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# Geometry Section 5-2 1112

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Medians and Altitudes of Triangles

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### Geometry Section 5-2 1112

1. 1. SECTION 5-2 Medians and Altitudes of TrianglesThursday, March 1, 2012
2. 2. ESSENTIAL QUESTIONS How do you identify and use medians in triangles? How do you identify and use altitudes in triangles?Thursday, March 1, 2012
3. 3. VOCABULARY 1. Median: 2. Centroid: 3. Altitude: 4. Orthocenter:Thursday, March 1, 2012
4. 4. VOCABULARY 1. Median: A segment in a triangle that connects a vertex to the midpoint of the opposite side 2. Centroid: 3. Altitude: 4. Orthocenter:Thursday, March 1, 2012
5. 5. VOCABULARY 1. Median: A segment in a triangle that connects a vertex to the midpoint of the opposite side 2. Centroid: The point of concurrency where the medians intersect in a triangle (always inside) 3. Altitude: 4. Orthocenter:Thursday, March 1, 2012
6. 6. VOCABULARY 1. Median: A segment in a triangle that connects a vertex to the midpoint of the opposite side 2. Centroid: The point of concurrency where the medians intersect in a triangle (always inside) 3. Altitude: A segment in a triangle that connects a vertex to the opposite side so that the side and altitude are perpendicular 4. Orthocenter:Thursday, March 1, 2012
7. 7. VOCABULARY 1. Median: A segment in a triangle that connects a vertex to the midpoint of the opposite side 2. Centroid: The point of concurrency where the medians intersect in a triangle (always inside) 3. Altitude: A segment in a triangle that connects a vertex to the opposite side so that the side and altitude are perpendicular 4. Orthocenter: The point of concurrency where the altitudes of a triangle intersectThursday, March 1, 2012
8. 8. 5.7 - CENTROID THEOREM The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite sideThursday, March 1, 2012
9. 9. 5.7 - CENTROID THEOREM The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side If G is the centroid of ∆ABC, then 2 2 2 AG = AF , BG = BD, and CG = CE 3 3 3Thursday, March 1, 2012
10. 10. Special Segments and Points in Triangles Point of Special Name Example Example Concurrency Property Perpendicular Circumcenter is Circumcenter equidistant from Bisector each vertex Angle In center is Incenter equidistant from Bisector each side Centroid is two- thirds the distance Median Centroid from vertex to opposite midpoint Altitudes are Altitude Orthocenter concurrent at orthocenterThursday, March 1, 2012
11. 11. EXAMPLE 1 In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.Thursday, March 1, 2012
12. 12. EXAMPLE 1 In ∆XYZ, P is the centroid and YV = 12. Find YP and PV. 2 YP = YV 3Thursday, March 1, 2012
13. 13. EXAMPLE 1 In ∆XYZ, P is the centroid and YV = 12. Find YP and PV. 2 YP = YV 3 2 YP = (12) 3Thursday, March 1, 2012
14. 14. EXAMPLE 1 In ∆XYZ, P is the centroid and YV = 12. Find YP and PV. 2 YP = YV 3 2 YP = (12) 3 YP = 8Thursday, March 1, 2012
15. 15. EXAMPLE 1 In ∆XYZ, P is the centroid and YV = 12. Find YP and PV. 2 YP = YV PV = YV − YP 3 2 YP = (12) 3 YP = 8Thursday, March 1, 2012
16. 16. EXAMPLE 1 In ∆XYZ, P is the centroid and YV = 12. Find YP and PV. 2 YP = YV PV = YV − YP 3 2 YP = (12) PV = 12 − 8 3 YP = 8Thursday, March 1, 2012
