Geometry ook 8, rt 2: Unit 3 IMRTNT NTE: For mny questions in this unit, there re multiple correct nswers, e.g. line segment cn e written s, RST is the sme s TSR, etc. Where pproprite, techers should e sure to check for similr equivlent nswers in their students work. ook G8-15 pge 57 1. nswers will vry 2. ) E, E ST, TS 3. ) Techer to check. Line segment 4. 6.9 cm = 69 mm 5. Techer to check. 6. ) For endpoint, circle: ii) S iii) iv) H ii) ST iii) J iv) HG 7. ) R M, Q, X 8. nswers will vry 9. ), E RS, FG, MN, JI, I 10. ) EF, UT, S H, F, IE, IG, I 11. nswers will vry Smple nswer: X Y ook G8-16 pge 59 1. Techer to check. 2. ircle: V UVW, WVU ircle: E FE, EF d) ircle: M LMN, NML 3. ) HIJ RM NUS There re 6 unique ngles: 2 cute: E, E 2 otuse: E, E 2 stright: E, E 4. ircle two: T or T, T or T 5. ) ircle:,,, ircle: L,, J LJ, LJ, JL 6. ) QR,, WX, XY, YW 7. MJL YVTZX d) EF 8. ) ircle three: LYX, LYX, XYL nswers will vry NUS nswers will vry techer to check. Smple nswers: ),,, F E,,, E, EF, F NUS ) ircle two:, None ook G8-17 pge 61 1. ) cute tuse cute d) tuse 2. ) cute; 30 tuse; 150 tuse; 130 d) cute; 40 3. ) 30 130 78 4. ) 65, 25; 90 30, 60; 90 5. Techer to check. ook G8-18 pge 63 1. Techer to check. Smple sketches: ) 2. ) ircle: left ircle: left 3. ) ircle: left ircle: right ircle: left 4. ) 2 cm 5. ) 2 cm 1 cm 1 cm V 120 6 cm T 2 cm U R 6. ) ircle: right ircle: left 7. ) 70 rectngulr gift ox is 15 cm long nd 10 cm wide. It hs trnsprent lid decorted with pper flower 5 cm in dimeter. Wht is the re of the ottom of the ox? 10 cm 8. ) ircle: right * not necessrily rectngle ircle: left * not necessrily right ngle tringle 9. Tringle 1: 3 cm Tringle 2: 10. Tringle 1: Tringle 2: 15 cm 45 11. ) ircle: left ircle: right Q 4 cm 4 cm 3 cm 70 40 5 cm 40 100 70 40 3 cm YRIGHT 2011 JUM MTH: NT T E IE V-14 nswer eys for ook 8.2
Geometry ook 8, rt 2: Unit 3 (continued) YRIGHT 2011 JUM MTH: NT T E IE 12. nswers will vry Smple nswers: ) d) J M S E J S L E 13. Techer to check dditionl informtion. ) c 2 = 2 + 2 = 3 2 + 4 2 = 25 c = 5 cm c 2 = 5 2 + 12 2 = 169 c = 13 cm 14. ) No; doesn t show tht the wll nd ground re perpendiculr. Sys: wll = verticl, ground = horizontl perpendiculr d) ; It s simpler, with only the essentil detils. e) x 2 + 5 2 = 13 2 ; x 2 = 13 2 5 2 = 169 25 = 144 x = 12 m f) No; 12.5 m Q M R Q R L ook G8-19 pge 66 1. ) EF J Q E d) 2. Techer to check. ) The 4 middle ngles re 90. The 4 corner ngles re 90. The 4 corner nd the 4 middle ngles re ll 90. 3. ; ; ; 4. Techer to check. 5. ) Techer to check. Smple explntion: The two lines re perpendiculr ecuse the ngle ws constructed to e 90. Techer to check. 6. ) Yes; Since its ngles re ll 90, ny pir of djcent sides will e perpendiculr. 7. ) 40; = + = (90 50) Techer to check. 8. ) 1, 2,, d) 1, 2; 3, 4 e) 1, 5; 5, 1 f) 3, 2; 3, 2 9. ) 45 53 50 10. 