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Anything greater than 24 cm. P = 2*L + 2*W, since the length is fixed at 24cm we have: 2*(24cm) +2*W > 96cm 2W > 96 - 48cm 2W > 48cm, or Width > 24cm. So any width greater than 24cm will make the perimeter >96cm. Obviously a width of 24cm wouldn't work since a polygon with all four sides equal would be a square and not a rectangle.

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Q: The length of a rectangle is fixed at 24 cm what widths will make the perimeter greater than 96 cm?

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Using algebra to solve this geometric problem, call the width w inches long.Therefore the length is 2w+2 inches (i.e. "2in more than twice its width").There are two widths and two lengths in the perimeter so now add them up. w+w+2w+2+2w+2=6w+4 The perimeter is equal to 16 inches, so 6w+4=16 6w=12 w=2 inches

length/width

Since the question asks about the perimeter, lengths and widths, it is not clear what the 30 feet measure, which is given in the question, refers to! Without that information, it is impossible to answer the question.

Algebra works great here. Let's use three variables: P for perimeter, L for length and W for width. Then we can rewrite the above in terms of math:W = (1/2)(L - 6)

The question is perhaps a bit confusing, because normally when we use the terms length and width in describing rectangles, the length refers to the longer dimension, and the width to the shorter. Because the question is worded so as to depart from that convention, it is reasonable to revert to the graph paper convention of making widths horizontal (along rows) and lengths vertical (along columns), even if the length is shorter than the width. So that means the answer is to draw along the perimeter of a group of 10 consecutive boxes in a single row on the graph paper.

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Add up (two lengths) plus (two widths) and you have the perimeter.

P (perimeter of a rectangle) = 2*l+2*w 2*24+2*w > 100 2*w > 52 w > 26 Any width greater than 26cm will cause the perimeter to be greater than 100cm.

You can't tell the dimensions from the perimeter. There are an infinite number of different rectangles, all with different lengths and widths, that all have the same perimeter.

x=width x+25=length 4x+50=130 4x=80 x=20 width is 20 length is 45

The perimeter of a rectangle is given by the formula P = 2(l + w). It is clear that as the length, l, increases, the perimeter, P, increases, as well. We say, therefore, that P is directly proportional to l. If l is the length and b is width of a rectangle then, the perimeter P of the rectangle is 2(l + b) units. P = 2(l + b) P = 2l + 2b If have b as a constant then, 2b will be a constant. Now l is the varying quantity. Say 2b = K P = 2l +K Perimeter changes if the length of the rectangle changes. In particular, if the length increases the perimeter of the rectangle increases. Similarly, if the length decreases the perimeter also decreases. So, the perimeter is directly proportional to the length of the rectangle. Source: www.icoachmath.com In the most simplest explanation, the sum of both lengths, and both widths of the rectangle, IS the perimeter. So obviously the perimeter is directly proportionate to its length (and its width).

To find the perimeter of the rectangle you add up all the sides. There are two lengths and widths for every rectangle, so you know the length is 40 meters x 2 = 80 meters. Perimeter is the total length of all the sides so you just minus 120 from the two lengths which is 80. So the width is 40, divide 2 (2 widths) = 20 meters. Width = 20 meters OR if they're asking for widthS then it'll be 40 meters, but they're not.

The perimeter of a rectangle, like the perimeter of any closed two-dimensional figure, is the distance around it. The perimeter of the rectangle is the sum of two lengths plus two widths.

Perimeter = 2 lengths and 2 widths In your case length + width = 25cm If length is 5cm more than width then length = 15cm and width = 10cm

Yes, it normally has opposite congruent lengths and opposite congruent widths. The length of a rectangle is normally greater than its width.

There will be Length/Width widths in 1 length. This will normally be a number that is greater than 1.

You can't find the widths of a rectangle with the lengths because the widths can be anything lower than the lengths. Like if your rectangle had a length of 7 the width can be 6, 5, 4, 3, 2, and so on.

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.

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