To find the length and width of a rectangle given the area and perimeter, we can use two simple formulas:

1. Area (A): $A = \text{length} \times \text{width}$
2. Perimeter (P): $P = 2(\text{length} + \text{width})$

We know:

• $A = 24 \, \text{cm}^2$
• $P = 24 \, \text{cm}$

Let's call the length $l$ and the width $w$.

From the perimeter formula, we can express one equation: $2(l + w) = 24 \quad \Rightarrow \quad l + w = 12$

From the area formula: $l \times w = 24$

Now, we have the system of equations:

1. $l + w = 12$
2. $l \times w = 24$

We can solve for one variable, say $w$, from the first equation: $w = 12 - l$

Substitute this into the second equation: $l \times (12 - l) = 24$ Expanding this: $12l - l^2 = 24$ Simplifying: $l^2 - 12l + 24 = 0$

This is a quadratic equation, but since you asked for no quadratic solving, you can use trial and error or estimation to find that the length $l$ is approximately 6 cm and the width $w$ is approximately 4 cm.

Would you like more details on this process or have any questions?

Here are 5 related questions:

1. How would the dimensions change if the perimeter were increased?
2. What if the area were doubled, but the perimeter stayed the same?
3. How do you solve for dimensions if the shape is a square?
4. Can you find the diagonal length of the rectangle using these dimensions?
5. How would this problem change if you were working with a triangle?

Tip: For problems like this, even if quadratic equations arise, simple estimation or factoring can help if you avoid formal quadratic solutions!