a rectangle has a perimeter of 38ft. length is 1ft less than 3 times the width. find the dimensions.?

Answer 1

2L + 2W = 38
3W - 1 = L

--> 6W -2 + 2W = 38

--> W = 5
--> L = 14

Answer 2

Let W = width of rectangle. Then, since the length of 1 ft less than 3 times the width,

L = 3W - 1

Recall that the perimeter of a rectangle is calculated by the following formula:

P = 2L + 2W

But we know L = 3W - 1, so we effectively replace it. We also know P = 38. Therefore,

38 = 2(3W - 1) + 2W

Expanding the brackets, we get

38 = 6W - 2 + 2W

Grouping like terms, and moving the -2 to the left hand side, we get:

40 = 8W

Therefore, W = 5.

Now that we've solved for W, we can actually get L.

L = 3W - 1 = 3(5) - 1 = 14

Therefore, L = 14 and W = 5, and the dimensions of the rectangle are 14 x 5

Answer 3

The dimensions of the rectangle is 5 x 14.

Width = x
Length = 3x-1

x + x + 3x -1 + 3x -1 = 38

Solve for x

x= 5

Answer 4

Let Width is b
So Length is 3b-1
Now 2(3b-1+b)=38
or, 8b=38+2=40
b=40/8=5 ft. (width)
Length=3b-1=3*5-1=14ft Ans.

Answer 5

W=5 ft
L=14 ft