11. x + 4x = 90 so x = 90 5 x = 18, 4x = 72 x 20 4x + ( 20) = 90 2 = 110 = 55, ( 20) = 35 ook G8-20 pge 69 1. ) ; ; ; 1, 5; 3, 7; 1, 5; 3, 7 ; + E; E + E; + E 2. ) 60 115 125 d) 112, 72 e) 65, 121 f) 52 g) 90 h) 65, 65 3. ) 1, 3; 4, 2; 4, 1; 3, 2 180, 180 3; 180, 180 3; They re equl. s stright ngles, 2 + 3 = 180 nd 2 + 4 = 180, which mens tht: 2 + 3 = 2 + 4 3 = 4 4. ) ; = ; = 4, 3; 8, 7 d) 2 = 5, 1 = 4; 3 = 6 e) E = ; = E f) = ; = 5. ) 135 50, 105 60, 90, 30 d) 105, 75 e) 51, 138 f) 55, 115, 115 ook G8-21 pge 71 1. Techer to check line extensions; the lines will intersect. 2.,, 3. nswers will vry 4. ) d) 5. ) XY FG RS LM GH 6. = E = F = 1.35 cm; They re ll equl. ook G8-22 pge 72 1. 9, 3 4, 10 13, 7 8, 14 3, 9 4, 10 13, 7 8, 14 INVESTIGTIN 1. 3, 4, 7, 8. 100, 80 100, 80 80, 100 80, 100. equl nswer eys for ook 8.2 V-15
Geometry ook 8, rt 2: Unit 3 (continued) 1. ) 5, 7; 6, 8 3, 4; 6 11, 13; 12, 2 INVESTIGTIN 2. 3, 4. 45, 135, 75, 105. No (since the rys ren t prllel) 2. ) 115 111; 69 78 d) 128; 52 e) 105; 95 f) 60; 45 3. ) 85; prllel 95; not prllel 63; prllel d) 99; not prllel 4. none none d) k l e) none f) r s g) x w h) y z, ook G8-23 pge 74 1. ) 6; 7 5; 6 nswers my vry NTE: Students should give 6 unique ngle pirs. 8, 5; 11, 2; 3; 10 INVESTIGTIN. 6, 5. NTE: ngle mesures my vry slightly ut ech pir of supplementry ngles must dd to 180. 107, 73, 73, 107; equl. prllel; corresponding; opposite, 7; 6. nswers my vry Smple proof: Since m n, we know 4 = 8 (corresponding) nd 8 = 5 (opposite) 4 = 5. 2. ) 48 72, 68 92, 88 3. ) = 125 = 55 = 95 = 122 = 85 = 58 = 125 = 55 4. L, L; L; prllel; MN L 5. ) c d g (given), e f p k (given) 6. nswers my vry Smple explntion: Since, we know tht 1 = 5 (corresponding) nd 1 + 3 = 180 (supplementry) 5 + 3 = 180. 7. ) Techer to check., ;, nswers my vry Smple explntion: y extending the length of the sides, you cn use the presence of equl corresponding ngles to prove tht the pirs of opposite sides re prllel. ook G8-24 pge 76 1. 180, 180; 180, 150; 30 2. ) 70 30, 30; 120 180, 90, 40; 50 d) 35; 60; 85 e) 55; 35 f) 43; 43; 69 3. ) 90 90 60 4. ) ; equl; E; equl, E = + + E = 180 5. ) 93 55, 125 45 d) 70, 60 e) 120 f) 120, 120 g) 54, 50, 83 h) 132, 92, 44 i) 45, 75, 60 6. ) + + + c = 180 (form tringle) x + c = 180 (supplementry) d) + + c = x + c x = + ook G8-25 pge 78 INVESTIGTIN. G, F,,, E, G, E, F,., F;, E, G;,, E, F;,, ; G. isosceles; equilterl 1. ) 120; otuse 90; right 80; cute 2. ) no (or 0); sclene 2; isosceles 3; equilterl 3. 4 m or 7 m (see G8-18 #9 for detils) 4. Techer to check. 5. ) right cute otuse sclene isosceles equilterl Neither n otuse or right equilterl tringle is possile; n equilterl tringle lwys hs 3 equl ngles, ech one mesuring 60 (which is cute). 6. : right, sclene EF, GEF: right, sclene EG: otuse, isosceles HJI: right, isosceles YRIGHT 2011 JUM MTH: NT T E IE V-16 nswer eys for ook 8.2
