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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. $\dfrac{\sqrt{21}a}{7}.$
B. $\dfrac{\sqrt{2}a}{2}.$
C. $\dfrac{\sqrt{2}a}{4}.$
D. $\dfrac{\sqrt{21}a}{14}.$
B. $\dfrac{\sqrt{2}a}{2}.$
C. $\dfrac{\sqrt{2}a}{4}.$
D. $\dfrac{\sqrt{21}a}{14}.$
Cách 1:
Ta có: ${C}'M\cap \left( {A}'BC \right)=\left\{ C \right\}$ nên $\dfrac{d\left( M,\left( {A}'BC \right) \right)}{d\left( {C}',\left( {A}'BC \right) \right)}=\dfrac{MC}{{C}'C}=\dfrac{1}{2}$
Lại có $A{C}'\cap \left( {A}'BC \right)=I$ là trung điểm của $A{C}'$ nên:
$\dfrac{d\left( {C}',\left( {A}'BC \right) \right)}{d\left( A,\left( {A}'BC \right) \right)}=\dfrac{{C}'I}{AI}=1\Rightarrow d\left( {C}',\left( {A}'BC \right) \right)=d\left( A,\left( {A}'BC \right) \right)$
Gọi $N$ là trung điểm của $BC$ thì $AN\bot BC$.
Mà $BC\bot {A}'A$ nên $BC\bot \left( {A}'AN \right)$
Kẻ $AH\bot {A}'N$.
Do $BC\bot \left( {A}'AN \right)$ nên $BC\bot AH$.
Vậy $AH\bot \left( {A}'BC \right)$ hay $d\left( A,\left( {A}'BC \right) \right)=AH$.
Ta có: $AN=\dfrac{a\sqrt{3}}{2},{A}'A=a$.
$\Rightarrow \dfrac{1}{A{{H}^{2}}}=\dfrac{1}{{A}'{{A}^{2}}}+\dfrac{1}{A{{N}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{\left( \dfrac{a\sqrt{3}}{2} \right)}^{2}}}=\dfrac{7}{3{{a}^{2}}}\Rightarrow A{{H}^{2}}=\dfrac{3{{a}^{2}}}{7}\Rightarrow AH=\dfrac{\sqrt{21}a}{7}$
$\Rightarrow d\left( M,\left( {A}'BC \right) \right)=\dfrac{1}{2}AH=\dfrac{\sqrt{21}a}{14}$.
Cách 2:
Do ${C}'M\cap \left( {A}'BC \right)=C$, 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 ${{V}_{{C}'.{A}'BC}}=\dfrac{1}{3}{{V}_{ABC\cdot {A}'{B}'{C}'}}=\dfrac{1}{3}.{C}'C.{{S}_{\Delta ABC}}=\dfrac{1}{3}.a.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{{{a}^{3}}\sqrt{3}}{12}$.
Lai 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 $d\left( {C}',\left( {A}'BC \right) \right)=\dfrac{3{{V}_{{{C}'}}}{A}'BC}{{{S}_{\Delta {A}'BC}}}=\dfrac{3\cdot \dfrac{{{a}^{3}}\sqrt{3}}{12}}{\dfrac{{{a}^{2}}\sqrt{7}}{4}}=\dfrac{a\sqrt{21}}{7}$
vậy $d\left( M,\left( {A}'BC \right) \right)=\dfrac{1}{2}d\left( {C}',\left( {A}'BC \right) \right)=\dfrac{1}{2}\cdot \dfrac{a\sqrt{21}}{7}=\dfrac{a\sqrt{21}}{14}$.
Lại có $A{C}'\cap \left( {A}'BC \right)=I$ là trung điểm của $A{C}'$ nên:
$\dfrac{d\left( {C}',\left( {A}'BC \right) \right)}{d\left( A,\left( {A}'BC \right) \right)}=\dfrac{{C}'I}{AI}=1\Rightarrow d\left( {C}',\left( {A}'BC \right) \right)=d\left( A,\left( {A}'BC \right) \right)$
Gọi $N$ là trung điểm của $BC$ thì $AN\bot BC$.
Mà $BC\bot {A}'A$ nên $BC\bot \left( {A}'AN \right)$
Kẻ $AH\bot {A}'N$.
Do $BC\bot \left( {A}'AN \right)$ nên $BC\bot AH$.
Vậy $AH\bot \left( {A}'BC \right)$ hay $d\left( A,\left( {A}'BC \right) \right)=AH$.
Ta có: $AN=\dfrac{a\sqrt{3}}{2},{A}'A=a$.
$\Rightarrow \dfrac{1}{A{{H}^{2}}}=\dfrac{1}{{A}'{{A}^{2}}}+\dfrac{1}{A{{N}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{\left( \dfrac{a\sqrt{3}}{2} \right)}^{2}}}=\dfrac{7}{3{{a}^{2}}}\Rightarrow A{{H}^{2}}=\dfrac{3{{a}^{2}}}{7}\Rightarrow AH=\dfrac{\sqrt{21}a}{7}$
$\Rightarrow d\left( M,\left( {A}'BC \right) \right)=\dfrac{1}{2}AH=\dfrac{\sqrt{21}a}{14}$.
Cách 2:
Ta có ${{V}_{{C}'.{A}'BC}}=\dfrac{1}{3}{{V}_{ABC\cdot {A}'{B}'{C}'}}=\dfrac{1}{3}.{C}'C.{{S}_{\Delta ABC}}=\dfrac{1}{3}.a.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{{{a}^{3}}\sqrt{3}}{12}$.
Lai 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 $d\left( {C}',\left( {A}'BC \right) \right)=\dfrac{3{{V}_{{{C}'}}}{A}'BC}{{{S}_{\Delta {A}'BC}}}=\dfrac{3\cdot \dfrac{{{a}^{3}}\sqrt{3}}{12}}{\dfrac{{{a}^{2}}\sqrt{7}}{4}}=\dfrac{a\sqrt{21}}{7}$
vậy $d\left( M,\left( {A}'BC \right) \right)=\dfrac{1}{2}d\left( {C}',\left( {A}'BC \right) \right)=\dfrac{1}{2}\cdot \dfrac{a\sqrt{21}}{7}=\dfrac{a\sqrt{21}}{14}$.
Đáp án D.
