Question

name alyonna bush
envision geometry
7-4 mathematical literacy and vocabulary
you can use a proportion to determine a geometric mean by multiplying and simplifying.
problem what is the geometric mean of 4 and 8?
\\(\frac{4}{x} = \frac{x}{8}\\) set up a proportion.
\\(32 = x^2\\) multiply each side of the equation by 8x.
\\(\sqrt{32} = \sqrt{x^2}\\) take the positive square root.
\\(\sqrt{16 \cdot 2} = \sqrt{x^2}\\) factor the perfect square 16.
\\(4\sqrt{2} = x\\) simplify
1 solve \\(\sqrt{324}\\)

1. circle the geometric mean of 6 and 12 \\(6\sqrt{2}\\) 9 \\(9\sqrt{2}\\)
2. circle the geometric mean of 10 and 40 \\(10\sqrt{3}\\) 20 25
3. find the geometric mean of 50 and 75

you can use the geometric mean to find the altitude of a right triangle.
problem what is the value of x?
\\(\frac{ad}{cd} = \frac{cd}{db}\\) set up a proportion.
\\(\frac{8}{x} = \frac{x}{18}\\) substitute.
\\(144 = x^2\\) multiply each side of the equation by 18x.
\\(\sqrt{144} = \sqrt{x^2}\\) take the positive square root.
\\(12 = x\\) simplify.
solve for x.
5.
6.

Explanation:

Step1: Solve $\sqrt{324}$

$\sqrt{324} = \sqrt{18^2} = 18$

Step2: Find geometric mean of 6 and 12

Set up proportion: $\frac{6}{x} = \frac{x}{12}$
Cross multiply: $x^2 = 6 \times 12 = 72$
Take positive root: $x = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

Step3: Find geometric mean of 10 and 40

Set up proportion: $\frac{10}{x} = \frac{x}{40}$
Cross multiply: $x^2 = 10 \times 40 = 400$
Take positive root: $x = \sqrt{400} = 20$

Step4: Find geometric mean of 50 and 75

Set up proportion: $\frac{50}{x} = \frac{x}{75}$
Cross multiply: $x^2 = 50 \times 75 = 3750$
Simplify root: $x = \sqrt{3750} = \sqrt{625 \times 6} = 25\sqrt{6}$

Answer:

1. $18$
2. $6\sqrt{2}$
3. $20$
4. $25\sqrt{6}$