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 minium perimeter of the quadrilateral is

Options

A. 4
B. 2
C. 1
D. none of these

Explanation

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

### Step 1: Understand the formula for the area of a cyclic quadrilateral
The area \( A \) of a cyclic quadrilateral is given by the formula:
\[ A^2 = (s - a)(s - b)(s - c)(s - d) \]
where \( s = \frac{a + b + c + d}{2} \) is the semi-perimeter and \( a, b, c, d \) are the sides of the quadrilateral.

### Step 2: Set the area equal to 1
Given that the area is 1 square unit, we have:
\[ 1^2 = (s - a)(s - b)(s - c)(s - d) \]
This simplifies to:
\[ 1 = (s - a)(s - b)(s - c)(s - d) \]

### Step 3: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
Using the AM-GM inequality, we know that:
\[ \frac{(s - a) + (s - b) + (s - c) + (s - d)}{4} \geq \sqrt[4]{(s - a)(s - b)(s - c)(s - d)} \]
Substituting our area condition:
\[ \frac{(s - a) + (s - b) + (s - c) + (s - d)}{4} \geq \sqrt[4]{1} = 1 \]

### Step 4: Simplify the left side
The left side can be rewritten as:
\[ \frac{4s - (a + b + c + d)}{4} = \frac{4s - 2s}{4} = \frac{2s}{4} = \frac{s}{2} \]
Thus, we have:
\[ \frac{s}{2} \geq 1 \implies s \geq 2 \]

### Step 5: Relate semi-perimeter to perimeter
Since \( s = \frac{a + b + c + d}{2} \), we can express the perimeter \( P \) as:
\[ P = a + b + c + d = 2s \]
From our previous result, we find:
\[ P \geq 2 \times 2 = 4 \]

### Conclusion
The minimum perimeter of the cyclic quadrilateral ABCD is:
\[ \boxed{4} \]