Từ giả thiết có tam giác ABC đều cạnh bằng a
Gọi \(O=AC\cap BD\Rightarrow BO=\frac{a\sqrt{3}}{2}\Rightarrow BD=a\sqrt{3}\Rightarrow HD=\frac{3}{4}BD=\frac{3}{4}a\sqrt{3}\)
\(SH^{2}=SD^{2}-HD^{2}=2a^{2}-\frac{27a^{2}}{16}=\frac{5a^{2}}{16}\Rightarrow SH=\frac{a\sqrt{5}}{4}\)
Diện tích tứ giác ABCD là \(S_{ABCD}=AB.BC.\sin \widehat{ABC}=a^{2}.\sin 60^{\circ}=\frac{a^{2}\sqrt{3}}{2}\)
Thể tích khối chóp S.ABCD là \(V_{S.ABCD}=\frac{1}{3}SH.S_{ABCD}=\frac{1}{3}.\frac{a\sqrt{5}}{4}.\frac{a^{2}\sqrt{3}}{2}=\frac{a^{3}\sqrt{15}}{24}\)
\(SB^{2}+SH^{2}+HB^{2}=\frac{5a^{2}}{16}+\frac{3a^{2}}{16}\Rightarrow SB=\frac{a\sqrt{2}}{2}\)
\(\left\{\begin{matrix} BD\perp AC\\ AC\perp SH \end{matrix}\right.AC\perp (SBD)\Rightarrow AC\perp OM\)
Diện tích tam giác MAC là \(S_{MAC}=\frac{1}{2}OM.AC=\frac{1}{4}SB.AC=\frac{1}{4}.\frac{a\sqrt{2}}{2}.a=\frac{a^{2}\sqrt{2}}{8}\)
\(SB//OM\Rightarrow SB//(MAC)\Rightarrow d(SB,CM)=d(SB,(MAC))=d(S,(MAC))=d(D,(MAC))\)
\(V_{M.ACD}=\frac{1}{3}d(M,(ABCD)).S_{ACD}=\frac{1}{3}.\frac{1}{2}d(S,(ABCD))\frac{1}{2}S_{ABCD}=\frac{1}{4}V_{S.ABCD}-\frac{a^{3}\sqrt{15}}{96}\)
Mặt khác \(V_{M.ACD}=\frac{1}{3}d(D,(MAC)).S_{MAC}\Rightarrow d(D,(MAC))=\frac{3V_{M.ACD}}{S_{MAC}}=\frac{\frac{a^{3}\sqrt{15}}{32}}{\frac{a^{2}\sqrt{2}}{8}}=\frac{a\sqrt{30}}{8}\)
