To determine whose argument is correct, we need to analyze the volume formulas given for the cone and the square pyramid.
-
Volume of the Cone (W):
- The formula for the volume of a cone is given by: \[ V = \frac{1}{3} \times \text{(area of the base)} \times \text{height} \]
- The base of cone W has a radius \( r = 10 \) cm, so: \[ \text{Area of base} = \pi r^2 = \pi (10^2) = 100\pi \approx 314.16, \text{cm}^2 \]
- Hence, \[ V_W = \frac{1}{3} \times 100\pi \times 5 = \frac{500\pi}{3} \approx 523.33 , \text{cm}^3 \]
-
Volume of the Square Pyramid (X):
- The volume of a square pyramid is calculated using the formula: \[ V = \frac{1}{3} \times \text{(area of the base)} \times \text{height} \]
- However, the base of pyramid X is a square with the same area as the base of cone W. The area of the square base must also be 314.16 cm².
- Thus, we should find the side length of the square base: \[ \text{Area of base} = s^2 \implies s^2 = 100\pi \implies s = \sqrt{100\pi} = 10\sqrt{\pi}, \text{cm} \]
- Now calculating the volume: \[ V_X = \frac{1}{3} \times 100\pi \times 5 = \frac{500\pi}{3} \approx 523.33, \text{cm}^3 \]
From the calculations:
- The volume of cone W (V_W) is approximately 523.33 cm³.
- The volume of square pyramid X (V_X) as calculated is also approximately 523.33 cm³.
Conclusion:
Manuel’s claim that the volume of square pyramid X is equal to the volume of cone W is correct.
Correct Statement: Manuel's argument is correct; Paul used the incorrect formula to find the volume of square pyramid X.
This is because both volumes calculated were actually equal, while Paul mistakenly asserted that the volume of pyramid X was three times that of cone W.
1 answer