9.2 The Pythagorean Theorem Geometry Mrs. Spitz Spring 2005 Objectives/Assignment •Prove the Pythagorean Theorem •Use the Pythagorean Theorem to solve real-life problems such as determining how.

```9.2 The Pythagorean Theorem
Geometry
Mrs. Spitz
Spring 2005
Objectives/Assignment
•Prove the Pythagorean Theorem
•Use the Pythagorean Theorem to solve real-life
problems such as determining how far a ladder
will reach.
•Assignment: pp. 538-539 #1-31 all
•Assignment due today: 9.1 pp. 531-532 #1-34
all
History Lesson
•Around the 6th century BC, the Greek
mathematician Pythagorus founded a school for
the study of philosophy, mathematics and science.
Many people believe that an early proof of the
Pythagorean Theorem came from this school.
•Today, the Pythagorean Theorem is one of the
most famous theorems in geometry. Over 100
different proofs now exist.
Proving the Pythagorean Theorem
•In this lesson, you will study one of the
most famous theorems in mathematics—the
Pythagorean Theorem. The relationship it
describes has been known for thousands of
years.
Theorem 9.4: Pythagorean Theorem
•In a right triangle,
the square of the
length of the
hypotenuse is equal to
the sum of the
squares of the legs.
c
a
b
c2 = a2 + b2
Proving the Pythagorean Theorem
•There are many different proofs of the
Pythagorean Theorem. One is shown below.
Other proofs are found in Exercises 37 and
38 on page 540 and in the Math & History
feature on page 557.
Given: In ∆ABC, BCA is a right angle.
Proof: Prove: c = a + b
2
2
2
Plan for proof: Draw altitude CD to the
hypotenuse just like in 9.1. Then apply
Geometric Mean Theorem 9.3 which states that
when the altitude is drawn to the hypotenuse of a
right triangle, each leg of the right triangle is
the geometric mean of the hypotenuse and the
segment of the hypotenuse that is adjacent to
that leg.
Proof
Statements:
Reasons:
1. Draw a perpendicular from C to AB.
2.
and
c = a
a
e
c = b
b
f
1. Perpendicular Postulate
2. Geometric Mean Thm.
3. ce = a2 and cf = b2
3. Cross Product Property
4. ce + cf = a2 + b2
4. Addition Property of =
5. c(e + f) = a2 + b2
5. Distributive Property
6. e + f = c
6. Segment Add. Postulate
7. c2 = a2 + b2
7. Substitution Property of =
Using the Pythagorean Theorem
•A Pythagorean triple is a set of three
positive integers a, b, and c that satisfy the
2
2
2
equation c = a + b For example, the
integers 3, 4 and 5 form a Pythagorean
Triple because 52 = 32 + 42.
Ex. 1: Finding the length of the hypotenuse.
•Find the length of
the hypotenuse of the
right triangle. Tell
whether the sides
lengths form a
Pythagorean Triple.
12
5
x
Solution:
(hypotenuse)2 = (leg)2 + (leg)2
x2 = 52 + 122
x2 = 25 + 144
x2 = 169
x = 13
 Because the side lengths
5, 12 and 13 are integers,
they form a Pythagorean
Triple. Many right triangles
have side lengths that do
not form a Pythagorean
Triple as shown next slide.
12
5
x
Pythagorean Theorem
Substitute values.
Multiply
Add
Find the positive square
root.
Note: There are no
negative square roots
until you get to Algebra
II and introduced to
“imaginary numbers.”
Ex. 2: Finding the Length of a Leg
•Find the length of
the leg of the right
triangle.
x
7
14
Solution:
(hypotenuse)2 = (leg)2 + (leg)2
142 = 72 + x2
196 = 49 + x2
147 = x2
√147 = x
√49 ∙ √3 = x
7√3 = x
x
7
14
Pythagorean Theorem
Substitute values.
Multiply
Subtract 49 from each side
Find the positive square root.
Use Product property
Simplify the radical.
In example 2, the side length was written as a radical in the
simplest form. In real-life problems, it is often more
convenient to use a calculator to write a decimal
approximation of the side length. For instance, in Example 2,
x = 7 ∙√3 ≈ 12.1
Ex. 3: Finding the area of a triangle
•Find the area of the
triangle to the nearest
tenth of a meter.
•You are given that the
base of the triangle is
10 meters, but you do
not know the height.
7m
7m
h
10 m
Because the triangle is isosceles,
it can be divided into two congruent
triangles with the given dimensions.
Use the Pythagorean Theorem to
find the value of h.
Solution:
7m
7m
h
10 m
Steps:
Reason:
(hypotenuse)2 = (leg)2 + (leg)2
72 = 52 + h2
49 = 25 + h2
24 = h2
√24 = h
Pythagorean Theorem
Substitute values.
Multiply
Subtract 25 both sides
Find the positive square
root.
Now find the area of the original triangle.
Area of a Triangle
7m
7m
h
10 m
Area = ½ bh
= ½ (10)(√24)
≈ 24.5 m2
2
The area of the triangle is about 24.5 m
Ex. 4: Indirect Measurement
•Support Beam: The skyscrapers shown on page 535
are connected by a skywalk with support beams.
You can use the Pythagorean Theorem to find the
approximate length of each support beam.
•Each support beam forms the hypotenuse of
a right triangle. The right triangles are
congruent, so the support beams are the same
length. Use the Pythagorean Theorem to
show the length of each support beam (x).
Solution:
(hypotenuse)2 = (leg)2 + (leg)2
x2 = (23.26)2 + (47.57)2
x2 = √ (23.26)2 + (47.57)2
x ≈ 13
Pythagorean Theorem
Substitute values.
Multiply and find the
positive square root.
Use a calculator to
approximate.
Reminder:
•Friday is the last day to turn in assignments for
credit for the quarter.
•Quiz after 9.3 on Monday 3/10/05.
•Quiz after 9.5 probably Thursday (5th) and
Friday (2nd and 6th).
•Test on Chapter 9 will be before you go on
Spring Break. Do yourself a favor and take it
before you go.
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