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.
Rectangles are related to the distributive property because you can divide a rectangle into smaller rectangles. The sum of the areas of the smaller rectangles will equal the area of the larger rectangle.
Yes. Say there are two rectangles, both with perimeter of 20. One of the rectangles is a 2 by 8 rectangle. The area of this rectangle is 2 x 8 which is 16. The other rectangle is a 4 by 6 rectangle. It has an area of 4 x 6 which is 24.
i am doing my homework right now and I am stuck on that problem
because it was estimation, the lengths were different and the rectangles are not the same
What is an "oddly shaped rectangle"? Rectangles have four sides, with two pairs of sides that are equal in length and parallel to each other, and four right angles. Anything that fits this definition is a rectangle, period. There's nothing "odd" about any of them. But the area of any rectangle can be found by multiplying the lengths of any two adjacent sides.
wht u hve to do is to cut the shape into rectangles and then times the length and width together on each rectangle. then add up all the rectangles areas and add them alll up. ta da
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
For the first rectangle, (L x W) = (5 x W) = 15, so W = 3 cm.The second rectangle is included in the question just to confuse you.
There is no relationship between the perimeter and area of a rectangle. Knowing the perimeter, it's not possible to find the area. If you pick a number for the perimeter, there are an infinite number of rectangles with different areas that all have that perimeter. Knowing the area, it's not possible to find the perimeter. If you pick a number for the area, there are an infinite number of rectangles with different perimeters that all have that area.
There are only two distinct areas of a rectangle - the inside and the outside. Anything else would require the rectangle to be partitioned.
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.
The areas are proportional to the square of the scale factor.