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.
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
they dont
divide that angle iron in to 2 rectangles and solve it according to the farmulas of rectangles
A = 24cm2
The width is 3cm.
Restricting yourself to whole numbers, it's 1 x 24 2 x 12 3 x 8 4 x 6
26 cm
24cm2
square circle rectangle diamond
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 smallest perimeter is 4*sqrt(24) = approx 19.6 cm There is no largest perimeter.
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