Answer:
4 units
Step-by-step explanation:
according to question:
perimeter of square = 16 units
4a = 16
where a is the side length
a = 16/4
= 4 units
Step-by-step explanation:
that r = P/(2π) so A = π(P/(2π))2 = P2/(4π). Any positive area less than this is also possible. So in this problem the largest area possible is (16 in.)2/(4π) = 64/π sq.
