How do you find the perimeter of a rectangle when given area?

Answer 1

You cannot. There are infinitely many possible answers.

Given Area = A square units, select any value of L > sqrt(A) units and let B = A/L units.

Then for every one of the infinitely many values of L, the rectangle with length L and breadth B, the area = L*B = L*(A/L) = A square units.

The reason for selecting L > sqrt(A) is simply to ensure that each different value of L gives a different rectangle and you do not have the length and breadth of one rectangle being the breadth and length of another.

You cannot. There are infinitely many possible answers.

Given Area = A square units, select any value of L > sqrt(A) units and let B = A/L units.

Then for every one of the infinitely many values of L, the rectangle with length L and breadth B, the area = L*B = L*(A/L) = A square units.

The reason for selecting L > sqrt(A) is simply to ensure that each different value of L gives a different rectangle and you do not have the length and breadth of one rectangle being the breadth and length of another.

You cannot. There are infinitely many possible answers.

Given Area = A square units, select any value of L > sqrt(A) units and let B = A/L units.

Then for every one of the infinitely many values of L, the rectangle with length L and breadth B, the area = L*B = L*(A/L) = A square units.

The reason for selecting L > sqrt(A) is simply to ensure that each different value of L gives a different rectangle and you do not have the length and breadth of one rectangle being the breadth and length of another.

You cannot. There are infinitely many possible answers.

Given Area = A square units, select any value of L > sqrt(A) units and let B = A/L units.

Then for every one of the infinitely many values of L, the rectangle with length L and breadth B, the area = L*B = L*(A/L) = A square units.

The reason for selecting L > sqrt(A) is simply to ensure that each different value of L gives a different rectangle and you do not have the length and breadth of one rectangle being the breadth and length of another.

Answer 2

You cannot. There are infinitely many possible answers.

Given Area = A square units, select any value of L > sqrt(A) units and let B = A/L units.

Then for every one of the infinitely many values of L, the rectangle with length L and breadth B, the area = L*B = L*(A/L) = A square units.

The reason for selecting L > sqrt(A) is simply to ensure that each different value of L gives a different rectangle and you do not have the length and breadth of one rectangle being the breadth and length of another.