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the length of a rectangle is four times its width. if the area of the r…

Question

the length of a rectangle is four times its width. if the area of the rectangle is 324 ft², find its perimeter.

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) feet. Then the length \( l \) is \( 4w \) feet (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 \( A = 324 \, \text{ft}^2 \), so substitute \( l = 4w \) into the area formula:

$$ 324 = 4w \times w $$
$$ 324 = 4w^2 $$

Step3: Solve for \( w \)

Divide both sides by 4:

$$ w^2 = \frac{324}{4} = 81 $$

Take the square root of both sides:

$$ w = \sqrt{81} = 9 \, \text{ft} \quad (\text{we take the positive root since width can't be negative}) $$

Step4: Find length \( l \)

Since \( l = 4w \), substitute \( w = 9 \):

$$ l = 4 \times 9 = 36 \, \text{ft} $$

Step5: Calculate perimeter

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). Substitute \( l = 36 \) and \( w = 9 \):

$$ P = 2(36 + 9) = 2 \times 45 = 90 \, \text{ft} $$

Answer:

\( 90 \)