QUESTION IMAGE
Question
find the geometric mean of each pair of numbers. 1. 12 and 50 2. 24 and 45 find the value of x. 3. triangle image 4. triangle image 5. triangle image 6. triangle image
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For 1. Geometric mean of 12 and 50
Step1: Recall geometric mean formula
The geometric mean of two numbers $a$ and $b$ is $\sqrt{a \times b}$.
Step2: Substitute values and calculate
$$\sqrt{12 \times 50} = \sqrt{600} = \sqrt{100 \times 6} = 10\sqrt{6} \approx 24.49$$
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For 2. Geometric mean of 24 and 45
Step1: Use geometric mean formula
Geometric mean = $\sqrt{a \times b}$
Step2: Substitute and simplify
$$\sqrt{24 \times 45} = \sqrt{1080} = \sqrt{36 \times 30} = 6\sqrt{30} \approx 32.86$$
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For 3. Right triangle altitude theorem
Step1: Identify similar triangle segments
In a right triangle, $\frac{x}{5} = \frac{21}{x}$ (altitude/leg proportionality)
Step2: Cross-multiply and solve for $x$
$$x^2 = 5 \times 21 = 105$$
$$x = \sqrt{105} \approx 10.25$$
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For 4. Right triangle altitude theorem
Step1: Set up proportionality equation
$\frac{x}{10} = \frac{10}{6}$ (altitude/leg proportionality)
Step2: Solve for $x$
$$6x = 100$$
$$x = \frac{100}{6} = \frac{50}{3} \approx 16.67$$
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For 5. Right triangle altitude theorem
Step1: Set up proportionality equation
$\frac{x}{4} = \frac{24}{x}$ (altitude/leg proportionality)
Step2: Solve for $x$
$$x^2 = 4 \times 24 = 96$$
$$x = \sqrt{96} = 4\sqrt{6} \approx 9.80$$
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For 6. Right triangle altitude theorem
Step1: Set up proportionality equation
$\frac{24}{18} = \frac{x}{24}$ (leg/hypotenuse proportionality)
Step2: Solve for $x$
$$18x = 24 \times 24 = 576$$
$$x = \frac{576}{18} = 32$$
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- $10\sqrt{6}$ (or ~24.49)
- $6\sqrt{30}$ (or ~32.86)
- $\sqrt{105}$ (or ~10.25)
- $\frac{50}{3}$ (or ~16.67)
- $4\sqrt{6}$ (or ~9.80)
- 32