123x123=123
3
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.
(24 cm)² = 576 sq cm, the area of a square with sides 24 cm 24 cm² = an area of 24 sq cm
I can give the width of one of the rectangles. The first rectangle of area 15 cm2 and length of 5 cm has width of 3 cm. It is impossible to know the width of the other rectangle of area 60 cm2. However, if you had said that the two rectangles were similar, then the dimensions of the second rectangle would be 10 cm X 6 cm. But you didn't say that the two rectangles were similar; so there are infinite possibilities of what the dimensions of the second rectangle might be.
The area of rectangle is : 24.0
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In order to get a rectangle with an area of 24 centimeters, the length and width multiplied need to equal 24. On top of that, length and width may not be equal, or the shape would be a square instead of a rectangle. Examples of rectangles with 24cm areas: 1x24 cm 2x12 cm 3x8 cm 4x6 cm
8:32
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. So, there are 5 rectangles with an area of 36 cm^2 is 5.
Infinitely many. Suppose the length of the rectangle is L cm where L >= sqrt(24) = 4.899 cm (approx). Let the breadth of the rectangle be B = 24/L. Then area = L*B = L*24/L = 24 Since there are infinitely many choices for L, there are infinitely many rectangles. For example, 5 * 4.8 6 * 4 8 * 3 50* 0.48 60 * 0.4 600 * 0.04 6000 * 0.004 60000 * 0.0004 etc
Infinitely many. Select any number, L, such that 12 < L < 24. Let W = 24 - L. Then a rectangle with sides of length L cm and width W cm will have a perimeter of 48 cm. And since the choice of L was arbitrary, there are infinitely many possible values of L and thence infnitely many rectangles.
To form a square using rectangles measuring 6 cm by 15 cm, we need to find the least common multiple (LCM) of the rectangle's dimensions. The LCM of 6 and 15 is 30 cm. A square with an area of 900 cm² (30 cm x 30 cm) can be formed, requiring a total of 30 cm / 6 cm = 5 rectangles along one side and 30 cm / 15 cm = 2 rectangles along the other side, resulting in 5 x 2 = 10 rectangles needed in total.
330سم
To find the number of rectangles with an area of 36 cm², we need to consider the pairs of positive integer factors of 36. The pairs are (1, 36), (2, 18), (3, 12), (4, 9), and (6, 6), resulting in 5 unique combinations. Since the order of the dimensions does not matter (e.g., a rectangle of dimensions 2 cm by 18 cm is the same as 18 cm by 2 cm), there are 5 distinct rectangles that can be drawn with an area of 36 cm².
12
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.
5