If the shape measuring 8 centimeters by 5 centimeters is a rectangle, the perimeter of the rectangle is 25 centimeters (2w + 2l)
To find the width of the rectangle, we need to subtract twice the length from the perimeter. Given that the length is 25 centimeters and the perimeter is 84 centimeters, twice the length would be 50 centimeters. Subtracting 50 from 84 gives us a width of 34 centimeters for the rectangle.
The area of rectangle is : 100.0
P = 50 so L + W = 25; L = W + 7 so L = 16 cm and W = 9 cm.
The perimeter of a rectangle is equal to twice the length plus twice the width, so P = 2l + 2w P = (2x25) + (2x13) P = 50 + 26 P = 76 centimeters
58 cm
The perimeter of a rectangle is P = 2(l + w) where: l = length w = width Let: l = 15 cm w = 10 cm Then: P = 2(15 cm + 10 cm) = 2(25 cm) = 50 cm So the perimeter of a rectangle is 50 cm.
If the shape measuring 8 centimeters by 5 centimeters is a rectangle, the perimeter of the rectangle is 25 centimeters (2w + 2l)
To find the width of the rectangle, we need to subtract twice the length from the perimeter. Given that the length is 25 centimeters and the perimeter is 84 centimeters, twice the length would be 50 centimeters. Subtracting 50 from 84 gives us a width of 34 centimeters for the rectangle.
The maximum length is 25 cm.
The area of rectangle is : 100.0
P = 50 so L + W = 25; L = W + 7 so L = 16 cm and W = 9 cm.
The perimeter of a rectangle is equal to twice the length plus twice the width, so P = 2l + 2w P = (2x25) + (2x13) P = 50 + 26 P = 76 centimeters
P=(L+W)x2 Where P = Perimeter, L=Length and W=Width.
80 feet 375/25 = 15so 25 + 25 + 15 + 15 = 80
74 cm
Perimeter = 25+36+25+36 = 122 units of measurement Use Pythagoras' theorem to find the other side of the rectangle