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Câu hỏi: Cho hình chóp đều S.ABC có cạnh đáy bằng a, cạnh bên bằng 2a. Khoảng cách từ điểm A đến mặt phẳng $\left( SBC \right)$ bằng
A. $\dfrac{a\sqrt{165}}{30}.$
B. $\dfrac{a\sqrt{165}}{45}.$
C. $\dfrac{a\sqrt{165}}{15}.$
D. $\dfrac{2a\sqrt{165}}{15}.$
Kẻ $SH\bot \left( ABC \right)$, gọi $K=AH\cap BC$.
Kẻ $HP\bot \text{S}K\Rightarrow d\left( A;(SBC) \right)=\dfrac{3}{2}d\left( H;(SBC) \right)=\dfrac{3}{2}HP=d$.
Ta có $\dfrac{1}{H{{P}^{2}}}=\dfrac{1}{S{{H}^{2}}}+\dfrac{1}{H{{K}^{2}}}$. Cạnh $HK=\dfrac{AB}{2\sqrt{3}}=\dfrac{a}{2\sqrt{3}}$
$S{{H}^{2}}=S{{A}^{2}}-A{{H}^{2}}=4{{\text{a}}^{2}}-{{\left( \dfrac{AB}{\sqrt{3}} \right)}^{2}}=\dfrac{11{{\text{a}}^{2}}}{3}$
$\Rightarrow HP=a\sqrt{\dfrac{11}{135}}\Rightarrow d\left( A;(SBC) \right)=\dfrac{a\sqrt{165}}{15}$.
A. $\dfrac{a\sqrt{165}}{30}.$
B. $\dfrac{a\sqrt{165}}{45}.$
C. $\dfrac{a\sqrt{165}}{15}.$
D. $\dfrac{2a\sqrt{165}}{15}.$
Kẻ $SH\bot \left( ABC \right)$, gọi $K=AH\cap BC$.
Kẻ $HP\bot \text{S}K\Rightarrow d\left( A;(SBC) \right)=\dfrac{3}{2}d\left( H;(SBC) \right)=\dfrac{3}{2}HP=d$.
Ta có $\dfrac{1}{H{{P}^{2}}}=\dfrac{1}{S{{H}^{2}}}+\dfrac{1}{H{{K}^{2}}}$. Cạnh $HK=\dfrac{AB}{2\sqrt{3}}=\dfrac{a}{2\sqrt{3}}$
$S{{H}^{2}}=S{{A}^{2}}-A{{H}^{2}}=4{{\text{a}}^{2}}-{{\left( \dfrac{AB}{\sqrt{3}} \right)}^{2}}=\dfrac{11{{\text{a}}^{2}}}{3}$
$\Rightarrow HP=a\sqrt{\dfrac{11}{135}}\Rightarrow d\left( A;(SBC) \right)=\dfrac{a\sqrt{165}}{15}$.
Đáp án C.