Đáp án đúng: B
Giải chi tiết:
Áp dụng định lí Cosin trong \(\Delta ABC\) ta có:
\(\begin{array}{l}\cos \angle ABC = \dfrac{{A{B^2} + B{C^2} - A{C^2}}}{{2AB.BC}}\\ \Leftrightarrow \cos {135^0} = \dfrac{{{{\left( {3a} \right)}^2} + B{C^2} - {{\left( {3\sqrt 5 a} \right)}^2}}}{{2.3a.BC}}\\ \Leftrightarrow  - 3\sqrt 2 aBC = B{C^2} - 36{a^2}\\ \Leftrightarrow BC = 3\sqrt 2 a\end{array}\)
Ta có \({S_{\Delta ABC}} = \dfrac{1}{2}AB.BC.\sin \angle ABC = \dfrac{1}{2}.3a.3\sqrt 2 a.\sin {135^0} = \dfrac{9}{2}{a^2}\).
Gọi \(H\) là hình chiếu vuông góc của \(D\) lên \(\left( {ABC} \right)\) ta có:
\(\begin{array}{l}\left\{ \begin{array}{l}AB \bot DH\,\,\left( {DH \bot \left( {ABC} \right)} \right)\\AB \bot AD\,\,\left( {gt} \right)\end{array} \right. \Rightarrow AB \bot \left( {ADH} \right) \Rightarrow AB \bot AH\\\left\{ \begin{array}{l}BC \bot DH\,\,\left( {DH \bot \left( {ABC} \right)} \right)\\BC \bot BD\,\,\left( {gt} \right)\end{array} \right. \Rightarrow BC \bot \left( {BDH} \right) \Rightarrow BC \bot BH \Rightarrow \angle CBH = {90^0}\\ \Rightarrow \angle ABH = {135^0} - {90^0} = {45^0}\end{array}\)
\( \Rightarrow \Delta ABH\) vuông cân tại \(A \Rightarrow AB = AH = 3a\) vaf \(BH = 3a\sqrt 2 \).
Kẻ \(HM \bot AD\,\,\left( {M \in AD} \right),\,\,HN \bot BD\,\,\left( {N \in BD} \right)\) ta có:
\(\begin{array}{l}\left\{ \begin{array}{l}AB \bot \left( {ADH} \right)\,\,\left( {cmt} \right) \Rightarrow AB \bot HM\\HM \bot AD\end{array} \right. \Rightarrow HM \bot \left( {ABD} \right)\\\left\{ \begin{array}{l}BC \bot \left( {BDH} \right)\,\,\left( {cmt} \right) \Rightarrow BC \bot HN\\HN \bot BD\end{array} \right. \Rightarrow HN \bot \left( {BCD} \right)\end{array}\)
\( \Rightarrow \angle \left( {\left( {ABD} \right);\left( {BCD} \right)} \right) = \angle \left( {HM;HN} \right)\).
Mà \(HM \bot \left( {ABD} \right) \Rightarrow HM \bot MN \Rightarrow \Delta HMN\) vuông tại \(M\) \( \Rightarrow \angle \left( {HM;HN} \right) = \angle MHN\).
\( \Rightarrow \angle MHN = {30^0}\).
Đặt \(DH = x\,\,\left( {x > 0} \right)\), áp dụng hệ thức lượng trong tam giác vuông ta có:
\(\begin{array}{l}HM = \dfrac{{DH.AH}}{{\sqrt {D{H^2} + A{H^2}} }} = \dfrac{{x.3a}}{{\sqrt {{x^2} + 9{a^2}} }}\\HN = \dfrac{{DH.BH}}{{\sqrt {D{H^2} + B{H^2}} }} = \dfrac{{x.3a\sqrt 2 }}{{\sqrt {{x^2} + 18{a^2}} }}\end{array}\)
Xét tam giác vuông \(HMN\) ta có \(\cos \angle MHN = \dfrac{{HM}}{{HN}}\)
\(\begin{array}{l} \Leftrightarrow \cos {30^0} = \dfrac{{x.3a}}{{\sqrt {{x^2} + 9{a^2}} }}:\dfrac{{x.3a\sqrt 2 }}{{\sqrt {{x^2} + 18{a^2}} }}\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{{\sqrt {{x^2} + 18{a^2}} }}{{\sqrt 2 \sqrt {{x^2} + 9{a^2}} }}\\ \Leftrightarrow \dfrac{3}{4} = \dfrac{{{x^2} + 18{a^2}}}{{2{x^2} + 18{a^2}}}\\ \Leftrightarrow 6{x^2} + 54{a^2} = 4{x^2} + 72{a^2}\\ \Leftrightarrow x = 3a = DH\end{array}\)
Vậy \({V_{ABCD}} = \dfrac{1}{3}DH.{S_{\Delta ABC}} = \dfrac{1}{3}.3a.\dfrac{9}{2}{a^2} = \dfrac{{9{a^3}}}{2}\).
Chọn B