Geometry ook 8, rt 2: Unit 3 (continued) YRIGHT 2011 JUM MTH: NT T E IE NLM: right, sclene NL: right, isosceles NM: otuse, sclene since NM = 105 7. ) Techer to check. i) Yes; 2 options ii) No iii) Yes; Infinite options s s length chnges 8. Explntions nd sketches my vry ) No Smple nswer: oth nd re otuse ngles. In tringle, the ngles must dd to 180 so, if one is greter thn 90 (otuse), the sum of the other two must e less thn 90 mening they re oth cute. No Smple nswer: = = 90 gin, the ngles sum must e 180 so, if one is 90, the other two must dd to 90 so they must oth e cute. 9. ) Tringle 1: 30, 120 Tringle 2: 75, 75 35, 35 In ), we re given n cute ngle. In, we re given n otuse ngle. They hve different numer of solutions ecuse, if given tringle hs n otuse ngle, its other two ngles must e cute this is why hs only one solution. In ), we re given n cute ngle, which might either e one of the two equl ngles in the isosceles tringle, or the unequl ngle this is why it hs two possile solutions. ook G8-26 pge 80 1. 57 (T) 80 (STT) 55 (T) 65 (STT) d) 75 (T) 45 (T) 60 (ST or STT) e) 119 (ST) 119 (T) 119 (T) f) 55 (T) 95 (T) 30 (STT) g) 115 (T) 65 (ST) 115 (T) 65 (T or ST) h) 35 (STT) 35 (T) 55 (T or STT) 125 (ST) i) 81 (T) 39 (STT) 39 (T) 60 (STT) 2. Specific theorems used y students my vry lso techer to check ny numered helper ngles mrked y students. ) 1 = 67 (ET) 2 = 83 (STT) 1 = 32 (T) 2 = 58 (T) 1 = 75 (T/ST) 2 = 130 (ST) 3. gin, specific theorems used my vry techer to check. ) 1 = 2 = 37 (ITT) 1 + 2 = 60 (ET) 1 = 2 = 30 (ITT) We know tht the other two ngles in the tringle re ech 55 (ITT), so: 1 = 2 = 125 (ST) d) 1 = 80 (ET) 2 = 50 (ITT) 4. NTE: In nd d), the missing ngles cn e found using ITT only students don t necessrily hve to solve for x. y STT: 70 + (x + 20) + (3x + 10) = 180 4x + 100 = 180 4x = 80 x = 20 ngles re 70, 40 nd 70. y ITT: 6x + 5 = 35 6x = 30 x = 5 ngles re 35, 35 nd 110. d) y ITT, STT: 96 + 2x + 2x = 180 96 + 4x = 180 4x = 84 x = 21 ngles re 96, 42 nd 42. e) y STT: 90 + (2x + 15) + (4x + 45) = 180 6x + 150 = 180 6x = 30 x = 5 ngles re 90, 25 nd 65. f) y ST, STT: 45 + (5x + 15) + (4x + 12) = 180 9x + 72 = 180 9x = 108 x = 12 ngles re 45, 75 nd 60. g) y ET, STT: 4x 30 = (x + 5) + 40 3x = 75 x = 25 ngles re 40, 30 nd 110. ook G8-27 pge 82 1. Techer to check mrked ngles nd sides. ) rectngle prllelogrm squre d) (right) trpezoid 2. ) rhomus rectngle NUS trpezoid 3. ) ll no no d) some 4. Techer to check. 5. Similrities: oth re qudrilterls. ifferences: prllelogrm hs two pirs of prllel sides; trpezoid hs only one. prllelogrm hs equl opposite sides nd ngles; this isn t true for trpezoid. 6. ) It s prllelogrm with 4 right ngles. Their sides re not ll equl. trpezoid hs one pir of prllel sides; prllelogrm hs two. d) It s prllelogrm with ll sides equl. 7. Techer to check for proper sketches. ) ne squre plus mny different rhomuses nswer eys for ook 8.2 V-17
