To solve this problem, let's break it down using the provided information:

1. The perimeter $P$ of a rectangle is given by: $P = 2(l + w)$ where $l$ is the length and $w$ is the width. Here, $P = 18.6 \, \text{cm}$.

2. The area $A$ of a rectangle is given by: $A = l \cdot w$ Here, $A = 21.42 \, \text{cm}^2$.

From the perimeter formula: $l + w = \frac{P}{2} = \frac{18.6}{2} = 9.3 \, \text{cm}$

Now we have two equations:

1. $l + w = 9.3$
2. $l \cdot w = 21.42$

We can express $w$ in terms of $l$ from the first equation: $w = 9.3 - l$

Substitute $w = 9.3 - l$ into the area equation: $l \cdot (9.3 - l) = 21.42$ $9.3l - l^2 = 21.42$ $l^2 - 9.3l + 21.42 = 0$

This is a quadratic equation. Solve for $l$ using the quadratic formula: $l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -9.3$, and $c = 21.42$. Substituting these values: $l = \frac{-(-9.3) \pm \sqrt{(-9.3)^2 - 4(1)(21.42)}}{2(1)}$ $l = \frac{9.3 \pm \sqrt{86.49 - 85.68}}{2}$ $l = \frac{9.3 \pm \sqrt{0.81}}{2}$ $l = \frac{9.3 \pm 0.9}{2}$

This gives two solutions: $l = \frac{9.3 + 0.9}{2} = \frac{10.2}{2} = 5.1 \, \text{cm}$ $l = \frac{9.3 - 0.9}{2} = \frac{8.4}{2} = 4.2 \, \text{cm}$

Thus, the dimensions of the rectangle are: $l = 5.1 \, \text{cm}, \, w = 4.2 \, \text{cm} \quad \text{or} \quad l = 4.2 \, \text{cm}, \, w = 5.1 \, \text{cm}.$

Would you like further explanation or verification?

Questions to Explore Further:

1. How can the quadratic formula be derived step by step?
2. What are other methods to solve for the dimensions of a rectangle with given area and perimeter?
3. How do these results change if the rectangle is a square?
4. Can this approach be applied to non-rectangular shapes?
5. What are practical applications of finding dimensions with constraints?

Tip:

Always verify the solutions by substituting them back into the original equations to ensure accuracy.