2006-I-13

Note that $\mathrm{\Delta }VCD\sim \mathrm{\Delta }VAB$. Therefore, we have

$\begin{array}{rcl}{\displaystyle \frac{VC}{VA}}& =& {\displaystyle \frac{CD}{AB}}\\ {\displaystyle \frac{VC}{VC+8}}& =& {\displaystyle \frac{3}{6}}\\ VC& =& 8\text{ cm}\end{array}$

Therefore, the volume of solid $X$

$\begin{array}{cl}=& {\displaystyle \frac{1}{3}}\pi (6{)}^2(16)–{\displaystyle \frac{1}{3}}\pi (3{)}^2(8)+{\displaystyle \frac{2}{3}}\pi (6{)}^3\\ =& 312\pi {\text{ cm}}^3\end{array}$

Since solid $X$ is similar to solid $Y$, we have

$\begin{array}{rcl}{\displaystyle \frac{\text{the volume of solid }X}{\text{the volume of solid }Y}}& =& {\left({\displaystyle \frac{\text{the surface area of solid }X}{\text{the surface area of solid }Y}}\right)}^{\frac{3}{2}}\\ {\displaystyle \frac{312\pi }{\text{the volume of solid }Y}}& =& {\left({\displaystyle \frac{4}{9}}\right)}^{\frac{3}{2}}\\ {\displaystyle \frac{312\pi }{\text{the volume of solid }Y}}& =& {\displaystyle \frac{8}{27}}\\ \text{the volume of solid }Y& =& 1053\pi {\text{ cm}}^3\end{array}$

However,

$\begin{array}{rcl}{\displaystyle \frac{\text{the volume of solid }{X}^{\prime }}{\text{the volume of solid }{Y}^{\prime }}}& =& {\displaystyle \frac{312\pi +\frac{4}{3}\pi (1{)}^3}{1053\pi +\frac{4}{3}\pi (2{)}^3}}\\ & =& {\displaystyle \frac{940}{3191}}\\ & \ne & {\displaystyle \frac{1}{8}}\end{array}$

Therefore, they are not similar.