Câu hỏi 30

* Nhận xét: Chóp A’.ABC đều. Vẽ O là tâm \(\Delta ABC\Rightarrow A'O\bot \left( ABC \right)\).

* \(AO=\frac{2}{3}AM=\frac{2}{3}.\frac{a\sqrt{3}}{2}=\frac{a\sqrt{3}}{3}\)

* Tam giác vuông A’OA: \(A'O=\sqrt{{{a}^{2}}-\frac{3{{a}^{2}}}{9}}=\frac{a\sqrt{6}}{3}\)

*  \(d\left( N;\left( ABC \right) \right)=\frac{1}{2}A'O=\frac{a\sqrt{6}}{6}\).

* \({{V}_{NACM}}=\frac{1}{3}\left[ d\left( N;\left( ABC \right) \right) \right].{{S}_{\Delta AMC}}=\frac{1}{3}.\frac{a\sqrt{6}}{6}.\frac{1}{2}.\frac{a}{2}.\frac{a\sqrt{3}}{2}=\frac{{{a}^{3}}\sqrt{2}}{48}\)

* \(\Delta A'AB\) đều \(\Rightarrow AN=\frac{a\sqrt{3}}{2}=AM,\,\,MN=\frac{A'C}{2}=\frac{a}{2}\).

Vẽ \(AE\bot MN,\,\,AE=\sqrt{\frac{3{{a}^{2}}}{4}-\frac{{{a}^{2}}}{16}}=\frac{a\sqrt{11}}{4}\)

\(\Rightarrow {{S}_{\Delta AMN}}=\frac{1}{2}.\frac{a}{2}.\frac{a\sqrt{11}}{4}=\frac{{{a}^{2}}\sqrt{11}}{16}\)

* \(d\left( C;\left( AMN \right) \right)=\frac{3{{V}_{NACM}}}{{{S}_{\Delta AMN}}}=\frac{a\sqrt{22}}{11}\)

Chọn đáp án A.