Yes.
They will be both the same because the perimeter of the square is 32 units and the perimeter of the rectangle is also 32 units
There is insufficient information to answer the question. For a given area, the perimeter depends upon the shape. For a given area, the circle will have the smallest perimeter. For polygons, regular polygons will have a smaller perimeter than an irregular one of the same area. Also, for regular polygons, the greater the number of sides, the smaller the perimeter.
The perimeter also doubles.
The dimensions of a rectangle are the length and the width. With these two measurements , the area of the rectangle can be calculated : Area = length x width. The perimeter can also be found : Perimeter = (2 x length) + (2 x width).
Yes.
Of course, a rectangle can have a greater perimeter and a greater area. Simply double all the sides: the perimeter is doubled and the area is quadrupled - both bigger than they were.
They will be both the same because the perimeter of the square is 32 units and the perimeter of the rectangle is also 32 units
The perimeter of any rectangle is [ 2 x (length + width) ]. Since the length and width of a square are equal, the perimeter of a square is also [ 2 x (side + side) ] = (4 x side).
how do you find the area of a rectangle witha perimeter of 36 in You don't. You need more information For example a 1 x 17 rectangle has a perimeter of 36 and its area is 17. But a 2 x 16 rectangle also has a perimeter of 36 and its area is 32.
There is insufficient information to answer the question. For a given area, the perimeter depends upon the shape. For a given area, the circle will have the smallest perimeter. For polygons, regular polygons will have a smaller perimeter than an irregular one of the same area. Also, for regular polygons, the greater the number of sides, the smaller the perimeter.
Not at all. For example:A square of 2 x 2 will have a perimeter of 8, and an area of 4. A rectangle of 3 x 1 will also have a perimeter of 8, and an area of 3.A "rectangle" of 4 x 0 will also have a perimeter of 8, but the area has shrunk down to zero. The circle has the largest area for a given perimeter/circumference.
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.
The perimeter also doubles.
The dimensions of a rectangle are the length and the width. With these two measurements , the area of the rectangle can be calculated : Area = length x width. The perimeter can also be found : Perimeter = (2 x length) + (2 x width).
Yes. The perimeter is a measure of the combined length of all the sides. If you double the lengths of the sides then naturally this will also necessarilychange the perimeter (it will double the perimeter).
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).