Câu hỏi 31

Dễ thấy BD // B’D’, BC’ // AD’ nên (AB’D’) // (BC’D)

\(\Rightarrow d\left( \left( AB'D' \right);\left( BC'D \right) \right)=d\left( C';\left( AB'D' \right) \right)=\frac{3{{V}_{C'.AB'D'}}}{{{S}_{AB'D'}}}\)

Ta có :

\(\begin{align}&{{V}_{C'.AB'D'}}={{V}_{A.B'C'D'}}=\frac{1}{3}AA'.{{S}_{B'C'D'}}=\frac{1}{3}AA'.\frac{1}{2}B'C'.C'D' \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{6}{{V}_{ABCD.A'B'C'D'}}=\frac{1}{6}{{.2}^{3}}=\frac{4}{3} \\ \end{align}\)

Tam giác AB’D’ có \(AB'=AD'=B'D'=2\sqrt{2}\Rightarrow \Delta AB'D'\) là tam giác đều cạnh \(2\sqrt{2}\Rightarrow {{S}_{AB'D'}}=\frac{{{\left( 2\sqrt{2} \right)}^{2}}\sqrt{3}}{4}=2\sqrt{3}\)

Vậy \(d\left( C';\left( AB'D' \right) \right)=\frac{3.\frac{4}{3}}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\Rightarrow d\left( \left( AB'D' \right);\left( BC'D \right) \right)=\frac{2}{\sqrt{3}}\).

Chọn B.