Perimeter = 12+12+14+14 = 54 uits of measurement
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.
the formula for the perimeter of a rectangle is p = 2W + 2L, where L is the length and W is the width, so your first equation is 126 = 2W + 2L"twice as long as it is wide" means that the length L is 2 times the width W, so your second equation is L = 2WIn order to solve for the dimensions, you can substitute L for 2W in the first equation to get:126 = L + 2L126 = 3LL = 126/3L = 42and because L = 2W:42 = 2WW = 42/2W = 21
width:10,length;17 17*2+10*2=54
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