Anything that equals 27.
14 x 13
15 x 12
16 x 11
17 x 10
18 x 9
19 x 8
20 x 7
To find the possible dimensions of a rectangle with an area of 54, we can consider pairs of factors of 54. The factor pairs are (1, 54), (2, 27), (3, 18), (6, 9). Therefore, the possible dimensions of the rectangle are: 1 by 54, 2 by 27, 3 by 18, and 6 by 9.
Perimeter = 12+12+14+14 = 54 uits of measurement
Let the length be ( l ) and the width be ( w ). The perimeter is given by the formula ( P = 2(l + w) ), so ( l + w = 15 ). The area is given by ( A = l \times w = 54 ). Solving these equations, we find the dimensions are ( l = 18 ) inches and ( w = 3 ) inches.
Let x = width 3x - 6 = length Perimeter is 2(l + w) 2(x + 3x - 6) 2x + 6x - 12 8x - 12 = 148 8x = 160 x = 20 3x - 6 = 54 Rectangle is 20 x 54 Check it. 2(20 + 54) 2 x 74 = 148 It checks.
Perimeter = 2*(Length + Width) 54 = 2*(L + 9) = 2L + 18 So 2L = 54 - 18 = 36 and therefore L = 18 inches.
54
54 Meters
30
Perimeter = 12+12+14+14 = 54 uits of measurement
width=9 length=18
Let x = width 3x - 6 = length Perimeter is 2(l + w) 2(x + 3x - 6) 2x + 6x - 12 8x - 12 = 148 8x = 160 x = 20 3x - 6 = 54 Rectangle is 20 x 54 Check it. 2(20 + 54) 2 x 74 = 148 It checks.
This has several values.The perimeter is twice the sum of any two numbers whose product is 54.
Area: 54 square cm Perimeter: 30 cm
Perimeter = 2*(Length + Width) 54 = 2*(L + 9) = 2L + 18 So 2L = 54 - 18 = 36 and therefore L = 18 inches.
Let the length be x-12 and the width be x: 2(length+width) = perimeter 2(x-12+x) = 240 2x-24+2x = 240 4x = 240+24 4x = 264 x = 66 Therefore: length = 54 feet and width = 66 feet
No.Rectangle 5 x 10. Area = 50. Perimeter = 30.Rectangle 2 x 25. Area = 50. Perimeter = 54.
The perimeter of a rectangle = 2L + 2W So, from your problem we know that L = W + 27 Use this equation for the length and substitute in the perimeter equation, we get: 2(W+27) + 2W = 96 Now we rearrange and solve for W 2W + 54 + 2W = 96cm 4W + 54 -54 = 96 -54 4W = 42 W = 10.5cm, and using this value L = 10.5 + 27 = 37.5cm