Of course, this depends on the eraser. Generally 1 or 2 cm.
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Well, isn't that just a happy little question! The width of an eraser can vary depending on the brand and type, but typically they are around half an inch to an inch wide. Remember, erasers come in all shapes and sizes, so just choose the one that feels right for you and your creative journey!
The width of an eraser can vary depending on the brand, type, and size of the eraser. On average, a standard pencil eraser has a width of around 0.25 inches (6.35 millimeters). However, larger erasers, such as block erasers or eraser sticks, can have widths ranging from 0.5 inches (12.7 millimeters) to 1 inch (25.4 millimeters) or more. It is important to note that these measurements are approximate and can differ based on the specific eraser being used.
length*width = area 24.7*width = 550.81 Divide both sides by 24.7 to find the width: width = 22.3 meters
Let the width of rectangle is w cm (Length is 12 cm more) so length is w + 12 Perimeter = 16 times of width 2 (length + width) = 16 * width Rest of the homework, do yourself.
Length is 12ft, width is 4 ft. 2 x (length + width) = perimeter = 32 ft ⇒ length = 16ft - width [1] length x width = area = 48 sq ft Using [1] above: ⇒ (16ft - width) x width = 48 sq ft multiplying out and rearranging: ⇒ width2 - 16 x width + 48 = 0 factorizing: ⇒ (width - 4)x(width - 12) = 0 Thus width = 4 or 12, so using [1] above: ⇒ width = 4ft, length = 16 - 4 = 12ft or width = 12ft, length = 16 - 12 = 4ft Convention is that the length is larger than the width, thus rectangle is 12ft long by 4 ft wide.
Important to note are these formulae: Perimeter_of_rectangle = 2 x (length + width) Area_of_rectangle = length x width So if the perimeter and area are known, then: 2 x (length + width) = perimeter => length + width = perimeter / 2 => length = perimeter / 2 - width length x width = area => (perimeter / 2 - width) x width = area (substituting for length given above) => perimeter / 2 x width - width2 = area => width2 - perimeter / 2 x width + area = 0 which is a quadratic and can be solved either by factorization or by using the formula: width = (perimeter / 2 +/- sqrt(perimeter2 / 4 - 4 x area)) / 2 = (perimeter +/- sqrt(perimeter2 - 16 x area)) / 4 This will provide two values for the width. However, each of these values is the length for the other, so the larger value is the length and the smaller value is the width. Sometimes only 1 value will be found for the width above. In this case, the rectangle is actually a square which means that the length and width are both the same. Examples: 1. perimeter = 6, area = 2 width2 - perimeter / 2 x width + area = 0 => width2 - 6 / 2 x width + 2 = 0 => width2 - 3 x width + 2 = 0 => (width - 2) x (width - 1) = 0 => width = 2 or 1. So the length is 2 and the width is 1. 2. perimeter = 12, area = 9 width2 - perimeter / 2 x width + area = 0 => width2 - 12 / 2 x width + 9 = 0 => width2 - 6 x width + 9 = 0 => (width - 3)2 = 0 => width = 3 So the rectangle is a square with both length and width of 3.
perimeter 24 so length + width half of that ie 12. Length 3 times width must be 9 and width 3.