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Answers for the lesson “Use Median and Altitude” LESSON 5.4 11. Z Skill Practice 1. circumcenter: when it is an acute triangle, when it is a right triangle, when it is an obtuse triangle; incenter: always, never, never; centroid: always, never, never; orthocenter: when it is an acute triangle, when it is a right triangle, when it is an obtuse triangle. C Y X 12. O Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved. 2. Both are perpendicular to a side of the triangle, although the altitude contains the vertex opposite the side while a perpendicular bisector bisects the side but does not necessarily contain the opposite vertex; both bisect one side of a triangle, although the perpendicular bisector does not necessarily contain the opposite vertex while the median is not necessarily perpendicular to the side but does contain the opposite vertex. 3. 12 4. 9 6. 5 7. D 5. 10 8. a. (8, 1); (5, 1) b. (5, 21); SQ 5 4 and SR 5 6, 2 therefore SQ 5 }3SR. 9. (3, 2) 10. (3, 4) B A C 13. no; no; yes 14. yes; yes; yes 15. no; yes; no 16. T is the orthocenter, but the centroid is needed to reach the conclusion. 17. altitude 18. angle bisector 19. median 20. perpendicular bisector, angle bisector, median, altitude 21. perpendicular bisector, angle bisector, median, altitude Geometry Answer Transparencies for Checking Homework 152 22. perpendicular bisector, angle 31. bisector, median, altitude O 23. 6, 228; nABD > nCBD by HL, use corr. parts of > ns are >. 24. 908, 228; nABD > nCBD by SSS, use definition of a linear pair and corr. parts of > ns are >. 25. 3 3 27. } 2 26. 2 28. If the base angles of the isosceles triangle are placed at (2a, 0) and (0, a), the vertex angle will be on the y-axis. C Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved. 29. O C B A 32. Sample answer: The midpoint of } is L(3, 7), so the equation of FG the median from H(6, 1) to L(3, 7) is y 5 22x 1 13. P(4, 5) lies on } this median. The midpoint of GH is J(5, 5), so the equation of the median from F(2, 5) to J(5, 5) is y 5 5. P(4, 5) lies on this median, so all three medians intersect at the centroid. 5 33. } 2 34. 9 } B A O 30. 35. 4 } 9Ï 19 9Ï 19 36. }, }; yes; the height and 2 2 base of both triangles will always be the same. B Problem Solving 37. B; it is the centroid of the triangle. A Geometry Answer Transparencies for Checking Homework 153 4. ADB and CDB are right angles. (Definition of altitude) 38. Sample: y y (n, h) (0, h) x (2n, 0) 5. nABD and nCDB are right triangles. (Definition of right triangles) x (0, 0) (2n, 0) (n, 0) 39. 6.75 in.2; altitude 40. Right; the orthocenter is on the 41. (0, 2); y 8 3 y2 4 x 5 Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved. (0, 5) 2 x 3 y3 x 4 2 42. a. Statements (Reasons) } is an 1. nABC is equilateral, BD altitude of nABC. (Given) 2. } AB > } BC 8. AD 5 CD (Definition of congruent segments) b. Statements (Reasons) 2. BA 5 BC (2, 2) (0, 4) y1 3x 4 (Corr. parts s are >.) of > n 1. nABC is equilateral. (Given) (4, 2) (2, 1) 7. } AD > } CD (HL) 9. } BD is a perpendicular bisector of } AC. (Definition of perpendicular bisector) right angle. (4, 8) 6. nABD > nCBD (Definition of equilateral triangle) 3. } BD > } BD (Reflexive Property of Segment Congruence) 3. } BD > } AC (Definition of equilateral triangle) (Definition of altitude) 4. B is on the perpendicular bisector of } AC. (Converse of the Perpendicular Bisector Theorem) 5. Only one line exists through B perpendicular to } AC, so the perpendicular bisector and the altitude are contained in the same line. (Perpendicular Postulate) Geometry Answer Transparencies for Checking Homework 154 43. a. Check students’ work. b. Statements (Reasons) b. Their areas are the same. c. They weigh the same; it means the weight of nABC is evenly distributed around its centroid. 2 44. a. 2} 3 1 2 28 3 2 3 b. y 5 }x, x 5 8, y 5 2}x 1 }, (8, 4) c. Find the equation of each perpendicular bisector of each side and solve the system. 45. a. Statements (Reasons) Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved. 1. } LP and } MQ are medians of ]› scalene nLMN, R is on LP such that } LP > } PR, S is › ] on MQ such that } MQ > } QS. (Given) 2. } MP > } NP, } QL > } QN (Definition of median) 1. nMPL > nNPR, nMQL > nSQN (Exercise 45a) 2. MLP > NRP, MLQ > SQN (Corr. parts s are >.) of > n } (Converse } i RN }, LM } i SN 3. LM of Alternate Interior Angles Theorem) c. Statements (Reasons) 1. } LM i } RN, } LM i } SN (Exercise 45b) 2. There is exactly one line through N parallel to } LM, so ‹]› ‹]› RN and SN are the same line. (Parallel Postulate) 3. S, N, and R are collinear. (Definition of collinear points) 3. MPL > NPR, MQL > SQN (Vertical Angles Congruence Theorem) 4. nMPL > nNPR, nMQL > nSQN (SAS) 5. } ML > } NR, } ML > } NS s are >.) (Corr. parts of > n } (Transitive Property } > NS 6. NR of Segment Congruence) Geometry Answer Transparencies for Checking Homework 155