To find the dimensions of the rectangular kitchen, we can use the formulas for area and perimeter of a rectangle. Let’s denote the length as \( L \) and the width as \( W \).
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Area Equation: \[ L \times W = 220 \quad \text{(1)} \]
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Perimeter Equation: \[ 2L + 2W = 62 \quad \text{(2)} \] Simplifying equation (2): \[ L + W = 31 \quad \text{(3)} \]
Now, we can solve these two equations. From equation (3), we can express \( W \) in terms of \( L \): \[ W = 31 - L \quad \text{(4)} \]
Next, we substitute equation (4) into equation (1): \[ L \times (31 - L) = 220 \] Expanding this: \[ 31L - L^2 = 220 \] Rearranging gives us a standard quadratic equation: \[ L^2 - 31L + 220 = 0 \]
Now we can solve for \( L \) using the quadratic formula: \[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -31 \), and \( c = 220 \).
Calculating the discriminant: \[ b^2 - 4ac = (-31)^2 - 4 \times 1 \times 220 = 961 - 880 = 81 \]
Now substituting it into the formula: \[ L = \frac{31 \pm \sqrt{81}}{2} \] \[ L = \frac{31 \pm 9}{2} \] Calculating the two possible values for \( L \):
- \( L = \frac{40}{2} = 20 \)
- \( L = \frac{22}{2} = 11 \)
So the dimensions are \( L = 20 \) meters and \( W = 11 \) meters (using equation (4) to get \( W = 31 - L \)).
Thus, the kitchen dimensions are: \[ 20 \text{ meters by } 11 \text{ meters} \]
Final Answer
The dimensions of the kitchen are: 20 meters by 11 meters.