Cho hình lăng trụ  $ABC.{A_1}{B_1}{C_1}$ có diện...

A

- Hướng dẫn giải

Do $C{C_1}//A{A_1} \Rightarrow C{C_1}//\left( {AB{B_1}{A_1}} \right)$  nên $d\left( {C{C_1};\left( {AB{B_1}{A_1}} \right)} \right) = d\left( {C;\left( {AB{B_1}{A_1}} \right)} \right) = 6$

Nhận xét:

${V_{{A_1}.ABC}} = {V_{C.{A_1}{B_1}{C_1}}}$ (do ${S_{\Delta ABC}} = {S_{\Delta {A_1}{B_1}{C_1}}};d\left( {{A_1};\left( {ABC} \right)} \right) = d\left( {C;\left( {{A_1}{B_1}{C_1}} \right)} \right)$ )  (1).

${V_{{A_1}.{B_1}BC}} = {V_{{A_1}.{B_1}{C_1}C}} = {V_{C.{A_1}{B_1}{C_1}}}$ (do ${S_{\Delta {B_1}BC}} = {S_{\Delta C{B_1}{C_1}}};d\left( {{A_1};\left( {{B_1}BC} \right)} \right) = d\left( {{A_1};\left( {{B_1}C{C_1}} \right)} \right)$) (2)

Từ (1) và (2), ta có: ${V_{ABC.{A_1}{B_1}{C_1}}} = 3.{V_{C.{A_1}AB}} = 3.\frac{1}{3}.d\left( {C;\left( {AB{B_1}{A_1}} \right)} \right).{S_{\Delta AB{A_1}}} = 3.\frac{1}{3}.6.\frac{1}{2}.4 = 12$