Cone W has a radius of 8 cm and a height of 5 cm. Square pyramid X has the same base area and height as cone W. Paul and Manuel disagree on how the volumes of cone W and square pyramid X are related.

| Paul | Manuel |
|---|---|
| The volume of square pyramid X is equal to the volume of cone W. This can be proven by finding the base area and volume of cone W, along with the volume of square pyramid X. The base area of cone W is [tex]\(\pi(r^2)=\pi(8^2)=200.96 \, \text{cm}^2\)[/tex]. The volume of cone W is [tex]\(\frac{1}{3} (\text{area of base}) (h)=\frac{1}{3}(200.96)(5)=334.93 \, \text{cm}^3\)[/tex]. The volume of square pyramid X is [tex]\(\frac{1}{3} (\text{area of base}) (h)=\frac{1}{3}(200.96)(5)=334.93 \, \text{cm}^3\)[/tex]. | The volume of square pyramid X is three times the volume of cone W. This can be proven by finding the base area and volume of cone W, along with the volume of square pyramid X. The base area of cone W is [tex]\(\pi(r^2)=\pi(8^2)=200.96 \, \text{cm}^2\)[/tex]. The volume of cone W is [tex]\(\frac{1}{3} (\text{area of base}) (h)=\frac{1}{3}(200.96)(5)=334.93 \, \text{cm}^3\)[/tex]. The volume of square pyramid X is (\text{area of base}) (h)=(200.96)(5)=1,004.8 \, \text{cm}^3. |

Examine their arguments. Which statement explains whose argument is correct and why?

A. Paul's argument is correct; Manuel used the incorrect formula to find the volume of square pyramid X.

B. Paul's argument is correct; Manuel used the incorrect base area to find the volume of square pyramid X.

C. Manuel's argument is correct; Paul used the incorrect formula to find the volume of square pyramid X.

D. Manuel's argument is correct; Paul used the incorrect base area to find the volume of square pyramid X.



Answer :

To solve this problem, let's analyze the calculations and arguments made by both Paul and Manuel regarding the volumes of cone [tex]\( W \)[/tex] and square pyramid [tex]\( X \)[/tex].

### Step-by-Step Solution:

1. Base Area Calculation:

Both objects, cone [tex]\( W \)[/tex] and square pyramid [tex]\( X \)[/tex], have the same base area.

The base area of cone [tex]\( W \)[/tex], which is also the base area of square pyramid [tex]\( X \)[/tex], is given by:
[tex]\[ \text{Base Area} = \pi r^2 = \pi (8^2) = \pi \cdot 64 = 201.06 \, \text{cm}^2 \][/tex]

2. Volume of Cone [tex]\( W \)[/tex]:

The volume of cone [tex]\( W \)[/tex] is calculated using the formula for the volume of a cone:
[tex]\[ \text{Volume of cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \cdot 201.06 \cdot 5 = 335.10 \, \text{cm}^3 \][/tex]

3. Volume of Square Pyramid [tex]\( X \)[/tex]:

Both Paul and Manuel are calculating the volume of square pyramid [tex]\( X \)[/tex], but let's evaluate the correct approach.

The correct formula for the volume of a square pyramid is:
[tex]\[ \text{Volume of pyramid} = \frac{1}{3} \text{Base Area} \times \text{Height} \][/tex]

Applying the given values:
[tex]\[ \text{Volume of pyramid} = \frac{1}{3} \cdot 201.06 \cdot 5 = 335.10 \, \text{cm}^3 \][/tex]

### Analyzing the Arguments:

- Paul's Argument:
Paul correctly uses the formula [tex]\(\frac{1}{3} \text{Base Area} \times \text{Height}\)[/tex] to calculate the volume of the square pyramid [tex]\( X \)[/tex]. His calculation:
[tex]\[ \text{Volume of pyramid} = \frac{1}{3} \cdot 201.06 \cdot 5 = 335.10 \, \text{cm}^3 \][/tex]
matches the volume of the cone [tex]\( W \)[/tex].

- Manuel's Argument:
Manuel incorrectly multiplies the base area by the height without the [tex]\(\frac{1}{3}\)[/tex] factor:
[tex]\[ \text{Volume of pyramid} = 201.06 \cdot 5 = 1005.30 \, \text{cm}^3 \][/tex]
This result is incorrect because he did not use the correct formula for the volume of a pyramid.

### Conclusion:

Based on the correct application of geometric volume formulas:

- Paul's argument is correct; Manuel used the incorrect formula to find the volume of square pyramid [tex]\( X \)[/tex].

Thus, the correct statement is:

Paul's argument is correct; Manuel used the incorrect formula to find the volume of square pyramid [tex]\( X \)[/tex].