Geometric Mean Worksheet.pdf - Keller ISD Schools - Free Printable

Geometric Mean Worksheet Solution

The geometric mean of two numbers \( a \) and \( b \) is given by the formula:
\[
\text{Geometric Mean} = \sqrt{a \cdot b}
\]

We will solve each problem step by step.

---

#### 1. Find the geometric mean of 8 and 18.

\[
\text{Geometric Mean} = \sqrt{8 \cdot 18} = \sqrt{144} = 12
\]

Answer:
\[
\boxed{12}
\]

---

#### 2. Find the geometric mean of 20 and 25.

\[
\text{Geometric Mean} = \sqrt{20 \cdot 25} = \sqrt{500} \approx 22.36
\]

Rounding to the nearest tenth:
\[
\boxed{22.4}
\]

---

#### 3. 15 is the geometric mean of 25 and what other number?

Let the unknown number be \( x \). The geometric mean is given by:
\[
\sqrt{25 \cdot x} = 15
\]

Square both sides:
\[
25 \cdot x = 15^2
\]
\[
25 \cdot x = 225
\]

Solve for \( x \):
\[
x = \frac{225}{25} = 9
\]

Answer:
\[
\boxed{9}
\]

---

#### 4. Find the geometric mean of 3 and 7.

\[
\text{Geometric Mean} = \sqrt{3 \cdot 7} = \sqrt{21} \approx 4.58
\]

Rounding to the nearest tenth:
\[
\boxed{4.6}
\]

---

#### 5. 32 is the geometric mean of 16 and what other number?

Let the unknown number be \( x \). The geometric mean is given by:
\[
\sqrt{16 \cdot x} = 32
\]

Square both sides:
\[
16 \cdot x = 32^2
\]
\[
16 \cdot x = 1024
\]

Solve for \( x \):
\[
x = \frac{1024}{16} = 64
\]

Answer:
\[
\boxed{64}
\]

---

Solve for the missing variable

#### 6. \( x = \)
The triangle is divided into two smaller triangles by an altitude, forming similar triangles. The geometric mean relationship applies here:
\[
x = \sqrt{12 \cdot 8} = \sqrt{96} \approx 9.8
\]

Rounding to the nearest tenth:
\[
\boxed{9.8}
\]

#### 7. \( x = \)
Similarly, for this triangle:
\[
x = \sqrt{9 \cdot 3} = \sqrt{27} \approx 5.2
\]

Rounding to the nearest tenth:
\[
\boxed{5.2}
\]

#### 8. \( x = \), \( y = \)
Using the Pythagorean theorem in the right triangle:
\[
x^2 + 4^2 = 6^2
\]
\[
x^2 + 16 = 36
\]
\[
x^2 = 20
\]
\[
x = \sqrt{20} \approx 4.5
\]

For \( y \), use the geometric mean relationship:
\[
y = \sqrt{x \cdot 6} = \sqrt{4.5 \cdot 6} = \sqrt{27} \approx 5.2
\]

Rounding to the nearest tenth:
\[
x \approx 4.5, \quad y \approx 5.2
\]

Answer:
\[
\boxed{4.5, 5.2}
\]

#### 9. \( x = \), \( y = \)
Using the geometric mean relationship:
\[
y = \sqrt{10 \cdot 25} = \sqrt{250} \approx 15.8
\]

For \( x \), use the Pythagorean theorem:
\[
x^2 + 10^2 = 25^2
\]
\[
x^2 + 100 = 625
\]
\[
x^2 = 525
\]
\[
x = \sqrt{525} \approx 22.9
\]

Rounding to the nearest tenth:
\[
x \approx 22.9, \quad y \approx 15.8
\]

Answer:
\[
\boxed{22.9, 15.8}
\]

#### 10. \( x = \), \( y = \)
Using the geometric mean relationship:
\[
x = \sqrt{7 \cdot 28} = \sqrt{196} = 14
\]

For \( y \), use the Pythagorean theorem:
\[
y^2 = 7^2 + 14^2
\]
\[
y^2 = 49 + 196
\]
\[
y^2 = 245
\]
\[
y = \sqrt{245} \approx 15.7
\]

Rounding to the nearest tenth:
\[
x = 14, \quad y \approx 15.7
\]

Answer:
\[
\boxed{14, 15.7}
\]

#### 11. \( x = \)
Using the geometric mean relationship:
\[
x = \sqrt{5 \cdot 15} = \sqrt{75} \approx 8.7
\]

Rounding to the nearest tenth:
\[
\boxed{8.7}
\]

---

Final Answers:

1. \(\boxed{12}\)
2. \(\boxed{22.4}\)
3. \(\boxed{9}\)
4. \(\boxed{4.6}\)
5. \(\boxed{64}\)
6. \(\boxed{9.8}\)
7. \(\boxed{5.2}\)
8. \(\boxed{4.5, 5.2}\)
9. \(\boxed{22.9, 15.8}\)
10. \(\boxed{14, 15.7}\)
11. \(\boxed{8.7}\)