The length of a rectangle is 6 inches longer than it is wide. If the perimeter is 56 inches, what are the dimensions of the rectangle?

The width, or shorter side is  inches

The length, or longer side is  inches

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.

The length of a rectangle is 6 inches longer than it is wide. If the perimeter is 56 inches, what are the dimensions of the rectangle?

Learn how to calculate the dimensions of a rectangle when the length is 6 inches longer than the width and the perimeter is 56 inches. This problem uses algebra and geometry concepts, suitable for students in Grades 6-8.

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