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Câu hỏi: Cho khối chóp $S.ABC$ có $\widehat{ASB}=\widehat{BSC}=\widehat{CSA}=60{}^\circ ,$ $SA=a,$ $SB=2a,$ $SC=4a$. Tính thể tích khối chóp $S.ABC$ theo $a$.
A. $\dfrac{{{a}^{3}}\sqrt{2}}{3}$.
B. $\dfrac{2{{a}^{3}}\sqrt{2}}{3}$.
C. $\dfrac{4{{a}^{3}}\sqrt{2}}{3}$.
D. $\dfrac{8{{a}^{3}}\sqrt{2}}{3}$.
Gọi $B',C'$ lần lượt thuộc $SB,SC$ sao cho $SB'=a,SC'=a\Rightarrow \Delta AB'C'$ là tam giác đều cạnh $a.$
Xét $\Delta SOA$ có
$SO=\sqrt{S{{A}^{2}}-A{{C}^{2}}}=\sqrt{{{a}^{2}}-{{\left( \dfrac{a\sqrt{3}}{3} \right)}^{2}}}=\dfrac{a\sqrt{6}}{3}$
$\Rightarrow {{V}_{SAB'C'}}=\dfrac{1}{3}SO.{{S}_{\Delta AB'C'}}=\dfrac{1}{3}.\dfrac{a\sqrt{6}}{3}.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{{{a}^{3}}\sqrt{2}}{12}$
Lại có công thức Sin-San.
$\dfrac{{{V}_{SAB'C'}}}{{{V}_{SABC}}}=1.\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{8}$
$\Rightarrow {{V}_{SABC}}=8.{{V}_{S.A'B'C'}}=8.\dfrac{{{a}^{3}}\sqrt{2}}{12}=\dfrac{2\sqrt{2}{{a}^{3}}}{3}$.
A. $\dfrac{{{a}^{3}}\sqrt{2}}{3}$.
B. $\dfrac{2{{a}^{3}}\sqrt{2}}{3}$.
C. $\dfrac{4{{a}^{3}}\sqrt{2}}{3}$.
D. $\dfrac{8{{a}^{3}}\sqrt{2}}{3}$.
Xét $\Delta SOA$ có
$SO=\sqrt{S{{A}^{2}}-A{{C}^{2}}}=\sqrt{{{a}^{2}}-{{\left( \dfrac{a\sqrt{3}}{3} \right)}^{2}}}=\dfrac{a\sqrt{6}}{3}$
$\Rightarrow {{V}_{SAB'C'}}=\dfrac{1}{3}SO.{{S}_{\Delta AB'C'}}=\dfrac{1}{3}.\dfrac{a\sqrt{6}}{3}.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{{{a}^{3}}\sqrt{2}}{12}$
Lại có công thức Sin-San.
$\dfrac{{{V}_{SAB'C'}}}{{{V}_{SABC}}}=1.\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{8}$
$\Rightarrow {{V}_{SABC}}=8.{{V}_{S.A'B'C'}}=8.\dfrac{{{a}^{3}}\sqrt{2}}{12}=\dfrac{2\sqrt{2}{{a}^{3}}}{3}$.
Đáp án B.