Any length and width which add up to (4 x 8)/2 eg 1 & 15 or 2 & 14 or 3 & 13 etc etc
The length should be 6cm and the width is 4cm.
perimeter is the measure around the figure; area is the measure within the figure formula: perimeter: length+length+width+width=perimeter (for square or rectangle) area: length times width= area ( for square or rectangle)
Length: 4; Width: 5 or Length: 5; Width: 4
28 inches
Length = 9 Width = 9 Your rectangle is a square.
The perimeter of any rectangle is [ 2 x (length + width) ]. Since the length and width of a square are equal, the perimeter of a square is also [ 2 x (side + side) ] = (4 x side).
The length should be 6cm and the width is 4cm.
perimeter is the measure around the figure; area is the measure within the figure formula: perimeter: length+length+width+width=perimeter (for square or rectangle) area: length times width= area ( for square or rectangle)
If the shape is a rectangle (or square), then Perimeter = 2*(Length + Breadth) So Breadth = Perimeter/2 - Length
Perimeter is 36 mArea is 72 square m
Length: 4; Width: 5 or Length: 5; Width: 4
28 inches
98 square feet
Length = 9 Width = 9 Your rectangle is a square.
To find the least perimeter of a rectangle with a fixed area of 32 square feet, we can use the relationship between area and perimeter. For a rectangle, the area ( A = l \times w ) (length times width) and the perimeter ( P = 2(l + w) ). To minimize the perimeter while keeping the area constant, the rectangle should be a square. The side length of a square with an area of 32 ft² is ( \sqrt{32} ), which is approximately 5.66 ft. Thus, the least perimeter is ( 4 \times \sqrt{32} ), which is approximately 22.63 ft.
If the dimensions of the rectangle are in feet, the perimeter will be in feet as well. The area will be in square feet. The area is length x width. The perimeter is 2 x (length + width) or 2 x length + 2 x width.
They will be both the same because the perimeter of the square is 32 units and the perimeter of the rectangle is also 32 units