2 ft
46 ft.
18??
A 4 ft by 9 ft rectangle
The rectangle is 6-ft long and 4-ft wide.
11 ft. 11+11+6+6=34
30 ft. ************************ Method: (11×2)+(4×2) = 30ft
If you extend the perimeter by 3 feet in each direction, you will get a new rectangle 18 ft by 21 ft with a perimeter of 78 feet (18+21+18+21 = 78)
2 ft
If it is a rectangle then the perimeter is 78ft
The perimeter is 33 feet 2 inches
20ft
The rectangle with the smallest perimeter for a given area is the square. The rectangle with the greatestperimeter for a given area can't be specified. The longer and skinnier you make the rectangle, the greater its perimeter will become. No matter how great a perimeter you use to enclose 24 ft2, I can always specify a longer perimeter. Let me point you in that direction with a few examples: 6 ft x 4 ft = 24 ft2, perimeter = 20 ft 8 ft x 3 ft = 24 ft2, perimeter = 22 ft 12 ft x 2 ft = 24 ft2, perimeter = 28 ft 24 ft x 1 ft = 24 ft2, perimeter = 50 ft 48 ft x 6 inches = 24 ft2, perimeter = 97 ft 96 ft x 3 inches = 24 ft2, perimeter = 192.5 ft 288 ft x 1 inch = 24 ft2, perimeter = 576ft 2inches No matter how great a perimeter you find to enclose 24 ft2, I can always specify a rectangle with the same area and a longer perimeter.
Perimeter of [ 16 x 26 ] rectangle = 16 + 16 + 26 + 26 = 84-ft . Perimeter of a 20-ft square = 4 x 20 = 80-ft . The rectangle has the greater perimeter.
-3
46 ft.
Half an acre = 21780 sq feet.This could have a perimeter of any length greater than (or equal to) 523.16 feet.If the area is in the form of a circle of radius 83.26 feet, it will have the minimum perimeter of 523.16 feet.Or, you can have a square of 147.6 ft giving a perimeter of 590.3 ft,or a rectangle of 150 ft * 145.2 ft - perimeter = 590.4 ftor a rectangle of 100 ft * 217.8 ft - perimeter = 635.6 ftor a rectangle of 10 ft * 2178 ft - perimeter = 4376 ftor a rectangle of 1 ft * 21780 ft - perimeter of = 43562 ftetc, with the perimeter increasing without limit.Or, the area could be circular giving the minim590 feet.