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Câu hỏi: Cho lăng trụ $ABCD.{A}'{B}'{C}'{D}'$ có đáy ABCD là hình chữ nhật với $AB=\sqrt{6},AD=\sqrt{3},{A}'C=3$ và mặt phẳng $\left( AC{C}'{A}' \right)$ vuông góc với mặt đáy. Biết hai mặt phẳng $\left( AC{C}'{A}' \right)$, $\left( AB{B}'{A}' \right)$ tạo với nhau góc $\alpha $ có $\tan \alpha =\dfrac{3}{4}.$ Thể tích của khối lăng trụ $ABCD.{A}'{B}'{C}'{D}'$ là
A. $V=12$
B. $V=6$
C. $V=8$
D. $V=10$
A. $V=12$
B. $V=6$
C. $V=8$
D. $V=10$
Cách giải:
Trong (ABCD) kẻ $BH\bot AC\left( H\in AC \right)$ ta có:
$\left\{ \begin{aligned}
& \left( ABCD \right)\bot \left( AC{C}'{A}' \right)=AC \\
& AH\subset \left( ABCD \right),AH\bot AC \\
\end{aligned} \right.\Rightarrow AH\bot \left( AC{C}'{A}' \right)$
Trong (ACC'A') kẻ $HK\bot A{A}'\left( K\in A{A}' \right)$ ta có: $\left\{ \begin{aligned}
& HK\bot A{A}' \\
& BH\bot A{A}' \\
\end{aligned} \right.$
$A{A}'\bot \left( BHK \right)\Rightarrow A{A}'\bot BK$
$\left\{ \begin{aligned}
& \left( AC{C}'{A}' \right)\cap \left( A{A}'{B}'B \right)=A{A}' \\
& HK\subset \left( AC{C}'{A}' \right),HK\bot A{A}' \\
& BK\subset \left( AC{C}'{A}' \right),BK\bot A{A}' \\
\end{aligned} \right.$
$\Rightarrow \angle \left( \left( AC{C}'{A}' \right),\left( \left( AB{B}'{A}' \right) \right) \right)=\angle \left( HK,BK \right)=\angle BKH=\alpha $
Ta có: $BH\bot \left( AC{C}'{A}' \right)\Rightarrow BH\bot HK\Rightarrow \Delta BHK$ vuông tại H
Ta có: $BH=\dfrac{AB.BC}{\sqrt{A{{B}^{2}}+B{{C}^{2}}}}=\dfrac{\sqrt{6}.\sqrt{3}}{\sqrt{6+3}}=\sqrt{2}$
$\Rightarrow HK=BH.\cot \alpha =\sqrt{2}.\dfrac{4}{3}=\dfrac{4\sqrt{2}}{3}$
Áp dụng hệ thức lượng trong tam giác vuông ABC ta có: $AH=\dfrac{A{{B}^{2}}}{AC}=\dfrac{6}{\sqrt{6+3}}=2$
Xét tam giác vuông AHK có: $\sin HAK=\dfrac{HK}{AH}=\dfrac{4\sqrt{2}}{3}:2=\dfrac{2\sqrt{2}}{3}\Rightarrow \cos HAK=\pm \sqrt{1-{{\sin }^{2}}HAK}=\pm \dfrac{1}{3}=\cos {A}'AC$
Áp dụng định lý Cosin trong tam giác A'AC có:
TH1: $\cos {A}'AC=\dfrac{1}{3}$
$\cos {A}'AC=\dfrac{A{{{{A}'}}^{2}}+A{{C}^{2}}-{A}'{{C}^{2}}}{2.{A}'A.AC}\Rightarrow \dfrac{1}{3}=\dfrac{A{{{{A}'}}^{2}}+9-9}{2.{A}'A.3}\Leftrightarrow \dfrac{1}{3}=\dfrac{A{A}'}{6}\Leftrightarrow A{A}'=2$
TH2: $\cos {A}'AC=-\dfrac{1}{3}$
$\cos {A}'AC=\dfrac{A{{{{A}'}}^{2}}+A{{C}^{2}}-{A}'{{C}^{2}}}{2A{A}'.AC}\Leftrightarrow \dfrac{1}{3}=\dfrac{A{{{{A}'}}^{2}}+9-9}{2.A{A}'.3}\Leftrightarrow \dfrac{1}{3}=\dfrac{{A}'A}{6}\left( ktm \right)$
$\Rightarrow {{S}_{\Delta A{A}'C}}=\dfrac{1}{2}A{A}'.AC.\sin {A}'AC=\dfrac{1}{2}.2.3.\dfrac{2\sqrt{2}}{3}=2\sqrt{2}$
$\Rightarrow {{S}_{AC{C}'{A}'}}=2{{S}_{\Delta A{A}'C}}=4\sqrt{2}$
