Question

1. the length of a rectangle is eight more than twice its width. the perimeter is 88 feet. find the dimensions of the rectangle.

variable:
equation:
solution:

Explanation:

Step1: Define the variable

Let \( w \) be the width of the rectangle (in feet). Then the length \( l \) is \( 2w + 8 \) (since length is eight more than twice the width).

Step2: Recall the perimeter formula for a rectangle

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). We know \( P = 88 \), so substitute \( l = 2w + 8 \) into the formula:
\[
88 = 2((2w + 8) + w)
\]

Step3: Simplify the equation

First, simplify inside the parentheses: \( (2w + 8) + w = 3w + 8 \). Then the equation becomes:
\[
88 = 2(3w + 8)
\]
Divide both sides by 2:
\[
44 = 3w + 8
\]

Step4: Solve for \( w \)

Subtract 8 from both sides:
\[
44 - 8 = 3w
\]
\[
36 = 3w
\]
Divide both sides by 3:
\[
w = 12
\]

Step5: Find the length

Now that we have \( w = 12 \), substitute into the length formula \( l = 2w + 8 \):
\[
l = 2(12) + 8 = 24 + 8 = 32
\]

Answer:

• Variable: Let \( w \) be the width (in feet), length \( l = 2w + 8 \)
• Equation: \( 88 = 2((2w + 8) + w) \)
• Solution: Width \( w = 12 \) feet, Length \( l = 32 \) feet