The perimeters of a square and a rectangle are equal. If their area be 'A' m² and 'B' m² then correct statement is

In this condition always square area is greater than any other quadrilateral
So, option (C) is correct.

Alternate Solution:-
Let perimeter = 16 cm
Perimeter of square = 4a = 16
                                      a = 4
Area (A) of square = 16 m²
Perimeter of rectangle = 2($$l$$ + b)
16 = 2($$l$$ + b)
$$l$$ + b = 8
Area (B) of rectangle = $$\left( {\frac{{4,\,4}}{{6,\,2}}} \right)$$
= 6 × 2
= 12 cm²