The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is 48 meters. Find the length and width of the rectangle.

Let the length and width of the rectangle be \( x \) and \( x+2 \), respectively, where both \( x \) and \( x+2 \) are consecutive odd integers.

The perimeter of a rectangle is given by the formula:

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]

Substituting the given values:

\[
48 = 2 \times (x + (x+2))
\]

Simplify the equation:

\[
48 = 2 \times (2x + 2)
\]

Divide both sides by 2:

\[
24 = 2x + 2
\]

Subtract 2 from both sides:

\[
22 = 2x
\]

Divide by 2 to solve for \( x \):

\[
x = 11
\]

Thus, the width is \( x = 11 \) meters, and the length is \( x+2 = 13 \) meters.

So, the length of the rectangle is **13 meters**, and the width is **11 meters**.

Would you like more details or have any questions?

Here are five related questions for further practice:
1. What would the area of the rectangle be with these dimensions?
2. If the perimeter was 60 meters, what would the dimensions of the rectangle be?
3. What if the length and width were consecutive even integers instead?
4. How would the dimensions change if the perimeter was doubled?
5. What would the diagonal of the rectangle be with the found dimensions?

**Tip:** For problems involving perimeter, remember that the perimeter of a rectangle is twice the sum of its length and width. This formula can be rearranged to solve for either dimension when given the perimeter.

Discover how to find the length and width of a rectangle with a perimeter of 48 meters. Step-by-step solution provided, including the use of the perimeter formula. Suitable for students in Grades 6-8.

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