Question:
The length of a rectangle is 8 cm more than its breadth. If the area of the rectangle is 209 \( cm^2 \), then the perimeter of the rectangle, in cm, is
The length of a rectangle is 8 cm more than its breadth. If the area of the rectangle is 209 \( cm^2 \), then the perimeter of the rectangle, in cm, is
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Factoring quadratic equations for dimensions requires selecting the positive root.
Updated On: Jun 15, 2026
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The Correct Option is C
Solution and Explanation
Concept:
The area is \( L \times B = 209 \), with \( L = B + 8 \). The perimeter is \( 2(L + B) \).
Step 1: Solve for breadth \( B \).
\[ B(B + 8) = 209 \implies B^2 + 8B - 209 = 0 \] Using the quadratic formula: \[ B = \frac{-8 + \sqrt{64 + 836}}{2} = \frac{-8 + 30}{2} = 11 \text{ cm} \]
Step 2: Find length and perimeter.
\( L = 11 + 8 = 19 \) cm. \[ \text{Perimeter} = 2(19 + 11) = 60 \text{ cm} \] \centerline{{60}}
Step 1: Solve for breadth \( B \).
\[ B(B + 8) = 209 \implies B^2 + 8B - 209 = 0 \] Using the quadratic formula: \[ B = \frac{-8 + \sqrt{64 + 836}}{2} = \frac{-8 + 30}{2} = 11 \text{ cm} \]
Step 2: Find length and perimeter.
\( L = 11 + 8 = 19 \) cm. \[ \text{Perimeter} = 2(19 + 11) = 60 \text{ cm} \] \centerline{{60}}
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