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Chapter 9
Geometry
Copyright © 2015, 2010, and 2007 Pearson Education, Inc.
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CHAPTER
9
Geometry
9.1 Perimeter
9.2 Area
9.3 Circles
9.4 Volume
9.5 Angles and Triangles
9.6 Square Roots and the Pythagorean Theorem
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9.6
Square Roots and the Pythagorean
Theorem
OBJECTIVES
a Simplify square roots of squares such as 25.
b Approximate square roots.
c Given the lengths of any two sides of a right triangle,
find the length of the third side.
d Solve applied problems involving right triangles.
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Square Root
If a number is a product of two identical factors,
then either factor is called a square root of the
number. (If a = c2, then c is a square root of a.) The
symbol
(called a radical sign) is used in naming
square roots.
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Example
Simplify.
49
169
324
Solution
49
49  7
169
169  13
324
324  18
Note that
72 = 49.
Note that
132 = 169.
Note that
182 = 324.
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Many square roots can’t be written as whole numbers
or fractions.
We can use a calculator to find a decimal
approximation.
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Example
Approximate to the nearest thousandth.
5
32
190
Solution
We use a calculator to find each square root. Since
more than three decimal places are given, we round
back to three places.
5
5  2.236
32
32  5.657
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190
190  13.784
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Recall that a right triangle is a triangle with a 90
angle, as shown here.
In a right triangle, the longest
Hypotenuse
a
c
side is called the hypotenuse.
b
It is also opposite the right angle.
Leg
The other two sides are called legs.
We generally use the letters a and b for
the lengths of the legs and c for the length of the
hypotenuse.
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Leg
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The Pythagorean Theorem
In any right triangle, if a and b are the lengths of the
legs and c is the length of the hypotenuse, then
a2 + b2 = c2, or
(Leg)2 + (Leg)2 = (Hypotenuse)2.
c
a
b
The equation a2 + b2 = c2 is called the Pythagorean
equation.
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Example
Find the length of the hypotenuse of this right triangle.
12
c
a b  c
2
16
2
2
12  16  c
2
2
2
144  256  c 2
400  c 2
c  400  20
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Example
Find the length b for the right triangle shown. Give an
exact answer and an approximation to three decimal
places.
a b  c
2
20
2
20  b  24
2
b
2
2
2
400  b 2  576
24
b 2  176
Exact answer:
Approximation:
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b  176
b  13.266
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Example
How long must a guy wire be to reach from the top of
a 15-m telephone pole to a point on the ground 10 m
from the foot of the pole?
Solution
1. Familiarize. We make a drawing and label the
known distances. We label the unknown length.
2. Translate. We use the
Pythagorean theorem. a 2  b 2  c 2
15  10  c
2
2
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2
15 m
c
10 m
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continued
3. Carry out.
152  102  c 2
2
225  100  c
2
325  c
325  c Exact answer:
18.028  c Approximation:
4. Check. a 2  b 2  c 2
152  102  18.0282
225 100  325.001
5. State. The guy wire should be about 18.028 m long.
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