To find the dimensions of the rectangular garage, we can use the formulas for perimeter and area. Let's define the length of the garage as \( l \) meters and the width as \( w \) meters.
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Perimeter Formula: The perimeter \( P \) of a rectangle is given by: \[ P = 2l + 2w \] Given that the perimeter is 36 meters, we have: \[ 2l + 2w = 36 \] Dividing the entire equation by 2: \[ l + w = 18 \quad \text{(1)} \]
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Area Formula: The area \( A \) of a rectangle is given by: \[ A = lw \] Given that the area is 80 square meters, we have: \[ lw = 80 \quad \text{(2)} \]
Now we have a system of two equations (1) and (2). We can express one variable in terms of the other using equation (1) and substitute it into equation (2).
From equation (1), we can express \( w \): \[ w = 18 - l \]
Now, substitute \( w \) into equation (2): \[ l(18 - l) = 80 \] This simplifies to: \[ 18l - l^2 = 80 \] Rearranging it gives us a quadratic equation: \[ l^2 - 18l + 80 = 0 \]
Now we can solve this quadratic equation using the quadratic formula: \[ l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -18 \), and \( c = 80 \).
Calculating the discriminant: \[ b^2 - 4ac = (-18)^2 - 4 \cdot 1 \cdot 80 = 324 - 320 = 4 \]
Now plug in the values: \[ l = \frac{18 \pm \sqrt{4}}{2 \cdot 1} = \frac{18 \pm 2}{2} \]
This gives us two potential solutions for \( l \): \[ l = \frac{20}{2} = 10 \quad \text{and} \quad l = \frac{16}{2} = 8 \]
So, the possible lengths are \( l = 10 \) and \( l = 8 \). We can find the corresponding widths using \( w = 18 - l \):
- If \( l = 10 \), then \( w = 18 - 10 = 8 \).
- If \( l = 8 \), then \( w = 18 - 8 = 10 \).
Therefore, the dimensions of the garage are: \[ \text{Length} = 10 \text{ meters}, \quad \text{Width} = 8 \text{ meters} \]
Thus, the dimensions of the garage are \( 10 \) meters by \( 8 \) meters.