Pearson ducation, Inc., publishing as Pearson Prentice all. ll rights reserved. hapter 4 nswers Practice 4-1 1. m 1 = 110; m 2 = 120 2. m 3 = 90; m 4 = 135 3. m 5 = 140; m 6 = 90; m 7 = 40; m 8 = 90 4. J,, J 5. J,, 6. WZ JM, WX JK, XY K, ZY M 7. W J, X K, Y, Z M 8. Yes; GJ IJ by heorem 4-1 and by the eflexive Property. herefore, GJ IJ by the definition triangles. 9. o; Q V because vertical angles are congruent, and Q V by heorem 4-1, but none of the sides are necessarily congruent. 10a. 10b. Vertical angles are. 10c. heorem 4-1 10d. 10e. efinition triangles Practice 4-2 1. B B by 2. not possible 3. not possible 4. U XWV by 5. not possible 6. GF by 7. MK KMJ by 8. P PQ by 9. not possible 10. 11. B and B 12. and B 13. 14a. 14b. eflexive Property of ongruence 14c. Postulate 15. tatements easons 1. F FG, F F 1. 2. F FG 2. Vertical s are. 3. F FG 3. Postulate Practice 4-3 1. not possible 2. Postulate 3. heorem 4. heorem 5. not possible 6. not possible 7. Postulate 8. not possible 9. heorem 10. tatements easons 1. K M, K M 1. 2. JK PM 2. Vertical s are. 3. JK PM 3. Postulate 11. Q Q eflexive Property 12. B F 13. KJ KG or KJ GK 14. M PQ Practice 4-4 Q heorem 1. B is a common side, so B B by, and by P. 2. F is a common side, so F FG by, and FG by P. 3. KJ PM by, so K P by P. 4. Q is a common side, so Q Q by. Q Q by P. 5. VX is a common side, so UVX WVX by, and U W by P. 6. ZY and B are vertical angles, so B YZ by. Z by P. 7. G is a common side, so G FG by, and FG G by P. 8. JK and KM are vertical angles, so JK MK by, and JK K by P. 9. P is a common side, so P QP by, and Q by P. 10. First, show that B and are vertical angles. hen, show B by. ast, show by P. 11. First, show F as a common side. hen, show JF GF by. ast, show FG J by P. Practice 4-5 1. x = 35; y = 35 2. x = 80; y = 90 3. t = 150 4. r = 45; s = 45 5. x = 55; y = 70; z = 125 6. a = 132; b = 36; c = 60 7. x = 6 8. a = 30; b = 30; c = 75 9. z = 120 10. ; F 11. G; G G 12. KJ; KIJ KJI 13. ; 14. B; BJ JB 15. B; B B 16. 130 17. 65 18. 130 19. 90 20. x = 70; y = 55 21. x = 70; y = 20 22. x = 45; y = 45 Practice 4-6 1. tatements easons 1. B # B, # F 1. 2. B, are right s. 2. Perpendicular lines 3. F, B 3. 4. B F 4. heorem 2. tatements easons 1. P, are right s. 1. 2. P Q 2. 3. Q Q 3. eflexive Property 4. PQ Q 4. heorem 3. MJ and MJK are right s. Perpendicular lines M MK MJ MJ eflexive Property MJ MJK heorem Geometry hapter 4 nswers 35
hapter 4 nswers (continued) 4. GI JI I I eflexive Property IG IJ heorem GI and JI GI JI are right s. heorem 2-5 m GI + m JI = 180 ngle ddition Postulate 5. VW 6. none 7. m and m F = 90 8. G J 9. P 10. UV or V U 11. m and m X = 90 12. m F and m = 90 13. GI # J Practice 4-7 1. ZWX YXW; 2. B B; 3. JG FK; 4. P M; 5. F BG; 6. UVY VUX; 7. B B common side: B 8. F G G F common side: FG 9. common angle: M K J 10. ample: tatements easons 1. X Y 1. 2. X # B, BY # 2. 3. m X and 3. Perpendicular lines m BY = 90 4. 4. eflexive Property 5. BY X 5. Postulate 11. ample: Because F G, FG GF, and FG GF, then FG GF by. hus, F G by P and, then G F by. eteaching 4-1 1. b 2. c 3. a 4. 117 5. 119 eteaching 4-2 1. 2. heck students work. 3. ; B B 4. ; MQ P 5. ; PQ VU 6. : JMK MK 7. ; QP Q 8. ; YX WX eteaching 4-3 1. 2. or 3. heck students work. 4. B B 5. JMK KM or JKM MK 6. UZ YZ 7. Y 8. P 9. Y Y eteaching 4-4 ) 1a. QK Q; QB bisects KQ 1b. definition of bisector 1c. BQ BQ 1d. Postulate 1e. P 2. tatements easons 1. M MP, 1. P 2. M M 2. eflexive Property 3. MP M 3. Postulate 4. P 4. P 3. tatements easons 1. bisects J, 1. J 2. J 2. efinition of bisector 3. 3. eflexive Property 4. J 4. heorem 5. J 5. P eteaching 4-5 1. ach angle is 60. 2. 120 3. 120 4. 50 5. 70 6. 60 7. 65 8. 115 9. 55 10. 120 11. 60 eteaching 4-6 1. ample: = 1.3 cm, = 1.6 cm, Q = 2.5 cm, Q = 2.3 cm; not congruent 2. ample: = 2.3 cm, G = 2.3 cm, = 1.9 cm; G 3. ample: = 3.3 cm, = 2.8 cm, M = 2.8 cm; M 4. heorem can be applied; B. 5. heorem cannot be applied. 6. heorem can be applied; MU M. 7. heorem can be applied; F F or F F. 8. heorem can be applied; K. 9. heorem cannot be applied. Pearson ducation, Inc., publishing as Pearson Prentice all. ll rights reserved. 36 nswers Geometry hapter 4
