Geometry 6.4 Geometric Mean

6.4 More Similar Triangles Objectives Explore the relationships created when an altitude is drawn to the hypotenuse of a right triangle. Prove the Right Triangle/Altitude Similarity Theorem. Use the geometric mean to solve for unknown lengths.

7𝑥=10𝑥−30 −3𝑥=−30 𝑥=10 16= 𝑥 2 16 = 𝑥 2 𝑥=±4 42= 𝑥 2 42 = 𝑥 2 𝑥=± 42

8𝑥=50 𝑥= 50 8 = 25 4 64=3𝑥 𝑥= 64 3 9=9𝑥 𝑥= 9 9 =1

Problem 1: Right Triangles Together 1-5

Problem 1: Right Triangles Hypotenuse of a Right Triangle The longest side Opposite the right angle Altitude “Height” of a triangle Drawn perpendicular from a vertex to the opposite side Altitude drawn to the Hypotenuse A perpendicular segment from the right angle to the hypotenuse

Problem 1: Right Triangles #1 Sketch an altitude to the hypotenuse rather than construct #2 ∆𝐴𝐵𝐶, ∆𝐴𝐶𝐷 𝑎𝑛𝑑 ∆𝐶𝐵𝐷 #3-5 We are not going to do the activity, lets look at the screen for the diagram of triangles. Draw all triangles below #3

Problem 1: Right Triangles Right Triangle/Altitude Similarity Theorem #4 ∆𝐴𝐵𝐶 ~ ∆𝐴𝐶𝐷 They both have right angles They both share ∠𝐴 AA~ #5 ∆𝐴𝐵𝐶 ~ ∆𝐴𝐶𝐷 𝐴𝐵 𝐴𝐶 = 𝐵𝐶 𝐶𝐷 = 𝐴𝐶 𝐴𝐷

Problem 1: Right Triangles Right Triangle/Altitude Similarity Theorem #4 ∆𝐴𝐵𝐶 ~ ∆𝐶𝐵𝐷 They both have right angles They both share ∠𝐵 AA~ #5 ∆𝐴𝐵𝐶 ~ ∆𝐶𝐵𝐷 𝐴𝐵 𝐶𝐵 = 𝐵𝐶 𝐵𝐷 = 𝐴𝐶 𝐶𝐷

Problem 1: Right Triangles Right Triangle/Altitude Similarity Theorem #4 ∆𝐴𝐶𝐷~ ∆𝐶𝐵𝐷 ∆𝐶𝐵𝐷~∆𝐴𝐵𝐶~∆𝐴𝐶𝐷 Transitive Property #5 ∆𝐴𝐶𝐷 ~ ∆𝐶𝐵𝐷 𝐴𝐶 𝐶𝐵 = 𝐶𝐷 𝐵𝐷 = 𝐴𝐷 𝐶𝐷

Problem 2: Geometric Mean The second and third terms in a proportion They are equal terms in the proportion 𝑎 𝑥 = 𝑥 𝑏 𝑥 2 =𝑎𝑏

Problem 2: Geometric Mean Right Triangle Altitude/Hypotenuse Theorem #1a The measure of the altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse. 𝐴𝐷 𝐶𝐷 = 𝐶𝐷 𝐷𝐵 = 𝐴𝐶 𝐵𝐶

Problem 2: Geometric Mean Right Triangle Altitude/Leg Theorem #1b If 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 adjacent to the leg. 𝐴𝐶 𝐴𝐷 = 𝐶𝐵 𝐷𝐶 = 𝐴𝐵 𝐴𝐶 𝐴𝐶 𝐶𝐷 = 𝐶𝐵 𝐷𝐵 = 𝐴𝐵 𝐶𝐵

Problem 2: Geometric Mean Together 2(a-d) 2. 4 𝑥 = 𝑥 9 𝑥 2 =36 𝑥= 36 =6

𝑥 8 = 8 4 4𝑥=64 𝑥=16

20 𝑥 = 𝑥 24 𝑥 2 =480 𝑥= 480 ≈21.9

Collaborate #3 (3 Minutes) 8 𝑥 = 𝑥 2 𝑥 2 =16 𝑥=4 Collaborate #3 (3 Minutes)

3. 10+5=15 The Altitude is the mean between the parts of the Hypotenuse 10 𝑥 = 𝑥 5 𝑥 2 =50 𝑥= 50 ≈7.1 The Leg is the mean between the Hypotenuse and the Part closest 15 𝑦 = 𝑦 10 𝑦 2 =150 𝑦= 150 ≈12.2 The Leg is the mean between the Hypotenuse and the Part closest 15 𝑧 = 𝑧 5 𝑧 2 =75 𝑧= 75 ≈8.7

Problem 3: Bridge Over the Canyon 45 yards 130 feet MUST COMPARE THE SAME UNITS The Altitude is the mean between the parts of the Hypotenuse 135 130 = 130 𝑥 Cross-Multiply Then Divide 135𝑥=16,900 𝑥=125.19 𝑓𝑡 130 feet 3x45 = 135 feet X

Formative Assessment Skills Practice 6.4 Pg. 537-544 (1-26) Vocabulary: Fill in Blanks Problem Set #’s 1-4 Sketch the altitude