To solve the problem, we start by letting the length of the rectangle be \( l \) and the width be \( w \).
From the problem, we know two things:
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The perimeter \( P \) of a rectangle is given by the formula: \[ P = 2l + 2w \] In this case, the perimeter is 32 cm, so: \[ 2l + 2w = 32 \] Dividing the entire equation by 2, we get: \[ l + w = 16 \quad \text{(1)} \]
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The width \( w \) is described as being 2 cm longer than half the length: \[ w = \frac{l}{2} + 2 \quad \text{(2)} \]
Now, we can substitute equation (2) into equation (1):
Substituting \( w \) in (1): \[ l + \left(\frac{l}{2} + 2\right) = 16 \]
Now simplify this: \[ l + \frac{l}{2} + 2 = 16 \] Combining the terms: \[ \frac{2l}{2} + \frac{l}{2} + 2 = 16 \] \[ \frac{3l}{2} + 2 = 16 \] Subtracting 2 from both sides: \[ \frac{3l}{2} = 14 \] Multiplying both sides by \(\frac{2}{3}\): \[ l = \frac{2 \times 14}{3} = \frac{28}{3} \approx 9.33 \text{ cm} \]
Now, we can find the width \( w \) using equation (2): \[ w = \frac{l}{2} + 2 = \frac{\frac{28}{3}}{2} + 2 \] \[ w = \frac{28}{6} + 2 = \frac{28}{6} + \frac{12}{6} = \frac{40}{6} = \frac{20}{3} \approx 6.67 \text{ cm} \]
Having determined the length and width, we can now find the area \( A \) of the rectangle: \[ A = l \times w = \left(\frac{28}{3}\right) \times \left(\frac{20}{3}\right) = \frac{28 \times 20}{9} = \frac{560}{9} \text{ cm}^2 \approx 62.22 \text{ cm}^2 \]
Thus, the area of the rectangle is: \[ \boxed{\frac{560}{9} \text{ cm}^2} \]