The dimensions would be 512 feet long x 431.346094 feet wide.
0.5 acres = 21,780 square feet.So you know that (length) times (width) must be 21,780 but there are noother constraints on what the dimensions must be. There are an infinite numberof possible length/width combinations.
An acre is a unit for area. Width x length. You can obtain the 2.54 acre area with a lot of length and width combinations. If you don't know the length, you cannot determine with any certainty what the width will be.
To determine the length and width of 1.5 acres, we need to know the shape of the area. Since acres are a unit of area and not length or width, we would need to define the shape of the area to calculate the dimensions. If we assume a rectangular shape, we can use the formula for the area of a rectangle (length x width = area) to find the dimensions. For 1.5 acres, we would need more information about the shape to provide specific dimensions.
To find the length and width of 20 acres of land, we first need to know the shape of the land. If the land is a perfect square, we can calculate the length and width by taking the square root of the total area (20 acres). The square root of 20 acres is approximately 4.47 acres. Therefore, the length and width of 20 acres of land would be 4.47 acres each.
The area doesn't tell you the shape or dimensions. There are an infinite numberof different shapes, and an infinite number of rectangles with different dimensions,that all have the same area.If the field is 2.5 acres, then all you know about it is that IF all 4 corners are square,then the length in feet, multiplied by the width in feet, must be 108,900 . (Or, if themeasurements are in yards, then the product must be 12,100 .) But you don't knowwhat the length or the width must be.
if your object is a rectangle then it is not possible to figure out the exact length of and width because we do not know the ratio of the length to width. if your object is a square then... the length is 29.08607914 the width is 29.08607914
To find the length and width of a rectangular area of 10,890 square feet, you need to know either the length or the width, as there are multiple combinations that can yield this area. For example, if the length is 90 feet, the width would be 121 feet (since 90 x 121 = 10,890). Alternatively, if the length is 100 feet, the width would be 108.9 feet. The specific dimensions depend on the desired shape of the area.
To find the perimeter of 150 acres, we need to know the shape of the land. Since acres measure area, we can't directly calculate the perimeter without additional information. If the land is a square, we would need to find the square root of 150 to determine the side length, then multiply by 4 to find the perimeter. If the land is a rectangle, we would need the length and width to calculate the perimeter using the formula 2(length + width).
One acre equals:0.0015625 square miles,4,840 square yards,43,560 square feet orabout 4,047 square metres (0.405 hectares)So, an acre is a unit of area. To know the length and width, the heometry sholud be known.For example it could be:a rectangle of length 1000 meters and width 4.047 meters, ora rectangle of length 500 meters and width 8.094 metersa square of length = width = 63.616 meters
If this is for a square or rectangle... the formula is: Area=Length x Width so, if you know the Area and the Length, the Width would be found this way: Area/Length = Width, or written out: Area divided by Length = Width Example: A rectangle where the known dimensions are: Area = 30 feet squared Length = 6 feet 30 feet squared / 6 feet = 5 feet Answer: Width = 5 feet Special Note: The units for the Area, meaning feet or inches or centimeters or whatever unit you're measuring in, the "Area" answer is always in Square Units. This is because as you multiply the length units times the width units, the units are multiplied and are thus "squared units." Hope this helps!
You need to know the width too.
A width of a pool is how much feet of water its can provide you with,you just need to know what you are capable of.