Geometry Section 5-2 1112

2,138 views

Published on

Medians and Altitudes of Triangles

0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
2,138
On SlideShare
0
From Embeds
0
Number of Embeds
1,319
Actions
Shares
0
Downloads
23
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

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

×