There is not enough information here to solve this
PERIMETER = 2L + 2W = 27
Thereare two unknowns and only one equation, so there are an infinite number of possibilities. You need more info, like the area, to solve this
Do-it-in-your-head method: Length + width is half the perimeter which is 27 inches; 27 minus 6 is 21 and one-third of 21 is 7 which is the width. (Length = 2 x 7 + 6 = 20)
The length of the rectangle is 27 meters and the width is 19 meters.
If that's a rectangle, the perimeter is 24 inches and the area is 27 square inches.
243 = 27 x 9 so perimeter is 2 x 36 ie 72.
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
Do-it-in-your-head method: Length + width is half the perimeter which is 27 inches; 27 minus 6 is 21 and one-third of 21 is 7 which is the width. (Length = 2 x 7 + 6 = 20)
22+27+22+27
The length of the rectangle is 27 meters and the width is 19 meters.
The rectangle is 9 inches wide X 27 inches long. Clue: width is X; length is 3 times width (3X). Perimeter is width plus length times two (X + 3X times 2 = 8X). 8X = 72; X = 9. 3X = 27. Proof: 9 + 9 (2sides) +27 + 27 (2 lengths) = 72.
Let X represent the length, then X-4 represents the width. Perimeter of the rectangle is length plus width multiplied by two. Hence, 2(X + X - 4) = 100; 4X - 8 = 100; 4X = 108; X = 27. The length is 27 mm and the width is 27 - 4 = 23 mm. width = w length = w + 4 perimeter = w + w + w + 4 + w + 4 = 100 4w + 8 = 100 4w = 92 w = 23 width = 23 length = 27 area = 23 x 27 = 621
It is: 2*(9.5+4) = 27 yards
If that's a rectangle, the perimeter is 24 inches and the area is 27 square inches.
243 = 27 x 9 so perimeter is 2 x 36 ie 72.
Suppose width is x metres. Then length is x+5 metres. Now, perimeter = 2*(width + length) so 54 = 2*[x + (x+5)] = 2*(2x + 5) So 27 = 2x + 5 => 22 = 2x => x = 11 and so x+5 = 16. So the length and width are 16 and 11 metres, respectively.
Let the length of the rectangle be 2x+3 and the width be x So: 2(2x+3+x) = 78 => 4x+6+2x = 78 => 6x = 78-6 => 6x = 72 => x = 12 Therefore: length = 27 cm Check for perimeter: 2(27+12) = 78 cm
Area = (length) x (width)72 = (24) x (width)Width = 72/24 = 3Perimeter = 2 x (length + width) = 2 x (24 + 3) = 2 x (27) = 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