Let's denote the length of the rectangle by $l$ and the width by $w$.

Step 1: Expressing the perimeter

The formula for the perimeter $P$ of a rectangle is: $P = 2(l + w)$ We are given that the perimeter is 116 centimeters, so: $2(l + w) = 116$ Dividing both sides by 2: l + w = 58 \tag{1}

Step 2: Expressing the width in terms of the length

We are told that the length is an even integer, and the width is 5 times the next consecutive even integer. If the length is $l$, the next consecutive even integer is $l + 2$. Thus, the width $w$ can be written as: w = 5(l + 2) \tag{2}

Step 3: Substitute the width into the perimeter equation

Now substitute equation (2) into equation (1): $l + 5(l + 2) = 58$ Simplify the equation: $l + 5l + 10 = 58$ $6l + 10 = 58$ Subtract 10 from both sides: $6l = 48$ Divide by 6: $l = 8$

Step 4: Find the width

Now that we have $l = 8$, substitute it back into equation (2) to find the width: $w = 5(8 + 2) = 5 \times 10 = 50$

Step 5: Conclusion

The length of the rectangle is $8$ centimeters, and the width is $50$ centimeters.

Would you like more details on this solution?

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Here are some follow-up questions:

1. How would the formula change if the width were expressed as a function of the previous even integer?
2. What would the dimensions be if the perimeter were 120 cm instead of 116 cm?
3. Can you find the area of the rectangle with the given dimensions?
4. How does the perimeter formula change if we consider shapes other than rectangles?
5. What would the result be if both the length and width were odd integers?

Tip: Always check if you can substitute values directly into the equation before solving for unknowns—it simplifies the process.