To find the dimensions of a rectangle where the length is three times its width and the perimeter is 80 cm, you can set up the equation based on the formula for the perimeter of a rectangle: $P = 2L + 2W$.

Let $W$ be the width of the rectangle. Given that the length $L$ is three times the width, we have $L = 3W$.

Using the perimeter formula: $P = 2L + 2W = 2(3W) + 2W = 6W + 2W = 8W$

Given $P = 80$ cm: $8W = 80$

Solving for $W$: $W = \frac{80}{8} = 10 \text{ cm}$

Now, calculate $L$: $L = 3W = 3 \times 10 = 30 \text{ cm}$

Thus, the width of the rectangle is 10 cm and the length is 30 cm.

Would you like further details or help with another question? Here are some related topics to explore:

1. How would you find the area of the rectangle with these dimensions?
2. Can you explain why the relationship between perimeter and area is important?
3. How would the solution change if the perimeter were different?
4. What is the significance of the ratio between length and width in various real-world applications?
5. How can you verify if a rectangle is a square based on its perimeter and length-width ratio?

Tip: When solving problems with multiple unknowns related by a constant, such as the length being three times the width, it's often helpful to express all unknowns in terms of one variable to simplify the equations.