Cho khối chóp đều `S.ABC` có cạnh đáy bằng `a`, hai mặt phẳng `(SAC)` và `(SBC)` vuông góc với nhau. Thể tích khối chóp đã cho bằng? `\bb A.` `\frac{a^{3}\sqrt{2}}{24}`. `\bb B.` `\frac{a^{3}\sqrt{2}}{8}`. `\bb C.` `\frac{a^{3}\sqrt{5}}{15}`. `\bb D.` `\frac{a^{3}\sqrt{5}}{4}`.

Kẻ $ AI \bot SC \Rightarrow AI \bot (SBC) \\ \left\{ {\begin{array}{*{20}{c}} {IA \bot IB} \\ {IA \bot IK} \end{array}} \right. \\ $

$\Rightarrow \Delta IAB$ vuông cân tại $I$

$ \Rightarrow IA = IB = \dfrac{{a\sqrt 2 }}{2} \\ \left\{ {\begin{array}{*{20}{c}} {AI \bot BC} \\ {AK \bot BC} \end{array}} \right. \\ \Rightarrow IK \bot BC \\ \Delta IKB \bot K \\ I{B^2} = I{K^2} + B{K^2} \\ \Rightarrow I{K^2} = I{B^2} - B{K^2} \\ = {\left( {\dfrac{{a\sqrt 2 }}{2}} \right)^2} - {\left( {\dfrac{a}{2}} \right)^2} \\ = \dfrac{{{a^2}}}{4} \\ \Rightarrow IK = \sqrt {\dfrac{{{a^2}}}{4}} = \dfrac{a}{2} \\ \Delta IKC \bot K \\ \Rightarrow I{C^2} = I{K^2} + K{C^2} \\ = {\left( {\dfrac{a}{2}} \right)^2} + {\left( {\dfrac{a}{2}} \right)^2} \\ = \dfrac{{{a^2}}}{2} \\ \Rightarrow IC = \dfrac{{a\sqrt 2 }}{2} \\ \Delta SAI \bot I \\ \Rightarrow S{A^2} = S{I^2} + A{I^2} \\ \Rightarrow S{I^2} = S{A^2} - A{I^2} \\ = S{A^2} - {\left( {\dfrac{{a\sqrt 2 }}{2}} \right)^2} \\ \Rightarrow SI = \sqrt {S{A^2} - {{\left( {\dfrac{{a\sqrt 2 }}{2}} \right)}^2}} \\ \Rightarrow SC = SI + IC \\ = \sqrt {S{A^2} - {{\left( {\dfrac{{a\sqrt 2 }}{2}} \right)}^2}} + \dfrac{{a\sqrt 2 }}{2} \\ SC = SA \\ \Rightarrow \sqrt {S{A^2} - {{\left( {\dfrac{{a\sqrt 2 }}{2}} \right)}^2}} + \dfrac{{a\sqrt 2 }}{2} = SA \\ \Leftrightarrow \sqrt {S{A^2} - {{\left( {\dfrac{{a\sqrt 2 }}{2}} \right)}^2}} = SA - \dfrac{{a\sqrt 2 }}{2} \\ \Rightarrow S{A^2} - {\left( {\dfrac{{a\sqrt 2 }}{2}} \right)^2} = {\left( {SA - \dfrac{{a\sqrt 2 }}{2}} \right)^2} \\ \Leftrightarrow S{A^2} - \dfrac{{{a^2}}}{2} = S{A^2} - a\sqrt 2 SA + \dfrac{{{a^2}}}{2} \\ \Rightarrow a\sqrt 2 SA = {a^2} \\ \Rightarrow SA = \dfrac{{{a^2}}}{{a\sqrt 2 }} = \dfrac{a}{{\sqrt 2 }} \\ AO = \dfrac{2}{3}AK = \dfrac{2}{3}.\dfrac{{a\sqrt 3 }}{2} = \dfrac{{a\sqrt 3 }}{3} \\ \Delta SAO \bot O \\ \Rightarrow S{A^2} = S{O^2} + O{A^2} \\ \Rightarrow S{O^2} = S{A^2} - O{A^2} \\ = {\left( {\dfrac{a}{{\sqrt 2 }}} \right)^2} - {\left( {\dfrac{{a\sqrt 3 }}{3}} \right)^2} \\ = \dfrac{{{a^2}}}{6} \\ \Rightarrow SO = \sqrt {\dfrac{{{a^2}}}{6}} = \dfrac{{a\sqrt 6 }}{6} \\ {S_{ABC}} = \dfrac{{{a^2}\sqrt 3 }}{4} \\ \Rightarrow {V_{SABC}} = \dfrac{1}{3}.SO.{S_{ABC}} \\ = \dfrac{1}{3}.\dfrac{{a\sqrt 6 }}{6}.\dfrac{{{a^2}\sqrt 3 }}{4} \\ = \dfrac{{{a^3}\sqrt 2 }}{{24}} \\ $

 Đáp án đúng: $A$