The length of a rectangle is five inches more than twice the width. The perimeter is 34 inches. Find the length and width.

Let the width of the rectangle be denoted by \( w \), and the length be denoted by \( l \).

### Step 1: Express the relationship between length and width
We are given that the length is five inches more than twice the width, so:
\[
l = 2w + 5
\]

### Step 2: Express the perimeter formula
The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2l + 2w
\]
We are told that the perimeter is 34 inches, so:
\[
2l + 2w = 34
\]

### Step 3: Substitute the expression for \( l \)
From step 1, we have \( l = 2w + 5 \). Substitute this into the perimeter equation:
\[
2(2w + 5) + 2w = 34
\]

### Step 4: Simplify and solve for \( w \)
Expand the equation:
\[
4w + 10 + 2w = 34
\]
Combine like terms:
\[
6w + 10 = 34
\]
Subtract 10 from both sides:
\[
6w = 24
\]
Divide both sides by 6:
\[
w = 4
\]

### Step 5: Solve for \( l \)
Now that we know \( w = 4 \), substitute this back into the expression for \( l \):
\[
l = 2(4) + 5 = 8 + 5 = 13
\]

### Final Answer:
- Width = 4 inches
- Length = 13 inches

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

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Here are 5 related questions you might want to explore:
1. What would the dimensions be if the perimeter were 40 inches?
2. How would the problem change if the length was three times the width instead?
3. What is the area of the rectangle with these dimensions?
4. Can you find the diagonal of the rectangle using the Pythagorean theorem?
5. What would happen if the perimeter were halved?

**Tip:** For any rectangle, once you know the length and width, you can easily compute its area by multiplying the two values together.

Learn how to find the dimensions of a rectangle where the length is five inches more than twice the width, and the perimeter is 34 inches. This problem involves key algebraic and geometric concepts and is suitable for students in Grades 6-8.

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