Geometry ook 8, rt 2: Unit 3 (continued) ne rectngle plus mny different kites, prllelogrms, nd qudrilterls with n indenttion The kites nd ny shpes with n indenttion ren t prllelogrms. 8. Trpezoid, prllelogrm nd rhomus Techer to check tht drwings re correct. 9. ) Two correct options: nd or nd Yes; Yes Two correct options: nd or nd d) No (185); No e) Trpezoid 10. ) Techer to check. roperty No e/s 1 pir e/s I Shpes F, G, J 2 pirs e/s,, E, H 4 e/s, No p/s, F, H 1 pir p/s G, I, J 2 pirs p/s,,, E No r/ 1 r/ F 2 r/ H, J 4 r/, No e/,, E, G, I F, G 1 pir e/, H, J 2 pirs e/, E, I 4 e/, ook G8-28 pge 84 INVESTIGTIN. Techer to check. In the top prllelogrm, the opposite ngles re 130 nd 50. In the other, the opposite ngles re 110 nd 70.. 2; 2. 180 ; 180 ; ; 180 ; 1. Explntions nd (counter) exmples my vry ) True; Smple explntion: In prllelogrm, the opposite sides re lwys prllel. d So, from the reminder ox on pge 83, we know tht ech pir of djcent ngles (/, /c, c/d nd d/) dds to 180. Flse; Smple counterexmple: Isosceles trpezoids hve 2 pirs of equl ngles ut re not prllelogrms. Flse; Smple counterexmple: In the ove trpezoid, we see tht + 180. d) Flse; Smple counterexmple: Here, the two sets of opposite ngles dd to 180 ut this qudrilterl is kite, not trpezoid or prllelogrm. 2. ) It doesn t tlk out rectngle eing prllelogrm. c Yes, this will lwys e true. ( squre is lso rectngle.) Using the following sketch, nd the reminder t the top of pge 83: + = 180 + = 180 So must e prllelogrm. Since is prllelogrm nd it hs 4 right ngles, we know is rectngle. Jen is correct. 3.,, E NUS If qudrilterl hs two pirs of equl ngles (let s cll them pir nd pir, then only two possile cses exist: 1. the pirs of equl ngles re opposite ech other; 2. the pirs of equl ngles re djcent to ech other. se 1: pposite Looking clockwise, the ngle order is,,,. s qudrilterl, we know: + + + = 360 2( + = 360 + = 180 From this, s in #2 ove, we see tht ll four djcent ngle pirs dd to 180, so opposite sides will e prllel it s prllelogrm. se 2: djcent Looking clockwise, the ngle order here is,,,. s qudrilterl, + = 180 lso pplies here. So, with two djcent (nd distinct) ngle pirs dding to 180, this shpe will hve one set of prllel sides it s trpezoid. 4. ) = 2 2 + 2 2 = 8 = 2 2 + 5 2 = 29 c = 2 2 + 5 2 = 29 d = 2 2 + 2 2 = 8 = d nd = c so this is kite. d = 5 = 2 2 + 6 2 = 40 c = 2 2 + 6 2 = 40 d = 4 2 + 3 2 = 25 = 5 = d nd = c so this is kite. d d = 4 2 + 3 2 = 25 = 5 = 4 2 + 5 2 = 41 c = 3 2 + 6 2 = 45 d = 2 2 + 3 2 = 13 this is not kite. 5. Techer to check drwn digonls. The kites hve digonls tht intersect t right ngle. 6. ) = 5 units c = 2 units = d = 3 units orrect prediction: kite s digonls re perpendiculr nd one of the digonls isects the other. Techer to check drwings of the two other kites. c c c YRIGHT 2011 JUM MTH: NT T E IE V-18 nswer eys for ook 8.2
