2021-I-14

$\begin{array}{rcl}
\dfrac{1}{3} \pi r^2 (24) & = & 800\pi \\
r^2 & = & 100 \\
r & = & 10
\end{array}$

$\therefore$ the base radius of $Y$ is $10\text{ cm}$.

$\begin{array}{cl}
= & \pi (10)^2(20) \\
= & 2000\pi \text{ cm}^3
\end{array}$

$\therefore$ the volume of $Z$

$\begin{array}{cl}
= & 2000\pi +800\pi \\
= & 2800\pi \text{ cm}^3
\end{array}$

Consider the radio of the volumes of $Y$ and $Z$, we have

$\begin{array}{rcll}
\dfrac{\text{The volume of $Y$}}{\text{The volume of $Z$}} & = & \dfrac{800\pi}{2800\pi} & = & \dfrac{2}{7}
\end{array}$

Consider the radio of the base radii of $Y$ and $Z$, we have

$\begin{array}{rcll}
\left(\dfrac{\text{The base radius of $Y$}}{\text{The base radius of $Z$}}\right)^3 & = & \left(\dfrac{10}{20} \right)^3 & = & \dfrac{1}{8} \\
\end{array}$

$\begin{array}{rcl}
\because \dfrac{\text{The volume of $Y$}}{\text{The volume of $Z$}} & \neq & \left(\dfrac{\text{The base radius of $Y$}}{\text{The base radius of $Z$}}\right)^3
\end{array}$

$\therefore Y$ and $Z$ are not similar.

$\begin{array}{cl}
= & 2 \pi (10) (20) \\
= & 400\pi\text{ cm}^2
\end{array}$

The slant height of $Y$

$\begin{array}{cl}
= & \sqrt{10^2 +24^2} \\
= & 26\text{ cm}
\end{array}$

$\therefore$ the curved surface area of $Y$

$\begin{array}{cl}
= & \pi (10) (26) \\
= & 260\pi\text{ cm}^2
\end{array}$

Hence, the sum of the curved surface areas of $X$ and $Y$

$\begin{array}{cl}
= & 400\pi +260\pi \\
= & 660\pi\text{ cm}^2
\end{array}$

Let $h\text{ cm}$ be the height of $Z$.

$\begin{array}{rcl}
\dfrac{1}{3} \pi (20)^2 h & = & 2800\pi \\
h & = & 21
\end{array}$

$\therefore$ the height of $Z$ is $21\text{ cm}$.

The slant height of $Z$

$\begin{array}{cl}
= & \sqrt{20^2 +21^2} \\
= & 29\text{ cm}
\end{array}$

Hence, the curved surface area of $Z$

$\begin{array}{cl}
= & \pi (20)(29) \\
= & 580 \pi \text{ cm}^2
\end{array}$

Since the sum of the curved surface areas of $X$ and $Y$ are greater then the curved surface area of $Z$,

then the claim is agreed.