Cho lăng trụ tam giác đều ABC.A'B'C' có tất cả các...

B

- Hướng dẫn giải

Ta có:

\({V_{A.A'BC}} = \frac{1}{3}.{S_{A'BC}}.d\left[ {A,\left( {A'BC} \right)} \right]\)

* Tính \({S_{\Delta A'BC}}\)

\(A'B = A'C = a\sqrt 2 \)

\(\begin{array}{l}
A'H = \sqrt {A'{B^2} - B{H^2}} \\
 = \sqrt {2{a^2} - \frac{{{a^2}}}{4}}  = \frac{{a\sqrt 7 }}{2}
\end{array}\)

\(\begin{array}{l}
{S_{\Delta A'BC}} = \frac{1}{2}.AH.BC\\
 = \frac{1}{2}.\frac{{a\sqrt 7 }}{2}.a = \frac{{{a^2}\sqrt 7 }}{4}
\end{array}\)

* Ta lại có: \({V_{A.A'BC}} = {V_{A'.ABC}}\)

\(\begin{array}{*{20}{l}}
{{V_{A'.ABC}} = \frac{1}{3}{S_{\Delta ABC}}.AA'}\\
{ = \frac{1}{3}.\frac{{{a^2}\sqrt 3 }}{4}.a = \frac{{{a^3}\sqrt 3 }}{{12}}}
\end{array}\)

\(\begin{array}{l}
 \Rightarrow d\left[ {A,\left( {A'BC} \right)} \right] = \frac{{3V}}{{{S_{\Delta A'BC}}}}\\
 = \frac{{a\sqrt {21} }}{7}
\end{array}\)

Chọn B.