17. 17. EXAMPLE 1 In ∆XYZ, P is the centroid and YV = 12. Find YP and PV. 2 YP = YV PV = YV − YP 3 2 YP = (12) PV = 12 − 8 3 YP = 8 PV = 4Thursday, March 1, 2012
18. 18. EXAMPLE 2 In ∆ABC, CG = 4. Find GE.Thursday, March 1, 2012
19. 19. EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE 3Thursday, March 1, 2012
20. 20. EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE 3 2 4 = CE 3Thursday, March 1, 2012
21. 21. EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE 3 2 4 = CE 3 CE = 6Thursday, March 1, 2012
22. 22. EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE GE = CE − CG 3 2 4 = CE 3 CE = 6Thursday, March 1, 2012
23. 23. EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE GE = CE − CG 3 2 4 = CE GE = 6 − 4 3 CE = 6Thursday, March 1, 2012
24. 24. EXAMPLE 2 In ∆ABC, CG = 4. Find GE. 2 CG = CE GE = CE − CG 3 2 4 = CE GE = 6 − 4 3 CE = 6 GE = 2Thursday, March 1, 2012
25. 25. EXAMPLE 3 An artist is designing a sculpture that balances a triangle on tip of a pole. In the artistʼs design on the coordinate plane, the vertices are located at (1, 4), (3, 0), and (3, 8). What are the coordinates of the point where the artist should place the pole under the triangle so that is will balance?Thursday, March 1, 2012
26. 26. EXAMPLE 3 (1, 4), (3, 0), (3, 8)Thursday, March 1, 2012
27. 27. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) xThursday, March 1, 2012
28. 28. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) xThursday, March 1, 2012
29. 29. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) xThursday, March 1, 2012
30. 30. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) We need to ﬁnd the centroid, so we start by ﬁnding the midpoint of our vertical side. xThursday, March 1, 2012
31. 31. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) We need to ﬁnd the centroid, so we start by ﬁnding the midpoint of our vertical side. x ⎛ 3 + 3 0 + 8⎞ M =⎜ , ⎟ ⎝ 2 2 ⎠Thursday, March 1, 2012
32. 32. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) We need to ﬁnd the centroid, so we start by ﬁnding the midpoint of our vertical side. x ⎛ 3 + 3 0 + 8⎞ ⎛ 6 8⎞ M =⎜ , ⎟ = ⎜ 2 , 2⎟ ⎝ 2 2 ⎠ ⎝ ⎠Thursday, March 1, 2012
33. 33. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) We need to ﬁnd the centroid, so we start by ﬁnding the midpoint of our vertical side. x ⎛ 3 + 3 0 + 8⎞ ⎛ 6 8⎞ M =⎜ , ⎟ = ⎜ 2 , 2⎟ ⎝ 2 2 ⎠ ⎝ ⎠ ( ) = 3,4Thursday, March 1, 2012
34. 34. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) We need to ﬁnd the centroid, so we start by ﬁnding the midpoint of our vertical side. x ⎛ 3 + 3 0 + 8⎞ ⎛ 6 8⎞ M =⎜ , ⎟ = ⎜ 2 , 2⎟ ⎝ 2 2 ⎠ ⎝ ⎠ ( ) = 3,4Thursday, March 1, 2012
35. 35. EXAMPLE 3 y (1, 4), (3, 0), (3, 8) We need to ﬁnd the centroid, so we start by ﬁnding the midpoint of our vertical side. x ⎛ 3 + 3 0 + 8⎞ ⎛ 6 8⎞ M =⎜ , ⎟ = ⎜ 2 , 2⎟ ⎝ 2 2 ⎠ ⎝ ⎠ ( ) = 3,4Thursday, March 1, 2012
36. 36. EXAMPLE 3 y xThursday, March 1, 2012
37. 37. EXAMPLE 3 y Next, we need the distance from the opposite vertex to this midpoint. xThursday, March 1, 2012
38. 38. EXAMPLE 3 y Next, we need the distance from the opposite vertex to this midpoint. d = (1 − 3)2 + (4 − 4)2 xThursday, March 1, 2012
39. 39. EXAMPLE 3 y Next, we need the distance from the opposite vertex to this midpoint. d = (1 − 3)2 + (4 − 4)2 x = (−2) + 0 2 2Thursday, March 1, 2012
