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Câu hỏi: Cho hình chóp $S.ABCD$ có đáy $ABCD$ là hình chữ nhật, $AB=a,AD=2a$ và $SA$ vuông góc với đáy. Gọi $M$ là trung điểm của $SC,$ biết khoảng cách từ $M$ đến mặt phẳng $\left( SBD \right)$ bằng $\dfrac{a}{4}.$ Tính thể tích khối chóp $S.ABM$
A. $\dfrac{{{a}^{3}}\sqrt{11}}{66}.$
B. $\dfrac{4{{a}^{3}}\sqrt{11}}{33}.$
C. $\dfrac{2{{a}^{3}}\sqrt{11}}{33}.$
D. $\dfrac{{{a}^{3}}\sqrt{11}}{33}.$
Hạ $AE\bot BD$ mà $BD\bot SA\Rightarrow BD\bot \left( SAE \right)$.
Hạ $AH\bot SE$ thì $BD\bot AH$ (Do $BD\bot \left( SAE \right)$ ), suy ra $AH\bot \left( SBD \right)$ hay $d\left( A,\left( SBD \right) \right)=AH.$
Ta có: $CM\cap \left( SBD \right)=\left\{ S \right\}\Rightarrow \dfrac{d\left( M,\left( SBD \right) \right)}{d\left( C,\left( SBD \right) \right)}=\dfrac{MS}{CS}=\dfrac{1}{2}$, mà $d\left( C,\left( SBD \right) \right)=d\left( A,\left( SBD \right) \right)$
Nên: $d\left( C,\left( SBD \right) \right)=d\left( A,\left( SBD \right) \right)$ nên $AH=d\left( A,\left( SBD \right) \right)=2d\left( M,\left( SBD \right) \right)=\dfrac{a}{2}$
Xét tam giác $ABD:$ $\dfrac{1}{A{{E}^{2}}}=\dfrac{1}{A{{B}^{2}}}+\dfrac{1}{A{{D}^{2}}}\Rightarrow AE=\dfrac{2\sqrt{5}}{5}a$
Xét tam giác $SAE:$ $\dfrac{1}{A{{H}^{2}}}=\dfrac{1}{A{{E}^{2}}}+\dfrac{1}{S{{A}^{2}}}\Rightarrow SA=\dfrac{2\sqrt{11}}{11}a\Rightarrow {{V}_{S.ABCD}}=\dfrac{1}{3}SA.AB.AD=\dfrac{4{{a}^{3}}\sqrt{11}}{33}.$
Ta có ${{V}_{S.ABC}}=\dfrac{1}{2}{{V}_{S.ABCD}}=\dfrac{1}{2}\dfrac{4{{a}^{3}}\sqrt{11}}{33}=\dfrac{2{{a}^{3}}\sqrt{11}}{33}$
Mặt khác, ta có:
$\dfrac{{{V}_{S.ABM}}}{{{V}_{S.ABC}}}=\dfrac{SA.SB.SM}{SA.SB.SC}=\dfrac{1}{2}\Rightarrow {{V}_{S.ABM}}=\dfrac{1}{2}{{V}_{S.ABC}}=\dfrac{1}{2}\dfrac{2{{a}^{3}}\sqrt{11}}{33}=\dfrac{{{a}^{3}}\sqrt{11}}{33}$.
A. $\dfrac{{{a}^{3}}\sqrt{11}}{66}.$
B. $\dfrac{4{{a}^{3}}\sqrt{11}}{33}.$
C. $\dfrac{2{{a}^{3}}\sqrt{11}}{33}.$
D. $\dfrac{{{a}^{3}}\sqrt{11}}{33}.$
Hạ $AH\bot SE$ thì $BD\bot AH$ (Do $BD\bot \left( SAE \right)$ ), suy ra $AH\bot \left( SBD \right)$ hay $d\left( A,\left( SBD \right) \right)=AH.$
Ta có: $CM\cap \left( SBD \right)=\left\{ S \right\}\Rightarrow \dfrac{d\left( M,\left( SBD \right) \right)}{d\left( C,\left( SBD \right) \right)}=\dfrac{MS}{CS}=\dfrac{1}{2}$, mà $d\left( C,\left( SBD \right) \right)=d\left( A,\left( SBD \right) \right)$
Nên: $d\left( C,\left( SBD \right) \right)=d\left( A,\left( SBD \right) \right)$ nên $AH=d\left( A,\left( SBD \right) \right)=2d\left( M,\left( SBD \right) \right)=\dfrac{a}{2}$
Xét tam giác $ABD:$ $\dfrac{1}{A{{E}^{2}}}=\dfrac{1}{A{{B}^{2}}}+\dfrac{1}{A{{D}^{2}}}\Rightarrow AE=\dfrac{2\sqrt{5}}{5}a$
Xét tam giác $SAE:$ $\dfrac{1}{A{{H}^{2}}}=\dfrac{1}{A{{E}^{2}}}+\dfrac{1}{S{{A}^{2}}}\Rightarrow SA=\dfrac{2\sqrt{11}}{11}a\Rightarrow {{V}_{S.ABCD}}=\dfrac{1}{3}SA.AB.AD=\dfrac{4{{a}^{3}}\sqrt{11}}{33}.$
Ta có ${{V}_{S.ABC}}=\dfrac{1}{2}{{V}_{S.ABCD}}=\dfrac{1}{2}\dfrac{4{{a}^{3}}\sqrt{11}}{33}=\dfrac{2{{a}^{3}}\sqrt{11}}{33}$
Mặt khác, ta có:
$\dfrac{{{V}_{S.ABM}}}{{{V}_{S.ABC}}}=\dfrac{SA.SB.SM}{SA.SB.SC}=\dfrac{1}{2}\Rightarrow {{V}_{S.ABM}}=\dfrac{1}{2}{{V}_{S.ABC}}=\dfrac{1}{2}\dfrac{2{{a}^{3}}\sqrt{11}}{33}=\dfrac{{{a}^{3}}\sqrt{11}}{33}$.
Đáp án D.