12.3 The Pythagorean

TheoremCORDMrs. 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.

ca

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.

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.

x

12

5

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.

Pythagorean TheoremSubstitute values.MultiplyAddFind the positive square

root.

Note: There are no negative square roots until you get to Algebra II and introduced to “imaginary numbers.”

x

12

5

Ex. 2: Finding the Length of a Leg

•Find the length of the leg of the right triangle.

x

14

7

Solution:(hypotenuse)2 = (leg)2 + (leg)2

142 = 72 + x2

196 = 49 + x2

147 = x2

√147 = x

√49 ∙ √3 = x

7√3 = x

Pythagorean Theorem

Substitute values.

Multiply

Subtract 49 from each side

Find the positive square root.

Use Product property

Simplify the radical.

x

14

7

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

c2 < a2 + b2 Acute ∆

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

c2 = √72 + √92

c2 =7 + 9c2 = 16c = 4

Note:

7

49

7772

Area of a rectangle

•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.

40 = bhb = 2h

+ 2

?

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.

h

10 m

7 m7 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:Steps:

(hypotenuse)2 = (leg)2 + (leg)2

72 = 52 + h2

49 = 25 + h2

24 = h2

√24 = h

Reason:

Pythagorean Theorem

Substitute values.

Multiply

Subtract 25 both sides

Find the positive square root.

h

10 m

7 m7 m

Now find the area of the original triangle.

Area of a TriangleArea = ½ bh = ½ (10)(√24)

≈ 24.5 m2

The area of the triangle is about 24.5 m2

h

10 m

7 m7 m