Cho hình lăng trụ đứng ABC.A’B’C’ có

D

- Hướng dẫn giải

Đáp án D.

Tam giác ABC có

$$\begin{gathered} \widehat{BAC}={120}^0 \\ \Rightarrow S_{\Delta ABC}=\frac{1}{2}.AB.AC.\sin \widehat{BAC}=\frac{\sqrt{3}}{2} \end{gathered}$$

và $BC=\sqrt{7}.$

Ta có

$$\begin{gathered} \frac{S_{\Delta A\text{'}BM}}{S_{\Delta A\text{'}BB\text{'}}}=\frac{BM}{BB\text{'}}=\frac{3}{4} \\ \Rightarrow \frac{V_{N.A\text{'}BM}}{V_{N.A\text{'}B\text{'}B}}=\frac{3}{4} \end{gathered}$$

mà 

$$\begin{gathered} V_{N.A\text{'}B\text{'}B}=V_{C\text{'}.A\text{'}B\text{'}B} \\ =\frac{1}{2}{V}_{C\text{'}.ABB\text{'}A\text{'}}=\frac{1}{3}{V}_{ABC.A\text{'}B\text{'}C\text{'}} \end{gathered}$$

Suy ra:

$$\begin{gathered} V_{N.A\text{'}BM}=\frac{3}{4}{V}_{N.A\text{'}B\text{'}B} \\ =\frac{1}{4}{V}_{ABC.A\text{'}B\text{'}C\text{'}} \\ =\frac{1}{4}\text{.AA}\text{'}.S_{\Delta ABC}=\frac{3\sqrt{3}}{8}. \end{gathered}$$

Tam giác $A\text{'}BN$ có $A\text{'}B=\sqrt{10},BN=\sqrt{11}$ và $A\text{'}N=\sqrt{5}\stackrel{}{\to }{S}_{\Delta A\text{'}BN}=\frac{\sqrt{46}}{2}.$

Khi đó:

$$\begin{gathered} V_{N.A\text{'}BM}=\frac{1}{3}.d\left(M;\left(A\text{'}BN\right)\right).S_{\Delta A\text{'}BN} \\ \Rightarrow d\left(M;\left(A\text{'}BN\right)\right) \\ =\frac{9\sqrt{3}}{8}:\frac{\sqrt{46}}{2}=\frac{9\sqrt{138}}{46}. \end{gathered}$$