proof is inorret euse the student ssumes the result tht is to e proved. The student ssumes tht the Pthgoren Theorem is true in order to write tht 2 = e 2 + f 2 nd 2 = f 2 + d 2. Module Lesson 4
8 Elorte 12. How would ou eplin to friend how to find the geometri men of two numers? Possile nswers: Write proportion nd multipl oth sides the produt of the denomintors, then simplif nd tke the squre root of oth sides; multipl the two numers nd then tke the squre root. 13. XYZ is n isoseles right tringle nd the right ngle is Y. Suppose the ltitude to hpotenuse _ XZ intersets _ XZ t point P. Desrie the reltionships mong tringles XYZ, YPZ nd XPY. Possile nswer: ll three tringles re similr. In ddition, YPZ XPY. 14. n two different pirs of numers hve the sme geometri men? If so, give n emple. If not, eplin wh not. Yes; possile nswer: the geometri men of 4 nd 16 is 8, nd the geometri men of 2 nd 32 is lso Essentil Question hek-in How is the ltitude to the hpotenuse of right tringle relted to the segments of the hpotenuse it retes? The length of the ltitude is the geometri men of the lengths of the segments of the hpotenuse. Module Lesson 4
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