$\Rightarrow {{V}_{B.AC{C}'{A}'}}=\dfrac{1}{3}BH.{{S}_{AC{C}'{A}'}}=\dfrac{1}{3}.\sqrt{2}.4\sqrt{2}=\dfrac{8}{3}$
$\Rightarrow {{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{3}{2}{{V}_{B.AC{C}'{A}'}}=4$
$\Rightarrow {{V}_{ABCD.{A}'{B}'{C}'{D}'}}=2{{V}_{ABC.{A}'{B}'{C}'}}=8$
Trong (ABCD) kẻ $BH\bot AC\left( H\in AC \right)$ ta có:
$\left\{ \begin{aligned}
& \left( ABCD \right)\bot \left( AC{C}'{A}' \right)=AC \\
& AH\subset \left( ABCD \right),AH\bot AC \\
\end{aligned} \right.\Rightarrow AH\bot \left( AC{C}'{A}' \right)$
Trong (ACC'A') kẻ $HK\bot A{A}'\left( K\in A{A}' \right)$ ta có: $\left\{ \begin{aligned}
& HK\bot A{A}' \\
& BH\bot A{A}' \\
\end{aligned} \right.$
$A{A}'\bot \left( BHK \right)\Rightarrow A{A}'\bot BK$
$\left\{ \begin{aligned}
& \left( AC{C}'{A}' \right)\cap \left( A{A}'{B}'B \right)=A{A}' \\
& HK\subset \left( AC{C}'{A}' \right),HK\bot A{A}' \\
& BK\subset \left( AC{C}'{A}' \right),BK\bot A{A}' \\
\end{aligned} \right.$
$\Rightarrow \angle \left( \left( AC{C}'{A}' \right),\left( \left( AB{B}'{A}' \right) \right) \right)=\angle \left( HK,BK \right)=\angle BKH=\alpha $
Ta có: $BH\bot \left( AC{C}'{A}' \right)\Rightarrow BH\bot HK\Rightarrow \Delta BHK$ vuông tại H
Ta có: $BH=\dfrac{AB.BC}{\sqrt{A{{B}^{2}}+B{{C}^{2}}}}=\dfrac{\sqrt{6}.\sqrt{3}}{\sqrt{6+3}}=\sqrt{2}$
$\Rightarrow HK=BH.\cot \alpha =\sqrt{2}.\dfrac{4}{3}=\dfrac{4\sqrt{2}}{3}$
Áp dụng hệ thức lượng trong tam giác vuông ABC ta có: $AH=\dfrac{A{{B}^{2}}}{AC}=\dfrac{6}{\sqrt{6+3}}=2$
Xét tam giác vuông AHK có: $\sin HAK=\dfrac{HK}{AH}=\dfrac{4\sqrt{2}}{3}:2=\dfrac{2\sqrt{2}}{3}\Rightarrow \cos HAK=\pm \sqrt{1-{{\sin }^{2}}HAK}=\pm \dfrac{1}{3}=\cos {A}'AC$
Áp dụng định lý Cosin trong tam giác A'AC có:
TH1: $\cos {A}'AC=\dfrac{1}{3}$
$\cos {A}'AC=\dfrac{A{{{{A}'}}^{2}}+A{{C}^{2}}-{A}'{{C}^{2}}}{2.{A}'A.AC}\Rightarrow \dfrac{1}{3}=\dfrac{A{{{{A}'}}^{2}}+9-9}{2.{A}'A.3}\Leftrightarrow \dfrac{1}{3}=\dfrac{A{A}'}{6}\Leftrightarrow A{A}'=2$
TH2: $\cos {A}'AC=-\dfrac{1}{3}$
$\cos {A}'AC=\dfrac{A{{{{A}'}}^{2}}+A{{C}^{2}}-{A}'{{C}^{2}}}{2A{A}'.AC}\Leftrightarrow \dfrac{1}{3}=\dfrac{A{{{{A}'}}^{2}}+9-9}{2.A{A}'.3}\Leftrightarrow \dfrac{1}{3}=\dfrac{{A}'A}{6}\left( ktm \right)$
$\Rightarrow {{S}_{\Delta A{A}'C}}=\dfrac{1}{2}A{A}'.AC.\sin {A}'AC=\dfrac{1}{2}.2.3.\dfrac{2\sqrt{2}}{3}=2\sqrt{2}$
$\Rightarrow {{S}_{AC{C}'{A}'}}=2{{S}_{\Delta A{A}'C}}=4\sqrt{2}$
$\Rightarrow {{V}_{B.AC{C}'{A}'}}=\dfrac{1}{3}BH.{{S}_{AC{C}'{A}'}}=\dfrac{1}{3}.\sqrt{2}.4\sqrt{2}=\dfrac{8}{3}$
$\Rightarrow {{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{3}{2}{{V}_{B.AC{C}'{A}'}}=4$
$\Rightarrow {{V}_{ABCD.{A}'{B}'{C}'{D}'}}=2{{V}_{ABC.{A}'{B}'{C}'}}=8$
Đáp án C.