confidential1 geometry medians and altitudes of triangle
TRANSCRIPT
CONFIDENTIAL 1
Geometry
Medians and Altitudes of
Triangle
CONFIDENTIAL 2
Warm up
1) NQ, QP, and QM are perpendicular bisectors of JKL. Find each Measure.
a) KL b) QJ c) m JQL
K
PN
M
36˚
40˚ 9.1
L
Q
7
J
CONFIDENTIAL 3
A median of a triangle is a segment whose endpoints are a vertex of the triangle and
the midpoint of the opposite side.
Medians and Altitudes of Triangle
median
A B
C
DEvery triangle has three medians, and the median are
concurrent, as shown in the next slide.
CONFIDENTIAL 4
Construction Centroid of a Triangle
A
B
C
X
Z
Y
A
B
C
X
Z
Y
Next page :
Draw ABC. Construct themidpoint of AB, BC, and AC.Label the midpoints of the sidesX, Y, and Z, respectively.
Darw AY, BZ, and CX. Theseare the three median of ABC.
CONFIDENTIAL 5
A
B
C
X
Z
YP
Label the point where AY, BZ,and CX intersect as P.
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.
AP = 2
3AY BP =
2
3BZ CP =
2
3CX
A
B
C
X
Z
YP
CONFIDENTIAL 8
A
B
C
E
D
F
G
Using the Centroid to Find Segment Lengths
In ∆ABC, AF = 9, and GE = 2.4. Find each length.
A) AG
Centroid Thm.
Substitute 9 for AF
Simplify.
AG = 2
3AF
AG = 2
3(9)
AG = 6
Next page :
CONFIDENTIAL 9
B) CE
CG = 2
3CE
CG + GE = CE
2
3CE + GE = CE
GE = 1
3CE
2.4 = 1
3CE
7.2 = CE
Centroid Thm.Seg. Add. Post.Substitute 2 CE for CG. 3Subtract 2CE from both sides. 3
Substitute 2.4 for GE.Multiply both sides by 3.
A
B
C
E
D
F
G
CONFIDENTIAL 10
Now you try!
1) In ∆JKL, ZW =7,and LX = 8.1.
Find each length.
a) KW
b) LZ W
K
Z
L
JX
CONFIDENTIAL 11
R(6, 4)
y
xP(3, 0)
Q(0, 8)
0
2
4
6
8
2 4 6 8
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?
1) Understand the Problem
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).
Next page :
CONFIDENTIAL 12
R(6, 4)
y
xP(3, 0)
Q(0, 8)
0
2
4
6
8
2 4 6 8
2) Make a Plan
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.
CONFIDENTIAL 13
R(6, 4)
y
xP(3, 0)
Q(0, 8)
0
2
4
6
8
2 4 6 8
Let M be the midpoint of QR and N Be the midpoint of QP.
m = 0+6
2,8+4
2 = (3, 6) n =
0+3
2,8+0
2 = (1.5, 4)
PM is vertical. Its equation is x = 3. RN is horizontal.Its equation is y =4. The coordinates of the centroid are S(3, 4).
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.
R(6, 4)
y
xP(3, 0)
Q(0, 8)
0
2
4
6
8
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.
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.
Q
R
SP Z
X
Y
CONFIDENTIAL 16
Finding the Orthocenter
Next page :
(-2, 4)
K
LJ
X = -2
x
y
0-4Y = x + 6
7
2
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.
CONFIDENTIAL 17
Next page :
(-2, 4)
K
LJ
X = -2
x
y
0-4Y = x + 6
7
2
Step 3 Find an equation of the line containing the altitude from J to KL.
slope of KL = 2 - 6
2- (-2) = -1
The slope of a line
perpendicular to KL is 1.
This line must pass
through J(-4, 2).
CONFIDENTIAL 18
(-2, 4)
K
LJ
X = -2
x
y
0-4Y = x + 6
7
2
y - y1 = m(x - x1)
y - 2 = 1[x-(-4)]
y - 2 = x + 4
y = x + 6
Point-slope form
Substitute 2 for y1. 1 for m, and -4 for x1.
Distribute 1.
Add 2 to both sides.
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).
