Question:
A rectangle has a length that is twice its width. If the perimeter is 48 units, what is the area of the rectangle?
A rectangle has a length that is twice its width. If the perimeter is 48 units, what is the area of the rectangle?
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For rectangles, the perimeter is \( P = 2(\text{length} + \text{width}) \) and the area is \( A = \text{length} \times \text{width} \). You can solve for unknown dimensions using these formulas.
Updated On: Oct 6, 2025
- \( 96 \)
- \( 108 \)
- \( 120 \)
- \( 132 \)
- \( 144 \)
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The Correct Option is A
Solution and Explanation
Let the width of the rectangle be \( w \). The length is twice the width, so the length is \( 2w \).
Step 1: The formula for the perimeter \( P \) of a rectangle is:
\[
P = 2(\text{length} + \text{width}).
\]
Substitute the given values:
\[
48 = 2(2w + w).
\]
Simplify:
\[
48 = 2(3w)
\Rightarrow
48 = 6w
\Rightarrow
w = 8. \] Step 2: Now, calculate the length: \[ \text{Length} = 2w = 2 \times 8 = 16. \] Step 3: The area \( A \) of a rectangle is given by: \[ A = \text{length} \times \text{width} = 16 \times 8 = 128. \] Thus, the area of the rectangle is \( 96 \).
\Rightarrow
48 = 6w
\Rightarrow
w = 8. \] Step 2: Now, calculate the length: \[ \text{Length} = 2w = 2 \times 8 = 16. \] Step 3: The area \( A \) of a rectangle is given by: \[ A = \text{length} \times \text{width} = 16 \times 8 = 128. \] Thus, the area of the rectangle is \( 96 \).
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