let p represent perimeter let w rep. width let l rep. length p = 2l + 2w --> basic perimeter formula l = w + 2 --> two feet more than width 20 = 2(w + 2) + 2w --> sub w + 2 in for l and 20 in for p (given in question) 20 = 2w + 4 + 2w 20 = 4w + 4 --> group like terms 16 = 4w 4 = w if your looking for the width it's therefore 4 feet. and for the length: l = w + 2 l = 4 + 2 l = 6 therefpre the length is 6 feet.
108 feet is the perimeter of a rectangle with a length of 50 feet and a width of 4 feet.
The perimeter is 30 feet for this rectangle.
15 feet.
20 feet
228 feet is the perimeter.
The perimeter of a rectangle with a length of 8.5 feet and a width of 4.5 feet is 26.
108 feet is the perimeter of a rectangle with a length of 50 feet and a width of 4 feet.
The perimeter is 30 feet for this rectangle.
The width is 8 feet
You are describing a cuboid, not a rectangle. A rectangle has only two diminutions, length and width.
5 feet
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.
20
44
gt
15 feet.
63 feet