To find a shape with a perimeter of 30 feet and an area of 35 square feet, we can consider a rectangle. If we denote the length as ( l ) and the width as ( w ), we have the equations ( 2l + 2w = 30 ) for the perimeter and ( l \times w = 35 ) for the area. Solving these equations, we can find specific dimensions, but there are multiple combinations that could satisfy these conditions, such as a rectangle or other shapes. However, achieving both the exact area and perimeter might require specific dimensions that may not correspond to a simple geometric figure.
Anything from the smallest possible width to just under 35 feet. ie 0<width<35 feet
Perimeter = 15+15+35+35 = 100 cm
Let the length be x and the width is 20:- 2(x+20) = 110 2x+40 = 110 Subtract 40 from both sides and then divide both sides by 2 to find the width:- x = 35 feet which is the length Check: 2(35+20) = 110 feet which is the perimeter
220 ft
The width of the rectangle is 7 feet. The lengthe of the rectangle is 35 feet.
To find a shape with a perimeter of 30 feet and an area of 35 square feet, we can consider a rectangle. If we denote the length as ( l ) and the width as ( w ), we have the equations ( 2l + 2w = 30 ) for the perimeter and ( l \times w = 35 ) for the area. Solving these equations, we can find specific dimensions, but there are multiple combinations that could satisfy these conditions, such as a rectangle or other shapes. However, achieving both the exact area and perimeter might require specific dimensions that may not correspond to a simple geometric figure.
Anything from the smallest possible width to just under 35 feet. ie 0<width<35 feet
Perimeter = 15+15+35+35 = 100 cm
35 mm
35 and 28
Perimeter:20 inches Area:35
Perimeter = 24 inches Area = 35 square inches
the answer is 11 sq cm
The rectangle is 35 square feet, which would be 5,040 square inches
Let the length be x and the width is 20:- 2(x+20) = 110 2x+40 = 110 Subtract 40 from both sides and then divide both sides by 2 to find the width:- x = 35 feet which is the length Check: 2(35+20) = 110 feet which is the perimeter
220 ft