Question

question 2
2 pts 1 details
the length of a rectangle is eight inches more than nine times the width. the perimeter is 56 inches. find the length and width.
the length is
inches, and the width is
inches.
check answer

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) inches. Then the length \( l \) is \( 9w + 8 \) inches (since length is eight inches more than nine times the width).

Step2: Use perimeter formula

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). We know \( P = 56 \), so substitute \( l = 9w + 8 \) into the formula:
\[
56 = 2((9w + 8) + w)
\]

Step3: Simplify the equation

First, simplify inside the parentheses:
\[
56 = 2(10w + 8)
\]
Then divide both sides by 2:
\[
28 = 10w + 8
\]

Step4: Solve for \( w \)

Subtract 8 from both sides:
\[
20 = 10w
\]
Divide both sides by 10:
\[
w = 2
\]

Step5: Find the length

Substitute \( w = 2 \) into \( l = 9w + 8 \):
\[
l = 9(2) + 8 = 18 + 8 = 26
\]

Answer:

The length is \( 26 \) inches, and the width is \( 2 \) inches.