Download 9.2 The Pythagorean Theorem

9.2 The Pythagorean Theorem
Geometry
Objectives:
• Prove the Pythagorean Theorem
• Use the Pythagorean Theorem to solve real-life problems such as
determining how far a ladder will reach.
History Lesson
• Around the 6th century BC, the Greek mathematician Pythagoras
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.
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 equation c2 = a2 + b2 For example, the integers 3, 4
and 5 form a Pythagorean Triple because 52 = 32 + 42.
Example 1:
• 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.”
Example 2:
•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
Example 3: Finding the area of a triangle
• Find the area of the triangle
to the nearest tenth of a
meter.
7m
7m
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.
h
10 m
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
Area = ½ bh
= ½ (10)(√24)
≈ 24.5 m2
The area of the triangle is about 24.5 m2
7m
7m
h
10 m