Suppose that the width of a certain rectangle is 1 inch more than one-fourth of its length. The perimeter of the rectangle is 32 inches. Find the length and width of the rectangle.
If the length is x, the width is 1 + x/4
So, 2(x + 1 + x/4) = 32
Now just solve for x and get the width.
12
31.25
To solve this problem, we need to use the given information to set up equations and then solve them.
Let's start by assigning variables to the unknowns. Let's say:
- L represents the length of the rectangle
- W represents the width of the rectangle
We are given two pieces of information:
1. The width of the rectangle is 1 inch more than one-fourth of its length:
W = (1/4)L + 1
2. The perimeter of the rectangle is 32 inches:
2L + 2W = 32
Now, we can use these two equations to solve for L and W.
Let's substitute the value of W from the first equation into the second equation:
2L + 2((1/4)L + 1) = 32
We can simplify this equation:
2L + (1/2)L + 2 = 32
(5/2)L + 2 = 32
Next, let's solve for L by isolating it:
(5/2)L = 32 - 2
(5/2)L = 30
Divide both sides of the equation by 5/2 to solve for L:
L = (30) / (5/2)
L = 30 * (2/5)
L = 60/5
L = 12
Now that we have the value of L, we can substitute it back into the first equation to find W:
W = (1/4)L + 1
W = (1/4)(12) + 1
W = 3 + 1
W = 4
So, the length of the rectangle is 12 inches and the width of the rectangle is 4 inches.