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Cone W has a radius of 10 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 [tex]$W$[/tex] and square pyramid [tex]$X$[/tex] are related. Examine their arguments. Which statement explains whose argument is correct, and why?

\begin{tabular}{|c|c|}
\hline
Paul & Manuel \\
\hline
\begin{tabular}{l}
The volume of square pyramid [tex]$X$[/tex] is three times the volume of cone [tex]$W$[/tex]. This can \\
be proven by finding the base area and volume of cone [tex]$W$[/tex], along with the volume \\
of square pyramid [tex]$X$[/tex]. The base area of cone [tex]$W$[/tex] is [tex]$\pi(r^2) = \pi(10^2) = 314 \, \text{cm}^2$[/tex]. The \\
volume of cone [tex]$W$[/tex] is [tex]$\frac{1}{3} (\text{area of base}) (h) = \frac{1}{3}(314)(5) = 523.33 \, \text{cm}^3$[/tex]. The volume \\
of square pyramid [tex]$X$[/tex] is (\text{area of base}) (h) = (314)(5) = 1,570 \, \text{cm}^3.
\end{tabular} &
\begin{tabular}{l}
The volume of square pyramid [tex]$X$[/tex] is equal to the volume of cone [tex]$W$[/tex]. This can be \\
proven by finding the base area and volume of cone [tex]$W$[/tex], along with the volume of \\
square pyramid [tex]$X$[/tex]. The base area of cone [tex]$W$[/tex] is [tex]$\pi(r^2) = \pi(10^2) = 314 \, \text{cm}^2$[/tex]. The \\
volume of cone [tex]$W$[/tex] is [tex]$\frac{1}{3} (\text{area of base}) (h) = \frac{1}{3}(314)(5) = 523.33 \, \text{cm}^3$[/tex]. The volume \\
of square pyramid [tex]$X$[/tex] is [tex]$\frac{1}{3} (\text{area of base}) (h) = \frac{1}{3}(314)(5) = 523.33 \, \text{cm}^3$[/tex].
\end{tabular} \\
\hline
\end{tabular}

\begin{itemize}
\item Paul's argument is correct. Manuel used the incorrect base area to find the volume of square pyramid [tex]$X$[/tex].
\item Manuel's argument is correct. Paul used the incorrect formula to find the volume of square pyramid [tex]$X$[/tex].
\end{itemize}

\begin{tabular}{|c|}
\hline
[tex]\(\square\)[/tex] \\
\hline
\end{tabular}



Answer :

GinnyAnswer
Let us examine the arguments of Paul and Manuel.

First, we need to determine the correct base area and volumes for both the cone [tex]\(W\)[/tex] and the square pyramid [tex]\(X\)[/tex].

Step 1: Base Area of Cone [tex]\(W\)[/tex]

- The base of cone [tex]\(W\)[/tex] is a circle with a radius [tex]\(r = 10 \text{ cm}\)[/tex].
- The formula for the area of a circle is [tex]\( \text{Area} = \pi r^2 \)[/tex].
- Substituting the radius [tex]\( r = 10 \text{ cm} \)[/tex]:

[tex]\[ \text{Base area of cone W} = \pi (10)^2 = 314.159 \text{ cm}^2 \][/tex]

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

- The formula for the volume of a cone is [tex]\( V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \)[/tex].
- The height [tex]\( h = 5 \text{ cm} \)[/tex].
- Substituting the base area and height into the volume formula:

[tex]\[ \text{Volume of cone W} = \frac{1}{3} \times 314.159 \times 5 = 523.5988 \text{ cm}^3 \][/tex]

Step 3: Base Area of Square Pyramid [tex]\(X\)[/tex]

- According to the problem, the base area of square pyramid [tex]\(X\)[/tex] is the same as the base area of cone [tex]\(W\)[/tex], which we have already calculated:

[tex]\[ \text{Base area of square pyramid X} = 314.159 \text{ cm}^2 \][/tex]

Step 4: Volume of Square Pyramid [tex]\(X\)[/tex]

- The formula for the volume of a square pyramid is also [tex]\( V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \)[/tex].
- The height is the same as cone [tex]\(W\)[/tex], [tex]\( h = 5 \text{ cm} \)[/tex].
- Substituting the base area and height into the volume formula:

[tex]\[ \text{Volume of square pyramid X} = \frac{1}{3} \times 314.159 \times 5 = 523.5988 \text{ cm}^3 \][/tex]

From these calculations, we see that both the volume of cone [tex]\(W\)[/tex] and the volume of square pyramid [tex]\(X\)[/tex] are equal, specifically:

[tex]\[ \text{Volume of cone W} = 523.5988 \text{ cm}^3 \][/tex]
[tex]\[ \text{Volume of square pyramid X} = 523.5988 \text{ cm}^3 \][/tex]

Conclusion:

Between Paul and Manuel, Manuel's argument is correct. The volume of square pyramid [tex]\(X\)[/tex] is indeed equal to the volume of cone [tex]\(W\)[/tex]. Paul's calculation that the volume of the square pyramid is three times the volume of the cone is incorrect. Therefore, the correct statement is:

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

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