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Câu hỏi: Cho hình chóp đều $S.ABC$ có cạnh đáy bằng $a$, cạnh bên $2a$, $M$ là trung điểm của $SA$. Tính khoảng cách từ điểm $M$ đến mặt phẳng $\left( SBC \right)$.
A. $\dfrac{a\sqrt{165}}{45}$.
B. $\dfrac{a\sqrt{165}}{30}$.
C. $\dfrac{a\sqrt{165}}{15}$.
D. $\dfrac{a\sqrt{165}}{20}$.
Gọi $N$ là trung điểm $BC$. Khi đó, ta có: $\left\{ \begin{aligned}
& BC\bot AN \\
& BC\bot SG \\
\end{aligned} \right.\Rightarrow BC\bot \left( SAN \right)\Rightarrow \left( SBC \right)\bot \left( SAN \right)$
Lại có: $\left( SBC \right)\cap \left( SAN \right)=SN$. Do đó, kẻ $\left\{ \begin{aligned}
& MK\bot SN\Rightarrow MK\bot \left( SBC \right) \\
& AH\bot SN\Rightarrow AH\bot \left( SBC \right) \\
\end{aligned} \right.\Rightarrow MK//AH$.
Ta có: $d\left( M,\left( SBC \right) \right)=MK=\dfrac{1}{2}AH$ (vì $\dfrac{MK}{AH}=\dfrac{SM}{SA}=\dfrac{1}{2}$ ).
Ta có: $\sin \widehat{ANH}=\dfrac{AH}{AN}\Rightarrow AH=\sin \widehat{ANH}.AN=\dfrac{SG}{SN}.AN=\dfrac{\sqrt{S{{A}^{2}}-A{{G}^{2}}}.AN}{\sqrt{S{{B}^{2}}-N{{B}^{2}}}}$
$=\dfrac{\sqrt{{{\left( 2a \right)}^{2}}-{{\left( \dfrac{2}{3}.\dfrac{a\sqrt{3}}{2} \right)}^{2}}}.\dfrac{a\sqrt{3}}{2}}{\sqrt{{{\left( 2a \right)}^{2}}-{{\left( \dfrac{1}{2}a \right)}^{2}}}}=\dfrac{a\sqrt{165}}{15}$.
Vậy $d\left( M,\left( SBC \right) \right)=MK=\dfrac{1}{2}AH=\dfrac{a\sqrt{165}}{30}$.
A. $\dfrac{a\sqrt{165}}{45}$.
B. $\dfrac{a\sqrt{165}}{30}$.
C. $\dfrac{a\sqrt{165}}{15}$.
D. $\dfrac{a\sqrt{165}}{20}$.
& BC\bot AN \\
& BC\bot SG \\
\end{aligned} \right.\Rightarrow BC\bot \left( SAN \right)\Rightarrow \left( SBC \right)\bot \left( SAN \right)$
Lại có: $\left( SBC \right)\cap \left( SAN \right)=SN$. Do đó, kẻ $\left\{ \begin{aligned}
& MK\bot SN\Rightarrow MK\bot \left( SBC \right) \\
& AH\bot SN\Rightarrow AH\bot \left( SBC \right) \\
\end{aligned} \right.\Rightarrow MK//AH$.
Ta có: $d\left( M,\left( SBC \right) \right)=MK=\dfrac{1}{2}AH$ (vì $\dfrac{MK}{AH}=\dfrac{SM}{SA}=\dfrac{1}{2}$ ).
Ta có: $\sin \widehat{ANH}=\dfrac{AH}{AN}\Rightarrow AH=\sin \widehat{ANH}.AN=\dfrac{SG}{SN}.AN=\dfrac{\sqrt{S{{A}^{2}}-A{{G}^{2}}}.AN}{\sqrt{S{{B}^{2}}-N{{B}^{2}}}}$
$=\dfrac{\sqrt{{{\left( 2a \right)}^{2}}-{{\left( \dfrac{2}{3}.\dfrac{a\sqrt{3}}{2} \right)}^{2}}}.\dfrac{a\sqrt{3}}{2}}{\sqrt{{{\left( 2a \right)}^{2}}-{{\left( \dfrac{1}{2}a \right)}^{2}}}}=\dfrac{a\sqrt{165}}{15}$.
Vậy $d\left( M,\left( SBC \right) \right)=MK=\dfrac{1}{2}AH=\dfrac{a\sqrt{165}}{30}$.
Đáp án B.
