Cho hình lăng trụ ABC.A′B′C′ có đáy ABC là tam giác cân

* Hướng dẫn giải

Trong (A’B’C’) kẻ\[B'K \bot A'C'\,\,\left( {K \in A'C'} \right)\]

Ta có:

\(\left. {\begin{array}{*{20}{c}}{AB\prime \bot A\prime C\prime (AB\prime \bot (A\prime B\prime C\prime ))}\\{B\prime K \bot A\prime C\prime }\end{array}} \right\} \Rightarrow A\prime C\prime \bot (AB\prime K)\)

\[ \Rightarrow A\prime C\prime \bot AK\]

\(\left. {\begin{array}{*{20}{c}}{(AA\prime C\prime ) \cap (A\prime B\prime C\prime ) = A\prime C\prime }\\{(AA\prime C\prime ) \supset AK \bot A\prime C\prime }\\{(A\prime B\prime C\prime ) \supset B\prime K \bot A\prime C\prime }\end{array}} \right\} \Rightarrow ((AA\prime \widehat {C\prime );(A\prime }B\prime C\prime )) = (A\widehat {K;B}\prime K) = \widehat {AKB\prime } = {30^0}\)

Ta có:

\[\begin{array}{*{20}{l}}{{S_{A'B'C'}} = \frac{1}{2}A'B'.A'C'.\sin 120 = \frac{1}{2}{a^2}.\frac{{\sqrt 3 }}{2} = \frac{{{a^2}\sqrt 3 }}{4} = \frac{1}{2}B'K.A'C'}\\{ \Rightarrow B'K = \frac{{2{S_{A'B'C'}}}}{{A'C'}} = \frac{{\frac{{{a^2}\sqrt 3 }}{2}}}{a} = \frac{{a\sqrt 3 }}{2}}\end{array}\]

\[AB' \bot \left( {A'B'C'} \right) \Rightarrow AB' \bot B'K \Rightarrow {\rm{\Delta }}AB'K\] vuông tại B’

\[ \Rightarrow AB' = B'K.tan30 = \frac{{a\sqrt 3 }}{2}.\frac{{\sqrt 3 }}{3} = \frac{a}{2}\]

Vậy\[{V_{ABC.A'B'C'}} = AB'.{S_{A'B'C'}} = \frac{a}{2}.\frac{{{a^2}\sqrt 3 }}{4} = \frac{{{a^3}\sqrt 3 }}{8}\]

Đáp án cần chọn là: C