Let the width of the rectangle be $w$ inches. Since the length is 6 inches longer than the width, the length can be expressed as $w + 6$ inches.

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

Substituting the given perimeter and the expressions for length and width: $56 = 2[(w + 6) + w]$

Simplifying this equation: $56 = 2(2w + 6)$

Divide both sides by 2: $28 = 2w + 6$

Subtract 6 from both sides: $22 = 2w$

Divide by 2: $w = 11$

So, the width of the rectangle is 11 inches. Now, substitute this back to find the length: $\text{Length} = w + 6 = 11 + 6 = 17$

Thus, the dimensions of the rectangle are:

• Width: 11 inches
• Length: 17 inches

Would you like a detailed explanation or have any questions?

Here are 5 related questions:

1. How would the dimensions change if the perimeter was 60 inches?
2. What is the area of this rectangle?
3. How would you express the perimeter if the width was $x$ and the length $x + 6$?
4. What would happen to the perimeter if both the length and width were doubled?
5. Can you find the diagonal length of this rectangle using the Pythagorean theorem?

Tip: Always break down word problems into smaller algebraic expressions to simplify solving.