Let's denote the length of the rectangular lawn as \( l \) meters and the width as \( w \) meters.
We have two equations based on the information provided:
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The area of the rectangle: \[ l \times w = 187 \]
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The perimeter of the rectangle: \[ 2l + 2w = 56 \] This simplifies to: \[ l + w = 28 \]
Now we have a system of equations:
\[ \begin{align*}
- & \quad lw = 187 \quad \text{(1)} \
- & \quad l + w = 28 \quad \text{(2)} \end{align*} \]
From equation (2), we can express \( w \) in terms of \( l \): \[ w = 28 - l \]
Now we substitute \( w \) into equation (1): \[ l(28 - l) = 187 \] Expanding this gives: \[ 28l - l^2 = 187 \] Rearranging the equation, we get: \[ l^2 - 28l + 187 = 0 \]
Next, we'll use the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -28 \), and \( c = 187 \). Calculating the discriminant: \[ b^2 - 4ac = (-28)^2 - 4 \cdot 1 \cdot 187 = 784 - 748 = 36 \]
Now we can find \( l \): \[ l = \frac{28 \pm \sqrt{36}}{2} = \frac{28 \pm 6}{2} \]
Calculating the two possible values for \( l \): \[ l = \frac{34}{2} = 17 \quad \text{and} \quad l = \frac{22}{2} = 11 \]
So, we have two possible dimensions for \( l \):
- If \( l = 17 \), then \( w = 28 - 17 = 11 \)
- If \( l = 11 \), then \( w = 28 - 11 = 17 \)
Thus, the dimensions of the lawn are: \[ \text{11 meters} \times \text{17 meters} \]
So the answer is: 11 meters by 17 meters.