1. Medians and Altitudes of Triangle
Geometry
Medians and Altitudes of Triangle
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2. Warm up K 7 N Q P
9.1
40˚
36˚ L J M
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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.
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4. Construction Centroid of a TriangleB B X X Y Y A C A C Z Z
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5. B X P Y C A Z
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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.
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7. Theorem 3.1 Centroid Theorem
The centroid of a triangle is located 2 of the distance from each vertex to the midpoint of the opposite side. B X Y P A C Z
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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
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9. Subtract 2CE from both sides.
B) CE E F G
Centroid Thm.
Seg. Add. Post.
Substitute 2 CE for CG. 3
Subtract 2CE from both sides.
Substitute 2.4 for GE.
Multiply both sides by 3. A C D
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10. Now you try! 1) In ∆JKL, ZW =7,and LX = 8.1. Find each length. KW LZ X
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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 2 4 6 8
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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 2 4 6 8
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13. y
Q(0, 8) 8 6
R(6, 4) 4 2
P(3, 0) x 2 4 6 8
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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 2 4 6 8
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15. In ∆QRS, altitude QY is inside the triangle, but RX and SZ are not.
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
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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 -4 2
Y = x + 6
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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 -4 2
Y = x + 6
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18. Step 4 Solve the system to find the coordinates of the orthocenter.
X = -2
Step 4 Solve the system to find the coordinates of the orthocenter.
x = -2
y = x + 2
y = = 4
The coordinates of the orthocenter are (-2,4). K 7
(-2, 4) J L x -4 2
Y = x + 6
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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 -4 2
Y = x + 6
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20. Now some problems for you to practice !
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21. VX = 205, and RW = 104. Find each length.
Assessment
VX = 205, and RW = 104. Find each length.
1) VW ) WX
3) RY ) WY T Y X W V R
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22. 5) The diagram shows a plan for a piece of a mobile
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 2
C(5, 0) 8
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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)
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24. Medians and Altitudes of Triangle
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.
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25. Construction Centroid of a TriangleB B X X Y Y A C A C Z Z
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26. B X P Y C A Z
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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.
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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
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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
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30. Subtract 2/3 CE from both sides.
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
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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 2 4 6 8
Next page :
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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 2 4 6 8
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33. y
Q(0, 8) 8 6
R(6, 4) 4 2
P(3, 0) x 2 4 6 8
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34. In ∆QRS, altitude QY is inside the triangle, but RX and SZ are not.
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
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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 -4 2
Y = x + 6
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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 -4 2
Y = x + 6
Next page :
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37. Step 4 Solve the system to find the coordinates of the orthocenter.
X = -2
Step 4 Solve the system to find the coordinates of the orthocenter.
x = -2
y = x + 2
y = = 4
The coordinates of the orthocenter are (-2,4). K 7
(-2, 4) J L x -4 2
Y = x + 6
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38. You did a great job today!
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