52 ft
52 (13•4)
For a fixed perimeter, the area will always be the same, regardless of how you describe the rectangle.
47
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.
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.
52 (13•4)
If we restrict ourselves to whole numbers, then 1 x 14 will have the least possible area.
what is the value of x so that the perimeter of the rectangle shown is at least 92 centimeters
7 x 19 cm
square
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
For a fixed perimeter, the area will always be the same, regardless of how you describe the rectangle.
47
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.
14 cm^3 (14+14+1+1=30 and 14*1=14cm^3)
0.000...1cm length on two sides. 19.999...9cm length on the other two.
16 and 12