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Câu hỏi: Tính thể tích khối chóp tam giác đều S.ABC biết cạnh đáy bằng $a\sqrt{3}$, cạnh bên bẳng $2a$.
A. $\dfrac{3}{4}{{a}^{3}}$.
B. $\dfrac{\sqrt{11}}{4}{{a}^{3}}$.
C. $\dfrac{\sqrt{11}}{12}{{a}^{3}}$.
D. $\dfrac{9}{4}{{a}^{3}}$.
$SH\bot \left( ABC \right)\Rightarrow AH=\dfrac{AB}{\sqrt{3}}=a\Rightarrow SH=\sqrt{S{{A}^{2}}-A{{H}^{2}}}=a\sqrt{3}$
${{V}_{S.ABC}}=\dfrac{1}{3}SH.{{S}_{ABC}}=\dfrac{1}{3}a\sqrt{3}.\dfrac{{{\left( a\sqrt{3} \right)}^{2}}\sqrt{3}}{4}=\dfrac{3}{4}{{a}^{3}}$.
A. $\dfrac{3}{4}{{a}^{3}}$.
B. $\dfrac{\sqrt{11}}{4}{{a}^{3}}$.
C. $\dfrac{\sqrt{11}}{12}{{a}^{3}}$.
D. $\dfrac{9}{4}{{a}^{3}}$.
$SH\bot \left( ABC \right)\Rightarrow AH=\dfrac{AB}{\sqrt{3}}=a\Rightarrow SH=\sqrt{S{{A}^{2}}-A{{H}^{2}}}=a\sqrt{3}$
${{V}_{S.ABC}}=\dfrac{1}{3}SH.{{S}_{ABC}}=\dfrac{1}{3}a\sqrt{3}.\dfrac{{{\left( a\sqrt{3} \right)}^{2}}\sqrt{3}}{4}=\dfrac{3}{4}{{a}^{3}}$.
Đáp án A.