Let the length of the shingle be \( l \) inches and the width be \( w \) inches. We are given the following two equations based on the area and perimeter:
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Area equation: \[ l \times w = 60 \]
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Perimeter equation: \[ 2l + 2w = 32 \] Simplifying the perimeter equation, we can divide everything by 2: \[ l + w = 16 \]
Now we have a system of equations:
- \( lw = 60 \)
- \( l + w = 16 \)
From the second equation, we can express \( w \) in terms of \( l \): \[ w = 16 - l \]
Now we can substitute this expression for \( w \) into the area equation: \[ l(16 - l) = 60 \] Expanding this, we get: \[ 16l - l^2 = 60 \] Rearranging to form a standard quadratic equation: \[ l^2 - 16l + 60 = 0 \]
Next, we can use the quadratic formula to find the values of \( l \): \[ l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -16 \), and \( c = 60 \): \[ l = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 60}}{2 \cdot 1} \] \[ l = \frac{16 \pm \sqrt{256 - 240}}{2} \] \[ l = \frac{16 \pm \sqrt{16}}{2} \] \[ l = \frac{16 \pm 4}{2} \]
This gives us two potential solutions for \( l \): \[ l = \frac{20}{2} = 10 \quad \text{or} \quad l = \frac{12}{2} = 6 \]
Now, we can find \( w \) for both cases:
- If \( l = 10 \): \[ w = 16 - 10 = 6 \]
- If \( l = 6 \): \[ w = 16 - 6 = 10 \]
Thus, the dimensions of the shingle are \( 10 \) inches by \( 6 \) inches.
The final answer is: 10 inches by 6 inches.
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