Day 4 - 8.1 Geometric Mean Worksheet | PDF | Teaching Mathematics - Free Printable

Worksheet on Geometric Mean

Let's solve the problems step by step.

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#### I. Simplify each radical

1. Simplify \(\sqrt{3}\):
- \(\sqrt{3}\) is already in its simplest form.
- Answer: \(\sqrt{3}\)

2. Simplify \(\sqrt{48}\):
- Factorize 48: \(48 = 16 \times 3\)
- \(\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}\)
- Answer: \(4\sqrt{3}\)

3. Simplify \(\frac{1}{\sqrt{2}}\):
- Rationalize the denominator:
\[
\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}
\]
- Answer: \(\frac{\sqrt{2}}{2}\)

4. Simplify \(\sqrt{\frac{3}{5}}\):
- Separate the square root of the numerator and the denominator:
\[
\sqrt{\frac{3}{5}} = \frac{\sqrt{3}}{\sqrt{5}}
\]
- Rationalize the denominator:
\[
\frac{\sqrt{3}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{15}}{5}
\]
- Answer: \(\frac{\sqrt{15}}{5}\)

5. Simplify \(\frac{2}{3\sqrt{5}}\):
- Rationalize the denominator:
\[
\frac{2}{3\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{3 \cdot 5} = \frac{2\sqrt{5}}{15}
\]
- Answer: \(\frac{2\sqrt{5}}{15}\)

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#### II. Find the geometric mean between each pair of numbers

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

6. Find the geometric mean between 2 and 8:
\[
\sqrt{2 \cdot 8} = \sqrt{16} = 4
\]
- Answer: \(4\)

7. Find the geometric mean between 9 and 16:
\[
\sqrt{9 \cdot 16} = \sqrt{144} = 12
\]
- Answer: \(12\)

8. Find the geometric mean between 4 and 5:
\[
\sqrt{4 \cdot 5} = \sqrt{20} = 2\sqrt{5}
\]
- Answer: \(2\sqrt{5}\)

9. Find the geometric mean between \(\sqrt{3}\) and \(\sqrt{5}\):
\[
\sqrt{\sqrt{3} \cdot \sqrt{5}} = \sqrt{\sqrt{15}} = \sqrt[4]{15}
\]
- Answer: \(\sqrt[4]{15}\)

10. Find the geometric mean between 5 and 1.25:
\[
\sqrt{5 \cdot 1.25} = \sqrt{6.25} = 2.5
\]
- Answer: \(2.5\)

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#### III. NOTES

The diagram shows a right triangle with an altitude drawn to the hypotenuse, dividing it into two smaller right triangles. The relationships are as follows:

- \(h\) is the altitude to the hypotenuse.
- \(a\) is one leg of the original triangle.
- \(b\) is the other leg of the original triangle.
- \(x\) and \(y\) are the segments of the hypotenuse created by the altitude.
- \(z\) is the hypotenuse of the original triangle.

The geometric mean relationships are:
1. \(h\) is the geometric mean between \(x\) and \(y\):
\[
h = \sqrt{x \cdot y}
\]
2. \(a\) is the geometric mean between \(z\) and \(x\):
\[
a = \sqrt{z \cdot x}
\]
3. \(b\) is the geometric mean between \(z\) and \(y\):
\[
b = \sqrt{z \cdot y}
\]

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#### IV. Solve for the variables in the diagrams

11. Diagram with \(x\), \(y\), and \(z\):
- Given: \(x = 6\) and \(z = 24\)
- Use the Pythagorean theorem in the smaller triangles:
\[
y^2 + x^2 = z^2
\]
\[
y^2 + 6^2 = 24^2
\]
\[
y^2 + 36 = 576
\]
\[
y^2 = 540
\]
\[
y = \sqrt{540} = \sqrt{36 \cdot 15} = 6\sqrt{15}
\]
- Use the geometric mean relationship for \(x\):
\[
x = \sqrt{z \cdot y}
\]
\[
6 = \sqrt{24 \cdot y}
\]
\[
36 = 24y
\]
\[
y = \frac{36}{24} = \frac{3}{2}
\]
- Answers: \(x = 6\), \(y = 6\sqrt{15}\), \(z = 24\)

12. Diagram with \(x\), \(y\), and \(z\):
- Given: \(z = 15\) and the hypotenuse of the larger triangle is 25.
- Use the Pythagorean theorem in the larger triangle:
\[
z^2 + y^2 = 25^2
\]
\[
15^2 + y^2 = 625
\]
\[
225 + y^2 = 625
\]
\[
y^2 = 400
\]
\[
y = \sqrt{400} = 20
\]
- Use the geometric mean relationship for \(x\):
\[
x = \sqrt{z \cdot y}
\]
\[
x = \sqrt{15 \cdot 20} = \sqrt{300} = 10\sqrt{3}
\]
- Answers: \(x = 10\sqrt{3}\), \(y = 20\), \(z = 15\)

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Final Answers:

\[
\boxed{
\begin{array}{l}
\text{I.} \\
1. \sqrt{3} \\
2. 4\sqrt{3} \\
3. \frac{\sqrt{2}}{2} \\
4. \frac{\sqrt{15}}{5} \\
5. \frac{2\sqrt{5}}{15} \\
\text{II.} \\
6. 4 \\
7. 12 \\
8. 2\sqrt{5} \\
9. \sqrt[4]{15} \\
10. 2.5 \\
\text{IV.} \\
11. x = 6, y = 6\sqrt{15}, z = 24 \\
12. x = 10\sqrt{3}, y = 20, z = 15
\end{array}
}
\]