Cho hình lập phương  ABCD...

B

- Hướng dẫn giải

Đáp án B

Ta có $A\text{'}D//B\text{'}C\Rightarrow A\text{'}D//\left(B\text{'}KC\right)\Rightarrow d\left(CK;A\text{'}D\right)=d\left(D;\left(B"KC\right)\right)$ 

$=d\left(D\text{'};\left(B"KC\right)\right)=\frac{1}{2}xd\left(C\text{'};\left(B"KC\right)\right)=\frac{3x{V}_{K.B\text{'}C\text{'}C}}{2S_{\Delta B\text{'}KC}}.$ 

Thể tích khối chóp KB'C'C là $V=\frac{1}{3}.d\left(K;\left(B\text{'}C\text{'}C\right)\right).S_{\Delta B\text{'}C\text{'}C}=\frac{a^2}{6}$

Tam giác B’KCcó $CK=\frac{a\sqrt{5}}{2};B\text{'}C=a\sqrt{2};B\text{'}K=\frac{3a}{2}$ 

=>Diện tích $\Delta B\text{'}KC$là $S_{\Delta B\text{'}KC}=\frac{3a^2}{4}.$Vậy $d\left(CK;A\text{'}D\right)=\frac{a}{3}$