Given that the side length of a rhombus is the geometric mean of the length of its diagonals. The degree measure of the acute angle of the rhombus is `15^0` (b) `30^0` (c) `45^0` (d) `60^0`

Options

A. `15^@`
B. `30^@`
C. `45^@`
D. `60^@`

Explanation

To solve the problem, we need to find the acute angle of a rhombus given that the side length is the geometric mean of the lengths of its diagonals. Let's denote the diagonals as \(d_1\) and \(d_2\) and the side length as \(s\).

### Step-by-step Solution:

1. **Understanding the relationship**:
The side length \(s\) of the rhombus is given to be the geometric mean of the lengths of the diagonals \(d_1\) and \(d_2\). This can be expressed mathematically as:
\[ s = \sqrt{d_1 \cdot d_2} \]

2. **Using the properties of the rhombus**:
In a rhombus, the diagonals bisect each other at right angles. Therefore, each diagonal can be split into two halves. If we let \(d_1 = a\) and \(d_2 = b\), then:
\[ \text{Half of } d_1 = \frac{a}{2}, \quad \text{Half of } d_2 = \frac{b}{2} \]

3. **Applying the Pythagorean theorem**:
The side length \(s\) can also be expressed using the halves of the diagonals:
\[ s = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2} \]
Simplifying this gives:
\[ s = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \sqrt{\frac{a^2 + b^2}{4}} = \frac{\sqrt{a^2 + b^2}}{2} \]

4. **Equating the two expressions for \(s\)**:
From the geometric mean, we have:
\[ s = \sqrt{ab} \]
From the Pythagorean theorem, we have:
\[ s = \frac{\sqrt{a^2 + b^2}}{2} \]
Setting these equal gives:
\[ \sqrt{ab} = \frac{\sqrt{a^2 + b^2}}{2} \]

5. **Squaring both sides**:
Squaring both sides to eliminate the square roots:
\[ ab = \frac{a^2 + b^2}{4} \]
Multiplying through by 4 results in:
\[ 4ab = a^2 + b^2 \]

6. **Rearranging the equation**:
Rearranging gives:
\[ a^2 - 4ab + b^2 = 0 \]

7. **Using the quadratic formula**:
This is a quadratic equation in terms of \(a\) (or \(b\)). We can use the quadratic formula:
\[ a = \frac{4b \pm \sqrt{(4b)^2 - 4 \cdot 1 \cdot b^2}}{2 \cdot 1} \]
Simplifying gives:
\[ a = \frac{4b \pm \sqrt{16b^2 - 4b^2}}{2} = \frac{4b \pm \sqrt{12b^2}}{2} = \frac{4b \pm 2\sqrt{3}b}{2} = 2b \pm \sqrt{3}b \]

8. **Finding the angle**:
The acute angle \(\theta\) can be found using the sine function:
\[ \sin \theta = \frac{1}{2} \]
This gives:
\[ \theta = 30^\circ \]

### Final Answer:
The degree measure of the acute angle of the rhombus is \(30^\circ\).