Geometry ook 8, rt 2: Unit 3 (continued) YRIGHT 2011 JUM MTH: NT T E IE 7. Techer to check. Students should notice tht kites lwys hve one pir of equl, opposite ngles: 1. 2. F G H E 8. ) Squre Rhomus Squre d) Rhomus ook G8-29 pge 86 1. 6 cm; 3 cm; Techer to check tht proper midpoint is mrked. 2. Midpoint t 4 cm; Techer to check. 3. ), ; N, No, there is nothing to indicte tht FE = E. NUS No, it cn t. y definition, line is infinite nd hs no endpoints it cn hve no midpoint. 4. Techer to check. 5. ) M1, M2 M1, M2 M1 d), 6. Techer to check. M2 M1 M2 7. ), Rhomus 8. ), Squre 9. ), Q Rectngle 10. ) X W L ite NUS X 2 + Y 2 = Z 2 + Y 2 2 XY = ZY XY = ZY X 2 + W 2 = Z 2 2 + W 2 2 XW = ZW XW = ZW ook G8-30 pge 88 1. Techer to check. 2. EF, FG, EG 3., R, Q 4. = S, E = L, F = M 5. W = R, X = T, Y = S; WY = RS, XW = TR, XY = TS 6. No; she forgot to check if the corresponding sides were equl. Z 2 S N M R Y 7. The third ngle in ny tringle will e equl to 180 1 2. In this cse, = S nd = T, so: = 180 = 180 S T = R = R, even though they ren t mrked. 8. 5: WXY RTS 7: RTS 9. ) i) = F These four ngles re equl. ii) iii) = = E Smple: FE Smple: EF i) G = J, ll six ngles re equl. H =, I = L ii) Smple: IHG LJ iii) Smple: GHI JL 10. orresponding sides Q R L M Q R L M orresponding ngles Q R L M Q R L M 11. VW = YZ V = Y UW = XZ W = Z 12. None of Tom s sttements re correct; XVW ook G8-31 pge 91 1. ) SS; EF S; RST ZXY SSS; MN FGH 2. ) No; They ren t equl. Yes No; They re equl ut re positioned differently in reltion to the corresponding ngles. 3. orresponding? Equl? ) No Yes Yes Yes No Yes d) Yes No 4. Techer to check. 5. Techer to check. 6. 5 cm G 36 54 6.4 cm 5 cm 3 cm 3.7 cm 36 54 4 cm H Three equl ngles: =, = G, = H ne equl side: = G No, they ren t congrent their equl sides re opposite different ngles. 7. Techer to check tringle sketches. ) No Yes (SS) No d) Yes (SSS) 8. Techer to check tringle sketches. F Smple sketch: H G NTE: nd FGH hve two equl ngles so they re similr y. ) No, this is just further evidence of their similrity. Yes (S) Yes (S) d) No nswer eys for ook 8.2 V-19
Geometry ook 8, rt 2: Unit 3 (continued) 9. ) 10. No, it depends where the equl side nd ngle re positioned within the tringles. se 1: E Here, EF. se 2: F Here, EF. INVESTIGTIN. =, = EF nd = E. No: the corresponding ngle isn t etween the corresponding sides.. No. No, neither hs its corresponding ngle etween its corresponding sides only SS is proper congruence rule. 11. Yes, y SSS nd the ythgoren Theorem; The third side of oth tringles is 13 2 5 2 = 12 cm long. 12. Yes, y SSS nd ythgoren; oth tringles hve sides tht re 9 cm, 12 cm nd 15 cm in length. F E 13. ) No, y SSS nd ythgoren; ne s sides tht re 1, 2 nd 5, nd the other s sides re 1, 2 nd 3. No, y STT/ITT; ne hs ngles tht re 80, 50 nd 50. The other s re 70, 55 nd 55. ook G8-32 pge 94 1. ) 53; 53 + 53 = 106 75; 75 + 75 = 150 20; 20 + 20 = 40 2. ) No; In this rectngle, the digonl is t 20/70 ngle to the vertex. Yes; squre s digonl is lwys t 45 ngle to the vertex. (Techer to check drwing.) 