Let's denote the length of the rectangle as $L$ and the width as $W$.

Given:

1. The perimeter of the rectangle is 40 inches.
2. The width is 2 times the next consecutive integer after the length.

Step 1: Set up perimeter equation

The formula for the perimeter $P$ of a rectangle is: $P = 2L + 2W$ Since $P = 40$: $2L + 2W = 40$ Dividing both sides by 2: $L + W = 20$

Step 2: Define $W$ in terms of $L$

According to the problem, the width $W$ is "2 times the next consecutive integer." This means $W = 2(L + 1)$.

Step 3: Substitute $W$ into the perimeter equation

Substitute $W = 2(L + 1)$ into $L + W = 20$: $L + 2(L + 1) = 20$ Expanding and simplifying: $L + 2L + 2 = 20$ $3L = 18$ $L = 6$

Step 4: Find $W$

Now, substitute $L = 6$ back into $W = 2(L + 1)$: $W = 2(6 + 1)$ $W = 2 \times 7 = 14$

Solution

The length $L$ is 6 inches, and the width $W$ is 14 inches.

Verification

To confirm, check the perimeter: $2L + 2W = 2(6) + 2(14) = 12 + 28 = 40$ Thus, the solution is correct.

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Would you like further details or have any questions about this problem?

Related Questions

1. What would the length and width be if the perimeter were 60 inches instead of 40?
2. If the width were 3 times the next consecutive integer after the length, how would the dimensions change?
3. How would you find the dimensions if the length is given as twice the width?
4. What are the possible integer pairs for length and width if only the perimeter is known?
5. How does this solution approach change if both length and width are unknown?

Tip: For perimeter problems, setting up clear equations based on the definitions can simplify solving for unknowns.