10,764 feet
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For a three-acre square-shaped lot, you'd need a minimum of about 1,446 feet of fencing.
It depends on what kind of animal you are trying to fence in (or out). For an average cattle pasture, I prefer setting them 10 feet apart, so you would need about 150 T-posts. I still prefer setting three six-inch treated wooden posts for each corner; they will hold up better to the tensioning of the fence/barb wire.
That depends on the shape of the lot. But, assuming the lot is a perfect square, you would need 10,560 feet (two miles) of fence to surround it.
To determine the amount of fence needed for a perimeter fence around 8 acres, we first need to convert the area from acres to square feet. Since 1 acre is equivalent to 43,560 square feet, 8 acres would be 8 x 43,560 = 348,480 square feet. The perimeter of a rectangular area can be calculated by using the formula P = 2(l + w), where P is the perimeter, l is the length, and w is the width. If we assume the area is a square (which maximizes the perimeter for a given area), we can find the length of one side by taking the square root of the total area, which gives us approximately 590.61 feet. Therefore, the total amount of fence needed for the perimeter would be 4 x 590.61 = 2,362.44 feet.
You can't tell the linear dimensions from the area. There are an infinite number of shapes that all have the same area. Even if you only consider rectangles, there are still an infinite number of rectangles, all with different dimensions, that all have areas of 2.5 acres. The only thing you know for sure from the area is the area. 2.5 acres = 108,900 square ft. The shortest possible length of fence to enclose 2.5 acres occurs when the plot is a square, 330-ft = 1/16th of a mile on a side. For the square, you need 1,320-ft of fence. For 2.5 acres in any rectangular shape: -- Pick one dimension. Call it 'D' in feet. -- The other dimension is (108,900 / D) feet. -- You'll need (2D + 217,800/D) feet of fence to enclose it.