3. ) S SS SSS d) SSS 4., ; ;, SSS; ;, 90º; ; 5. ) SSS Techer to check ngles re mrked. d) Techer to check lel for E. E E (SS) e) Explntions my vry techer to check. Smple explntion: Since E E (SS), we know tht: E = E. ut E is stright ngle, so: E = E = 90 nd. From d), we know E E nd E = E. This proves tht is the perpendiculr isector of. f) i) T ii) F iii) T iv) T 6. ) F T F d) T e) T NUS Explntions will vry 7. They re equl. ) ; = ; = (S) d) =, = ; Yes 8. ) Mesurements my vry slightly ut reltionships must remin to e correct. E = E = 2.8 cm E = E = 1.5 cm 9. orrect prediction: The digonls of prllelogrm will isect ech other. Techer to check test shpes. E = E (T) E = E (T) nd =, we know tht E E (S) d) From, we know tht E = E nd E = E the prllologrms digonls ( nd ) isect ech other t E. ook G8-33 pge 97 INVESTIGTIN 1. =. They will e equl (think of n isosceles tringle).. M M y SS: M = M (given), M = M (given), nd side M is common. This congruency mens tht corresponding sides nd re equl. INVESTIGTIN 2.. Isosceles. E. No, nd would hve to e perpendiculr; (SSS) =, ut this cn only hppen if oth ngles equl 90, mening. YRIGHT 2011 JUM MTH: NT T E IE V-20 nswer eys for ook 8.2
Geometry ook 8, rt 2: Unit 3 (continued) YRIGHT 2011 JUM MTH: NT T E IE 1. ) In ll cses, the centre lies on the perpendiculr isector of. This hppens ecuse = (oth re rdii) nd ll points equidistnt from nd lie on their perpendiculr isector. 2. ) In ll cses, the three perpendiculr isectors go through centre tht is, they intersect t. This hppens ecuse = = (ll rdii) nd ll points equidistnt from two endpoints lies on their line segment s perpendiculr isector. 3. ) ll three intersect t the sme point. 4. She should choose this intersection point, which we ll lel s ; rdius = = = 3.2 cm M NTE: y drwing LM nd its three sides perpendiculr isectors, you identify the centre of the circle through, L nd M. NUS Techer to check drwings. NTE: The centre of right tringle s circumcentre is t the midpoint of the tringle s hypoteneuse. ook G8-34 pge 99 1. ) : EF = 2 : 6 = 3 : 1 : FG = 1 : 3 : GH = 2 : 6 = 3 : 1 : EH = 1 : 3 Yes y definition, ll of rectngle s ngles re 90. 2. Side rtio = 1 : 2 = 2 :? So the length of the unknown side, indicted y? = 2 2 cm = 4 cm. 3. ) 6 cm 12 cm 15 cm 4. No; squre hs ll sides equl, while rectngle only needs its opposite sides to e equl. 5. No; squre lwys hs two pirs of prllel sides ut, y definition, trpezoid hs exctly one pir of prllel sides. 