To evaluate Paul and Manuel's arguments regarding the volumes of cone W and square pyramid X, we first need to correctly calculate the volumes based on the formulas for each shape.
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Volume of Cone W:
- The formula for the volume of a cone is: \[ V = \frac{1}{3} \text{(Area of Base)} \cdot h \]
- The base area \( A \) of cone W, with a radius \( r = 10 \text{ cm} \), is: \[ A = \pi r^2 = \pi (10^2) = \pi (100) \approx 314 \text{ cm}^2 \]
- The height \( h \) of cone W is \( 5 \text{ cm} \).
- Therefore, the volume \( V \) of cone W is: \[ V = \frac{1}{3} \cdot 314 \cdot 5 \approx 523.33 \text{ cm}^3 \]
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Volume of Square Pyramid X:
- Square pyramid X has the same base area and height as cone W. Given that it has the same base area of approximately \( 314 \text{ cm}^2 \) and a height \( h \) of \( 5 \text{ cm} \), its volume is calculated using the formula: \[ V = \frac{1}{3} \text{(Area of Base)} \cdot h \]
- Thus, the volume of square pyramid X is: \[ V = \frac{1}{3} \cdot 314 \cdot 5 \approx 523.33 \text{ cm}^3 \]
From the calculations above, we find that both the volume of cone W and the volume of square pyramid X are approximately \( 523.33 \text{ cm}^3 \).
Analyzing the Arguments:
- Paul's Argument: He stated that the volume of square pyramid X is three times the volume of cone W. This is incorrect since both volumes are equal to \( 523.33 \text{ cm}^3 \).
- Manuel's Argument: He argued that the volume of square pyramid X is equal to the volume of cone W, which is correct.
Conclusion:
Manuel's argument is correct. The correct explanation is that Paul used the incorrect formula for finding the volume of square pyramid X, effectively stating that square pyramid X's volume is equal to that of cone W, when the correct calculation shows that they share the same volume. Thus, the correct choice is:
Manuel's argument is correct; Paul used the incorrect formula to find the volume of square pyramid X.
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