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Answers for 5.2 For use with pages 306–309 5.2 Skill Practice 19. Sample: 1. circumcenter 2. all points on the perpendicular bisector of } AB 3. 15 4. 30 5. 55 6. yes 7. yes 8. no 9. B 10. You don’t know that EC 5 DC. 11. 35 12. 43 13. 50 14. 50 15. Yes; the Converse of the Perpendicular Bisector Theorem ‹]› guarantees L is on JP . Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 16. 9 17. 11 18. Sample answer: In the 20. Sometimes; a scalene triangle can be acute, right, or obtuse. 21. Always; congruent sides are created. 22. Sometimes; consider an equilateral triangle and a scalene triangle. construction of the segment bisector four congruent triangles are created. In the process four pairs of congruent angles are formed which are right angles making the bisector perpendicular to the segment. Geometry Answer Transparencies for Checking Homework ngat-0500.indd 6 A6 3/27/06 1:53:14 PM Answers for 5.2 For use with pages 306–309 continued 26. Statements (Reasons) 23. a– c. Statements (Reasons) 1. P is a perpendicular bisector of } XZ at Y; W and V lie in plane P. (Given) 2. AP 5 BP, m∠CPA 5 m∠CPB 5 908 (Definition of perpendicular bisector) 2. XW 5 ZW, XV 5 ZV (Perpendicular Bisector Theorem) AP > } BP (Definition of 3. } segment congruence) XW > } ZW, } XV > } ZV 3. } (Definition of segment congruence) 4. ∠CPA > ∠CPB (Definition of angle congruence) WV > } WV (Reflexive Property 4. } of Segment Congruence) 5. nVXW > nVZW Copyright © by McDougal Littell, a division of Houghton Mifflin Company. ‹]› is the perpendicular 1. PC bisector of } AB. (Given) (SSS) 6. ∠VXW > ∠VZW (Corr. Parts s are >.) of > n 5.2 Problem Solving CP > } CP (Reflexive Property 5. } of Segment Congruence) 6. nCPA > nCPB CA > } CB 7. } (SAS) (Corr. parts s are >.) of > n (Definition of 8. CA 5 CB segment congruence) 24. 59.6 m, 195.5 m; Perpendicular Bisector Theorem 25. Theorem 5.4 shows you that you can find a point equidistant from three points by using the perpendicular bisectors of the sides of the triangle formed by the three points. Geometry Answer Transparencies for Checking Homework ngat-0500.indd 7 A7 3/27/06 1:53:15 PM Answers for 5.2 For use with pages 306–309 continued 27. Statements (Reasons) 30. Midpoint of the hypotenuse. (Given) 1. CA 5 CB ‹]› } ⊥ AB through 2. Draw PC point C. (Perpendicular Postulate) 3. } CA > } CB (Definition of segment congruence) CP > } CP (Reflexive Property 4. } of Segment Congruence) 5. ∠CPA and ∠CPB right angles. (Definition of perpendicular lines) Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 6. nCPA and nCPB are right triangles. (Definition of right triangle) 7. nCPA > nCPB PA > } PB 8. } (HL) (Corr. parts s are >.) of > n 9. C is on the perpendicular bisector of } AB. (Definition of perpendicular bisector) 28. a. Find the intersection of the perpendicular bisectors of the triangle formed by the three points. b. approximately (7, 6.5) 29. Check students’ work; Sample answer: Right triangle with vertices A(2a, 0), B(0, 2b), and C(0, 0); the midpoint of } AC is (a, 0) and midpoint of } BC is (0, b). The equation of the perpendicular bisectors of } AC } and BC are x 5 a and y 5 b. These two lines intersect in the point (a, b) which is the midpoint of } AB. 31. Case 1: Given: D, E, and B are collinear. AB > } BC Prove: } Statements (Reasons) 1. D, E, and B are collinear; } AD > } CD, } AE > } CE. (Given) ‹]› containing point E. 2. Draw DB (Two points determine a line.) DE > } DE (Reflexive Property 3. } of Segment Congruence) 4. nDAE > nDCE (SSS) 5. ∠AED > ∠CED (Corr. parts s are >.) of > n 6. ∠AED and ∠AEB, ∠CED and ∠CEB are linear pairs. (Definition of linear pair) Perpendicular Bisector Theorem. Geometry Answer Transparencies for Checking Homework ngat-0500.indd 8 A8 3/27/06 1:53:16 PM Answers for 5.2 For use with pages 306–309 continued 31. (cont.) 7. ∠AED and ∠AEB are supplementary, ∠CED and ∠CEB are supplementary. (Linear Pair Postulate) 8. ∠AEB > ∠CEB (Congruent Supplements Theorem) EB > } EB (Reflexive Property 9. } of Segment Congruence) 10. nAEB > nCEB AB > } BC 11. } (SAS) (Corr. parts s are >.) of > n AB > } BC Case 2: Given: } Prove: D, E, and B are collinear. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. Statements (Reasons) DE, } EB, and } AC. 1. Draw } (Two point determine a line.) AB > } BC, } AD > } CD, 2. } } } AE > CE (Given) 3. AB 5 CB, AD 5 CD, AE 5 CE (Definition of congruent segments) 4. B is on the perpendicular bisector of } AC, D is on the perpendicular bisector of } AC, E is on the perpendicular bisector of } AC. (Converse of the Perpendicular Bisector Theorem) 5. There exists only one line that is the perpendicular bisector of } AC, so B, D, and C are all on the same line. (Through a point not on a line, there exists only one line through the point perpendicular to the given line.) 6. D, E, and B are collinear. (Definition of collinear) 32. Statements (Reasons) 1. PQRST is a regular polygon, } SV > } RV. (Given) TP > } QP, } TS > } QR (Definition 2. } of regular polygon) PW > } PW, } PV > } PV 3. } (Reflexive Property of Segment Congruence) 4. VPTS > VPQR (Definition of congruent polygons) 5. ∠TPW > ∠QPW (Corr. parts of > polygons are >.) 6. nPWT > nPWQ TW > } QW 7. } (SAS) (Corr. parts s are >.) of > n 8. ∠PWT > ∠PWQ are a linear pair. (Definition of linear pair) Geometry Answer Transparencies for Checking Homework ngat-0500.indd 9 A9 3/27/06 1:53:17 PM Answers for 5.2 For use with pages 000–000 continued 32. (cont.) 41. Starting with 3, zero is added to PV ⊥ } TQ (If two lines intersect 9. } to form a linear pair of congruent angles, the lines are perpendicular.) get 3, then one is added to get 4, then 2 is added to get 6, next add 3; 9. PV is the perpendicular 10. } bisector of } TQ. (Definition of perpendicular bisector) 33. No; unless the four points determine a rectangle there is no single point to locate the school so that it is equidistant from the four towns. Cedar Falls Copyright © by McDougal Littell, a division of Houghton Mifflin Company. Shady Hills Lake City Willow Valley 5.2 Mixed Review 34. 612 35. 68 37. 18; 1588 } 36. 62Ï 7 38. 9; 908 39. The numbers are decreasing by 5; 1. 40. The next number in the sequence is the previous one multiplied by 3; 162. Geometry Answer Transparencies for Checking Homework ngat-0500.indd 10 A10 3/27/06 1:53:19 PM