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
To find the least perimeter of a rectangle with a fixed area of 32 square feet, we can use the relationship between area and perimeter. For a rectangle, the area ( A = l \times w ) (length times width) and the perimeter ( P = 2(l + w) ). To minimize the perimeter while keeping the area constant, the rectangle should be a square. The side length of a square with an area of 32 ft² is ( \sqrt{32} ), which is approximately 5.66 ft. Thus, the least perimeter is ( 4 \times \sqrt{32} ), which is approximately 22.63 ft.
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.