If you assume that the width is not greater than the length, it can have any value if the range (0, 9] yards.
If you assume that the width is not greater than the length, it can have any value if the range (0, 9] yards.
If you assume that the width is not greater than the length, it can have any value if the range (0, 9] yards.
If you assume that the width is not greater than the length, it can have any value if the range (0, 9] yards.
Perimeter is the length round the outside. A basketball court is a rectangle, so: perimeter = 2 x (length + width) = 2 x (30 yd + 40 yd) = 2 x 70 yd = 140 yd
To calculate the perimeter of a rectangle simply add the length and width then double your answer. In this case P = (9 + 11) x 2 = 20 x 2 = 40 yds
Assuming that you have a rectangle shaped area, 20 yards long by 15 yards wide, and you want to completely enclose the area with the fencing, then you have a problem where you need to find the perimeter of a rectangle. Perimeter = 2 * Length + 2 * Width Perimeter = 2 * (20 yd) + 2 * (15 yd) = 40 yd + 30 yd = 70 yards
The diagonal of a rectangle with the length of 30 yd and the width of 30 yd is approximately 42.43 yd
3 ft = 1 yd 6 ft = 2 yd 9 ft = 3 yd . . 36 ft = 12 yd
49
Well, L=6 and W=18If you do the math you get (6*2)+(18*2)=12+36=48Your final answer 48 yd. The perimeter is 48 yd.
Perimeter is the length round the outside. A basketball court is a rectangle, so: perimeter = 2 x (length + width) = 2 x (30 yd + 40 yd) = 2 x 70 yd = 140 yd
Perimeter = 8+8+10+10 = 36 yards
Perimeter = 2*(L+W) = 2*(3+7) = 2*10 = 20 yards. By the way, conventionally, the length is the bigger dimension.
To calculate the perimeter of a rectangle simply add the length and width then double your answer. In this case P = (9 + 11) x 2 = 20 x 2 = 40 yds
Assuming that you have a rectangle shaped area, 20 yards long by 15 yards wide, and you want to completely enclose the area with the fencing, then you have a problem where you need to find the perimeter of a rectangle. Perimeter = 2 * Length + 2 * Width Perimeter = 2 * (20 yd) + 2 * (15 yd) = 40 yd + 30 yd = 70 yards
44 sq yd = 396 sq ft, so width = 396/36 = 11 ft
The ratio of their perimeters is also 45/35 = 9/7. The ratio of their areas is (9/7)2 = 81/63
1 ft = (1/3) yd 1 in = (1/12) ft = (1/36) yd So 1yd 7ft 14in = 1 + (7/3) + (14/36) yd = (36/36) + (84/36) + (14/36) yd = (134/36) yd. (134/36) = 3.7222 yd (rounded)
The diagonal of a rectangle with the length of 30 yd and the width of 30 yd is approximately 42.43 yd
It depends on the width of the carpet.... I've never come across a carpet that isn't over 12" wide, so on that basis, yes. A sq yd is 36" by 36" so should have plenty spare with a sq yd.