2007-I-16

$\begin{array}{rcl}s& =& {\displaystyle \frac{AB+BC+AC}{2}}\\ & =& {\displaystyle \frac{9+5+6}{2}}\\ & =& 10\text{ cm}\end{array}$

Therefore, the area of triangular base $ABC$

$\begin{array}{cl}=& \sqrt{s(s-AB)(s-BC)(s-AC)}\\ =& \sqrt{10(10-9)(10-5)(10-6)}\\ =& \sqrt{200}\\ =& 10\sqrt{2}{\text{ cm}}^2\end{array}$

Hence, the volume of the souvenir $ABCDEF$

$\begin{array}{cl}=& 10\sqrt{2}\times 20+{\displaystyle \frac{1}{3}}\times 10\sqrt{2}(23-20)\\ =& 200\sqrt{2}+10\sqrt{2}\\ =& 210\sqrt{2}{\text{ cm}}^3\end{array}$

Note that $FG=BC=5\text{ cm}$, $GE=CA=6\text{ cm}$ and $EF=AB=9\text{ cm}$. In $\mathrm{\Delta }DFG$,

$\begin{array}{rcl}D{F}^2& =& D{G}^2+F{G}^2\\ DF& =& \sqrt{3^2+5^2}\\ DF& =& \sqrt{34}\text{ cm}\end{array}$

In $\mathrm{\Delta }DEG$,

$\begin{array}{rcl}D{E}^2& =& D{G}^2+G{E}^2\\ DE& =& \sqrt{3^2+6^2}\\ DE& =& \sqrt{45}\text{ cm}\end{array}$

By applying the cosine law to $\mathrm{\Delta }DEF$, we have

$\begin{array}{rcl}\cos \mathrm{\angle }DFE& =& {\displaystyle \frac{D{F}^2+E{F}^2–D{E}^2}{2(DF)(EF)}}\\ \cos \mathrm{\angle }DFE& =& 0.666938943\\ \mathrm{\angle }DFE& =& 48.168751{76}^{\circ }\\ \mathrm{\angle }DFE& \approx & {48.2}^{\circ }\end{array}$

Note that the shortest distance from $D$ to $EF$ is $DH$. In $\mathrm{\Delta }DFH$,

$\begin{array}{rcl}\sin \mathrm{\angle }DFH& =& {\displaystyle \frac{DH}{DF}}\\ DH& =& DF\sin \mathrm{\angle }DFH\\ DH& =& 4.344714399\\ DH& =& 4.34\text{ cm}\end{array}$

$\begin{array}{cl}=& 5\times 4\\ =& 20{\text{ cm}}^2\end{array}$

The area of $\mathrm{\Delta }DEF$

$\begin{array}{cl}=& {\displaystyle \frac{1}{2}}\times EF\times DH\\ =& 19.5512148{\text{ cm}}^2\\ <& 20{\text{ cm}}^2\end{array}$

Therefore, the thin rectangular metal plate cannot be fixed onto the triangular surface $DEF$.