Đáp án:
\({V_{ABC.A'B'C'}} = {{9{a^3}} \over 4}\)
Giải thích các bước giải:
$$\eqalign{
& Goi\,\,H\,\,la\,\,TD\,\,cua\,\,AB \Rightarrow A'H \bot \left( {ABC} \right) \cr
& Goi\,\,I = AB' \cap AH \cr
& \Rightarrow \angle \left( {AB';\left( {ABC} \right)} \right) = \angle \left( {AI;\left( {ABC} \right)} \right) = \angle IAH = {60^0} \cr
& Xet\,\,{\Delta _v}AHI:\,\,IH = AH.\tan {60^0} = {{a\sqrt 3 } \over 2} \cr
& Ap\,dung\,dinh\,\,li\,\,Ta - let: \cr
& {{IH} \over {IA'}} = {{AH} \over {A'B'}} = {1 \over 2} \Rightarrow IA' = 2IH = a\sqrt 3 \cr
& \Rightarrow A'H = IA' + IH = a\sqrt 3 + {{a\sqrt 3 } \over 2} = {{3a\sqrt 3 } \over 2} \cr
& {S_{\Delta ABC}} = {1 \over 2}AB.BC = {1 \over 2}.a.a\sqrt 3 = {{{a^2}\sqrt 3 } \over 2} \cr
& Vay\,\,{V_{ABC.A'B'C'}} = A'H.{S_{ABC}} = {{3a\sqrt 3 } \over 2}.{{{a^2}\sqrt 3 } \over 2} = {{9{a^3}} \over 4} \cr} $$
