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Câu hỏi: Cho hình chóp $S.ABC$ có $SA\bot \left( ABC \right),SA=a$, tam giác $ABC$ đều có cạnh $2a.$ Tính thể tích khối chóp $S.ABC.$
A. ${{a}^{3}}\sqrt{3}.$
B. $\dfrac{{{a}^{3}}\sqrt{3}}{3}.$
C. $\dfrac{{{a}^{3}}\sqrt{3}}{2}.$
D. $\dfrac{{{a}^{3}}\sqrt{3}}{6}.$
Ta có: ${{S}_{\Delta ABC}}=\dfrac{\sqrt{3}}{4}.\left( 2{{a}^{2}} \right)={{a}^{2}}\sqrt{3}$
${{V}_{S.ABC}}=\dfrac{1}{3}{{S}_{\Delta ABC}}.SA=\dfrac{1}{3}{{a}^{2}}\sqrt{3}.a=\dfrac{{{a}^{3}}\sqrt{3}}{3}.$
A. ${{a}^{3}}\sqrt{3}.$
B. $\dfrac{{{a}^{3}}\sqrt{3}}{3}.$
C. $\dfrac{{{a}^{3}}\sqrt{3}}{2}.$
D. $\dfrac{{{a}^{3}}\sqrt{3}}{6}.$
Ta có: ${{S}_{\Delta ABC}}=\dfrac{\sqrt{3}}{4}.\left( 2{{a}^{2}} \right)={{a}^{2}}\sqrt{3}$
${{V}_{S.ABC}}=\dfrac{1}{3}{{S}_{\Delta ABC}}.SA=\dfrac{1}{3}{{a}^{2}}\sqrt{3}.a=\dfrac{{{a}^{3}}\sqrt{3}}{3}.$
Đáp án B.