Question

the length of a rectangle is four times its width. if the area of the rectangle is 144 ft², find its perimeter.

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) (in feet). Then the length \( l \) is \( 4w \) (since length is four times the width).

Step2: Use area formula

The area \( A \) of a rectangle is given by \( A = l \times w \). We know \( A = 144 \, \text{ft}^2 \), so substitute \( l = 4w \) into the area formula:
\[
144 = 4w \times w
\]
\[
144 = 4w^2
\]

Step3: Solve for width

Divide both sides by 4:
\[
w^2 = \frac{144}{4} = 36
\]
Take the square root of both sides (we consider positive value since width can't be negative):
\[
w = \sqrt{36} = 6
\]

Step4: Find length

Since \( l = 4w \), substitute \( w = 6 \):
\[
l = 4 \times 6 = 24
\]

Step5: Calculate perimeter

The perimeter \( P \) of a rectangle is \( P = 2(l + w) \). Substitute \( l = 24 \) and \( w = 6 \):
\[
P = 2(24 + 6) = 2 \times 30 = 60
\]

Answer:

The perimeter of the rectangle is \( 60 \) feet.