Cho hình chóp S.ABCD có đáy là...

A

- Hướng dẫn giải

Ta có $\left(SCD\right)\cap \left(ABCD\right)=CD$

$$\begin{gathered} \left\{\begin{array}{l}CD\perp SA\\ CD\perp AC\end{array}\right. \\ \Rightarrow CD\perp \left(SAC\right) \\ \Rightarrow SC\perp CD \end{gathered}$$

Vì $\left\{\begin{array}{l}SC\perp CD,SC\subset \left(SCD\right)\\ AC\perp CD,AC\subset \left(ABCD\right)\end{array}\right.$

Nên $\left(\widehat{\left(SCD\right),\left(ABCD\right)}\right)=\widehat{SCA}={45}^o$

Dễ thấy $∆SAC$ vuông cân tại A

Suy ra SA = AC = $a\sqrt{2}$

Lại có

 $$\begin{gathered} S_{MCD}=\frac{1}{2}MC.MD \\ =\frac{1}{2}a.a=\frac{a^2}{2} \end{gathered}$$

Do đó

 $$\begin{gathered} V=V_{S.MCD}=\frac{1}{3}{S}_{MCD}SA \\ =\frac{1}{3}.\frac{a^2}{2}.a\sqrt{2}=\frac{a^3\sqrt{2}}{6} \end{gathered}$$

Ta có

 $$\begin{gathered} \left\{\begin{array}{l}BD\parallel MN\\ MN\subset \left(SMN\right)\end{array}\right. \\ \Rightarrow BD\parallel \left(SMN\right) \end{gathered}$$

Khi đó d( SM,BD ) = d( SM, (SMN) ) = d( D, (SMN) ) = d( A, ( SMN) )

Kẻ $\left\{\begin{array}{l}AP\perp MN,P\in MN\\ AH\perp SP,H\in SP\end{array}\right.$

Suy ra $AH\perp \left(SMN\right)\Rightarrow d\left(A\left(SMN\right)\right)=AH$

$∆SAP$ vuông tại A có

$$\begin{gathered} \frac{1}{A{H}^2}=\frac{1}{S{A}^2}+\frac{1}{A{P}^2} \\ =\frac{1}{S{A}^2}+\frac{1}{A{N}^2}+\frac{1}{A{M}^2} \\ =\frac{1}{2a^2}+\frac{1}{{\displaystyle \frac{a^2}{4}}}+\frac{1}{a^2} \\ =\frac{11}{2a^2} \end{gathered}$$

Do đó d = d( SM, BD ) = AH = $\frac{a\sqrt{22}}{11}$

Đáp án A