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Perimeter = 2 lengths + 2 widths 1/2 perimeter = length + width Length = 1/2 perimeter - width Area = length x width That's (1/2 perimeter - width) x (width) .
Do-it-in-your-head method: 1. Length + width is half the perimeter ie 116 in 2. 116 minus 68 is 48, half of 48 is 24 which is the width, making the length 92 in
A rectangle with a perimeter of 48 and 1 side equal to 5 would have its other sides lengths of 5, 19, 19. Such a rectangle would have an area of 95 sq/m. Your question is somewhat strange, as the "length" of a rectangle generally refers to the longer side - in this case the length would be 19 and the width 5.
Any length greater than 1 mile. The area of a rectangle is not sufficient to determine its perimeter.
This question has no unique answer. A (3 x 2) rectangle has a perimeter = 10, its area = 6 A (4 x 1) rectangle also has a perimeter = 10, but its area = 4 A (4.5 x 0.5) rectangle also has a perimeter = 10, but its area = 2.25. The greatest possible area for a rectangle with perimeter=10 occurs if the rectangle is a square, with all sides = 2.5. Then the area = 6.25. You can keep the same perimeter = 10 and make the area anything you want between zero and 6.25, by picking different lengths and widths, just as long as (length+width)=5.
Length = (1/2 of perimeter) minus (Width) Diagonal = square root of [ (Length)2 + (Width)2 ]
Perimeter = 2(length + width) → width = perimeter ÷ 2 - length length = 1 foot → width = 70 ft ÷ 2 - 1 ft = 34 ft; length = 2 feet → width = 70 ft ÷ 2 - 2 ft = 33 ft.
34 feet
YOu add a # to the width and then you subtract the same # from the length! If you want to go all the way and make the area as big as possible, then you want to make the length and width both 1/4 of the perimeter.
that's easy.....it's 1 ;)
If a rectangle had a length of 2 and a perimeter of 2, its width would need to be negative 1. However, width, by definition is non-negative and so a width of -1 is impossible. As a result, such a rectangle cannot exist. And since it cannot exist, it cannot have an area.
The answer is centimetersImproved Answer:-The length works out as 9 and 1/3 cm
The length could be 3 cm (width = 1 cm), with a perimeter of 8 cm, which is not more than 72 cm. Or it could be 6 cm (w = 2 cm, perimeter = 16 cm).
Perimeter = 2*(Length + Width) = 2*(4+1) metres = 10 metres.
Perimeter = 2 lengths + 2 widths 1/2 perimeter = length + width Length = 1/2 perimeter - width Area = length x width That's (1/2 perimeter - width) x (width) .
What do we know about the perimeter of a rectangle? perimeter = 2 × (length + width) → 2 × (length + width) = 20 in → length + width = 10 in → length = 10 in - width What do we know about the area of a rectangle: area = length × width → length × width = 24.4524 in² But from the perimeter we know the length in terms of the width and can substitute it in: → (10 in - width) × width = 24.4524 in² → 10 in × width - width² = 24.4524 in² → width² - 10 in × width + 24.4524 in² = 0 This is a quadratic which can be solved by using the formula: ax² + bx + c → x = (-b ±√(b² - 4ac)) / (2a) → width = (-10 ±√(10² - 4 × 1 × 24.4524)) / (2 × 1) in → width = -5 ± ½√(100 - 97.8096) in → width = -5 ±½√2.1904 in → width = -5 ± 0.74 in → width = 4.26 in or 5.74 in → length = 10 in - 4.26 in = 5.74 in or 10 in - 5.74 in = 4.26 in (respectively) By convention the width is the shorter length (though it doesn't have to be) making the width 4.26 in and the length 5.74 in. Thus the rectangle is 5.74 in by 4.26 in
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.