The length of a rectangle is 6 inches more than twice its width. If the perimeter of the rectangle is 54 inches, find the width of the rectangle.
Let's start by assigning variables to the unknowns.
Let:
- L be the length of the rectangle,
- W be the width of the rectangle.
According to the problem, we know that:
1. The length of the rectangle is 6 inches more than twice its width, so we can write an equation: L = 2W + 6.
2. The perimeter of the rectangle is 54 inches, so we can also write another equation: 2L + 2W = 54.
Now we can solve the system of equations to find the value of W.
Substituting the value of L from equation 1 into equation 2, we get:
2(2W + 6) + 2W = 54
4W + 12 + 2W = 54
6W + 12 = 54
6W = 54 - 12
6W = 42
Dividing both sides by 6, we find:
W = 42 / 6
W = 7
Therefore, the width of the rectangle is 7 inches.
To find the width of the rectangle, we can set up an equation based on the given information.
Let's assume the width of the rectangle is "x" inches.
According to the given information, the length of the rectangle is "6 inches more than twice its width". So, the length would be "2x + 6" inches.
The formula for the perimeter of a rectangle is: P = 2(length + width).
Plugging in the values from the given information, we can create the equation: 54 = 2(2x + 6 + x).
Now we can solve the equation to find the value of "x".
54 = 2(3x + 6)
54 = 6x + 12
42 = 6x
x = 42/6
x = 7
Therefore, the width of the rectangle is 7 inches.
l =24
w= 3
From the information above, you should be able to develop the folowing two equations.
L = 2W + 6
2L + 2W = 54
Substitute the value of L from the first equation into the second and solve for W.
I hope this helps. Thanks for asking.