The Pythagorean Theorem: Finding Sides and Classifying Triangles

Page 1: The Pythagorean Theorem Title: The Pythagorean Theorem Theorems: Theorem 9.1: Pythagorean Theorem "In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs." Formula: c² = a² + b² Diagram included with a, b, and c marked and labeled leg, leg, and hypotenuse respectively. Common Pythagorean Triples and Some of Their Multiples: Bolded (Most Common): 3, 4, 5 5, 12, 13 8, 15, 17 7, 24, 25 Multiples of the above: 6, 8, 10 9, 12, 15 12, 16, 20 15, 20, 25 10, 24, 26 15, 36, 39 20, 48, 52 16, 30, 34 24, 45, 51 21, 72, 75 14, 48, 50 3x, 4x, 5x 5x, 12x, 13x 8x, 15x, 17x 7x, 24x, 25x "The most common Pythagorean triples are in bold. The other triples are the result of multiplying each integer in a bold-faced triple by the same factor." Theorem 9.2: Converse of the Pythagorean Theorem "If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle." If c² = a² + b², then ∆ABC is a right triangle sides follow this rule -> then the triangle is a right triangle Diagram included showing a, b and c. Theorem 9.3: Pythagorean Inequalities Theorem "For any ∆ABC where c is the length of the longest side, the following statements are true." If c² < a² + b², then ∆ABC is acute. * smaller leads to a smaller angle than 90° * Diagram included showing a, b, and c. If c² > a² + b², then ∆ABC is obtuse. * bigger leads to a bigger angle than 90° * Diagram included showing a, b, and c. Page 2: 9.1 Extra Practice (Exercises 1-6) Heading: 9.1 Extra Practice "In Exercises 1-6, find the value of x. Then tell whether the side lengths form a Pythagorean triple." Diagram of a right triangle, one leg is 9, one is 81, and c is x a^2 + b^2 = c^2 81^2 + 108^2 = x^2 6561 + 11664 = x^2 18225 = x^2 135=x Diagram of a right triangle, one leg is 4, one is 8, and c is x a^2 + b^2 = c^2 4^2 + 8^2 = c^2 16 + 64 = c^2 80 = c^2 = 4 * 5 = 4 5^1/2 Find your hypotenuse and the rest is easy! even 108^2 + 81^2 = x^2 will also work Diagram of a right triangle, one leg is 15, one is 20, and c is x a^2 + b^2 = c^2 20^2 + 15^2 = x^2 400 + 225 = x^2 625= x^2 25=x Diagram of a right triangle, one leg is 10, one is 6, and c is x a^2 + b^2 = c^2 10^2 + 6^2 = x^2 100 + 36= x^2 136=x^2 136=268 = 234 = 2*17 = 2 34^1/2 always take out pairs Diagram of a right triangle, one leg is 55, one is 77, and c is x x^2 + 55^2 = 77^2 x^2 + 3025 = 5929 -3025 - 3025 x^2= 2904 x = 2 11 3 = 22 6^1/2 Diagram of a right triangle, one leg is 48, one is 90, and c is x a^2 + b^2 = c^2 48^2 + 90^2 = x^2 2304+8100= x^2 10404=x^2 102=x Page 3: 9.1 Extra Practice (Exercises 7-9) Exercise 7: "From school, you biked 1.2 miles due south and then 0.5 mile due east to your house. If you had biked home on the street that runs directly diagonal from your school to your house, how many fewer miles would you have biked?" Diagram included showing the path. If x^2 = 1.2^2 + .5^2 x^2 = 1.44 + .25 = x^2 = 1.69, x=1.3 old route 1.2 + .5 = 1.7 miles .4 miles fewer had you biked on the diagonal Exercises 8-9: "In Exercises 8 and 9, verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?" 90, 216, and 234 small + middle > large 90 + 216 > 234 306 > 234 true we have a triangle * a^2 + b^2 = c^2 90^2+ 216^2 = 234^2 8100 + 46656 = 54756 54756 = 54756 since we have a right triangle 1, 1, and √3 ≈ 1.7 1 + 1 > √3 2 > 1.7 yup ... ∆ 1^2+ 1^2 = (√3)^2 1 + 1 = 3 2 = 3 no since hyp is longer we have an obtuse triangle