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Câu hỏi: Cho hình lăng trụ đứng $ABC.{A}'{B}'{C}'$ có tất cả các cạnh bằng $a$. Gọi $M$ là trung điểm của $C{C}'$ (tham khảo hình bên). Khoảng cách từ $M$ đến mặt phẳng $\left( {A}'BC \right)$ bằng
A. $\dfrac{\sqrt{21}a}{14}$.
B. $\dfrac{\sqrt{2}a}{2}$.
C. $\dfrac{\sqrt{21}a}{7}$.
D. $\dfrac{\sqrt{2}a}{4}$.
A. $\dfrac{\sqrt{21}a}{14}$.
B. $\dfrac{\sqrt{2}a}{2}$.
C. $\dfrac{\sqrt{21}a}{7}$.
D. $\dfrac{\sqrt{2}a}{4}$.
${C}'M\cap \left( {A}'BC \right)=C$, suy ra $\dfrac{d\left( M,\left( {A}'BC \right) \right)}{d\left( {C}',\left( {A}'BC \right) \right)}=\dfrac{{C}'M}{{C}'C}=\dfrac{1}{2}$.
Ta có ${{V}_{{C}'.{A}'BC}}=\dfrac{1}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{1}{3}.{C}'C.{{S}_{\Delta ABC}}=\dfrac{1}{3}.a.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{{{a}^{3}}\sqrt{3}}{12}$.
Lại có ${A}'B=a\sqrt{2}$, $CB=a$, ${A}'C=a\sqrt{2}$ $\Rightarrow {{S}_{{A}'BC}}=\dfrac{{{a}^{2}}\sqrt{7}}{4}$.
Suy ra $d\left( {C}',\left( {A}'BC \right) \right)=\dfrac{3{{V}_{{C}'.{A}'BC}}}{{{S}_{\Delta {A}'BC}}}=\dfrac{3.\dfrac{{{a}^{3}}\sqrt{3}}{12}}{\dfrac{{{a}^{2}}\sqrt{7}}{4}}=\dfrac{a\sqrt{21}}{7}$.
Vậy $d\left( M,\left( {A}'BC \right) \right)=\dfrac{1}{2}d\left( {C}',\left( {A}'BC \right) \right)=\dfrac{1}{2}.\dfrac{a\sqrt{21}}{7}=\dfrac{a\sqrt{21}}{14}$.
Ta có ${{V}_{{C}'.{A}'BC}}=\dfrac{1}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{1}{3}.{C}'C.{{S}_{\Delta ABC}}=\dfrac{1}{3}.a.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{{{a}^{3}}\sqrt{3}}{12}$.
Lại có ${A}'B=a\sqrt{2}$, $CB=a$, ${A}'C=a\sqrt{2}$ $\Rightarrow {{S}_{{A}'BC}}=\dfrac{{{a}^{2}}\sqrt{7}}{4}$.
Suy ra $d\left( {C}',\left( {A}'BC \right) \right)=\dfrac{3{{V}_{{C}'.{A}'BC}}}{{{S}_{\Delta {A}'BC}}}=\dfrac{3.\dfrac{{{a}^{3}}\sqrt{3}}{12}}{\dfrac{{{a}^{2}}\sqrt{7}}{4}}=\dfrac{a\sqrt{21}}{7}$.
Vậy $d\left( M,\left( {A}'BC \right) \right)=\dfrac{1}{2}d\left( {C}',\left( {A}'BC \right) \right)=\dfrac{1}{2}.\dfrac{a\sqrt{21}}{7}=\dfrac{a\sqrt{21}}{14}$.
Đáp án A.