1. Geometry B Chapter 8
Geometric Mean

2. Your Math Goal Today is…
Use geometric mean to find segment lengths in right triangles.
Apply similarity relationships in right triangles to solve problems.

3. Warm Up 1. Write a similarity statement comparing the two triangles.
Simplify.
Solve each equation.
x2 = 50
∆ADB ~ ∆EDC ±5

4. Vocabulary
geometric mean

5. In a right triangle, an altitude drawn from the vertex of the right angle to the hypotenuse forms two right triangles.

6. Please turn to page 538 Theorem 8.1

7. Example 1: Identifying Similar Right Triangles
Write a similarity statement comparing the three triangles.
Sketch the three right triangles with the angles of the triangles in corresponding positions. Z W
By Theorem 8.1, ∆UVW ~ ∆UWZ ~ ∆WVZ.

8. Note:
Writing similarity statements for triangles,
be sure to name the vertices in the corresponding
order in each triangle.

9. In Your Notes! Example 1
Write a similarity statement comparing the three triangles.
Sketch the three right triangles with the angles of the triangles in corresponding positions.
By Theorem 8.1, ∆LJK ~ ∆JMK ~ ∆LMJ.

10. Consider the proportion
Consider the proportion In this case, the means of the proportion are the same number, and that number is the geometric mean of the extremes.
The geometric mean of two positive numbers is the positive square root of their product. So the geometric mean of a and b is the positive number x such
that , or x2 = ab.

11. Example 2A: Finding Geometric Means
Find the geometric mean of each pair of numbers. If necessary, give the answer in simplest radical form.
4 and 25
Let x be the geometric mean.
x2 = (4)(25) = 100
Def. of geometric mean
x = 10
Find the positive square root.

12. Example 2B: Finding Geometric Means
Find the geometric mean of each pair of numbers. If necessary, give the answer in simplest radical form.
5 and 30
Let x be the geometric mean.
x2 = (5)(30) = 150
Def. of geometric mean
Find the positive square root.

13. In Your Notes! Example 2a
Find the geometric mean of each pair of numbers. If necessary, give the answer in simplest radical form.
2 and 8
Let x be the geometric mean.
x2 = (2)(8) = 16
Def. of geometric mean
x = 4
Find the positive square root.

14. In Your Notes! Example 2b
Find the geometric mean of each pair of numbers. If necessary, give the answer in simplest radical form.
10 and 30
Let x be the geometric mean.
x2 = (10)(30) = 300
Def. of geometric mean
Find the positive square root.

15. In Your Notes! Example 2c
Find the geometric mean of each pair of numbers. If necessary, give the answer in simplest radical form.
8 and 9
Let x be the geometric mean.
x2 = (8)(9) = 72
Def. of geometric mean
Find the positive square root.

16. You can use Theorem 8.1 to write proportions comparing the side lengths of the triangles formed by the altitude to the hypotenuse of a right triangle.
All the relationships in red involve geometric means.

17. Please turn to page 539

18. Example 3: Finding Side Lengths in Right Triangles
Find x, y, and z.
62 = (9)(x)
6 is the geometric mean of 9 and x.
x = 4
Divide both sides by 9.
y2 = (4)(13) = 52
y is the geometric mean of 4 and 13.
Find the positive square root.
z2 = (9)(13) = 117
z is the geometric mean of 9 and 13.
Find the positive square root.

19. Once you’ve found the unknown side lengths, you can use the Pythagorean Theorem to check your answers.
Helpful Hint

20. 92 = (3)(u) 9 is the geometric mean of
Check It Out! Example 3
Find u, v, and w.
92 = (3)(u) is the geometric mean of
u and 3.
u = Divide both sides by 3.
w2 = (27 + 3)(27) w is the geometric mean of
u + 3 and 27.
Find the positive square root.
v2 = (27 + 3)(3) v is the geometric mean of
u + 3 and 3.
Find the positive square root.

21. Example 4: Measurement Application
To estimate the height of a Douglas fir, Jan positions herself so that her lines of sight to the top and bottom of the tree form a 90º angle. Her eyes are about 1.6 m above the ground, and she is standing 7.8 m from the tree. What is the height of the tree to the nearest meter?

22. Let x be the height of the tree above eye level.
Example 4 Continued
Let x be the height of the tree above eye level.
7.8 is the geometric mean of 1.6 and x.
(7.8)2 = 1.6x
x = ≈ 38
Solve for x and round.
The tree is about = 39.6, or 40 m tall.

23. In Your Notes! Example 4
A surveyor positions himself so that his line of sight to the top of a cliff and his line of sight to the bottom form a right angle as shown.
What is the height of the cliff to the nearest foot?

24. Let x be the height of cliff above eye level.
Continued
Let x be the height of cliff above eye level.
(28)2 = 5.5x
28 is the geometric mean of 5.5 and x.
x  142.5
Divide both sides by 5.5.
The cliff is about , or 148 ft high.