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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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DevinYou 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 height of a rectangle with the base and perimeter given?
Height = (Perimeter/2) - Base
What is the relationship for perimeter and area for rectangle?
There is no relationship between the perimeter and area of a rectangle. Knowing the perimeter, it's not possible to find the area. If you pick a number for the perimeter, there are an infinite number of rectangles with different areas that all have that perimeter. Knowing the area, it's not possible to find the perimeter. If you pick a number for the area, there are an infinite number of rectangles with different perimeters that all have that area.
What is the formula to find the length of a rectangle when the perimeter and width is already given?
Length = (Perimeter - twice width) / 2
How do you find area when given perimeter and width?
If you are given the width and the perimeter, then figure out what the length is then calculate the area... hope this helps :)
He length of a rectangle is three times its width. If the area of the rectangle is 192 m2 find its perimeter.?
64 meters