6. Techer to check. L 7. ) Techer to check sketch. Yes, they re congruent; The ngles in ny equilterl tringle re ll 60 so, since one corresponding side is equl, we know (y S) tht EFG QRS. No; y, equilterl tringles will lwys e similr, ut they re only congrent if they hve t lest one equl, corresponding side. 8. To e similr, tringles must hve proportionte corresponding sides nd ll their corresponding ngles must e equl. To e congruent, ll their corresponding ngles nd sides must e equl. Since the corresponding sides in pir of congruent tringles re equl, they will hve 1:1 proportion which mens they lso meet the requirement for similrity. 9. ) No, 1 : 2 3 : 3 Yes, 1 : 3 = 2 : 6 Yes, 1 : 4 = 1 : 4 d) Yes, 2 : 3 = 4 : 6 e) No, 1 : 2.5 2 : 5.5 10. Techer to check drwing; new figure = 2 units high 11. nd re similr: their corresponding sides ll hve rtio of 1 : 2. NTE: Students will need to use the ythgoren Theorem in this question. 12. sed on the reltionship etween corresponding sides/ngles, we find: ongruent nd H nd I Similr E nd J nd G 13. ) NTE: Mesurements my vry slightly ut the 2 : 3 proportion must remin. 20.5 4.8 cm 6.6 cm Yes; ngles = = E = F Sides : F = : E = : EF = 2 : 3 14. ) x = 20, y = 25 x = 16, y = 12 15. ) No: they hve only one equl ngle. No: not ll of the corresponding sides re proportionte. ook G8-35 pge 101 4.4 cm 39 20.5 120.5 1.8 cm 3.2 cm 120.5 2.7 cm INVESTIGTIN. Techer to check. In ll cses, the resulting tringles should e similr with scle fctor of 2. Smple nswer: ) = 3 cm = 5 cm = 5.7 cm '' = 6 cm '' = 10 cm '' = 11.4 cm = ' = 86 = ' = 62 = ' = 32 ; Yes d) Scle fctor = 2. ) = 4.8 cm = 2.2 cm = 5.4 cm E 39 F nswer eys for ook 8.2 V-21
Geometry ook 8, rt 2: Unit 3 (continued) Techer to check '''. '' = 14.4 cm '' = 6.6 cm '' = 16.2 cm d) No, since they meet the SSS rule in ove, we know they re similr. e) Techer to check.. ) Techer to check. Techer to check. Yes: M = M' = 80 d) 'M' M = 10.8 cm 2.7 cm M'L' ML = 8 cm 2 cm 'M' M = 10.8 cm 2.7 cm M'L' ML = 8 cm 2 cm L'' L = 12 cm 3 cm their scle fctor is 4.. ) Techer to check. While *L*M* side lengths will vry, ll three rtios will e equl if correct. M = 80, M* = 80 (sum of the tringles ngles must e 180 ) d) Yes; Scle fctor will vry e) Techer to check. 1. ) SS SSS 2. ) '' '' = 1.65 cm 3.3 cm = 2.35 cm 4.7 cm nce reduced, oth equl 0.5. = ' = 19 No d) No 3. No; In Question 2, we showed tht SS isn t sufficient so S certinly isn t. 4. Techer to check. Smple exmple: ' 4 cm 6 cm ' 3cm 43 ' 2cm Here we cn see tht: '' = '' = 2 ut oviously ' so the tringles ren t similr. 5. ) is common = E = E They re similr. 6. ) 2 : 1; 2 : 1 They re similr, y SS. d) N = 2 N = 1 2 e) Two correct options: L nd M or L nd N; Ech corresponding pir is equl. f) MN L (T); LMN is trpezoid. E 7. nswers my vry NTE: Techers should see n exmple used in ech student explntion. Smple nswer - SSSS: This squre nd rhomus hve side proportion of 1 : 2 ut their ngles re different. NTE: 2 = 60 120 120 60 Smple nswer - : oth squre nd rectngle hve four right ngles ut there is no proportion etween their sides. ook G8-36 pge 104 INVESTIGTIN 1. Techer to check drwn prllelogrms; Yes; 3. 