Cho hình chóp SABC có đáy ABC là tam giác vuông tại A, AB=4cm

* Hướng dẫn giải

Đáp án là D.

Gọi I là điểm thuộc SA   sao cho $\frac{SI}{SA}=\frac{1}{3}\Rightarrow IM\text{//}AC$ .

Gọi  H là trung điểm của .AB  $\left.\begin{array}{l}\left(SAB\right)\perp \left(ABC\right)\\ \left(SAB\right)\cap \left(ABC\right)=AB\\ SH\perp AB\end{array}\right\}\Rightarrow SH\perp \left(ABC\right)$

$$\begin{gathered} \left.\begin{array}{l}AC\perp AB\\ AC\perp SH\end{array}\right\}\Rightarrow AC\perp \left(SAB\right)\Rightarrow IM\perp \left(SAB\right) \\ \Rightarrow IM\perp BI\Rightarrow \Delta BIM \end{gathered}$$

$$\begin{gathered} \frac{V_{SBAM}}{V_{SBAC}}=\frac{SM}{SC}=\frac{1}{3} \\ \Rightarrow V_{SBAM}=\frac{1}{3}{V}_{SBAC} \\ =\frac{1}{3}.\frac{1}{3}SH.S_{△ABC} \\ =\frac{1}{9}.\frac{4\sqrt{3}}{2}\frac{1}{2}AB.AC=\frac{4\sqrt{3}}{9}AC \end{gathered}$$

$$\begin{gathered} \frac{V_{ABIM}}{V_{ABSM}}=\frac{AI}{AS}=\frac{2}{3} \\ \Rightarrow V_{ABIM}=\frac{2}{3}{V}_{ABSM} \\ =\frac{2}{3}.\frac{4\sqrt{3}}{9}AC=\frac{8\sqrt{3}}{27}AC \end{gathered}$$

$$\begin{gathered} B{I}^2=A{B}^2+A{I}^2-2AB.AI.c\text{os}{60}^0 \\ =4^2+{\left(\frac{8}{3}\right)}^2-\mathrm{2.4.}\frac{8}{3}.c\text{os}{60}^0 \\ =\frac{112}{9}\Rightarrow BI=\frac{4\sqrt{7}}{3} \end{gathered}$$

$$\begin{gathered} S_{\Delta BIM}=\frac{1}{2}BI.IM \\ =\frac{1}{2}.\frac{4\sqrt{7}}{3}.\frac{1}{3}AC \\ =\frac{2\sqrt{7}}{9}AC \end{gathered}$$

$$\begin{gathered} V_{ABIM}=\frac{1}{3}{S}_{△BIM}.d\left(A,\left(BIM\right)\right) \\ \Rightarrow d\left(A,\left(BIM\right)\right)=\frac{3V_{ABIM}}{S_{△BIM}} \\ =\frac{3.\frac{8\sqrt{3}}{27}AC}{\frac{2\sqrt{7}}{9}AC}=\frac{4\sqrt{21}}{7} \end{gathered}$$