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Q: What is perimeter of a rectangle if the length is 4 and the width is 1?
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How do you get the area if you have the width and the perimeter?

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) .


The perimeter of a rectangle is 232 inches The length exceeds it by 68 inches Find the length and 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


What is the area of a rectangle with a perimeter of 48 and a length of 5?

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.


What is the perimeter of a 40 acre rectangle of land?

Any length greater than 1 mile. The area of a rectangle is not sufficient to determine its perimeter.


A rectangle has a perimeter of 10 ft Write the area A of the rectangle as a function of the length of one side x of the rectangle?

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.

Related questions

How do you find diagonal of rectangle when perimeter and width are known?

Length = (1/2 of perimeter) minus (Width) Diagonal = square root of [ (Length)2 + (Width)2 ]


What is the width of a rectangle with a perimeter of 70 feet if its length is 1 feet 2 feet?

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.


What is the width of a rectangle with a perimeter of 70 feet if its length is 1 foot?

34 feet


What do you do to the length and width of a rectangle if you want the perimeter to be the same as another rectangle but have a larger area?

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.


What is the length and width of a rectangle whose area is 1 and the perimeter is 4?

that's easy.....it's 1 ;)


Find Area of Rectangle Having perimeter 2 and length 2?

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 width of a rectangle is fourteen centimeters the perimeter of the rectangle is five times more than the length of a rectangle find then length?

The answer is centimetersImproved Answer:-The length works out as 9 and 1/3 cm


Length of rectangle is 3 times as great as its width. if the perimeter of the rectangle is no more than 72 cm. What is the length of the rectangle?

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).


What is the perimeter of a 4 meter by 1 meter rectangle?

Perimeter = 2*(Length + Width) = 2*(4+1) metres = 10 metres.


How do you get the area if you have the width and the perimeter?

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 is the length and width of a rectangle that has a perimeter of 20 inches and an area of 24.4524 square inches showing work with final answers?

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


How do you calculate the top dimensions of a rectangle if you know its perimeter and bottom area?

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.