Cho khối tứ diện $ABCD$ có thể tích $V=\dfrac{1}{6}$, góc...

\)">V=\dfrac{1}{3}.{{S}_{ABC}}.d\left( D,\left( ABC \right) \right)=\dfrac{1}{3}.\dfrac{1}{2}.CA.CB.\sin {{45}^{\circ }}.d\left( D,\left( ABC \right) \right)=\dfrac{1}{6}.\dfrac{1}{\sqrt{2}}CA.CB.d\left( D,\left( ABC \right) \right)\le \dfrac{1}{6}\dfrac{CA.CB.AD}{\sqrt{2}}\left( 1 \right)AD,BC,\dfrac{AC}{\sqrt{2}},\dfrac{AC}{\sqrt{2}}.BC.AD\le {{\left( \dfrac{\dfrac{AC}{\sqrt{2}}+BC+AD}{3} \right)}^{3}}.V\le \dfrac{1}{6}.{{\left( \dfrac{\dfrac{AC}{\sqrt{2}}+BC+AD}{3} \right)}^{3}}=\dfrac{1}{6}\left( 2 \right)V=\dfrac{1}{6}\left( 1 \right)\left( 2 \right)\left\{ \begin{matrix}
DA\bot \left( ABC \right) \\
\dfrac{AC}{\sqrt{2}}=BC=AD=1 \\
\end{matrix} \right.\Rightarrow \left\{ \begin{matrix}
CD=\sqrt{A{{C}^{2}}+D{{A}^{2}}} \\
BC=1,AD=1,AC=\sqrt{2} \\
\end{matrix} \right.\Rightarrow CD=\sqrt{3}$.