Let w be the width, then 2(7 + w) = 42
or 14 + 2w = 42
or 2w = 42 - 14
or 2w = 28
w = 14
L + W = P/2 = 42 ie L = 42 - W; L = 4W + 2 so 42 - W = 4W + 2 ie 5W = 40 so width is 8 cm making length 34 cm
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
Largest = 86, Smallest 26
42/7 = 6 units of length.
Write a a series of equations as follows p=2l+2w (formula for perimeter) a=l*w (Formula for area) l=3w (fact given in the question) a=147 (fact given in hte question) We don't currently have enough information to solve the perimeter formula for this rectangle, so we begin with the area formula to get more information. By substituteing 147 for the variable a and 3w for the variable l, we arrive at this equation: 147=w*3w With only one variable, we can solve this equation by isolating w on one side as follows 147=3w2(Multiply w by w to combine like terms) 147/3=3w2/3 (divide both sides by 3) 49=w2 sqrt49=sqrtw2(square root of both sides) 7=w We now know that the width of the rectangle is 7. We also know that the length is three times that, which is 21. We can substitute these values into the perimeter eequation as follows: p=(2*7)+(2*21) p=14+42 p=56 The perimeter of the rectangle is 56
A rectangle has two dimensions, length and width. You haven't said what the width is. The perimeter is the distance around the rectangle. Imagine going round it. You would go 42 inches, then a width, then another 42 inches, then another width. In general, the perimeter of a rectangle is twice its length plus twice its width.
The perimeter is 42 metres.
2 (x + 2x) = 42 x + 2x = 21 3x = 21 Therefore, x = 7 The width of the rectangle is 7 inches. The length of the rectangle is 14 inches.
The perimeter of a rectangle is 42. Meters. The length of the rectangle is threemeter less than twice the width.Mar
Perimeter = width*2 + length *2 112 = 14 * 2 + length * 2 56 = 14 + length length = 56 - 14= 42
When the linear dimensions of a plane figure are quadrupled, its perimeter is quadrupled, and its area is multiplied by 42 = 16 .
width = x and length = x+3 the perimeter = x + x +( x+3) +(x + 3) = 48, solve for x 4x + 6 = 48m 4x = 42 x = 10.5m (width) x+3 = 13.5m (length)
The perimeter of a 9.5 x 11.5 inch rectangle is 42 inches. To find the perimeter of a rectangle we use the formula Length * 2 + Width * 2 That gives us the following expression. 9.5 * 2 + 11.5 * 2 19 + 23 42
None. 42 yards is a length with no width. Therefore it cannot have a perimeter.
Rectangular perimeter = 2(Length + Width) Width = a Length = a+5 Then, 84 = 2(a+a+5) 42 = 2a+5 (42-5)/2 = a = 18.5m = Width Length = a+5 = 23.5m
42
There is insufficient information to answer this question. A rectangle with length L and width W has perimeter 2(L+W) So 2(L+W) = 42 which means L+W=21 So, given that the length should be greater than the width, any value of L such that 10.5 < L < 21 will meet the requirements. Obviously, for each value of L there will be a different value for W.