(a.k.a Geometry Mean)
Transcript of (a.k.a Geometry Mean)
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9.3Altitude-On-Hypotenuse
Theorems (a.k.a Geometry Mean)
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Warm Up
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a
x b
x
Given two numbers a and b, x is the geometric mean if
MEANS
EXTREMES
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Find the geometric mean of 9 and 16
a
x b
x
9
16
x
x
2 144x
12x
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Index Card: When an altitude is drawn from a vertex to a hypotenuse, then three similar triangles are formed.
**Identify the three similar triangles!Which theorem can they be proven similar?
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A
B
C
Altitudes!!!!What do you remember?
An Altitude*Starts at a vertex*Forms a right angleD
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A
B
C
The hypotenuse is split into two piecesBD and DA
D
CD (the altitude) is the geometric mean of those
two pieces
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A
B
C
D
If the altitude is in the means place, put the two segments of the hypotenuse into the extremes.
hyppiece 2nd
altitude
altitude
hyppiece1st
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A
B
C
D
Index CardTheorem: The altitude to the hypotenuse is the mean proportional (or geometric mean) between the segments of the hypotenuse.
y
h
h
x
h
x
ya
b
h = altitude
x and y are segments of the hypotenuse
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the LEG is the
geometric mean
A
BC
D
If the leg is in the means place, put the whole hyp. and segment adjacent to that leg into the extremes.
legtoadjside
leg
leg
hypwhole
AB
AC
AC
AD
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A
BC
D
elofhyp
elofleg
smallofhyp
smallofleg
arg
arg
x
b
b
cy
x
c
b
a
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the LEG is the
geometric mean
A
BC
D
AB
CB
CB
BD
legtocloseside
leg
leg
hypwhole
Using the other part of the hypotenuse:
Remember: You can use similar triangles…compare the hypotenuse
and the long leg using the large triangle and the medium-sized
triangle.
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A
B
C
D
9
16
x
EX: 1 Find x
x
x9
16
2 144x
12x
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EX: 2 Find the length of AB.
A
BC
D
32
15
102468.26
15x
x
15
32
32
x
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A
B
C
D
3
x
4
EX 3: Solve for x
33.53
16x
3
4
4
x
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EX: 4 Find the length of CB.
A
BC
D
x
32
50
x = 40
32
x 50
x
x2
= 1600
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A
B
C
D
x
12
5
EX: 5 Solve for x
08.212
25x
x
5 12
5
12x = 25
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Homework
• p. 379 #1-5, 16, 17