2, 2, 6, 6; se of = se of scle fctor. 1 4 36 2 : 1 6 : 1 2 : 6 = 1 : 3 4 : 1 36 : 1 4 : 36 = 1 : 9. If the se rtio of (originl) : (imge) is imge x : y, s = originl = y x ; re of re of = s 2 1. ) Techer to check. orrect rtio: 1 : 9 Techer to check re clcultions. 1 2 h : 1 2 (3(3h) 1 2 h : 9 2 h 1 : 9 2. Let s represent the scle fctor of the two tringles. ) 1 : s 2 = 125 : 5 1 s 2 = 125 5 s 2 = 5 125 = 1 25 s = 1 5 Q : UV = 5 : 1 (diltion) 1 : s 2 = 1 : 100 s = 10 Q : UV = 1 : 10 (enlrgement) 1 : s 2 = 50 : 25 1 s 2 = 50 25 s 2 = 25 50 = 1 2 s = 1 2 Q : UV = 2 : 1 (enlrgement) 3. ) M 2 cm 2 cm L 30 30 We know: M = 120 (STT) R 120 Q We know: = Q = 30 (ITT) LM nd QR re similr (). 1 : s 2 = 1 : 81 s = 9 R = QR = 9 2 = 18 cm INVESTIGTIN 2. : = 3.3 cm = 4.3 cm = 6.8 cm EF: F = 5.1 cm E = 3.2 cm FE = 2.5 cm YRIGHT 2011 JUM MTH: NT T E IE V-22 nswer eys for ook 8.2
Geometry ook 8, rt 2: Unit 3 (continued). F = 6.8 5.1 1.33 FE = 3.3 2.5 = 1.32 E = 4.3 3.2 1.34 yes, they re similr their corresponding sides re proportionte.. s 4 3 = 13. = (T) = (ITT) Since it s given tht = 2, we lso know tht =. nd re similr (). 5. ) 90 ; 90 ; ; They re similr (). i) QR Q = RS QS = QS S s = QS S = 6 8 = 0.75 j) 24 cm 2, multiply it y the squre of the scle fctor; 24(0.75) 2 = 13.5 cm 2 k) 24 + 13.5 = 37.5 cm 2 Qudrilterls Mny non-congruent Smple nswers:. 14.4 cm, 10.8 cm E. ( ) ( EF) = 14.4 10.8 = 13. = s ( EF) = ( ) s F. Techer to check pir of shpes drwn. erimeter rtio = er : er = 1 : 5 Smple nswer: 2 10 3 15 (originl) : (imge) = 2 : 10 = 3 : 15 = 1 : 5 their scle fctor is 5. erimeter (originl) = 10 erimeter (imge) = 50 = (common) = = 90 ; They re lso similr (). Yes: nd re similr, since they re oth similr to. 6. 2.88 m = 288 cm 7. ) QR = SQ = 90 (given) = (shred) QS nd RQ re similr (). QR = SQ, RQ = QS, RQ = QS QR QS = Q S = R Q d) Q, Q = 10 cm e) Q S = 10 8 = 1.25 f) 24 cm 2 8. Since LM nd LNM re similr, we know tht ML = MLN nd ML = MNL. ut, from LNM, we see tht MLN + MNL = 90 since NML = 90 (given). LN = MLN + ML = 90 9. NTE: In ) nd, there re mny possile nswers nd students cn t e expected to find them ll. The more they cn find, the etter. ) Tringles 8 non-congruent Mny non-similr nswers will vry the rtio etween their perimeters is lso 1 : 5. Generl rule: For similr shpes, the rtio etween the perimeters is the sme s the rtio etween the corresponding sides. g) (1.25) 2, 37.5 cm 2 h) Explntions my vry techer to check. SQ = QSR = 90 (given) QS + RQS = 90 (given) 6 non-similr 4. ) QS + QS = 90 (STT) YRIGHT 2011 JUM MTH: NT T E IE so QS = RQS QS nd QRS re similr (). From this similrity, we lso know tht SQ = SRQ. nswer eys for ook 8.2 V-23