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.
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.
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.
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.