hapter 4 nswers (continued) Pearson ducation, Inc., publishing as Pearson Prentice all. ll rights reserved. eteaching 4-7 1. tatements easons 1. P and PQ 1. are right s; QP and P 2. P and PQ 2. ight s are congruent. 3. P P 3. eflexive Property 4. QP P 4. heorem 5. QP 5. Vertical s are. 6. PQ 6. P 7. QP 7. heorem 2. ample: Prove MP QP by the heorem. hen use P and vertical angles to show M QP by the heorem. 3. ample: Prove B by the Postulate. hen use P and vertical angles to show BF F by the heorem. nrichment 4-1 heck students work. amples shown. 1. 2. 3. 4. 5. nrichment 4-2 1a. efinition of perpendicular lines 1b. KF G 1c. 2a. egment ddition Postulate 2b. + G = F + 2c. G 2d. lternate Interior ngles 2e. orresponding ngles 2f. JG 2g. nrichment 4-3 1. 11. heck students work. 2a. 2b. he top angles are congruent because the fold bisected the right angles formed by the folds in steps 1 and 3. he corners of the paper are right angles; therefore, those angles are congruent. he included sides are congruent because the fold in step 1 found the midpoint of the width of the paper, thus creating two equal segments. 3a. 3b. he top angles are congruent because the fold bisected the right angles formed by the folds in steps 1 and 2. he upper corners that became inside angles along the center line are right angles; therefore, those angles are congruent. he included sides are congruent because the fold in step 1 found the midpoint of the width of the paper, thus creating two equal segments. 4a. he top angles are congruent because the fold bisected the right angles formed by the fold in step 1. he inside angles along the center line are right angles because the horizontal fold that formed them is perpendicular to the original fold in step 1. 4b. he included sides are congruent because the fold in step 1 found the midpoint of the width of the paper, thus creating two equal segments. 8a. 8b. he top angles are congruent because the fold bisected the right angles formed by the fold in step 7. he inside angles along the center line are congruent because of the ngle ddition Postulate. he included sides are congruent because the fold in step 7 found the midpoint of the width of the paper, thus creating two equal segments. nrichment 4-4 1. B 2. 3. 45; 45; B; 4. 30; 30; B; 5. eflexive Property 6. heorem 7. P 8. efinition segments 9. efinition segments 10. efinition segments 11. Postulate 12. P 13. 60 nrichment 4-5 1. 60 2. 60 3. 60 4. 70 5. 70 6. 40 7. 72 8. 72 9. 36 10. 30 11. 30 12. 120 13. 80 14. 80 15. 20 16. 80 17. 80 18. 20 19. 41 20. 23 21. 116 22. 23 23. 41 24. 116 25. 80 26. 80 27. 20 28. 82 29. 82 30. 16 31. 78.5 32. 78.5 33. 37.5 34. 40 35. 40 36. 100 Geometry hapter 4 nswers 37
