Do \(\widehat{A}=60^0\Rightarrow\Delta ABD\) đều \(\Rightarrow BD=AB=a\); \(AC=2.\frac{a\sqrt{3}}{2}=a\sqrt{3}\)
\(\Rightarrow S_{ABCD}=\frac{1}{2}AC.BD=\frac{a^2\sqrt{3}}{2}\)
\(\Rightarrow SI=\frac{3V}{S_{ABCD}}=\frac{3a}{2}\)
Từ I kẻ \(IH\perp AB\Rightarrow AB\perp\left(SIH\right)\)
Từ \(I\) kẻ \(IK\perp SH\Rightarrow IK\perp\left(SAB\right)\Rightarrow IK=d\left(I;\left(SAB\right)\right)\)
\(IH=\frac{1}{2}.\frac{a\sqrt{3}}{2}=\frac{a\sqrt{3}}{4}\)
\(\frac{1}{IK^2}=\frac{1}{IH^2}+\frac{1}{SH^2}\Rightarrow IK=\frac{IH.SH}{\sqrt{IH^2+SH^2}}=\frac{3a}{8}\)
\(CI\) cắt (SAB) tại A mà \(CA=2CI\)
\(\Rightarrow d\left(C;\left(SAB\right)\right)=2d\left(I;\left(SAB\right)\right)=2IK=\frac{3a}{4}\)
