A rectangle with a length of 10 and a width of 24
You could consider the cross as two intersecting rectangles. Calculate the area of both rectangles and the area of the intersection (overlap). Then area of cross = sum of the areas of the rectangles minus the area of the overlap.
An L-shaped area can be divided into two rectangles. The total area is the sum of the areas of the two rectangles.
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.
When rectangles are inscribed, they lie entirely inside the area you're calculating. They never cross over the curve that bounds the area. Circumscribed rectangles cross over the curve and lie partially outside of the area. Circumscribed rectangles always yield a larger area than inscribed rectangles.
thare is only 1 differint rectangles
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.
they dont
Some rectangles don't have equal sides.
330سم
divide that angle iron in to 2 rectangles and solve it according to the farmulas of rectangles
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.
Area of a rectangle in square units = length*width