hapter 4 nswers (continued) nrichment 4-6 8 nrichment 4-7 1. ample: B 2. common angle: 3. ample: B 4. B common side: 12 V 2 3 4 I G 5 G U I B 16 5. 8 6. ample: GJ K 7. K G J 14 B 9 1 I U 15 8. ample: IM JIM 9. I I J common side: IM M M 10. 4 11. 4 12. PUV WV by ; PV QV by ; PVW VU by ; WQV UV by. 13. 14 14. heck students work. 15. heck students work. Y P 10 13 M P U M 7 6 F X I V I 11 V 17 hapter Project ctivity 1: Modeling triangle Yes; the brace makes two rigid triangles. ctivity 2: bserving heck students work. ctivity 3: Investigating tetrahedron ample: You could add in diagonals of the cube. heck students work. Finishing the Project heck students work. heckpoint Quiz 1 1. 2. 3. 4. not possible 5. 6. not possible 7. M Q, M Q,,, M Q, 8. lternate Interior ngles heorem 9. lternate Interior ngles heorem 10. heckpoint Quiz 2 1. B, B 2. ypotenuse-eg heorem 3. P 4. Q, Q, Q 5a. definition of a bisector 5b. eflexive 5c. 5d. P 5e. definition hapter est, Form 1. x = 50; y = 65 2. a = 118; b = 62; c = 59 3. 4. not possible 5. 6. 7. 8. 9. not possible 10. 11. not possible 12. heck students work; J P, K Q,, JK PQ, K Q, J P. 13. B 14. F 15. 16a. 16b. 16c. onverse of Isosceles riangle heorem 16d. Postulate 16e. P 16f. Isosceles riangle heorem 17. ample: tatements easons 1. B # ;is 1. midpoint of 2. B B 2. Perpendicular lines 3. 3. efinition of midpoint 4. B B 4. eflexive Property 5. B B 5. Postulate 6. B B 6. P 18. ample: that X is the midpoint of and B, X X and BX X by the definition of midpoint. XB X because all vertical angles are congruent. XB X by the Postulate, and therefore B by P. Pearson ducation, Inc., publishing as Pearson Prentice all. ll rights reserved. 38 nswers Geometry hapter 4
hapter 4 nswers (continued) Pearson ducation, Inc., publishing as Pearson Prentice all. ll rights reserved. hapter est, Form B 1. x = 58, y = 64 2. a = 40, b = 70, c = 70 3. 4. 5. 6. 7. not possible 8. not possible 9. or 10. or 11. 12. ; B ; F; B ; F; B F 13. B 14. B XY; B YZ; XZ 15. X; B Y; Z 16. 17. J 18a. 18b. 18c. ef. of equilateral 18d. eflexive Property 18e. Postulate lternative ssessment, Form K 1: coring Guide ample: : M : : P B B F kb kf P km kpm kb k kp kp 3 tudent s figures and information are clear and accurate. 2 tudent s figures and information contain minor errors or 1 tudent s figures and information contain significant errors or 0 tudent makes little or no attempt. K 2: coring Guide a. ample: 5 cm K 4 cm 4 cm M 5 cm b. ample: Using the Pythagorean theorem, show that K = M. hen K M by Postulate or Postulate. 3 tudent s figures and explanation are accurate and clear. 2 tudent s figures and explanation contain minor errors or 1 tudent s figures and explanation contain significant errors or 0 tudent makes little or no attempt. K 3: coring Guide ample: tatements easons 1., B 1. 2. 2. eflexive Property 3. B 3. heorem 4. B 4. P 5. B 5. egment ddition Postulate 3 tudent gives a proof that is accurate and complete. 2 tudent gives a proof that contains minor errors or 1 tudent gives a proof that contains significant errors or 0 tudent makes little or no attempt. K 4: coring Guide ample: he,, and Postulates are statements that are accepted as true without proof. he and heorems, on the other hand, can be proved true, using postulates, definitions, and previously proved theorems. 3 tudent gives an explanation that is thorough and correct. 2 tudent gives an explanation that is partially correct. 1 tudent gives an explanation that lacks demonstrated understanding of the difference between a theorem and a postulate. 0 tudent makes little or no attempt. Geometry hapter 4 nswers 39
hapter 4 nswers (continued) umulative eview 1. 2. G 3. 4. 5. 6. G 7. 8. G 9. 10. G 11. 12. J 13. 14. x = 102; y = 102 15. c, e, a, b, d or e, c, a, b, d 16. heorem P eflexive Property 17. ample: he alarm sounds if and only if there is smoke. If the alarm sounds, then there is smoke. If there is smoke, then the alarm sounds. 18., Pearson ducation, Inc., publishing as Pearson Prentice all. ll rights reserved. 40 nswers Geometry hapter 4