The area of any cyclic quadrilateral ABCD is given by `A^(2) = (s -a) (s-b) (s-c) (s-d)`, where `2s = a + b ++ c + d, a, b, c and d` are the sides of the quadrilateral
Now consider a cyclic quadrilateral ABCD of area 1 sq. unit and answer the following question
The minimum value of the sum of the lenghts of diagonals is

Options

A. `2sqrt2`
B. 2
C. `sqrt2`
D. none of these

Explanation

To find the minimum value of the sum of the lengths of the diagonals of a cyclic quadrilateral ABCD with an area of 1 square unit, we can follow these steps:

### Step 1: Understand the relationship between area and diagonals
The area \( A \) of a cyclic quadrilateral can be expressed in terms of its diagonals \( d_1 \) and \( d_2 \) and the angle \( \theta \) between them:
\[ A = \frac{1}{2} d_1 d_2 \sin \theta \]
Given that the area \( A = 1 \) square unit, we can set up the equation:
\[ 1 = \frac{1}{2} d_1 d_2 \sin \theta \]

### Step 2: Rearranging the equation
From the equation above, we can express the product of the diagonals in terms of \( \sin \theta \):
\[ d_1 d_2 = \frac{2}{\sin \theta} \]

### Step 3: Analyze the maximum value of \( d_1 d_2 \)
Since \( \sin \theta \) has a maximum value of 1, we can deduce that:
\[ d_1 d_2 \leq 2 \]
This implies that the product of the diagonals is maximized when \( \sin \theta = 1 \).

### Step 4: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
Using the AM-GM inequality, we know that:
\[ \frac{d_1 + d_2}{2} \geq \sqrt{d_1 d_2} \]
Thus, we can express the sum of the diagonals:
\[ d_1 + d_2 \geq 2 \sqrt{d_1 d_2} \]

### Step 5: Substitute the maximum value of \( d_1 d_2 \)
Substituting \( d_1 d_2 = 2 \) into the AM-GM inequality, we get:
\[ d_1 + d_2 \geq 2 \sqrt{2} \]

### Step 6: Conclusion
The minimum value of the sum of the lengths of the diagonals \( d_1 + d_2 \) occurs when \( d_1 d_2 = 2 \) and is given by:
\[ d_1 + d_2 = 2 \sqrt{2} \]

Thus, the minimum value of the sum of the lengths of the diagonals of the cyclic quadrilateral ABCD is:
\[ \boxed{2\sqrt{2}} \]