Đáp án:
$${{{a^3}} \over 4}$$
Giải thích các bước giải:
$$\eqalign{
& \Delta ABC:\,\,\left\{ \matrix{
AB = AD \hfill \cr
\angle BAD = {60^0} \hfill \cr} \right. \Rightarrow \Delta ABC\,\,deu \cr
& Goi\,\,O = AC \cap BD \Rightarrow AO = {{a\sqrt 3 } \over 2} \Rightarrow AC = a\sqrt 3 \cr
& AA' \bot \left( {ABCD} \right) \cr
& \Rightarrow \angle \left( {A'C;\left( {ABCD} \right)} \right) = \angle \left( {A'C;A'A} \right) = \angle A'CA = {45^0} \cr
& \Rightarrow \Delta A'AC\,\,vuong\,\,can\,\,tai\,\,A \cr
& \Rightarrow AA' = AC = a\sqrt 3 \cr
& {S_{\Delta ABC}} = {{{a^2}\sqrt 3 } \over 4} \cr
& \Rightarrow {V_{A'.ABC}} = {1 \over 3}AA'.{S_{ABC}} = {1 \over 3}.a\sqrt 3 .{{{a^2}\sqrt 3 } \over 4} = {{{a^3}} \over 4} \cr} $$
