Let the length of the lawn be \( l \) meters and the width be \( w \) meters.
We know the formulas for the perimeter and area of a rectangle:
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The perimeter \( P \) is given by: \[ P = 2l + 2w = 66 \]
Simplifying this gives: \[ l + w = 33 \quad \text{(1)} \]
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The area \( A \) is given by: \[ A = lw = 260 \quad \text{(2)} \]
Now, we can solve these two equations.
From equation (1), we can express \( l \) in terms of \( w \): \[ l = 33 - w \]
Now, substitute \( l \) in equation (2): \[ (33 - w)w = 260 \] Expanding this gives: \[ 33w - w^2 = 260 \] Rearranging to form a standard quadratic equation: \[ w^2 - 33w + 260 = 0 \]
Now we can use the quadratic formula to solve for \( w \): \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -33 \), and \( c = 260 \).
Calculating the discriminant: \[ b^2 - 4ac = (-33)^2 - 4 \cdot 1 \cdot 260 = 1089 - 1040 = 49 \]
Now we can find \( w \): \[ w = \frac{33 \pm \sqrt{49}}{2} = \frac{33 \pm 7}{2} \]
Calculating the two possible values:
- \( w = \frac{40}{2} = 20 \)
- \( w = \frac{26}{2} = 13 \)
Thus, the possible dimensions of the lawn are:
- If \( w = 20 \), then \( l = 33 - 20 = 13 \).
- If \( w = 13 \), then \( l = 33 - 13 = 20 \).
Therefore, the dimensions of the lawn are:
\[ \text{20 meters by 13 meters} \]
Final answer: 20 meters by 13 meters