Question

1. the length of a rectangle is five less than twice its width. if the perimeter of the rectangle is 56 inches, find its dimensions.

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) inches. Then the length \( l \) is \( 2w - 5 \) inches (since length is five less than twice the width).

Step2: Recall perimeter formula

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). We know \( P = 56 \), so substitute \( l = 2w - 5 \) into the formula:
\[
56 = 2((2w - 5) + w)
\]

Step3: Simplify the equation

First, simplify inside the parentheses: \( (2w - 5) + w = 3w - 5 \). Then the equation becomes:
\[
56 = 2(3w - 5)
\]
Divide both sides by 2:
\[
28 = 3w - 5
\]

Step4: Solve for \( w \)

Add 5 to both sides:
\[
3w = 28 + 5 = 33
\]
Divide both sides by 3:
\[
w = \frac{33}{3} = 11
\]

Step5: Find the length

Substitute \( w = 11 \) into the expression for length: \( l = 2(11) - 5 = 22 - 5 = 17 \).

Answer:

The width of the rectangle is 11 inches and the length is 17 inches.