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Geometry Medians and Altitudes of Triangle

Geometry Medians and Altitudes of Triangle. Warm up. K. 7. N. Q. P. 9.1. 40˚. 36˚. L. J. M. Medians and Altitudes of Triangle. A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. C. median. A. B. D.

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Geometry Medians and Altitudes of Triangle

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  1. Geometry Medians and Altitudes of Triangle CONFIDENTIAL

  2. Warm up K 7 N Q P 9.1 40˚ 36˚ L J M CONFIDENTIAL

  3. Medians and Altitudes of Triangle A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. C median A B D Every triangle has three medians, and the median are concurrent, as shown in the next slide. CONFIDENTIAL

  4. Construction Centroid of a Triangle B B X X Y Y A C A C Z Z Next page : CONFIDENTIAL

  5. B X P Y C A Z CONFIDENTIAL

  6. The point of concurrency of the medians of a triangle is the centroid of the triangle. The centroid is always inside the triangle. The centroid is also called the center of gravity because it is the point where a triangular region will balance. CONFIDENTIAL

  7. Theorem 3.1 Centroid Theorem The centroid of a triangle is located 2 of the distance from 3 each vertex to the midpoint of the opposite side. B X Y P A C Z CONFIDENTIAL

  8. Using the Centroid to Find Segment Lengths In ∆ABC, AF = 9, and GE = 2.4. Find each length. B A) AG E F G Centroid Thm. Substitute 9 for AF Simplify. A C D Next page : CONFIDENTIAL

  9. B B) CE E F G Centroid Thm. Seg. Add. Post. Substitute 2 CE for CG. 3 Subtract 2CE from both sides. 3 Substitute 2.4 for GE. Multiply both sides by 3. A C D CONFIDENTIAL

  10. Now you try! • 1) In ∆JKL, ZW =7,and LX = 8.1. • Find each length. • KW • LZ X K J Z W L CONFIDENTIAL

  11. Problem-Solving Application The diagram shows the plan for a triangular piece of a mobile. Where should the sculptor attach the support so that the triangle is balanced? y Q(0, 8) 8 6 1) Understand the Problem R(6, 4) 4 The answer will be the coordinates of the centroid of ∆PQR. The important information is the location of the vertices, P(3,0),Q(0,8), and R(6,4). 2 P(3, 0) x 0 2 4 6 8 Next page : CONFIDENTIAL

  12. 2) Make a Plan y Q(0, 8) The centroid of the triangle is the point of intersection of the three medians. So write the equations for two medians and find their point of intersection. 8 6 R(6, 4) 4 2 P(3, 0) x 0 2 4 6 8 CONFIDENTIAL

  13. y Q(0, 8) 8 6 R(6, 4) 4 2 P(3, 0) x 0 2 4 6 8 CONFIDENTIAL

  14. Now you try! 2) Find the average of the x-coordinates and the average of the y-coordinates of the vertices of ∆PQR. Make a conjecture about the centroid of a triangle. y Q(0, 8) 8 6 R(6, 4) 4 2 P(3, 0) x 0 2 4 6 8 CONFIDENTIAL

  15. An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. Every triangle has three altitudes. An altitude can be inside, outside, or on the triangle. R In ∆QRS, altitude QY is inside the triangle, but RX and SZ are not. Notice that the lines containing the altitudes are concurrent at P. This point of concurrency is the orthocenter of the triangle. Y Q X P Z S CONFIDENTIAL

  16. Finding the Orthocenter Find the orthocenter of ∆JKL with vertices J(-4,2), K(-2,6), and L(2,2). Step 1 Graph the triangle. Step 2 Find an equation of the line containing the altitude from K to JL. Since JL is horizontal, the altitude is vertical. The line containing it must pass through K(-2,6), so the equation of the line is x = -2. y X = -2 K 7 (-2, 4) J L x 0 -4 2 Y = x + 6 Next page : CONFIDENTIAL

  17. Step 3 Find an equation of the line containing the altitude from J to KL. y X = -2 The slope of a line perpendicular to KL is 1. This line must pass through J(-4, 2). K 7 (-2, 4) J L x 0 -4 2 Y = x + 6 Next page : CONFIDENTIAL

  18. y X = -2 Step 4 Solve the system to find the coordinates of the orthocenter. x = -2 y = x + 2 y = -2 + 6 = 4 The coordinates of the orthocenter are (-2,4). K 7 (-2, 4) J L x 0 -4 2 Y = x + 6 CONFIDENTIAL

