Lời giải:.png)
Ta có : \(AC = \sqrt {A{B^2} + B{C^2}} = 2a\), mà \(\Delta SAC\) vuông tại S \( \Rightarrow SI = \frac{{AB}}{2} = a\)
\( \Rightarrow SH = \sqrt {S{I^2} - H{I^2}} = \sqrt {{a^2} - \frac{{{a^2}}}{4}} = \frac{{a\sqrt 3 }}{2}\)
Kẻ \(HK \bot AB;AB \bot SH \Rightarrow AB \bot \left( {KHS} \right) \Rightarrow \left( {SAB} \right) \bot (KHS)\)
Mà \(\left( {SAB} \right) \cap \left( {KHS} \right) = SK\).
Kẻ \(HE \bot SK \Rightarrow HE \bot \left( {SAB} \right) \Rightarrow d(H,(SCD)) = HE\)
\(\begin{array}{l} A = HC \cap \left( {SAB} \right)\\ \Rightarrow \frac{{d\left( {C,\left( {SAB} \right)} \right)}}{{d\left( {H,(SAB)} \right)}} = \frac{{CA}}{{HA}} = 4\\ \Rightarrow d\left( {C,(SAB)} \right) = 4d(H,(SAB)) = 4HE \end{array}\)
\(\begin{array}{l} HE = \frac{{HK.SH}}{{\sqrt {H{K^2} + S{H^2}} }} = \frac{{\frac{{a\sqrt 3 }}{4}.\frac{{a\sqrt 3 }}{2}}}{{\sqrt {\frac{{3{a^2}}}{{16}} + \frac{{3{a^2}}}{4}} }} = \frac{{a\sqrt {15} }}{{10}}\\ \Rightarrow d\left( {C,(SAB)} \right) = \frac{{2a\sqrt {15} }}{5} \end{array}\)
