1. Geometry
6.4 Geometric Mean

2. 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.

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

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

5. Problem 1: Right Triangles
Together 1-5

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

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

15. 𝑥 8 = 8 4
4𝑥=64
𝑥=16

16. 20 𝑥 = 𝑥 24
𝑥 2 =480
𝑥= 480 ≈21.9

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

18. 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

19. 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
= 130 𝑥
Cross-Multiply
Then Divide
135𝑥=16,900
𝑥= 𝑓𝑡
130 feet
3x45 = 135 feet X

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