40. 40. EXAMPLE 3 y Next, we need the distance from the opposite vertex to this midpoint. d = (1 − 3)2 + (4 − 4)2 x = (−2) + 0 2 2 = 4Thursday, March 1, 2012
41. 41. EXAMPLE 3 y Next, we need the distance from the opposite vertex to this midpoint. d = (1 − 3)2 + (4 − 4)2 x = (−2) + 0 2 2 = 4 =2Thursday, March 1, 2012
42. 42. EXAMPLE 3 y xThursday, March 1, 2012
43. 43. EXAMPLE 3 y The centroid P is 2/3 of this distance from the vertex. xThursday, March 1, 2012
44. 44. EXAMPLE 3 y The centroid P is 2/3 of this distance from the vertex. ⎛ 2 ⎞ P = ⎜1 + (2),4 ⎟ ⎝ 3 ⎠ xThursday, March 1, 2012
45. 45. EXAMPLE 3 y The centroid P is 2/3 of this distance from the vertex. ⎛ 2 ⎞ P = ⎜1 + (2),4 ⎟ ⎝ 3 ⎠ x ⎛ 4 ⎞ P = ⎜1 + ,4 ⎟ ⎝ 3 ⎠Thursday, March 1, 2012
46. 46. EXAMPLE 3 y The centroid P is 2/3 of this distance from the vertex. ⎛ 2 ⎞ P = ⎜1 + (2),4 ⎟ ⎝ 3 ⎠ x ⎛ 4 ⎞ P = ⎜1 + ,4 ⎟ ⎝ 3 ⎠ ⎛7 ⎞ P = ⎜ ,4 ⎟ ⎝3 ⎠Thursday, March 1, 2012
47. 47. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ.Thursday, March 1, 2012
48. 48. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y xThursday, March 1, 2012
49. 49. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y H J x IThursday, March 1, 2012
50. 50. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y H J x IThursday, March 1, 2012
51. 51. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y H J x IThursday, March 1, 2012
52. 52. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J x IThursday, March 1, 2012
53. 53. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J Let’s ﬁnd the equations for the altitudes coming from I and H. x IThursday, March 1, 2012
54. 54. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J Let’s ﬁnd the equations for the altitudes coming from I and H. x I ( ) m JI = 1+ 3 −5 + 3Thursday, March 1, 2012
55. 55. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J Let’s ﬁnd the equations for the altitudes coming from I and H. x I ( ) m JI = 1+ 3 = 4 −5 + 3 −2Thursday, March 1, 2012
56. 56. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J Let’s ﬁnd the equations for the altitudes coming from I and H. x I ( ) m JI = 1+ 3 = 4 −5 + 3 −2 = −2Thursday, March 1, 2012
57. 57. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J Let’s ﬁnd the equations for the altitudes coming from I and H. x I ( ) m JI = 1+ 3 = 4 −5 + 3 −2 = −2 ⊥ m = 1 2Thursday, March 1, 2012
58. 58. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J Let’s ﬁnd the equations for the altitudes coming from I and H. x I ( ) m JI = 1+ 3 = 4 −5 + 3 −2 = −2 ⊥ m = 1 2 ( ) m HJ = 2 −1 1+ 5Thursday, March 1, 2012
59. 59. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J Let’s ﬁnd the equations for the altitudes coming from I and H. x I ( ) m JI = 1+ 3 = −5 + 3 −2 4 = −2 ⊥ m = 1 2 ( ) m HJ = 2 −1 1 1+ 5 6 =Thursday, March 1, 2012
60. 60. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y To ﬁnd the orthocenter, ﬁnd the intersection of two altitudes. H J Let’s ﬁnd the equations for the altitudes coming from I and H. x I ( ) m JI = 1+ 3 = −5 + 3 −2 4 = −2 ⊥ m = 1 2 ( ) m HJ = 2 −1 1 1+ 5 6 = ⊥ m = −6Thursday, March 1, 2012
61. 61. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through H H J x IThursday, March 1, 2012
62. 62. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through H 1 ⊥ m = , (1,2) H 2 J x IThursday, March 1, 2012
