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
When the perimeter is minimum, the quadrilateral is necessarily

Options

A. a square
B. a rectangle but not a square
C. a rhombus but not a square
D. none of these

Explanation

To solve the problem, we need to analyze the conditions under which the perimeter of a cyclic quadrilateral ABCD is minimized, given that its area is 1 square unit.

### Step-by-Step Solution:

1. **Understanding the Area Formula**:
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 lengths of the sides of the quadrilateral.

2. **Given Area**:
We know that the area \( A = 1 \) square unit. Therefore, we have:
\[ 1^2 = (s - a)(s - b)(s - c)(s - d) \]
This simplifies to:
\[ 1 = (s - a)(s - b)(s - c)(s - d) \]

3. **Minimizing the Perimeter**:
The perimeter \( P \) of the quadrilateral is given by:
\[ P = a + b + c + d = 2s \]
To minimize the perimeter, we need to minimize \( s \).

4. **Condition for Minimum Perimeter**:
The perimeter is minimized when the differences \( s - a, s - b, s - c, s - d \) are equal. This means:
\[ s - a = s - b = s - c = s - d \]
Let’s denote this common value as \( k \). Thus, we have:
\[ s - a = k \implies a = s - k \]
Similarly, we get:
\[ b = s - k, \quad c = s - k, \quad d = s - k \]
This implies:
\[ a = b = c = d \]

5. **Identifying the Quadrilateral**:
Since all sides are equal, the quadrilateral is a regular quadrilateral. The only regular quadrilateral is a square.

6. **Conclusion**:
Therefore, when the perimeter of the cyclic quadrilateral ABCD is minimized, it is necessarily a square.

### Final Answer:
The quadrilateral is necessarily a **square**.