Question

1 solve 32²=16
2 circle the geometric mean of 6 and 12
3 circle the geometric mean of 10 and 40
4 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?
set up a proportion.
substitute.
multiply each side of the equation by 18x.
144 = x²
144 = √x²
12 = x
simplify.

1. solve for x

6.

Explanation:

Step1: Solve $\sqrt{324}$

$\sqrt{324} = 18$

Step2: Find geometric mean of 6 and 12

Geometric mean formula: $\sqrt{a \times b}$
$\sqrt{6 \times 12} = \sqrt{72} = 6\sqrt{2}$

Step3: Find geometric mean of 10 and 40

$\sqrt{10 \times 40} = \sqrt{400} = 20$

Step4: Find geometric mean of 50 and 75

$\sqrt{50 \times 75} = \sqrt{3750} = \sqrt{625 \times 6} = 25\sqrt{6}$

Step5: Solve for $x$ (segment)

Geometric mean for segments: $x = \sqrt{7 \times 28}$
$x = \sqrt{196} = 14$

Step6: Solve for $x$ (altitude)

Geometric mean for altitude: $x = \sqrt{9 \times 27}$
$x = \sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3}$

Answer:

1. $18$
2. $\boldsymbol{6\sqrt{2}}$
3. $\boldsymbol{20}$
4. $25\sqrt{6}$
5. $14$
6. $9\sqrt{3}$