- By applying the cosine law on $\Delta ABC$, we have
$\begin{array}{rcl}
\cos \angle BAD & = & \dfrac{AB^2 + AC^2 -BC^2}{2(AB)(AC)} \\
& = & \dfrac{40^2 +90^2 – 60^2}{2 (40)(90)} \\
& = & \dfrac{61}{72}
\end{array}$Consider $\Delta ABD$,
$\begin{array}{rcl}
\cos \angle BAD & = & \dfrac{AD}{AB} \\
AD & = & 40 \times \dfrac{61}{72} \\
& = & \dfrac{305}{9} \text{ cm}
\end{array}$ -
-
- By applying the cosine law on $\Delta ACD$ and the result of (a), we have
$\begin{array}{rcl}
AD^2 + AC^2 -2(AD)(AC) \cos \angle DAC & = & CD^2 \\
AC^2 – 31.819~739~26 AC – 2000 & = & 0 \\
\end{array}$Therefore, $AC=63.376~952~45 \text{ cm}$ or $AC=-31.55721319\text{ cm}$ (rejected).
- By applying the Heron’s formula on $\Delta ABC$, we have
$\begin{array}{rcl}
s & = & \dfrac{1}{2} ( AB + AC + BC) \\
& = & 81.688~472~23 \text{ cm}
\end{array}$Hence, the area of $\Delta ABC$
$\begin{array}{cl}
= & \sqrt{s(s-AB)(s-AC)(s-BC)} \\
= & 1~162.961~054 \text{ cm}^2
\end{array}$ - The volume of the tetrahedron $ABCD$ with base $\Delta ACD$
$\begin{array}{cl}
= & \dfrac{1}{3} \times BD \times \text{the area of $\Delta ACD$} \\
= & \dfrac{1}{3} \times BD \times \dfrac{1}{2} \times AD \times AC \times \sin \angle DAC \\
= & 6~716.175~187 \text{ cm}^3
\end{array}$Let $h\text{ cm}$ be the height of the tetrahedron $ABCD$ from the vertex $D$ to the base $\Delta ABC$. The volume of the tetrahedron $ABCD$ with base $\Delta ABC$
$\begin{array}{cl}
= & \dfrac{1}{3} \times h \times \text{the area of $\Delta ABC$} \\
= & 387.653~684~7 h \text{ cm}^3
\end{array}$Hence, we have
$\begin{array}{rcl}
387.653~684~7 h & = & 6~716.175~187 \\
h & = & 17.325~193~73 \text{ cm}
\end{array}$Therefore, the required height is $17.325~193~73\text{ cm}$.
- By applying the cosine law on $\Delta ACD$ and the result of (a), we have
- Refer to part (b)(i)(3), the volume of tetrahedron $ABCD$
$\begin{array}{cl}
= & \dfrac{(AD)(AC)(BD)\sin \angle DAC}{6}
\end{array}$Since $AD$, $AC$, $BD$ and $\dfrac{1}{6}$ are constants, then the volume of tetrahedron $ABCD$ varies directly as $\sin \angle DAC$.
Hence, we have the following conclusion:
When $\angle ADC$ increases from $30^\circ$ to $90^\circ$, the value of $\sin \angle DAC$ increases and so the volume of tetrahedron $ABCD$ increases.
When $\angle ADC$ increases from $90^\circ$ to $150^\circ$, the value of $\sin \angle DAC$ decreases and so the volume of tetrahedron $ABCD$ decreases.
-
2006-I-17
Ans: (a) $\dfrac{305}{9}\text{ cm}$ (b) (i) (1) $63.4\text{ cm}$ (2) $1\ 160\text{ cm}^2$ (3) $17.3\text{ cm}$ (ii) $\angle ADC$ increases from $30^\circ$ to $90^\circ$, the volume increases. $\angle ADC$ increases from $90^\circ$ to $150^\circ$, the volume decreases.