12.3 The Pythagorean Theorem
CORD
Mrs. Spitz
Spring 2007
Objectives/Assignment
•Use Pythagorean Theorem
•Assignment: pp. 484-485 #4-39 all
•Assignment due today: 12.2 p. 479-480
#5-51
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
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
Note:
•Determine if the following lengths can
represent the sides of a right triangle.
c2 = a2 + b2
Right ∆
c2 < a2 + b2
Acute ∆
c2 > a2 + b2
Obtuse ∆
#32. 12, 11, 15
•The measures of the sides of a triangle are
given. Determine whether each triangle is a
right triangle.
c2 = a2 + b2
152 = 112 + 122
225 = 121 +144?
225 ≠ 265 Not a right Triangle
#14 a = √7 b =√9 c=?
c2 = a2 + b2
2
2
2
c = √7 + √9
2
c =7 + 9
c2 = 16
c=4
Note:
2
7  7 7
 49
7
Area of a rectangle
?
40 = bh
b = 2h + 2
•The area of a rectangle
is 40 square meters.
Find the length of a
diagonal of the
rectangle if its length
is 2 meters less than
twice its width.
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