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Câu hỏi: Cho hình chóp $S.ABC$ có $\widehat{ASB}=\widehat{ASC}=\widehat{BSC}={{60}^{o}},SA=5a,SB=6a,SC=3a.$ Tính thể tích khối chóp $S.ABC$ theo $a$.
A. $\dfrac{15{{a}^{3}}\sqrt{2}}{2}$.
B. $\dfrac{15{{a}^{3}}\sqrt{2}}{8}$.
C. $\dfrac{15{{a}^{3}}\sqrt{2}}{4}$.
D. $\dfrac{15{{a}^{3}}\sqrt{2}}{7}$.
A. $\dfrac{15{{a}^{3}}\sqrt{2}}{2}$.
B. $\dfrac{15{{a}^{3}}\sqrt{2}}{8}$.
C. $\dfrac{15{{a}^{3}}\sqrt{2}}{4}$.
D. $\dfrac{15{{a}^{3}}\sqrt{2}}{7}$.
Cách 1: Ta có
$\begin{aligned}
& {{V}_{S.ABC}}=\dfrac{SA.SB.SC}{6}\cdot \sqrt{1-{{\cos }^{2}}\widehat{ASB}-{{\cos }^{2}}\widehat{BSC}-{{\cos }^{2}}\widehat{CSB}+2\cos \widehat{ASB}.\cos \widehat{BSC}.\cos \widehat{CSB}} \\
& {{V}_{S.ABC}}=\dfrac{5a.6a.3a}{6}\sqrt{1-\dfrac{3}{4}+2.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}}=\dfrac{15{{a}^{3}}\sqrt{2}}{2} \\
\end{aligned}$
Cách 2: Ta có:
Chọn $A'\in SA:SA'=3a;B'\in SB:SB'=3a\Rightarrow \dfrac{{{V}_{S.ABC}}}{{{V}_{S.A'B'C}}}=\dfrac{SA}{SA'}.\dfrac{SB}{SB'}.\dfrac{SC}{SC'}=\dfrac{3a.3a.3a}{3a.5a.6a}=\dfrac{3}{10}$
Do tứ diện đều $S.ABC$ nên ${{V}_{S.A'B'C}}=\dfrac{{{\left( 3a \right)}^{3}}\sqrt{2}}{12}$ = $\dfrac{9\sqrt{2}}{4}{{a}^{3}}\Rightarrow {{V}_{S.ABC}}=\dfrac{15{{a}^{3}}\sqrt{2}}{2}$.
$\begin{aligned}
& {{V}_{S.ABC}}=\dfrac{SA.SB.SC}{6}\cdot \sqrt{1-{{\cos }^{2}}\widehat{ASB}-{{\cos }^{2}}\widehat{BSC}-{{\cos }^{2}}\widehat{CSB}+2\cos \widehat{ASB}.\cos \widehat{BSC}.\cos \widehat{CSB}} \\
& {{V}_{S.ABC}}=\dfrac{5a.6a.3a}{6}\sqrt{1-\dfrac{3}{4}+2.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}}=\dfrac{15{{a}^{3}}\sqrt{2}}{2} \\
\end{aligned}$
Cách 2: Ta có:
Chọn $A'\in SA:SA'=3a;B'\in SB:SB'=3a\Rightarrow \dfrac{{{V}_{S.ABC}}}{{{V}_{S.A'B'C}}}=\dfrac{SA}{SA'}.\dfrac{SB}{SB'}.\dfrac{SC}{SC'}=\dfrac{3a.3a.3a}{3a.5a.6a}=\dfrac{3}{10}$
Do tứ diện đều $S.ABC$ nên ${{V}_{S.A'B'C}}=\dfrac{{{\left( 3a \right)}^{3}}\sqrt{2}}{12}$ = $\dfrac{9\sqrt{2}}{4}{{a}^{3}}\Rightarrow {{V}_{S.ABC}}=\dfrac{15{{a}^{3}}\sqrt{2}}{2}$.
Đáp án A.