The increase in area of a rectangle having a perimeter of 62 cm is 66 cm2 when each dimension is increased by 2 cm.
For proof, after dividing the perimeter by 2 to obtain length plus width:
30 cm x 1 cm = 30 cm2 and 32 cm x 3 cm = 96 cm2;
16 cm x 15 cm = 240 cm2; 18 cm x 17 cm = 306 cm2.
Comparing both examples, the difference is 66 cm2.
absolutely no. length and width are independent parameters .
Hi, If you're looking for a straight numerical answer, then you will need to supply more information, for example a ratio or a width. EG: What is the length of a rectangle with a perimeter of 96 with a height to length ratio of 1:2 (height is 1, length is double that) or EG: What is the length of a rectangle with a perimeter of 96 and a height of 10. There's no way to tell what the length of a rectangle is if you don't know its height. however, if height is not relivent then the answer is anything that can be accuratel used in the definition of a rectangle. So...if the answer must be a whole number then we'll assume the minimum height of the rectangle is 1, therefor the maxium length is 47. If we continue down this track then each time we increase the height by 1 we reduce the length by 1. The So the table would go (using a ratio of Height:Length) {| |- | 1:47 2:46 3:45 4:44 5:43 6:42 7:41 8:40 9:39 10:38 11:37 12:36 13:35 14:34 15:33 16:32 17:31 18:30 19:29 20:28 21:27 22:26 23:25 24:24 would be a square so 23:25 is last option. |}
30%
Perimeter is a length, and a length cannot be the same as an area.Ignoring the units, all rectangles that have the same numerical perimeter as area are those that satisfy:2 x (length + width) = length x widthwhich can be rearranged to give:width = 2 x length/(length - 2)Meaning that given any length over 2, a width can be found to give a rectangle that meets the requirement that its numerical perimeter is the same as its numerical area. (For a length greater than 0 and less than 2, the width would be negative and not possible; similarly for a length less than 0, the length is negative and not possible. When the length is 2, the width is undefined and so not possible. When the length is 0 the width is 0 and it is not a rectangle.)At some stage as the length increases, the length will equal the width and as the length continues to increase the rectangle then given will match the previous rectangles with the length and widths swapped. This occurs when:length = 2 x length/(length - 2)⇒ length x (length - 4) = 0⇒ length = 4.So, as long as the length is greater than or equal 4 it will be the longer side - the length, by convention, is the longer side. Thus all rectangles satisfy:width = 2 x length/(length - 2)with length ≥ 4, will have the numerical value of their perimeter the same as the numerical value of their area.For example:4 cm x 4 cm: perimeter = 16 cm, area = 16 cm25 cm x 31/3 cm: perimeter = 162/3 cm, area = 162/3 cm26 cm x 3 cm: perimeter = 18 cm, area = 18 cm241/2 cm x 33/5 cm: perimeter = 161/5 cm, area = 161/5 cm2etcNote: the first example is a square which is a rectangle with all the sides the same length.
Any value greater than or equal to 7934.102 feet. The most compact shape would be a circular plot with radius 1262.752 feet and that would give a circumference (or perimeter) of 7934.102 feet. By flattening the circle you can increase the perimeter without limit. Similarly, 115 acres could be a square plot with sides of 2238.169 feet, giving the minimum perimeter (for a quadrilateral) of 8952.676 feet. But you could consider a rectangular plot of 22381.69 ft * 223.8169 ft with a perimeter of 45211.01 feet or 223816.9 ft * 22.38169 ft giving a perimeter of 447678.5 ft. or 2238169 ft * 2.238169 ft and a perimeter of 4476342 ft. etc.
66
As written, that's confusing. The length and width of a triangle wouldn't have any bearing on the perimeter and area of a rectangle unless they overlap in some drawing that only you are looking at. Let's assume you meant rectangle all along. If the dimensions of a rectangle increased 4 times the perimeter would also increase 4 times. The area would increase 16 times. Try it out. A 2 x 3 rectangle has perimeter 10 and area 6. An 8 x 12 rectangle has perimeter 40 and area 96.
If you increase the rectangle's length by a value, its perimeter increases by twice that value. If you increase the rectangle's width by a value, its perimeter increases by twice that value. (A rectangle is defined by its length and width, and opposite sides of a rectangle are the same length. The lines always meet at their endpoints at 90° angles.)
increase
If you increase the length then the width must decrease by the same amount if the perimeter is to remain the same.
400
It is also halved. In general, if you increase any linear measurement of a figure by a certain factor (i.e., stretching the figure so that you obtain a geometrically similar figure), then all linear measurements will increase by the same factor. In this case, all linear measurements of the rectangle are increased by a factor of 0.5 - that includes the length of any side, the perimeter, the half-perimeter, the diagonal, the sum of the diagonals, the length of half the shorter side, or any other linear measurement you can think of.
It was an acher
If the length and width of a rectangle is doubled, it means that both dimensions have increased by a factor of 2. As a result, the area of the rectangle will increase by a factor of 4, because the area is calculated by multiplying the length and width together. Additionally, the perimeter of the rectangle will also increase by a factor of 2, since it is calculated by adding the lengths of all four sides.
Nice problem !At first, I thought you can't tell, because the perimeter doesn't tell you the area,so you can't tell what the original area is. But then I tried a few, and discoveredthat although you can't tell the original area or the new area, the increase is alwaysthe same ... 66 square centimeters.Let's see if I can prove it:Perimeter = 2L + 2W = 62L + W = 31W = 31 - LOld area = L x W = L (31 - L) = 31L - L2New area = (L + 2) x (W + 2) = (L + 2) x (33 - L) = 33L - L2 + 66 - 2L = 31L - L2 + 66 = Old area + 66Well that wasn't so bad.
18" is not a possible perimeter measurement. Assume the dimensions of the rectangle are so close to those of a square that the difference can be disregarded. This is the condition when the perimeter is at its minimum. When the rectangle measures approximately 6" x 6", its area = 36 sq ins, its perimeter = 24" For the area to remain constant then as the length increases by a factor n the width must decrease by that same factor. Area = 6n x 6/n : perimeter = 12n + 12/n :so when n = 1, Perimeter = 12 + 12 = 24 As n increases, say n = 2, Perimeter = 24 + 6 = 30 : And the perimeter continues to increase as the rectangle becomes narrower. Eventually, it will become so narrow that for diagram purposes it will appear as a straight line.
This needs more information. Without some other factor, like a change in area, the width doesn't have to increase at all.