The minimum value of the sum of the lengths of diagonals of a cyclic quadrilateral of area `a^(2)` square units, is

Options

A. `sqrt(2)a`
B. `2sqrt(2)a`
C. 2a
D. none of these

Answer

To find the minimum value of the sum of the lengths of the diagonals of a cyclic quadrilateral with an area of \( a^2 \) square units, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the minimum value of the sum of the lengths of the diagonals \( AC + BD \) of a cyclic quadrilateral given its area \( A = a^2 \).

2. **Using the Area Formula**:
The area \( A \) of a cyclic quadrilateral can be expressed in terms of its diagonals \( AC \) and \( BD \) and the sine of the angle \( \theta \) between them:
\[ A = \frac{1}{2} \times AC \times BD \times \sin(\theta) \]
Here, \( \theta \) is the angle between the diagonals.

3. **Setting Up the Inequality**:
By applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
\[ AC + BD \geq 2 \sqrt{AC \cdot BD} \]

4. **Expressing \( AC \cdot BD \)**:
From the area formula, we can rearrange it to find \( AC \cdot BD \):
\[ a^2 = \frac{1}{2} \times AC \times BD \times \sin(\theta) \]
Rearranging gives:
\[ AC \cdot BD = \frac{2a^2}{\sin(\theta)} \]

5. **Substituting Back into the Inequality**:
Substitute \( AC \cdot BD \) into the AM-GM inequality:
\[ AC + BD \geq 2 \sqrt{\frac{2a^2}{\sin(\theta)}} \]

6. **Minimizing the Expression**:
To minimize \( AC + BD \), we need to maximize \( \sin(\theta) \). The maximum value of \( \sin(\theta) \) is 1 (when \( \theta = 90^\circ \)):
\[ AC + BD \geq 2 \sqrt{2a^2} \]

7. **Final Result**:
Therefore, the minimum value of the sum of the lengths of the diagonals is:
\[ AC + BD \geq 2\sqrt{2}a \]

### Conclusion:
The minimum value of the sum of the lengths of the diagonals of a cyclic quadrilateral with area \( a^2 \) square units is \( 2\sqrt{2}a \).