2008-I-13

$\begin{array}{rcl}\stackrel{⏜}{ABC}& =& 2\pi (20)\times {\displaystyle \frac{{216}^{\circ }}{{360}^{\circ }}}\\ & =& 240\pi \text{ cm}\end{array}$

Note that the arc length is equal to the base circumference of $X$. Therefore, we have

$\begin{array}{rcl}2\pi r& =& 24\pi \\ r& =& 12\end{array}$

Refer to the above figure, by applying the Pythagoras Theorem to $\mathrm{\Delta }OAD$, we have

$\begin{array}{rcl}O{A}^2& =& O{D}^2+D{A}^2\\ {20}^2& =& h^2+{12}^2\\ h& =& \sqrt{400-144}\\ & =& 16\end{array}$

Therefore, the base radius and the height of $X$ are $12\text{ cm}$ and $16\text{ cm}$.

$\begin{array}{cl}=& {\displaystyle \frac{1}{3}}\times \pi (12{)}^2\times 16\\ =& 768\pi {\text{ cm}}^3\end{array}$

$\begin{array}{rcl}2\pi y& =& 2\pi (10)\times {\displaystyle \frac{{108}^{\circ }}{{360}^{\circ }}}\\ y& =& 3\end{array}$

Hence, we have

$\begin{array}{rcl}{\displaystyle \frac{\text{the base radius of }X}{\text{the base radius of }Y}}& =& {\displaystyle \frac{12}{3}}\\ & =& 4\end{array}$

Also, we have

$\begin{array}{rcl}{\displaystyle \frac{OA}{PD}}& =& {\displaystyle \frac{20}{10}}\\ & =& 2\end{array}$

Since the above two ratios are not equal, then $X$ and $Y$ are not similar.