For a fixed perimeter, the area will always be the same, regardless of how you describe the rectangle.
The greatest area for a fixed perimeter will be when all the sides are equal or when the rectangle approaches the shape of a square.
A perimeter that will not change.
nope because if u have a square with a side length of 4 then the perimeter is 16 and the area is 16 and say if u have a rectangle with side lengths of 2 and 6 then the perimeter is 16 but the area is 12 so the answer is no
Anything greater than 24 cm. P = 2*L + 2*W, since the length is fixed at 24cm we have: 2*(24cm) +2*W > 96cm 2W > 96 - 48cm 2W > 48cm, or Width > 24cm. So any width greater than 24cm will make the perimeter >96cm. Obviously a width of 24cm wouldn't work since a polygon with all four sides equal would be a square and not a rectangle.
A fixed area of a rectangle is an area that doesn't change. An area is a quantity that measures the space of a shape.Consider this example:A = length x width, which is the formula of a rectangleIf A is fixed, then it depends on what values length and width are. Then, length is indirectly proportional to width in order for A to remain fixed.
You dont
The greatest area for a fixed perimeter will be when all the sides are equal or when the rectangle approaches the shape of a square.
The maximum area for a rectangle of fixed perimeter is that of the square that can be formed with the given perimeter. 136/4 = 34, so that the side of such a square will be 34 and its area 342 = 1156.
fixed perimeter is the perimeter being fixed
A perimeter that will not change.
P (perimeter of a rectangle) = 2*l+2*w 2*24+2*w > 100 2*w > 52 w > 26 Any width greater than 26cm will cause the perimeter to be greater than 100cm.
nope because if u have a square with a side length of 4 then the perimeter is 16 and the area is 16 and say if u have a rectangle with side lengths of 2 and 6 then the perimeter is 16 but the area is 12 so the answer is no
The clever person might realize that, though an infinite number of rectangles can be created with a fixed perimeter, there is a maximum and minimum area that any rectangle formed under the constriction can have. And we can work with that. The minimum area will be "near" zero. (With an area "at" zero, the rectangle will collapse and/or disappear.) The rectangle with "maximumized" area for a fixed perimeter will be a square. Its side (designated by "s") will be one fourth of the perimeter (designated by "p"). If s = p/4 and we use the formula for finding the area (As) of a square substituing our "p/4" for the side length "s" we will get the equation: As = (p/4)2 Our rectangle(s) will all have an area (Ar) within this range: Zero is less than Ar which is less than or equal to (p/4)2 Though we couldn't come up with a precise answer, we came up with the next best thing with the information supplied.
The rectangle with the most area for a given perimeter is the square.Build the room square, with sides of 94-ft. The area is 8,836 square ft.
Conventionally, length > width. With that assumption, if L is the length and W the width (both in metres) then 3.75 < L < 7.5 and W = 7.5 - L If L = 3.75 then W = L and the rectangle becomes a square.
For a fixed area, the perimeter is minimum for a circle, but has no maximum. Fractal figures (such as Koch snowflake) may have a finite area within an infinite perimeter.
L + W = P/2 = 49 so Length must be greater than 35 cm