  19. Now you try! 3) Show that the altitude to JK passes through the orthocenter of ∆JKL. y X = -2 K 7 (-2, 4) J L x 0 -4 2 Y = x + 6 CONFIDENTIAL

  20. Now some problems for you to practice ! CONFIDENTIAL

  21. Assessment VX = 205, and RW = 104. Find each length. 1) VW 2) WX 3) RY 4) WY T Y X W V R CONFIDENTIAL

  22. 5) The diagram shows a plan for a piece of a mobile. A chain will hang from the centroid of the triangle. At what coordinates should the artist attach the chain? y B(7, 4) 4 2 A(0, 2) x 0 2 C(5, 0) 8 CONFIDENTIAL

  23. Find the orthocenter of a triangle with the given vertices. K(2, -2), L(4, 6), M(8, -2) U(-4, -9), V(-4, 6),W(5, -3) P(-5, 8), Q(4, 5), R(-2, 5) C(-1, -3), D(-1, 2), E(9, 2) CONFIDENTIAL

  24. Let’s review Medians and Altitudes of Triangle A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. C median A B D Every triangle has three medians, and the median are concurrent, as shown in the construction below. CONFIDENTIAL

  25. Construction Centroid of a Triangle B B X X Y Y A C A C Z Z Next page : CONFIDENTIAL

  26. B X P Y C A Z CONFIDENTIAL

  27. The point of concurrency of the medians of a triangle is the centroid of the triangle. The centroid is always inside the triangle. The centroid is also called the center of gravity because it is the point where a triangular region will balance. CONFIDENTIAL

  28. Theorem 3.1 Centroid Theorem The centroid of a triangle is located 2/3 of the distance from each vertex to the midpoint of the opposite side. B X Y P A C Z CONFIDENTIAL

  29. Using the Centroid to Find Segment Lengths In ∆ABC, AF = 9, and GE = 2.4. Find each length. B A) AG E F G Centroid Thm. Substitute 9 for AF Simplify. A C D Next page : CONFIDENTIAL

  30. B B) CE E G Centroid Thm. Seg. Add. Post. Substitute 2/3 CE for CG. Subtract 2/3 CE from both sides. Substitute 2.4 for GE. Multiply both sides by 3. A D CONFIDENTIAL

  31. Problem-Solving Application The diagram shows the plan for a triangular piece of a mobile. Where should the sculptor attach the support so that the triangle is balanced? y Q(0, 8) 8 6 1) Understand the Problem R(6, 4) 4 The answer will be the coordinates of the centroid of ∆PQR. The important information is the location of the vertices, P(3,0),Q(0,8), and R(6,4). 2 P(3, 0) x 0 2 4 6 8 Next page : CONFIDENTIAL

  32. 2) Make a Plan y Q(0, 8) The centroid of the triangle is the point of intersection of the three medians. So write the equations for two medians and find their point of intersection. 8 6 R(6, 4) 4 2 P(3, 0) x 0 2 4 6 8 CONFIDENTIAL

  33. y Q(0, 8) 8 6 R(6, 4) 4 2 P(3, 0) x 0 2 4 6 8 CONFIDENTIAL

  34. An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. Every triangle has three altitudes. An altitude can be inside, outside, or on the triangle. R In ∆QRS, altitude QY is inside the triangle, but RX and SZ are not. Notice that the lines containing the altitudes are concurrent at P. This point of concurrency is the orthocenter of the triangle. Y Q X P Z S CONFIDENTIAL

  35. Finding the Orthocenter Find the orthocenter of ∆JKL with vertices J(-4,2), K(-2,6), and L(2,2). Step 1 Graph the triangle. Step 2 Find an equation of the line containing the altitude from K to JL. Since JL is horizontal, the altitude is vertical. The line containing it must pass through K(-2,6), so the equation of the line is x = -2. y X = -2 K 7 (-2, 4) J L x 0 -4 2 Y = x + 6 Next page : CONFIDENTIAL

  36. Step 3 Find an equation of the line containing the altitude from J to KL. y X = -2 The slope of a line perpendicular to KL is 1. This line must pass through J(-4, 2). K 7 (-2, 4) J L x 0 -4 2 Y = x + 6 Next page : CONFIDENTIAL

  37. y X = -2 Step 4 Solve the system to find the coordinates of the orthocenter. x = -2 y = x + 2 y = -2 + 6 = 4 The coordinates of the orthocenter are (-2,4). K 7 (-2, 4) J L x 0 -4 2 Y = x + 6 CONFIDENTIAL

  38. You did a great job today! CONFIDENTIAL

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