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Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 Pages Exercises 1. M 2. DF 3. XY 4. 5. 6. 7. 8. 9. 10. a. Given b. Reflexive Prop. of c. Given d. AAS e. CPCTC LQP PML; HL 4-7
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Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 RST UTS; SSS QDA UAD; SAS QPT RUS; AAS 15. TD RO if TDI ROE by AAS. TID REO if TEI RIE. TEI RIE by SSS. 16. AE DE if AEB DEC by AAS. AB DC and A D since they are corr. parts of ABC and DCB, which are by HL. QET QEU by SAS if QT QU. QT and QU are corr. parts of QTB and QUB which are by ASA. ADC EDG by ASA if A E. A and E are corr. parts in ADB and EDF, which are by SAS. 19–22. Answers may vary. Samples are given. 19. 20. 4-7
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Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 21. a. b. 22. a. ACE BCD by ASA; AC BC, A B (Given) C C (Reflexive Prop. of ) ACE BCD (ASA) WYX ZXY by HL; WY YX, ZX YX, WX ZY (Given) WYX and ZXY are rt. (Def. of ) XY XY (Reflexive Prop. of ) WYX ZXY (HL) 25. m 1 = 56; m 2 = 56; m 3 = 34; m 4 = 90; m 5 = 22; m 6 = 34; m 7 = 34; m 8 = 68; m 9 = 112 ABC FCG; ASA 27. a. Given b. Reflexive Prop. of c. Given d. ETI s 4-7
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Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 27. (continued) e. IRE f. CPCTC g. Given h. All rt. are . i. TDI j. ROE k. CPCTC 28. a. Given b. Def. of c. Def. of rt. d. Given e. BC BC f. Reflexive g. HL h. CPCTC 28. (continued) i. DEC j. Vert. are . k. AAS l. AE DE m. CPCTC s s 4-7
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Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 29–30. Proofs may vary. Samples are given. 29. It is given that 1 2 and Since QB QB by the Reflexive Prop. of , QTB QUB by ASA. So QT QU by CPCTC. Since QE QE by the Reflexive Prop. of , then QET QEU by SAS. AD ED (Given) 2. D is the midpt. of BF. (Given) 3. FD DB (Def. of midpt.) 4. FDE ADB (Vert. are .) 5. FDE BDA (SAS) 6. E A (CPCTC) s 30. (continued) 7. GDE CDA (Vert. are .) ADC EDG (ASA) 31. a. AD BC; AB DC; AE EC; DE EB 4-7
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Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 31. (continued) b. Use DB DB (refl.) and alt. int. to show ADB CBD (ASA). AB DC and AD BC (CPCTC). AEB CED (ASA) and AED CEB (ASA). Then AE EC and DE EB (CPCTC). s AC EC; CB CD (Given) 2. C C (Reflexive Prop. of ) 3. ACD ECB (SAS) 4. A E (CPCTC) 33. PQ RQ and PQT RQT by Def. of bisector. QT QT so PQT RQT by SAS. P R by CPCTC. QT bisects VQS so VQT SQT and PQT and RQT are both rt So VQP SQR since they are compl. of PQV RQS by ASA so QV QS by CPCTC. 34. C 35. F s 4-7
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Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 36. A 37. [2] a. HBC HED b. HB HE by CPCTC if HBC HED by ASA. Since BDC CED by AAS, then DBC CED by CPCTC and CHB DHE because vertical are . [1] one part correct s 38. [4] a. HL b. c. x = 30. In ADC m A + m ADC + m ACD = Substituting, 90 + x + x + x = Solving, x = 30. d. 120; it is suppl. to a 60° e. 6 m; DC = 2(AD) 4-7
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Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 1 3 38. (continued) [3] 4 parts answered correctly [2] 3 parts answered correctly [1] 2 parts answered correctly 39. a. right b. c. Reflexive d. HL 40. 41. 42. y + 6 = (x – 2) 43. y – 5 = 1(x – 0) 44. y – 6 = –2(x + 3) 45. y – 0 = – (x – 0) 46–48. Eqs. may vary, depending on pt. chosen. 46. y – 4 = 2(x – 1) 47. y + 5 = (x – 3) 48. y + 3 = – (x + 4) 5 3 5 6 1 2 4-7
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