Questions: Find the dimensions of a rectangle whose perimeter is 20 meters and whose area is 21 square meters. The sides of the rectangle measure square meters. (Use a comma to separate answers as needed.)

Solution Steps

To find the dimensions of a rectangle, we start with the formulas for perimeter and area. The perimeter \( P \) and area \( A \) are given by: \[ P = 2 \times (l + w) = 20 \] \[ A = l \times w = 21 \] where \( l \) is the length and \( w \) is the width of the rectangle.

From the perimeter equation, we can express the sum of the length and width: \[ l + w = \frac{20}{2} = 10 \] Now we have two equations:

1. \( l + w = 10 \)
2. \( l \times w = 21 \)

We can express \( w \) in terms of \( l \): \[ w = 10 - l \] Substituting this into the area equation: \[ l \times (10 - l) = 21 \] This simplifies to: \[ 10l - l^2 = 21 \] Rearranging gives us a quadratic equation: \[ l^2 - 10l + 21 = 0 \]

Factoring the quadratic equation: \[ (l - 3)(l - 7) = 0 \] This gives us the solutions: \[ l = 3 \quad \text{or} \quad l = 7 \]

Using \( l + w = 10 \):

• If \( l = 3 \), then \( w = 10 - 3 = 7 \).
• If \( l = 7 \), then \( w = 10 - 7 = 3 \).

Final Answer