P = perimeter, L = length, W = width
P = 2L + 2W = 2(8) + 2W = 2W + 16 = 46
2W = 30
W = 15
So as long as x is 15 inches or more, it will have a perimeter at least 46.
For a fixed perimeter, the area will always be the same, regardless of how you describe the rectangle.
52 ft
47
52 (13•4)
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.
what is the value of x so that the perimeter of the rectangle shown is at least 92 centimeters
square
For a fixed perimeter, the area will always be the same, regardless of how you describe the rectangle.
52 ft
47
52 (13•4)
If you restrict yourself to integers, the answer is 24 feet.
rectangle with 10" length and 2x-4 width 60" area
The dimensions of the rectangle will then work out as 14 cm by 10 cm because the perimeter is 14+10+14+10 = 48 cm
Whether that's the length, the width, the diagonal, or the perimeter, you don't have enough information. To get the area of a rectangle, you need at least two measurements.
The least perimeter is attained when the rectangle is, in fact a square. A square with an area of 32 square feet will have sides of sqrt(32) = 4*sqrt(2) ft. So the perimeter of the square will by 4*4*sqrt(2) = 16*sqrt(2) = 22.63 feet, approx.
To calculate the perimeter of a rectangle you need to add up the length of all four sides. Length + length + width + width = perimiter2L + 2W = PAssume L = 8, and P = 46, determine W.2 * 8 + 2W = 4616 + 2W = 46subtract 16 from both sides2W = 46 - 162W = 30divide both sides by 2W = 15.The other side is 15 inches long.