Substitute -2for x
CONFIDENTIAL 19
Now you try!
3) Show that the altitude to JK passes through the orthocenter of ∆JKL.
(-2, 4)
K
LJ
X = -2
x
y
0-4Y = x + 6
7
2
CONFIDENTIAL 20
Now some problems for you to practice !
CONFIDENTIAL 21
VX = 205, and RW = 104. Find each length.
1) VW 2) WX
3) RY 4) WY
WY
X
R
V
T
Assessment
CONFIDENTIAL 22
0
4
2
y
x
B(7, 4)
A(0, 2)
2 C(5, 0) 8
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?
CONFIDENTIAL 23
Find the orthocenter of a triangle with the given vertices.
6) K(2, -2), L(4, 6), M(8, -2)
7) U(-4, -9), V(-4, 6),W(5, -3)
8) P(-5, 8), Q(4, 5), R(-2, 5)
9) C(-1, -3), D(-1, 2), E(9, 2)
CONFIDENTIAL 24
Let’s review
A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side.
Medians and Altitudes of Triangle
median
A B
C
DEvery triangle has three medians, and the median are concurrent, as shown in the construction below.
CONFIDENTIAL 25
Construction Centroid of a Triangle
A
B
C
X
Z
Y
A
B
C
X
Z
Y
Next page :
Draw ABC. Construct themidpoint of AB, BC, and AC.Label the midpoints of the sidesX, Y, and Z, respectively.
Darw AY, BZ, and CX. Theseare the three median of ABC.
CONFIDENTIAL 26
A
B
C
X
Z
YP
Label the point where AY, BZ,and CX intersect as P.
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.
AP = 2
3AY BP =
2
3BZ CP =
2
3CX
A
B
C
X
Z
YP
CONFIDENTIAL 29
A
B
C
E
D
F
G
Using the Centroid to Find Segment Lengths
In ∆ABC, AF = 9, and GE = 2.4. Find each length.
A) AG
Centroid Thm.
Substitute 9 for AF
Simplify.
AG = 2
3AF
AG = 2
3(9)
AG = 6
Next page :
CONFIDENTIAL 30
B) CE
CG = 2
3CE
CG + GE = CE
2
3CE + GE = CE
GE = 1
3CE
2.4 = 1
3CE
7.2 = CE
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
B
E
D
G
CONFIDENTIAL 31
R(6, 4)
y
xP(3, 0)
Q(0, 8)
0
2
4
6
8
2 4 6 8
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?
1) Understand the Problem
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).
Next page :
CONFIDENTIAL 32
R(6, 4)
y
xP(3, 0)
Q(0, 8)
0
2
4
6
8
2 4 6 8
2) Make a Plan
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.
CONFIDENTIAL 33
R(6, 4)
y
xP(3, 0)
Q(0, 8)
0
2
4
6
8
2 4 6 8
Let M be the midpoint of QR and N Be the midpoint of QP.
m = 0+6
2,8+4
2 = (3, 6) n =
0+3
2,8+0
2 = (1.5, 4)
PM is vertical. Its equation is x = 3. RN is horizontal.Its equation is y =4. The coordinates of the centroid are S(3, 4).
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.
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.
Q
R
SP Z
X
Y
CONFIDENTIAL 35
Finding the Orthocenter
Next page :
(-2, 4)
K
LJ
X = -2
x
y
0-4Y = x + 6
7
2
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.
CONFIDENTIAL 36
Next page :
(-2, 4)
K
LJ
X = -2
x
y
0-4Y = x + 6
7
2
Step 3 Find an equation of the line containing the altitude from J to KL.
slope of KL = 2 - 6
2- (-2) = -1
The slope of a line
perpendicular to KL is 1.
This line must pass
through J(-4, 2).
CONFIDENTIAL 37
(-2, 4)
K
LJ
X = -2
x
y
0-4Y = x + 6
7
2
y - y1 = m(x - x1)
y - 2 = 1[x-(-4)]
y - 2 = x + 4
y = x + 6
Point-slope form
Substitute 2 for y1. 1 for m, and -4 for x1.
Distribute 1.
Add 2 to both sides.
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).
Substitute -2for x
CONFIDENTIAL 38
You did a great job today!