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.
Divide the room into two rectangles. Then calculate the area of each rectangle as length x width.
area = 144 square units perimeter = 48 units
Area of a rectangle in square units = length*width
Rectangles and squares.
Thee different rectangles with an area of 12 square units are 3 by 4, 2 by 6 and 1 by 12.
Treat it as two rectangles - calculate the area of each rectangle - then simply add the two figures together.
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.
Then its area will only be a quater of once it was.
Right triangle square rectangles