63. 63. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through H 1 ⊥ m = , (1,2) H 2 J 1 y − 2 = (x −1) 2 x IThursday, March 1, 2012
64. 64. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through H 1 ⊥ m = , (1,2) H 2 J 1 y − 2 = (x −1) 2 x 1 1 I y−2= x − 2 2Thursday, March 1, 2012
65. 65. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through H 1 ⊥ m = , (1,2) H 2 J 1 y − 2 = (x −1) 2 x 1 1 I y−2= x − 2 2 1 3 y= x+ 2 2Thursday, March 1, 2012
66. 66. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through I H J x IThursday, March 1, 2012
67. 67. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through I ⊥ m = −6, (−3,−3) H J x IThursday, March 1, 2012
68. 68. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through I ⊥ m = −6, (−3,−3) H y + 3 = −6(x + 3) J x IThursday, March 1, 2012
69. 69. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through I ⊥ m = −6, (−3,−3) H y + 3 = −6(x + 3) J x y + 3 = −6x −18 IThursday, March 1, 2012
70. 70. EXAMPLE 4 The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1). Find the coordinates of the orthocenter of ∆HIJ. y Altitude through I ⊥ m = −6, (−3,−3) H y + 3 = −6(x + 3) J x y + 3 = −6x −18 I y = −6x − 21Thursday, March 1, 2012
71. 71. EXAMPLE 4Thursday, March 1, 2012
72. 72. EXAMPLE 4 1 3 −6x − 21 = x + 2 2Thursday, March 1, 2012
73. 73. EXAMPLE 4 1 3 −6x − 21 = x + 2 2 +6x +6xThursday, March 1, 2012
74. 74. EXAMPLE 4 1 3 −6x − 21 = x + 2 2 +6x +6x 13 3 −21 = x + 2 2Thursday, March 1, 2012
75. 75. EXAMPLE 4 1 3 −6x − 21 = x + 2 2 +6x +6x 13 3 −21 = x + 2 2 3 3 − − 2 2Thursday, March 1, 2012
76. 76. EXAMPLE 4 1 3 −6x − 21 = x + 2 2 +6x +6x 13 3 −21 = x + 2 2 3 3 − − 2 2 45 13 − = x 2 2Thursday, March 1, 2012
77. 77. EXAMPLE 4 1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2 −6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13 2 2 13 ⎝ ⎠ ⎝ ⎠ +6x +6x 13 3 −21 = x + 2 2 3 3 − − 2 2 45 13 − = x 2 2Thursday, March 1, 2012
78. 78. EXAMPLE 4 1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2 −6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13 2 2 13 ⎝ ⎠ ⎝ ⎠ +6x +6x 45 13 3 x=− −21 = x + 13 2 2 3 3 − − 2 2 45 13 − = x 2 2Thursday, March 1, 2012
79. 79. EXAMPLE 4 1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2 −6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13 2 2 13 ⎝ ⎠ ⎝ ⎠ +6x +6x 45 13 3 x=− −21 = x + 13 2 2 ⎛ 45 ⎞ 1 ⎛ 45 ⎞ 3 3 3 −6 ⎜ − ⎟ − 21 = ⎜ − ⎟ + − − ⎝ 13 ⎠ 2 ⎝ 13 ⎠ 2 2 2 45 13 − = x 2 2Thursday, March 1, 2012
80. 80. EXAMPLE 4 1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2 −6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13 2 2 13 ⎝ ⎠ ⎝ ⎠ +6x +6x 45 13 3 x=− −21 = x + 13 2 2 ⎛ 45 ⎞ 1 ⎛ 45 ⎞ 3 3 3 −6 ⎜ − ⎟ − 21 = ⎜ − ⎟ + − − ⎝ 13 ⎠ 2 ⎝ 13 ⎠ 2 2 2 45 13 3 3 − = x − =− 2 2 13 13Thursday, March 1, 2012
81. 81. EXAMPLE 4 1 3 2 ⎛ 45 ⎞ ⎛ 13 ⎞ 2 −6x − 21 = x + ⎜ − 2 ⎟ = ⎜ 2 x ⎟ 13 2 2 13 ⎝ ⎠ ⎝ ⎠ +6x +6x 45 13 3 x=− −21 = x + 13 2 2 ⎛ 45 ⎞ 1 ⎛ 45 ⎞ 3 3 3 −6 ⎜ − ⎟ − 21 = ⎜ − ⎟ + − − ⎝ 13 ⎠ 2 ⎝ 13 ⎠ 2 2 2 45 13 3 3 − = x − =− 2 2 13 13 ⎛ 45 3 ⎞ ⎜ − 13 ,− 13 ⎟ ⎝ ⎠Thursday, March 1, 2012
82. 82. CHECK YOUR UNDERSTANDING p. 337 #1-4Thursday, March 1, 2012
83. 83. PROBLEM SETThursday, March 1, 2012
84. 84. PROBLEM SET p. 338 #5-31 odd, 49, 53 "Educations purpose is to replace an empty mind with an open one." – Malcolm ForbesThursday, March 1, 2012