Answered

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Cone W has a radius of [tex]$8 \, \text{cm}[tex]$[/tex] and a height of [tex]$[/tex]5 \, \text{cm}$[/tex]. 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.

\begin{tabular}{|c|c|}
\hline
Paul & Manuel \\
\hline
\begin{tabular}{l}
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.
\end{tabular}
&
\begin{tabular}{l}
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.
\end{tabular} \\
\hline
\begin{tabular}{l}
The base area of cone W is [tex]$\pi \left( r^2 \right) = \pi \left( 8^2 \right) = 200.96 \, \text{cm}^2$.
\end{tabular}
&
The base area of cone W is [tex]$\pi \left( r^2 \right) = \pi \left( 8^2 \right) = 200.96 \, \text{cm}^2$. \\
\hline
\begin{tabular}{l}
The volume of cone W is [tex]\frac{1}{3} \text{(area of base)} \cdot ( h ) = \frac{1}{3} (200.96) (5) = 334.93 \, \text{cm}^3[/tex].
\end{tabular}
&
\begin{tabular}{l}
The volume of cone W is [tex]\frac{1}{3} \text{(area of base)} \cdot ( h ) = \frac{1}{3} (200.96) (5) = 334.93 \, \text{cm}^3[/tex].
\end{tabular} \\
\hline
\begin{tabular}{l}
The volume of square pyramid X is [tex]\frac{1}{3} \text{(area of base)} \cdot ( h ) = \frac{1}{3} (200.96) (5) = 334.93 \, \text{cm}^3[/tex].
\end{tabular}
&
The volume of square pyramid X is [tex]\text{(area of base)} \cdot ( h ) = (200.96) (5) = 1,004.8 \, \text{cm}^3[/tex]. \\
\hline
\end{tabular}

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.

Sagot :

GinnyAnswer
Let's go through the problem step by step to determine who is correct.

### Step 1: Base Area of Cone \(W\)
Cone \(W\) has a radius \(r = 8\) cm. The base area of the cone is calculated using the formula for the area of a circle:
[tex]\[ \text{Base Area} = \pi \times r^2 \][/tex]
Substituting the given radius:
[tex]\[ \text{Base Area} = \pi \times (8^2) = \pi \times 64 \][/tex]
[tex]\[ \text{Base Area} \approx 201.06 \, \text{cm}^2 \][/tex]

### Step 2: Volume of Cone \(W\)
The height of cone \(W\) is \(h = 5\) cm. The volume of a cone is calculated using the formula:
[tex]\[ \text{Volume of Cone} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \][/tex]
Substituting the base area and height:
[tex]\[ \text{Volume of Cone} = \frac{1}{3} \times 201.06 \times 5 \][/tex]
[tex]\[ \text{Volume of Cone} \approx 335.10 \, \text{cm}^3 \][/tex]

### Step 3: Volume of Square Pyramid \(X\)
Square pyramid \(X\) has the same base area and height as cone \(W\). The volume of a pyramid is calculated using the formula:
[tex]\[ \text{Volume of Pyramid} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \][/tex]
Substituting the base area and height:
[tex]\[ \text{Volume of Pyramid} = \frac{1}{3} \times 201.06 \times 5 \][/tex]
[tex]\[ \text{Volume of Pyramid} \approx 335.10 \, \text{cm}^3 \][/tex]

### Examination of Arguments:
Now let’s analyze the arguments of Paul and Manuel:

#### Paul's Argument:
Paul states that 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 \(X\).

- Base Area of Cone \(W\): \( \approx 201.06 \, \text{cm}^2 \) (Correct)
- Volume of Cone \(W\): \( \approx 335.10 \, \text{cm}^3 \) (Correct)
- Volume of Pyramid \(X\): \( \approx 335.10 \, \text{cm}^3 \) (Correct)

Paul’s calculations are correct, and his conclusion is logically sound.

#### Manuel's Argument:
Manuel states that the volume of square pyramid \(X\) is three times the volume of cone \(W\). He uses the formula that incorrectly omits the \(\frac{1}{3}\) factor.

- Base Area of Cone \(W\): \( \approx 201.06 \, \text{cm}^2 \) (Correct)
- Volume of Cone \(W\): \( \approx 335.10 \, \text{cm}^3 \) (Correct)
- Volume of Pyramid \(X\): \( \approx 1005.31 \, \text{cm}^3 \) (Incorrect)

Manuel used an incorrect formula for the volume of the square pyramid \(X\).

### Conclusion:
Paul's argument is correct. Manuel used an incorrect formula to find the volume of square pyramid \(X\).

So the correct statement explaining the situation is:
"Paul's argument is correct, Manuel used the incorrect formula to find the volume of square pyramid [tex]$X$[/tex]."
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