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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
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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.
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the legs. Theorem 9.4: Pythagorean Theorem 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. Prove: c2 = a2 + b2 Proof: 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.
Statements: 1. Draw a perpendicular from C to AB. 2. and Reasons: Proof • Perpendicular Postulate • Geometric Mean Thm. • Cross Product Property c a c b = = a e b f 3. ce = a2 and cf = b2 4. ce + cf = a2 + b2 4. Addition Property of = 5. c(e + f) = a2 + b2 5. Distributive Property 6. Segment Add. Postulate 6. e + f = c 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 equation c2 = a2 + b2For example, the integers 3, 4 and 5 form a Pythagorean Triple because 52 = 32 + 42.
Find the length of the hypotenuse of the right triangle. Tell whether the sides lengths form a Pythagorean Triple. Ex. 1: Finding the length of the hypotenuse.
(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 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.” Solution:
Find the length of the leg of the right triangle. Ex. 2: Finding the Length of a Leg
(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. Solution: • 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
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. Ex. 3: Finding the area of a triangle 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.
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. Solution: 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
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. Ex. 4: Indirect Measurement
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).
(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. Solution:
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.