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Câu hỏi: Cho hình chóp $S.ABC$ có đáy là tam giác $ABC$ vuông tại $C$, $AB=2a$, $AC=a$ và $SA$ vuông góc với mặt phẳng $\left( ABC \right)$. Biết góc giữa hai mặt phẳng $\left( SAB \right)$ và $\left( SBC \right)$ bằng ${{60}^{o}}$. Tính thể tích khối chóp $S.ABC$.
A. $\dfrac{{{a}^{3}}\sqrt{6}}{4}$.
B. $\dfrac{{{a}^{3}}\sqrt{2}}{2}$.
C. $\dfrac{{{a}^{3}}\sqrt{2}}{6}$.
D. $\dfrac{{{a}^{3}}\sqrt{6}}{12}$.
Trong $\left( ABC \right)$ kẻ $CH\bot AB$ ta có: $\left\{ \begin{aligned}
& CH\bot AB \\
& CH\bot SA \\
\end{aligned} \right.\Rightarrow CH\bot \left( SAB \right)\Rightarrow CH\bot SB$
Trong $\left( SBC \right)$ kẻ $CK\bot SB$ ta có: $\left\{ \begin{aligned}
& CH\bot SB \\
& CK\bot SB \\
\end{aligned} \right.\Rightarrow SB\bot \left( CHK \right)\Rightarrow HK\bot SB$.
$\Rightarrow \left( \widehat{\left( SAB \right),\left( SBC \right)} \right)=\widehat{\left( HK,CK \right)}=\widehat{CKH}=60{}^\circ $.
Xét tam giác vuông ABC ta có: $BC=\sqrt{4{{a}^{2}}-{{a}^{2}}}=a\sqrt{3}$
$CH=\dfrac{AC.BC}{AB}=\dfrac{a\sqrt{3}.a}{2a}=\dfrac{a\sqrt{3}}{2}$.
Xét tam giác vuông $CHK$ có: $HK=HC.\cot 60{}^\circ =\dfrac{a\sqrt{3}}{2}.\dfrac{1}{\sqrt{3}}=\dfrac{a}{2}$.
$HB=\dfrac{B{{C}^{2}}}{AB}=\dfrac{3{{a}^{2}}}{2a}=\dfrac{3a}{2}$
Ta có $\Delta BHK\sim\Delta BSA\left( g.g \right)\Rightarrow \dfrac{HK}{SA}=\dfrac{HB}{SB}$ $\Rightarrow \dfrac{\dfrac{a}{2}}{SA}=\dfrac{\dfrac{3a}{2}}{\sqrt{S{{A}^{2}}+4{{a}^{2}}}}\Rightarrow 3SA=\sqrt{S{{A}^{2}}+4{{a}^{2}}}$
$\Leftrightarrow 9S{{A}^{2}}=S{{A}^{2}}+4{{a}^{2}}\Leftrightarrow 8S{{A}^{2}}=4{{a}^{2}}\Leftrightarrow SA=\dfrac{a\sqrt{2}}{2}$
Vậy ${{V}_{S.ABC}}=\dfrac{1}{3}SA.{{S}_{\Delta ABC}}=\dfrac{1}{3}.\dfrac{a\sqrt{2}}{2}.\dfrac{1}{2}.2a.a=\dfrac{{{a}^{3}}\sqrt{2}}{6}$.
A. $\dfrac{{{a}^{3}}\sqrt{6}}{4}$.
B. $\dfrac{{{a}^{3}}\sqrt{2}}{2}$.
C. $\dfrac{{{a}^{3}}\sqrt{2}}{6}$.
D. $\dfrac{{{a}^{3}}\sqrt{6}}{12}$.
Trong $\left( ABC \right)$ kẻ $CH\bot AB$ ta có: $\left\{ \begin{aligned}
& CH\bot AB \\
& CH\bot SA \\
\end{aligned} \right.\Rightarrow CH\bot \left( SAB \right)\Rightarrow CH\bot SB$
Trong $\left( SBC \right)$ kẻ $CK\bot SB$ ta có: $\left\{ \begin{aligned}
& CH\bot SB \\
& CK\bot SB \\
\end{aligned} \right.\Rightarrow SB\bot \left( CHK \right)\Rightarrow HK\bot SB$.
$\Rightarrow \left( \widehat{\left( SAB \right),\left( SBC \right)} \right)=\widehat{\left( HK,CK \right)}=\widehat{CKH}=60{}^\circ $.
Xét tam giác vuông ABC ta có: $BC=\sqrt{4{{a}^{2}}-{{a}^{2}}}=a\sqrt{3}$
$CH=\dfrac{AC.BC}{AB}=\dfrac{a\sqrt{3}.a}{2a}=\dfrac{a\sqrt{3}}{2}$.
Xét tam giác vuông $CHK$ có: $HK=HC.\cot 60{}^\circ =\dfrac{a\sqrt{3}}{2}.\dfrac{1}{\sqrt{3}}=\dfrac{a}{2}$.
$HB=\dfrac{B{{C}^{2}}}{AB}=\dfrac{3{{a}^{2}}}{2a}=\dfrac{3a}{2}$
Ta có $\Delta BHK\sim\Delta BSA\left( g.g \right)\Rightarrow \dfrac{HK}{SA}=\dfrac{HB}{SB}$ $\Rightarrow \dfrac{\dfrac{a}{2}}{SA}=\dfrac{\dfrac{3a}{2}}{\sqrt{S{{A}^{2}}+4{{a}^{2}}}}\Rightarrow 3SA=\sqrt{S{{A}^{2}}+4{{a}^{2}}}$
$\Leftrightarrow 9S{{A}^{2}}=S{{A}^{2}}+4{{a}^{2}}\Leftrightarrow 8S{{A}^{2}}=4{{a}^{2}}\Leftrightarrow SA=\dfrac{a\sqrt{2}}{2}$
Vậy ${{V}_{S.ABC}}=\dfrac{1}{3}SA.{{S}_{\Delta ABC}}=\dfrac{1}{3}.\dfrac{a\sqrt{2}}{2}.\dfrac{1}{2}.2a.a=\dfrac{{{a}^{3}}\sqrt{2}}{6}$.
Đáp án C.