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https://math.answers.com/algebra/How_do_you_find_the_perimeter_of_a_rectangle_when_given_area

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

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.



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

https://math.answers.com/algebra/How_do_you_find_the_perimeter_of_a_rectangle_when_given_area

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.



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https://math.answers.com/algebra/How_do_you_find_the_perimeter_of_a_rectangle_when_given_area

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

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.

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      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.

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