0.000...1cm length on two sides. 19.999...9cm length on the other two.
If we restrict ourselves to whole numbers, then 1 x 14 will have the least possible area.
Yes.
No, it is not possible for a rectangle to have a perimeter of 46 and an area of 42 simultaneously. For a rectangle, the perimeter ( P ) is given by ( P = 2(l + w) ), and the area ( A ) is ( A = l \times w ), where ( l ) is the length and ( w ) is the width. Solving these equations shows that the dimensions needed for these values are inconsistent, meaning no such rectangle exists.
42 square units.
You might want to investigate the rectangle that measures [ 6 by 8 ].
52 ft
52 (13•4)
If we restrict ourselves to whole numbers, then 1 x 14 will have the least possible area.
For a fixed perimeter, the area will always be the same, regardless of how you describe the rectangle.
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.
Assuming no fractional dimensions, least possible area would be a rectangle measuring 1cm x 9cm. Area increases to a maximum of 25 sq cm when shape is square, ie 5cm x 5cm.
47
Yes.
NO, because if you did it would be a square
(p/4)2, where p is the perimeter.
No, it is not possible for a rectangle to have a perimeter of 46 and an area of 42 simultaneously. For a rectangle, the perimeter ( P ) is given by ( P = 2(l + w) ), and the area ( A ) is ( A = l \times w ), where ( l ) is the length and ( w ) is the width. Solving these equations shows that the dimensions needed for these values are inconsistent, meaning no such rectangle exists.
You might want to investigate the rectangle that measures [ 6 by 8 ].