There are infinitely many rectangles. Let K = sqrt(11). Let L be any real number greater than M and let B = 11/L. Then, B < K so that for any two different values of L, the pair (L, B) are distinct even with a swap.The rectangle with length L and breadth B has an area = L*(11/L) = 11 cm2. Since there are infinitely many choices for L, there are infinitely many rectangles.
Infinite amounts.
16.
123x123=123
There are an infinite number of rectangles for any given area, while there is only one square for any given area. The number of integer-value rectangles depends on the area and the number of integer factors of a whole-number area. Example: a rectangular area of 6 square inches could be enclosed by rectangles that were 1x6, 2x3, 3x2, and 6x1. Non-integer dimensions would include 1.5x4 and 1.2x5 inches.
9
13
Infinite amounts.
technically the number is infinite
4
There are infinitely many rectangles. Let K = sqrt(11). Let L be any real number greater than M and let B = 11/L. Then, B < K so that for any two different values of L, the pair (L, B) are distinct even with a swap.The rectangle with length L and breadth B has an area = L*(11/L) = 11 cm2. Since there are infinitely many choices for L, there are infinitely many rectangles.
3 or 6, depending on whether rectangles rotated through 90 degrees are counted as different. The rectangles are 1x12, 2x6 3x4 and their rotated versions: 4x3, 6x2 and 12x1.
No. Many investigators have searched for such an example, but none have found it yet. According to all published research so far, two rectangles with the same area always have the same area. But the search goes on, in many great universities.
Infinitely many. Suppose the area of the rectangle is 100. We could create rectangles of different areas: 100x1 50x2 25x4 20x5 10x10 However, the side lengths need not be integers, which is why we can create infinitely many rectangles. Generally, if A is the area of the rectangle, and L, L/A are its dimensions, then the amount 2(L + (L/A)) can range from a given amount (min. occurs at L = sqrt(A), perimeter = 4sqrt(A)) to infinity.
thare is only 1 differint rectangles
5
The answer is Infinite...The rectangles can have an infinitely small area and therefore, without a minimum value to the area of the rectangles, there will be an uncountable amount (infinite) to be able to fit into that 10 sq.in.