1x16
2x8
and 4x4
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.
The area of a rectangle is worked out by the equation axb when a is length and b is width. Thus, if a and b are both integers, a good way to work out possible lengths would be to break down the area into its prime factors. These are 2, 3 and 3. Now any product of these primes for a, and the other prime for b, will work as a possible rectangle. For example a=2x3=6 and b=3 will work a=3x3=9 and b=2 will work Thus possible rectangles are 1x18, 2x9 and 3x6
Rectangles don't have height. They have length and width, one value for each. The area is the product of those two values.
You can't tell the area from knowing the perimeter. There are an infinite number of different rectangles, all with the same perimeter, that all have different areas. Here are a few rectangles that all have perimeters of 42. The last number after each one is its area: 1 cm by 20 cm . . . . . 20 square centimeters 2 x 19 . . . . . 38 3 x 18 . . . . . 54 4 x 17 . . . . . 68 5 x 16 . . . . . 80 10 x 11 . . . 110
A triangular prism.
To find all rectangles with an area of 51, we can use the formula ( \text{Area} = \text{length} \times \text{width} ). The pairs of factors of 51 are (1, 51) and (3, 17), which means the rectangles can have dimensions of either 1 by 51, 51 by 1, 3 by 17, or 17 by 3. These dimensions represent all possible rectangles with the specified area.
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.
There is no area because rectangles do not have equal sides.
Treat it as 3 rectangles.
No, it is not. I'll give you two examples of a rectangle with a perimeter of 1. The first rectangle has dimensions of 1/4x1/4. The area is 1/16. The second rectangle has dimensions of 3/8x1/8. The area is 3/64. You can clearly see that these two rectangles have the same perimeter, yet the area is different.
16 1x1 rectangles + 12 2x1 rectangles + 8 3x1 rectangles + 4 4x1 rectangles + 12 1x2 rectangles + 9 2x2 rectangles + 6 3x2 rectangles + 3 4x2 rectangles + 8 1x3 rectangles + 6 2x3 rectangles + 4 3x3 rectangles + 2 4x3 rectangles + 4 1x4 rectangles + 3 2x4 rectangles + 2 3x4 rectangles + 1 4x4 rectangle. A Grand Total of: 100 squares and rectangles. OR: A rectangle is formed by 2 horizontal lines and 2 vertical lines. There are 5 horizontal and 5 vertical lines so the number of rectangles is 5C2 * 5C2 = 10 * 10 = 100
3
3 X( Length x Width )
3*27 = 81 and 3+3+27+27 = a perimeter of 60 inches
Thee different rectangles with an area of 12 square units are 3 by 4, 2 by 6 and 1 by 12.
3
No, two rectangles with the same area do not necessarily have the same perimeter. For example, a rectangle with dimensions 2 x 6 has an area of 12 and a perimeter of 16, while a rectangle with dimensions 3 x 4 also has an area of 12 but a perimeter of 14. Thus, different combinations of length and width can yield the same area but different perimeters.