Đáp án:
a) Kẻ SH ⊥ BC
=> SH ⊥ (ABC) do (SBC) ⊥ (ABC)
$\begin{array}{l}
SH = SB.\sin \widehat {SBC} = 2a\sqrt 3 .\sin {30^0} = a\sqrt 3 \\
\Rightarrow {V_{SABC}} = \dfrac{1}{3}.SH.{S_{ABC}} = \dfrac{1}{3}.a\sqrt 3 .\dfrac{1}{2}.3a.4a = 2{a^3}\sqrt 3
\end{array}$
b)
Kẻ HN // BM; BM ⊥ AC
=> AC ⊥ (SHN)
Kẻ HK ⊥ SN
=> HK ⊥ (SAC)
=> HK là khoảng cách từ H xuống (SAC)
$\begin{array}{l}
\Rightarrow \dfrac{{{d_{B - \left( {SAC} \right)}}}}{{{d_{H - \left( {SAC} \right)}}}} = \dfrac{{BM}}{{HN}} = \dfrac{{BC}}{{CH}} = \dfrac{{3a}}{{BC - BH}}\\
= \dfrac{{3a}}{{3a - 2a\sqrt 3 .c{\rm{os3}}{{\rm{0}}^0}}} = \dfrac{{3a}}{{4a - 3a}} = 3\\
\Rightarrow HN = \dfrac{{BM}}{3} = \dfrac{{BA.BC}}{{AC.3}} = \dfrac{{3a.4a}}{{5a.3}} = \dfrac{{4a}}{5}\\
Do:\dfrac{1}{{H{K^2}}} = \dfrac{1}{{S{H^2}}} + \dfrac{1}{{H{N^2}}}\\
\Rightarrow HK = \dfrac{{4\sqrt {273} a}}{{91}}\\
\Rightarrow BM = 3HK = \dfrac{{12\sqrt {273} a}}{{91}}
\end{array}$
