Cho hình chóp \(S.ABC\) đáy \(ABC\) là tam giác vuông tại \(A,AB = a,AC = a\sqrt 3 \). Tam giác $SBC$ đều nằm trong mặt phẳng vuông góc với đáy. Tính thể tích khối chóp $S.ABC$

\(V = Sh\)

\(V = \dfrac{1}{2}Sh\)

\(V = \dfrac{1}{3}Sh\)

\(V = \dfrac{1}{6}Sh\) 

\(\dfrac{V}{{V'}} = k\)         

\(\dfrac{{V'}}{V} = {k^2}\)

\(\dfrac{V}{{V'}} = {k^3}\)

\(\dfrac{{V'}}{V} = {k^3}\)

\(\dfrac{{{V_{S.A'B'C'}}}}{{{V_{S.ABC}}}} = \dfrac{{SA'}}{{SA}} + \dfrac{{SB'}}{{SB}} + \dfrac{{SC'}}{{SC}}\)          

\(\dfrac{{{V_{S.ABC}}}}{{{V_{S.A'B'C'}}}} = \dfrac{{SA'}}{{SA}}.\dfrac{{SB'}}{{SB}}.\dfrac{{SC'}}{{SC}}\)

\(\dfrac{{{V_{S.A'B'C'}}}}{{{V_{S.ABC}}}} = \dfrac{{SA'}}{{SA}} = \dfrac{{SB'}}{{SB}} = \dfrac{{SC'}}{{SC}}\)         

\(\dfrac{{{V_{S.A'B'C'}}}}{{{V_{S.ABC}}}} = \dfrac{{SA'}}{{SA}}.\dfrac{{SB'}}{{SB}}.\dfrac{{SC'}}{{SC}}\)

\(\dfrac{{{a^3}}}{6}\)

\(\dfrac{{{a^3}}}{3}\)

\(\dfrac{{{a^3}}}{4}\)

\(\dfrac{{{a^3}}}{8}\)

\(\dfrac{{2{a^3}\sqrt 2 }}{3}\)         

\(\dfrac{{{a^3}\sqrt 2 }}{6}\)           

\(\dfrac{{2{a^3}}}{3}\)

\(\dfrac{{{a^3}\sqrt 2 }}{{12}}\)

\(\dfrac{{{a^3}\sqrt 6 }}{2}\)

\(\dfrac{{{a^3}\sqrt 6 }}{4}\)

\(\dfrac{{{a^3}\sqrt 6 }}{8}\)

\(\dfrac{{3{a^3}\sqrt 6 }}{8}\) 

\(\dfrac{1}{3}abc\)                

\(\dfrac{1}{9}abc\)

\(\dfrac{1}{6}abc\)    

\(\dfrac{2}{3}abc\)