2010-I-13

Therefore, the area of $\Delta ABC$

$\begin{array}{cl}
= & \sqrt{25(25-17)(25-17)(25-16)} \\
= & 120 \text{ cm}^2
\end{array}$

$\begin{array}{cl}
= & 120 \times 20 \\
= & 2~400\text{ cm}^3
\end{array}$

1. Since $\Delta ABC \sim \Delta APQ$, then we have

   $\begin{array}{rcl}
   \dfrac{\text{Area of }\Delta APQ}{\text{Area of }\Delta ABC} & = & \left(\dfrac{PQ}{BC}\right)^2 \\
   \dfrac{\text{Area of }\Delta APQ}{120} & = & \left(\dfrac{4}{16}\right)^2 \\
   \text{Area of }\Delta APQ & = & 7.5\text{ cm}^2
   \end{array}$

   Therefore, the volume of $APQRES$

   $\begin{array}{cl}
   = & 7.5 \times 20 \\
   = &150\text{ cm}^3
   \end{array}$

2. By the result of (b) and (c)(i), we have

   $\begin{array}{cl}
   & \dfrac{\text{Volume of }APQRES}{\text{Volume of }ABCDEF} \\
   = & \dfrac{2400}{150} \\
   = & 16
   \end{array}$

   However,

   $\begin{array}{cl}
   & \left(\dfrac{\text{Height of }APQRES}{\text{Height of }ABCDEF}\right)^3 \\
   = & \left(\dfrac{20}{20}\right)^3 \\
   = & 1 \\
   \neq & 16
   \end{array}$

   Therefore, the two wooden blocks are not similar.