Đáp án:
D
Giải thích các bước giải:
Ta có \(SB = \sqrt {S{A^2} + A{B^2}} = a\sqrt 5 \)
\(AC = \sqrt {A{B^2} + B{C^2}} = 2a,SC = \sqrt {S{A^2} + A{C^2}} = 2a\sqrt 2 \)
\[\begin{array}{l}SH.SB = S{A^2} \Leftrightarrow SH = \dfrac{{S{A^2}}}{{SB}} \Rightarrow \dfrac{{SH}}{{SB}} = {\left( {\dfrac{{SA}}{{SB}}} \right)^2} = \dfrac{4}{5}\\SK.SC = S{A^2} \Leftrightarrow SK = \dfrac{{S{A^2}}}{{SC}} \Rightarrow \dfrac{{SK}}{{SC}} = {\left( {\dfrac{{SA}}{{SC}}} \right)^2} = \dfrac{1}{2}\end{array}\]
\(\begin{array}{l} \Rightarrow \dfrac{{{V_{S.AHK}}}}{{{V_{S.ABC}}}} = \dfrac{{SA}}{{SA}}.\dfrac{{SH}}{{SB}}.\dfrac{{SK}}{{SC}} = 1.\dfrac{4}{5}.\dfrac{1}{2} = \dfrac{2}{5}\\ \Rightarrow \dfrac{{{V_{ABCHK}}}}{{{V_{S.ABC}}}} = 1 - \dfrac{2}{5} = \dfrac{3}{5}\\ \Rightarrow V = {V_{ABCHK}} = \dfrac{3}{5}{V_{S.ABC}} = \dfrac{3}{5}.\dfrac{1}{3}.2a.\dfrac{1}{2}a.a\sqrt 3 = \dfrac{{{a^3}\sqrt 3 }}{5}\\ \Rightarrow \dfrac{{{a^3}}}{V} = \dfrac{5}{{\sqrt 3 }} \approx 3\end{array}\)
