Let's denote the width of the rectangle as $w$ cm. According to the problem, the length of the rectangle is 4 cm longer than the width, so the length can be written as $w + 4$ cm.

The formula for the perimeter $P$ of a rectangle is given by: $P = 2 \times (\text{length} + \text{width})$

Substituting the given perimeter and the expressions for length and width into this formula: $28 = 2 \times ((w + 4) + w)$

Simplify the equation: $28 = 2 \times (2w + 4)$

Divide both sides by 2: $14 = 2w + 4$

Now, subtract 4 from both sides: $10 = 2w$

Divide both sides by 2: $w = 5$

Now that we have the width, we can find the length: $\text{length} = w + 4 = 5 + 4 = 9 \, \text{cm}$

The area $A$ of the rectangle is given by: $A = \text{length} \times \text{width} = 9 \times 5 = 45 \, \text{cm}^2$

Thus, the area of the rectangle is $45 \, \text{cm}^2$.

Would you like more details, or do you have any questions?

Here are 5 related questions to explore:

1. How would the area change if the perimeter were 40 cm?
2. What would happen to the area if the length were 6 cm longer than the width?
3. Can you find the length and width if the area is known and the perimeter is 28 cm?
4. How would the dimensions of the rectangle change if both the length and width were doubled?
5. Can you find the diagonal of this rectangle?

Tip: For a rectangle, the diagonal can be calculated using the Pythagorean theorem.