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.
In a 4 by 4 grid, there are 16 squares (1x1 squares), 9 rectangles that are 2x1, 6 rectangles that are 3x1, 4 rectangles that are 2x2, and 1 rectangle that is 4x4. Therefore, in total, there are 16 squares and 20 rectangles in a 4 by 4 grid.
16
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
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.
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.
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.
16.
An L-shaped area can be divided into two rectangles. The total area is the sum of the areas of the two rectangles.
* It is unclear if the question is asking about two rectangles, each with a perimeter of 16, or two rectangles whose perimeters sum to 16. This answer assumes the former.Other than the 4x4 square, which coincidentally has both a perimeter and area of 16, some examples would be:1 x 7 rectangle : perimeter 16 in. , area 7 sq. in2 x 6 rectangle : perimeter 16 in., area 12 sq. in3 x 5 rectangle: perimeter 16 in., area 15 sq. inYou can calculate that for a given perimeter, the largest area is found in the square with a side measurement of P/4, i.e. the length and the width are the same.
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.
16
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
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.