Câu hỏi 31

Ta có: \(A'M\parallel AC \Rightarrow \dfrac{{A'M}}{{AC}} = \dfrac{{A'I}}{{IC}} = \dfrac{1}{2} \Rightarrow \dfrac{{IC}}{{A'C}} = \dfrac{2}{3}.\)

Vì \(IA' \cap \left( {ABC} \right) = C\)\( \Rightarrow \dfrac{{d\left( {I;\left( {ABC} \right)} \right)}}{{d\left( {A';\left( {ABC} \right)} \right)}} = \dfrac{{IC}}{{A'C}} = \dfrac{2}{3}.\)

\(\begin{array}{l} \Rightarrow \dfrac{{{V_{I.ABC}}}}{{{V_{ABC.A'B'C'}}}} = \dfrac{{\dfrac{1}{3}d\left( {I;\left( {ABC} \right)} \right).{S_{ABC}}}}{{d\left( {A';\left( {ABC} \right)} \right).{S_{ABC}}}} = \dfrac{1}{3}.\dfrac{2}{3} = \dfrac{2}{9}\\ \Rightarrow {V_{I.ABC}} = \dfrac{2}{9}{V_{ABC.A'B'C'}}\end{array}\)

Ta có: \(AA' \bot \left( {ABC} \right) \Rightarrow AA' \bot AC \Rightarrow \Delta AA'C\) vuông tại \(A\).

\( \Rightarrow AC = \sqrt {A'{C^2} - AA{'^2}}  = \sqrt {9{a^2} - 4{a^2}}  = a\sqrt 5 .\)

Xét tam giác vuông ABC có: \(BC = \sqrt {A{C^2} - A{B^2}}  = \sqrt {5{a^2} - {a^2}}  = 2a.\)

\( \Rightarrow {S_{ABC}} = \dfrac{1}{2}AB.BC = \dfrac{1}{2}a.2a = {a^2}.\)

\( \Rightarrow {V_{ABC.A'B'C'}} = AA'.{S_{ABC}} = 2a.{a^2} = 2{a^3}.\)

Vậy \({V_{I.ABC}} = \dfrac{2}{9}{V_{ABC.A'B'C'}} = \dfrac{2}{9}.2{a^3} = \dfrac{{4{a^3}}}{9}.\)    

Chọn C.