Transcript Slide 1

Geometry
Medians and
Altitudes of Triangle
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1
Warm up
1)
NQ, QP, and QM are perpendicular
bisectors of JKL. Find each Measure.
K
a) KL
b) QJ
c) m
JQL
7
P
N
Q
40˚
L
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36˚
J
M
2
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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Construction
Centroid of a Triangle
B
B
X
A
X
Y
Z
C
Draw
ABC. Construc t the
midpoint of AB, BC, and AC.
Label the midpoints of the sides
X, Y, and Z, respectively.
A
Y
Z
C
Darw AY, BZ , and CX. These
are the three median of
ABC.
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B
X
A
P
Y
Z
C
Label the point where AY, BZ ,
and CX intersect as P.
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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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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
3
2
AY
BP = BZ
3
CP =
B
X
A
P
Z
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2
3
CX
Y
C
7
Using the Centroid to Find
Segment Lengths
In ∆ABC, AF = 9, and GE = 2.4. Find each length.
B
A)
AG
E
AG =
F
G
2
AF
3
2
AG = (9)
3
AG = 6
Centroid Thm.
Substitute 9 for AF
A
C
D
Simplify.
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B
B)
CE
E
2
CG = CE
3
CG + GE = CE
2
CE + GE = CE
3
1
GE = CE
3
1
2.4 = CE
3
7.2 = CE
F
G
Centroid Thm. A
D
Seg. Add. Post.
Substitute 2 CE for CG.
3
Subtract 2CE from both sides.
3
C
Substitute 2.4 for GE.
Multiply both sides by 3.
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Now you try!
1) In ∆JKL, ZW =7,and LX = 8.1.
Find each length.
a) KW
X
J
b) LZ
W
K
Z
L
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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).
y
8
Q(0, 8)
6
R(6, 4)
4
2
0
P(3, 0)
2 4 6
x
8
Next page :
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2)
Make a Plan
y
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
Q(0, 8)
6
R(6, 4)
4
2
0
P(3, 0)
2 4 6
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x
8
12
Let M be the midpoint of QR and N Be the midpoint of QP.
0+6 8+4
0+3 8+0
m=
,
= (3, 6) n =
,
= (1.5, 4)
2
2
2
2




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).
y
8
Q(0, 8)
6
R(6, 4)
4
2
0
P(3, 0)
2 4 6
x
8
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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
8
Q(0, 8)
6
R(6, 4)
4
2
0
P(3, 0)
2 4 6
x
8
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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.
CONFIDENTIAL
Y
X
P
Q
Z
S
15
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.
y
X = -2
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.
K 7
J
Y=x+6
L
-4
0
2
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x
Step 3 Find an equation of the line containing the
altitude from J to KL.
slope of KL =
2- 6
2- (-2)
= -1
y
X = -2
The slope of a line
perpendicular to KL is 1.
K 7
This line must pass
through J(-4, 2).
J
Y=x+6
L
-4
0
2
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x
y - y1 = m(x - x1)
Point-slope form
y - 2 = 1[x-(-4)]
Substitute 2 for y1. 1 for m, and -4 for x1.
y - 2 =x +4
Distribute 1.
y =x +6
Add 2 to both sides.
y
X = -2
Step 4 Solve the system
to find the coordinates of
the orthocenter.
x = -2
K 7
y=x+2
Substitute -2
for x
The coordinates of the
orthocenter are (-2,4).
y = -2 + 6 = 4
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J
Y=x+6
-4
L
0
2
18
x
Now you try!
3) Show that the altitude to JK passes through the
orthocenter of ∆JKL.
y
X = -2
K 7
J
Y=x+6
-4
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L
0
x
2
19
Now some problems for you to practice !
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Assessment
VX = 205, and RW = 104. Find each length.
1) VW
2) WX
3) RY
4) WY
T
X
W
Y
V
R
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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
0
A(0, 2)
2
x
C(5, 0) 8
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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)
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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
C
opposite side.
median
A
B
D
Every triangle has three medians, and the median are
concurrent, as shown in the construction below.
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24
Construction
Centroid of a Triangle
B
B
X
A
X
Y
Z
C
Draw
ABC. Construc t the
midpoint of AB, BC, and AC.
Label the midpoints of the sides
X, Y, and Z, respectively.
A
Y
Z
C
Darw AY, BZ , and CX. These
are the three median of
ABC.
Next page :
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B
X
A
P
Y
Z
C
Label the point where AY, BZ ,
and CX intersect as P.
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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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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
3
2
AY
BP = BZ
3
CP =
B
X
A
P
Z
CONFIDENTIAL
2
3
CX
Y
C
28
Using the Centroid to Find
Segment Lengths
In ∆ABC, AF = 9, and GE = 2.4.
Find each length.
A)
B
AG
E
AG =
F
G
2
AF
3
2
AG = (9)
3
AG = 6
Centroid Thm.
Substitute 9 for AF
A
C
D
Simplify.
Next page :
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B
B)
CE
E
2
CG = CE
3
CG + GE = CE
2
CE + GE = CE
3
1
GE = CE
3
1
2.4 = CE
3
7.2 = CE
Centroid Thm.
Seg. Add. Post.
G
A
D
Substitute 2/3 CE for
CG.
Subtract 2/3 CE from
both sides.
Substitute 2.4 for GE.
Multiply both sides by 3.
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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).
y
8
Q(0, 8)
6
R(6, 4)
4
2
0
P(3, 0)
2 4 6
x
8
Next page :
CONFIDENTIAL
31
2)
Make a Plan
y
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
Q(0, 8)
6
R(6, 4)
4
2
0
P(3, 0)
2 4 6
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x
8
32
Let M be the midpoint of QR and N Be the midpoint of QP.
0+6 8+4
0+3 8+0
m=
,
= (3, 6) n =
,
= (1.5, 4)
2
2
2
2




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).
y
8
Q(0, 8)
6
R(6, 4)
4
2
0
P(3, 0)
2 4 6
x
8
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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.
CONFIDENTIAL
Y
X
P
Q
Z
S
34
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.
y
X = -2
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.
K 7
J
Y=x+6
L
-4
0
2
Next page :
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x
Step 3 Find an equation of the line containing the
altitude from J to KL.
slope of KL =
2- 6
2- (-2)
= -1
y
X = -2
The slope of a line
perpendicular to KL is 1.
K 7
This line must pass
through J(-4, 2).
J
Y=x+6
L
-4
0
2
Next page :
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36
x
y - y1 = m(x - x1)
Point-slope form
y - 2 = 1[x-(-4)]
Substitute 2 for y1. 1 for m, and -4 for x1.
y - 2 =x +4
Distribute 1.
y =x +6
Add 2 to both sides.
y
X = -2
Step 4 Solve the system
to find the coordinates of
the orthocenter.
x = -2
K 7
y=x+2
Substitute -2
for x
The coordinates of the
orthocenter are (-2,4).
y = -2 + 6 = 4
CONFIDENTIAL
J
Y=x+6
-4
L
0
2
37
x
You did a great job
today!
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