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.
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.
Height = (Perimeter/2) - Base
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.
Length = (Perimeter - twice width) / 2
If you are given the width and the perimeter, then figure out what the length is then calculate the area... hope this helps :)
64 meters
You cannot.
Multiply length by width
If you are given the area you will have to think what do you times with the number you have to get it.
find the perimeter and area of a rectangle that is 15cm long and 5cm wide
The length of a rectangle is twice its width. If the perimeter of the rectangle is , find its area.
To find the perimeter you add and to find the area we multiply.
the length of a rectangle is 5 more then the width. Find the perimeter and the area of the rectangle
how do you find the area of a rectangle witha perimeter of 36 in You don't. You need more information For example a 1 x 17 rectangle has a perimeter of 36 and its area is 17. But a 2 x 16 rectangle also has a perimeter of 36 and its area is 32.
Length + width = half the perimeter, but more info eg area, is needed.
You cannot find the perimeter unless the rectangle is a regular rectangle (a square) in which case the perimeter is 4 times the square root of the area. With just the area the shape of the rectangle could be any number of shapes with different perimeter, for example, imagine 6 square units 1cm by 1cm arranged in a 1*6 configuration to give a long thin rectangle, the perimeter would be 6+6+1+1=14cm, the same 6 arranged in a 3*2 rectangle would have the same area, but a perimeter of 3+3+2+2=10cm, for this reason a rectangle's perimeter cannot be determined from the area alone.
Max rectangular area for a given perimeter is a square.P = 100S = 25Area = 252 = 625 square feet
Height = (Perimeter/2) - Base