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Transcript 
Slide #1
8.1 Geometric Mean and
Pythagorean Theorem
Geometry
Objectives/Assignment


Use Pythagorean
theorem to solve
problems
Use Geometric Mean
and Pythagorean
Theorem to solve reallife problems
Geometric Mean

The geometric mean of two positive numbers
a and b is the positive number x such that
a
x
=
x
b
If you solve this
proportion for x, you
find that
which is a positive
number.
Book Example Pg 397
Geometric Mean Example

For example, the geometric mean of 8 and 18
is 12, because
8
12
=
12
18
and also because x = √8 ∙ 18 = x = √144 = 12
Practice “Geometric Mean”
Geometric mean

PAPER SIZES.
International standard paper
sizes are commonly used
all over the world. The
various sizes all have the
same width-to-length ratios.
Two sizes of paper are
shown, called A4 and A3.
The distance labeled x is
the geometric mean of 210
mm and 420 mm. Find the
value of x.
A4
x
A3
210 mm
x
420 mm
A4
x
A3
420 mm
210 mm
x
Solution:
210
x
=
x
Write proportion
420
X2 = 210 ∙ 420
Cross product property
X = √210 ∙ 420
Simplify
X = 297mm
EXAMPLE 1
Find the length of a hypotenuse
Find the length of the hypotenuse of the right triangle.
(hypotenuse)2
= (leg)2 + (leg)2
Pythagorean Theorem
x2 = 62 + 82
Substitute.
x2 = 36 + 64
Square.
x2 = 100
Add.
x = 10
Find the positive square root.
GUIDED PRACTICE
for Example 1
1. Find the unknown side length of the right triangle. Write your answer
in simplest radical form.
ANSWER
4
GUIDED PRACTICE
2.
for Example 1
Find the unknown side length of the right triangle.
ANSWER
13
X = 3.2
Y=X+5
Y = 3.2 + 5
Y = 8.2
Y² = 32
Y = 5.7
5.7²+ 8² = x²
32.49 + 64 = x²
96.49 = x²
X = 9.8
x
4
x² = 20
x = 4.5
4.5²+ 5² = y²
20 + 25 = y²
45 = y²
X = 6.7
x² = 18
x = 4.2
6²+ 4.2² = y²
36 + 18 = y²
54 = y²
X = 7.3
x²+ 5² = 9²
x² + 25 = 81
x² = 81-25
X² = 56
X = 7.5
2²+ 2² = x²
4 + 4 = x²
x² = 8
X = 2.8
30²+ 16² = x²
900+256 = x²
x² = 1156
X = 34
x+ 60² = 65²
x² +3600 = 4225
x² = 4225-3600
x² = 625
X = 25
14² + 48² = 50²
50² + 75² = 85²
15² + 36² = 39²
45² + 60² = 80²
?
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